point-set-topology

[tender halo]

it is the combination of mappings

[tender halo]

^ this is right before 2.1.11

[rancid umbra]

there is a list of special symbols in the back of the book with references to the first time they appear

page 71

[tender halo]

regular triangle is the cartesian product of mappings

[rancid umbra]

not sure what the combination means in context of the sum of spaces

with regards to the second picture here

[tender halo]

you combine continuous functions on each summand

they are naturally defined on disjoint open subsets of the sum

so the combination is continuous

[neat mantle]

Thanks for finding it

[rancid umbra]

oh, yeah. true

[cosmic bone]

im back to an old problem about the space of countable ordinals with the order topology
I have to show its countably compact but not compact

In order to show its compact, I have to show that any arbitrary centered system of closed sets has a nonempty intersection. Consider $G_1$ = [1, 2, ..., $\omega_0$ ...], $G_2$ = [2, 3, ..., $\omega_0$, ...], ..., $G_n$ = [n, ..., $\omega_0$, ...]. Clearly any finite collection of G has a nonempty intersection. However, G must have an empty intersection. If not, there is some x $\in$ $\bigcap_{\alpha} G_\alpha$. However, x is not in $G_{x+1}$, and so it cannot be in this infinite intersection.

[gentle osprey]

stephenw

[cosmic bone]

To show its countably compact it suffices to show that any arbitrary countable centered systems of closed sets has a nonempty intersection.

This unfortunately requires a different argument (because we want to show any arbitrary closed blah blah has a nonempty intersection), but I'm kind of struggling to be careful with my different infinities

We have an uncountable topological space, and we want to show that any countable collection of closed sets with nonempty finite intersections must itself have a nonempty intersection.

[gaunt linden]

Uh, doesn't "compact" imply "countably compact"?

[rancid umbra]

surely they meant countably compact but not compact, right?

[cosmic bone]

oh oops

horrible completely inverted typo

apologies about that

[rancid umbra]

[0,w) for w countable doesn’t have a finite subcover…

[cosmic bone]

this was the problem you helped me with a few days ago c squared

[rancid umbra]

yea

[cosmic bone]

I can see why its not compact

[cosmic bone]

but Im struggling to directly see how we can invoke this to show its countable compact

[cosmic bone]

apologies about the typo, fixed it

[rancid umbra]

this condition implies that X is countably compact because you just take complements

like, if you know that for every countable collection of closed sets with empty intersection, it contains a finite subcollection with empty intersection, then for a given countable open cover of X, there is a finite subcollection of complements that have empty intersection, and so the finitely many open sets that they correspond to in the original cover in fact cover X

it sounded like there was general confusion tho. are you also asking how to show that every countable collection of closed subsets of X that has empty intersection contains a finite subcollection with empty intersection?

[cosmic bone]

yes

[cosmic bone]

precisely

In the uncountable case (for compactness), there was the system of sets we constructed $G_1$, $G_2$, ... which "obviously" had an empty intersection

[gentle osprey]

stephenw

[cosmic bone]

I dont see why a similar example couldnt be created with a countable set of such $G_n$'s

[gentle osprey]

stephenw

[cosmic bone]

If there were such a countable set of G_n's which have nonempty finite intersections but an empty intersection, then T is not countably compact. But I know T IS countably compact from the problem, so they cannot have an empty intersection, but I dont see how any one element could be in all of the sets

[rancid umbra]

hm? that doesn’t sound right

oh you meant has the finite intersection property

[cosmic bone]

yes

apologies about that

[rancid umbra]

all g

hmm. i have to go for a few. i’ll think about it while im gone and hopefully give an answer if nobody else has when i get back

[cosmic bone]

sounds good-- thank u a lot

[rancid umbra]

do you know that every sequence in X has an accumulation point in X?

i think last time i argued this

you take a monotone subsequence and look at its supremum, i.e., its union

[rancid umbra]

anyways, if you know this fact, consider some countable open cover with empty intersection, and assume for contradiction that every finite sub collection has non-empty intersection

after enumerating the closed sets however you like, take the smallest element in the first set, x_1. since the first and the second set intersect (by our faulty assumption), pick the smallest element in the intersection, x_2. continue in this way until you have exhausted all of the sets in the cover.

at each stage, the chosen x_n is in the intersection of the first n closed sets. the sequence x_n is monotone non-decreasing, so converges to its supremum x in X. notably, x is an accumulation point of the sequence of x_n’s

if x is not in the m-th closed set in the cover, say C_m, its complement is an open neighborhood of x, so by virtue of x being an accumulation point of x_n, there are infinitely many x_n in the complement of C_m. but for n >= m, x_n can’t be in the complement of C_m since it is in C_m by construction.

this x must be in the intersection of all C_m. contradiction.

[rancid umbra]

the wiki page on countably compact spaces
actually gives outlines for equivalent characterizations. i have just given 3 —> 4. i don’t think these characterizations are equivalent unless you assume some weak form of choice, though

[gentle osprey]

longboard kayak

[spare light]

tryna decimally expand all the reals
is that the hint?

[rancid umbra]

yea, but it requires some work for decimals with leading 0’s

[spare light]

or we just state the continum hypo and be done

[alpine nest]

I'd say this is easier by establishing two injections rather than an explicit bijection

(as is very often the case)

[rancid umbra]

it’s kinda funny, but you can find a bijection (1/10,1) <—> P(N) first. then compose with a bijection (0,1) <—>(1/10,1)

[alpine nest]

Cantor, Schroeder and Bernstein are lovely boys

[spare light]

alr

[prime elbow]

[tender halo]

finding an explicit bijection is miserable

for all the different variants of the continuum

[rancid umbra]

on the other hand, showing P(N) and P(N)^n have the same cardinality is quite fun

there is a nice bijection with interleaving

[gaunt linden]

That's only for one particular construction that you have in mind. There's a lot of other things one could do with decimal expansions that don't have this problem.
For example map 0.00729735256... to {10, 20, 37, 42, 59, 67, 73, 85, 92, 105, 116, ...}

[unreal stratus]

Real

[spare light]

cooked

i think we have to exclude the repeating cases and there are countably many of them, which upon deletion from uncountable cardinality, doesn't make difference?

[opaque scroll]

Yeah, you can also be very explicit.

Like pick some enumeration of the diadic rationals, then whenever you have something where the naive result would be a diadic rational you shuffle it around a little (like send it to the nth even or odd number on your enumeration depending on if it ends in repeating 1s or 0s).

Then you just have to tweak it a little bit so that since (0, 1) doesn't contain 0 and 1.

[spare light]

Is every countable topological space (i.e., one in which the underlying set is countable) separable? What about ccc? Can a countable space be first countable but not second countable?

Yes, every countable topological space is separable, because the whole space is countable and any dense subset must intersect every nonempty open set, so taking $X$ itself gives a countable dense subset.

Every countable topological space also satisfies the countable chain condition (ccc). If there were uncountably many pairwise disjoint nonempty open sets, then each would have to contain at least one distinct point of $X$, contradicting the countability of $X$.

i'm sure that the first one is correct, need confirmation for the 2nd one and hint for the 3rd one

also need hint for this

Construct a topological space $(X, \mathcal{T})$ which is countable (i.e., $X$ is countable) but not first countable.

[gentle osprey]

longboard kayak

[prime elbow]

You just take X, in case of countable topological space, X will be your countable dense set

[spare light]

yeh, which i mentioned

[spare light]

for the last one we need local(countable and neseted) basis but not global, if that makes sense

[gaunt linden]

My "as explicitly as possible" would go something like:

1. Reinterpret P(N) as the set of all binary sequences.
2. For each sequence that ends in repeating ones, negate the entire sequence and stick a 1 in front. Simultaneously, stick a 0 in front of every sequence that already ended in repeating zeroes.
3. Binary fractions to biject to [0,1)
4. Hilbert's hotel to shove 0 into (0,1)

[tender halo]

yea 2 is correct, for 3 consider that union of bases of all points is a base for the whole space

[spare light]

so

Let $X$ be a countable topological space.\\

If $X$ is first-countable then every $x \in X$ has a countable local basis $\mathcal{B}_x$. Taking $\mathcal{B} := \bigcup_{x \in X} \mathcal{B}_x$, we get a countable union of countable sets, hence countable. Now, for any open set $U$ and any $x \in U$, some $B \in \mathcal{B}_x$ exists with $x \in B \subseteq U$, so $\mathcal{B}$ is a basis. Hence $X$ is second-countable.\\

If $X$ is second-countable, then there exists a countable basis $\mathcal{B}$. For each $x \in X$, define $\mathcal{B}_x := \{B \in \mathcal{B} : x \in B\}$. Since $\mathcal{B}$ is countable, so is each $\mathcal{B}_x$, and itvaciouslyforms a local basis at $x$. So $X$ is first-countable.

[gentle osprey]

longboard kayak

[tender halo]

not sure why you included the second part but yeah

[gaunt linden]

Hmm, I think you can take something like Q with an extra point, and declare that a set containing the extra point is open iff its intersection with Q is dense open.

[spare light]

i got an idea tho i dont know if it would work

for non example of first countable m.SE was pointing out co-finite and particular point

if we imply that in Z

[prime elbow]

In 2, how it implies that each pairwise disjoint set contains at least one distinct point of dense set?

I got it, my bad

[gaunt linden]

Ah, you don't even need the extra point. Just Q with the topology

T = { Ø } cup { A intersect Q | A is an open and dense subset of R }

[tender halo]

Q with N collapsed to a point also works

[gaunt linden]

Ah, diagonalization!

[unreal stratus]

Nice question!

[gritty widget]

Single ultrafilter topology(I think that’s the name?) on omega works too

Coolest one fr 🥱

[gaunt linden]

Or the topology on N where a subset is closed iff it all of N or the sum of reciprocals of its elements converges.

[tender halo]

any uncountable ideal on N works i think

[gaunt linden]

Hmm, I don't think so? The ideal "all subsets that contain only even numbers" is uncountable, but if we make it (and N) into the closed sets of a topology, the space we get is first countable -- namely, every point n has a neighborhood basis consisting of the single set {n} cup {all odd numbers}.
Or am I misunderstanding you?

[tender halo]

uhh thats fair

i meant more essentialy-uncountable i guess

an ideal that does not have a ideal base in the form A1 \subseteq A2 \subseteq A3 ...

but i guess thats kinda circular

[tiny kindle]

hey, im taking topology next semester and we are using munkre's topolgoy

was wondering how much overlap does this have with real analysis?

[tender halo]

not that much for the basic course, there are a couple key results that will be used in real analysis in one variable but thats about it

namely connected and compact subsets of the real line

i guess the general notion of a limit

but for "meaty" theorems there is not much

[tiny kindle]

but like isn't compactness + pointwise convergence in real analysis?

similarly (Bolzano-Weierstrass) and Hausdorff Topological Spaces.

I've seen compactness + pointwise convergence in like 4 different coruses already in Complex analysis even

[tender halo]

sure but thats peanuts, its just vocabulary really

it starts mattering when you start talking about manifolds

[tiny kindle]

These are like the topics we covered in real analtis: "1. Sets and Cardinality
2. Metric Spaces
3. Sequences and Series of Functions
4. Topological Spaces
Hausdorff Topological Spaces
Bases for Topologies
Convergence in Topological Spaces
Nets
Homeomorphisms
5. Compactness"

[tiny kindle]

the topology course we are learning, doesn't even cover manifodls

It teaches, basic topology + some algebraic topology

This is like the course outline for topology: "Topics covered: Point set topology: metric spaces and topological spaces, compactness, connectedness, continuity, extension theorems, separation axioms, quotient spaces, topologies on function spaces, Tychonoff theorem. Algebraic topology: Fundamental groups and covering spaces, and related topics."

[tender halo]

idk there are a lot of ways to construct a real analysis course, some avoid talking about metric spaces altogether, living in R and C, some straight out presuppose you know the underlying topological notions, some try be inbetween

[tiny kindle]

I see, strange lol

[tender halo]

to answer your question differently: topology reframes real analysis results in a more general and transparent framework, but generally it is only used as a language, the heavy duty theorems are rarely invoked until you get to stuff like manifolds

[tiny kindle]

how hard is it to self learn manifolds?

[tender halo]

not particularly hard

as far as self study goes

[tiny kindle]

If i've done topology

i mean next semester

I'm also self learning differential geometry, made a wrong choice not taking it instead I chose math game theory

Also they call the toplogy course: "Geometry-Topology, Differential Geometry".

So weird its called differential geometry

[frail bane]

what does it mean for soemthing to be continuous in R^3

[tender halo]

you mean a continuous function from R^3 to somewhere?

[frail bane]

well basically

i was trying to explain what a surface in R3 is

but didn't know how to do that without the idea of a homeomorphism

which leads to me needing to know what is required for something to be continuous in R3

I'm iexplaining this to first year students so yeah

it needs to be brief but somewhat understandable (not too rigorous)

[opaque scroll]

I think "A surface is something that looks like the plane when you zoom in around a point. For example a sphere or a (the surface of a) donut" is a pretty good explanation.

It can be more rigorous of course if you need that

[frail bane]

could i say that for every point on the surface, there is a point which has a neighbourhood that is on the surface

[opaque scroll]

I'm not sure how to parse this sentence.

But you could say that every point on the surface has a neighborhood that looks like a disk.

[frail bane]

ah ok yeah

wait

but i mainly want a brief explanation on what makes f: R3 -> R3 continuous

[tender halo]

honestly i have real trouble marrying the image of a surface with the definition of a two dimensional manifold

[opaque scroll]

An explanation of continuity could be that small changes in the input result in small changes to the output.

But I don't really see the relevance of a function R3 -> R3 to surfaces. R2 -> R3 could be relevant

[frail bane]

oh

wait i'm kinda confused

is it fine if i can explain the context?

so that it's easier for me to write an explanation that makes sense

relavant to my context

[opaque scroll]

It's fine by me...

[frail bane]

sure

okay so

i was trying to derive the equation of a catenoid by consiering minimizing the area of a surface of revolution

however, one point of the proof was this:

which leads to me requiring to explain what a surface is

I wrote this

which led me to writing what a homemorphism is(which you did say was wrong so i'll change that after)

but then i need to briefly explain what continuity is for the homeomorphism

[opaque scroll]

It's not wrong. Just less general than the usual definition

[frail bane]

yeah i want to talk specifically about surfaces in R3 so isn't that what a homeomorphism for those surfaces are?

[opaque scroll]

It is yes

[frail bane]

so it's continuous between subsets of R3?

ok good

[frail bane]

so what was this about?

is it kinda like epsilon delta but in multiple directions?

[opaque scroll]

If your only interested in subsets of R3 it would make sense to define it for subsets of R3 sure

[frail bane]

ok sure

[opaque scroll]

You can define it with epsilon delta yes

[frail bane]

how can i explain that then

[opaque scroll]

f is continuous at x if for any epsilon > 0 there exists a delta > 0 such that if the distance between x and y is less than delta, then the distance between f(x) and f(y) is less than epsilon.

Or less rigorous, if you only change x a little, then f(x) also only changes a little

[frail bane]

ahhh but this time the distance can be 3 dimensional

as in

not 3 dimensional

but

the distance function in R3

[sick torrent]

Are there any applications of the generalized heine borel theorem? In particular the implication that totally bounded + complete => compact. Or is it just more like, a random curiosity? Like is there some application in which showing that something is totally bounded + complete is genuinely easier than just showing that a set is compact

[modest hornet]

I vaguely remember someone talking about the Extreme Value Theorem when we covered it in my class.

I think it's the first time in most topology courses that you get some sort of theoretical connection between the idea of being closed or bounded and the idea of compactness.

But I can't think of an application off the top of my head.

I need to go to sleep. 😴

[sick torrent]

Extreme Value Theorem and total boundedness? I mean you can just prove that with the fact that intervals are compact idk how total boundedness helps

[modest hornet]

I don't think it was a necessity to proving it.

[sick torrent]

Yeah I'm wondering more about times where proving total bounded + complete is actually easier than proving compactness directly (either by sequences or by covers)

[modest hornet]

Yeah, sorry. I cannot help you there. I haven't done topology in ages.

I was dusting off the cobwebs to remember that earlier bit. 😭

[ruby delta]

proving that Cantor set is compact, depending on your definition, might be easier using sequential compactness / Heine Borel

[sick torrent]

thanks! however, I'm asking for examples where proving totally bounded + complete is easier than proving the other two

[ruby delta]

yeah, I think that's an example that you're looking for

[sick torrent]

so that case is an example in which using covers, or sequences, is easier

[ruby delta]

sequences is much easier than covers here

[sick torrent]

yeah I know but I don't want that, I don't think you are understanding

I would like an example in which you prove totally bounded + complete, not that every sequence has a convergent subsequence

[ruby delta]

I mean, you can modify a sequential compactness proof into completeness proof

actually completeness is probably even easier here

since you can now assume convergence (in R)

[sick torrent]

well yeah but it would amount to saying something like: - R is complete and the cantor set is closed so the cantor set is complete. It is totally bounded because it is a closed subsets of a totally bounded set [0,1]

but this all seems really unnecessary as you can just say

it is a closed subset of [0,1]

[ruby delta]

well if you can straight up assume that the cantor set is closed then you're done, but that's not a trivial proof

[sick torrent]

which is compact

done

yeah I guess, I mean

in this case I would just use tychonoff's theorem with {0,1}^N and then give the explicit biyection with the set of real numbers, which I find much easier

so idk if I would count this as "an essential use" of Heine Borel

[tender halo]

from it you easily get that the completion of a space is compact iff the space is totally bounded

[prime elbow]

Maybe I asked this question earlier,

Every separable metric space has a Lindelof covering.

Say X has open covering {U_i}. Since X is separable metric space so it is second countable. So we can write U_i as a countable union of basis elements.

Now let D is a countable dense set in X, now for all d \in D, take the union of all basis elements B_d's such that d \in B_d and B_d \subset U_j, for all such j, and it is countable union.

Since D is countable, so countable union of countable union is countable union.

I think we have to add some extra basis elements in our collection to make U_i's, right?

[sudden flower]

If you take the "cross" |x| = |y| ≤ 1 in R^2, is this space locally contractible? It seems to be globally contractible as one can homotope it into (0,0)?

[rancid umbra]

yes, that is part of it. there is a basis for the topology on the cross in which every basis element is contractible

it consists of intervals and smaller crosses centered at the origin

each of which is contractible

[alpine nest]

Please don't spam this across multiple channels; #real-complex-analysis was the corect one.

[floral sage]

Ok

[sudden flower]

I mean in the subspace topology coming from the standard one on the plane

[heady skiff]

How can we apply Theorem 1.18 here? Don't we need C open, which we don't necessarily have?

[gritty widget]

Isn’t C open because you’re already assuming local path-connectedness(and hence local connectedness)?

[rancid umbra]

yes, the basis i mentioned will generate the subspace topology

[queen prism]

https://paracompact.space/en/definitions/ someone devoted an entire website to this?

[thorny agate]

that's fun

[unreal stratus]

Lmao "paracompact space" and the only reference is Lurie

[red yoke]

This is because "space" is ∞-topos

[pure shale]

Just sanity-checking. Can a basis ever fail to be a cover? My understanding is that it's no definitionally, but I'm reading some tutorials for one of my classes and it got me wondering.

[gaunt linden]

Right, that is no by definition. The whole space is open, and an open set must be the union of basis elements.

[pure shale]

sick, ty

When my answer is 3 lines and the tutor's is a page, one of us made a categorical error and I usually default to it being mine

[gaunt linden]

That might call for "please post a screenshot/photo of the problem", for verification...

[pure shale]

It's from Munkres. Given a metric space X, if X is second-countable, it is Lindelof.

Their argument is that since a basis {B_n} on X is countable, you can take a representative x_n in B_n.

Then, if we have some U = {u_n} that is an open cover of X, we can say each B_n is in u_n.

Then, U is a countable subcover of X.

My argument is that, if you have a basis, it is already a collection of open sets, and you neednt introduce U, because B is already countable (its already a countable subcover of itself).

If one defines bases to be ANY set that covers it makes sense, but in our course basis sets are open sets

[gaunt linden]

But for Lindelöf you don't get to decide which open cover you start with. It is given to you by the adversary, and it doesn't need to be related to the countable basis.

[pure shale]

love the use of 'adversary' very fun ty

[pure shale]

This is a great point, though it also invalidates the tutor's proof 😭 I think I need to show some relationship between 'the' countable basis and every other basis

Ok I see what I need to do. I need to put B_n in my arbitrary u_n and show that still covers the same space. I think that's probably what the tutor meant to write

[gaunt linden]

Heres a quote of a proof that seems to purport to be Munkres's own: https://math.stackexchange.com/questions/2257690/if-x-is-second-countable-then-x-is-lindelöf

[pure shale]

Oh ty

[gaunt linden]

The phrasing there looks a bit convoluted to me, though. I think I would separate out the core idea as a Lemma:
Let $X$ be a topological space, $\mathcal B$ any basis for it, and $\mathcal U$ an open cover of $X$. Then
$$ { b \in \mathcal B \mid \exists u \in \mathcal U : b \subseteq u } $$
is also a basis for $X$.

[gentle osprey]

Troposphere

[pure shale]

I think I get it

I re-read some notes and there's a little segment where our prof mentions that one way to define a basis is that, if u is an open set in our topology and B={b_i} is a basis, then we can always decompose u as a (possibly uncountable) union of basis sets

[gaunt linden]

Hmm, to me that is the definition of basis. How did you define it originally?

[pure shale]

Or symbolically:
$\forall u \in \tau_X \exists A \text{, (an interval), } s.t. \quad u = \bigcup_{\alpha \in A} B_\alpha$

[gentle osprey]

Attelis

[pure shale]

Yeah that's how I learnt it originally.

In this course we define $B$ as a basis if $B$ is an (open) cover of $X$, where $\forall x \in X$, and for any $B_1$ and $B_2$ that cover $x$, $\exists B_3 \in B$ such that $x \in B_3 \subseteq B_1 \cap B_2$.

[gentle osprey]

Attelis

[gaunt linden]

That looks rather indirect. Is one of B1 and B2 supposed to be in the basis, and the other an arbitrary open?

[pure shale]

B_1 and B_2 are in B, yeah. I miswrote

It's a weird definition but I get why you'd use it. EXTREMELY easy to check

[gaunt linden]

Ah, I see, that is a definition of "basis for some possible topology", whereas "basis for this already existing topology" would need additional definitions.

[pure shale]

really? how?

[gaunt linden]

As written, this implies that {X} is a basis.
(Which is true as long as we're not picky about which topology: it is a basis for the trivial topology).

[pure shale]

Sorry I should have added, $B \subseteq \tau$.

[gentle osprey]

Attelis

[gaunt linden]

{X} still qualifies. :-)

[pure shale]

Yeah true

I don't think it matters tho, bases are just a way to cover elements in a composable way

There's probably more sophisticated uses for them but I'm not a very sophisticated person so that's ok :)

[gaunt linden]

Getting the definitions right always matters. If we start with this definition, we cannot prove from it that every open set is a union of basis elements.

[pure shale]

I agree with you on principle and would normally use the open set definition, though I am sticking to the definitions provided in class because the tutors mark down for that kind of thing. Which I guess is fair but that's the decision calculus

Though curious, why do you say we can't prove the other definition?

[gaunt linden]

Because {X} satisfies the wrong definition, and it's not true that every open set is a union of elements of {X}.

An element-based definition that I think works would be

A "basis" for X means a family B of open sets with the property that for every open O and every x in O, there is a b in B such that x in b subseteq O.

[pure shale]

yeah we use that one too

[pure shale]

I could be hallucinating but I think this got covered as 'every open set is contained in a union of basis elements' which now I think about it consciously, is vacuously true

I think I will use the x in b subset O def tho

[pure shale]

I think I'm making another silly error. I'm trying to show that the lower-limit topology is not N_2. But surely [q, q+1) is a valid countable basis?

Or rather, [n,n+1) with n an integer. Then all x in R_l is covered by a countable basis

(Pg 190 of Munkres)

[pure shale]

...Ignore me, I was thinking 'cover' not basis. It fails the composability condition

Time for a break I think x_x

[prime elbow]

How can I show that GL(n,R) is dense in M(n,R)? I am thinking of doing this by showing some mapping?

[tardy carbon]

You can use a matrix decomposition like the SVD.

[gaunt linden]

Let A be some matrix that is not in GL(n,R). Consider the function f(t) = A+t(I-A) and g(t) = det(f(t)).
Then g is a polynomial in t (since the determinant is a polynomial in the matrix entries), and since g(0)=0, g(1)=1, it is not a constant polynomial.
Therefore g has only finitely many roots, so there are points arbitrarily close to 0 where g(t)!=0. Those points correspond to invertible matrices close to A.

[uncut minnow]

Hmmmmmmmmmm.... GL(n,R) is Zariski open in M(n,R). Is there a quick way to see that it is dense as well straight from the geometry?

[gaunt linden]

In general, nonempty Zariski opens are dense according to the Euclidean topology too -- but to see that, I think you need something akin to my argument with an auxiliary polynomial.

[uncut minnow]

I think so too... there is a slick argument if we were over complex numbers

but over the reals... mmm...

Okay maybe it still works. Let U be a zariski open set. We want to show this is dense in the euclidean topology.

Since U is zariski open, its complement is the zero locus of a bunch of polynomials, with at least one of them non-zero. Now let W be any euclidean open set. We want to show it meets U somewhere. If it did not, then W is in the complement and thus each of those poylnomials would vanish on W. But W is open and no non-zero real polynomial can vanish on an open set, so all of them have to be 0. This is a contradiction.

I think this does it?

[gaunt linden]

Yeah, I think that works.
The "no non-zero real polynomial can vanish on an open set" probably hides a reduction to a single-variable polynomial in the same vein as above.

[uncut minnow]

mmm... we have finitely many non-zero polynomials, they can vanish on only finitely many points, a non-empty open set in R^n cannot be finite. Does this seem like it works?

[gaunt linden]

With more than one variable, polynomials can and do vanish at infinitely many points.
(E.g. ax+by+c vanishes on an entire line).

[uncut minnow]

oh yes, these are multivariable polynomials

nvm

can we just throw in the identity theorem as justification?

[gaunt linden]

Yes, assuming you have an appropriate identity theorem available.

[uncut minnow]

there's one for real analytic functions iirc

[gaunt linden]

Sure.

[uncut minnow]

btw, if you don't mind.... can you help remove my helpers role? I don't really get much time to help in the help channels, so it is just causing my discord pings to go up 😭 i feel bad but...

[gaunt linden]

You can do that yourself by unchecking "Subscribe me to pings ..." in the Channels & Roles screen.
But I've done it for you now. (You can still help, of course, if you have time and notice a freshly opened help channel).

[uncut minnow]

Thank you! That was unchecked but somehow the role still remained...

And yes, if I do find the time I'll definitely help.

[unreal stratus]

ye i guess you can say y |-> f(x_1,...,x_n,y) vanishes at infinitely many points so it's zero but x_1,...,x_n are arbitrary so gg

[viral wing]

Can you help me with a proof, please? I’m not sure how to justify the highlighted step. I want to reprove this prop because Chevalley’s version is very messy

I guess I can take any open neighbourhood in the coset space and intersect it with the finite cover of the coset in G, would this work? I’m not sure if I got it right because it’s been a while since I touched topology

[uncut minnow]

Well, so essentially we have kind of a different proof for GL(n,R) being dense. Interesting

[unreal stratus]

You can also prove this more generally for arbitrary commutative rings if you use the language of affine schemes

Well okay maybe you can just still do the zriski top

[uncut minnow]

I believe it would be easier if you show that the natural surjection is perfect, and then note that G is the preimage of the projection so is compact because G/H is compact.

So two lemmas suffice :

Lemma 1 : the natural surjection is perfect, that is, continuous, surjective, closed and fibres are compact.

Lemma 2 : Preimages of compact sets by a perfect map are compact.

[elfin violet]

I'm getting into topology, I found Introduction to Topology and modern Analysis by George F. Simmons, recommended over Munkres by a friend of mine. Just for reference, would you agree?

[viral wing]

Thanks, I’ll give it a try. Just for my piece of mind, does my current version look sensible at least? 🙂

[uncut minnow]

The idea is correct, but some more work needs to be done (at least some more justifications).

If I am reading it correctly, you are essentially doing the same proof I mentioned, except working more hard

[prime elbow]

I got it

[upbeat scarab]

I've just started studying calculus. I'm currently studying Taylor expansions, but someday I'd like to learn differential topology. So I have a question: What fields should I complete at a minimum to learn differential topology? If possible, please let me know the detailed names of the fields I need to learn.

[uncut minnow]

Topology, real analysis and multivariable analysis and calculus at the very least. For multivariable analysis, you would need some amount of linear algebra too. A course on differential equations will not hurt but it is not strictly necessary unless we are going deep. This should be enough to start reading differential topology on your own.

[upbeat scarab]

Thank you for your reply! Is it correct to say that I don't need to delve too deeply into vector knowledge?

[uncut minnow]

oh you will have to... that is what the whole multivariable analysis and calculus part is for

and the linear algebra for that

[upbeat scarab]

That was helpful 🙂 Thanks

[viral wing]

Thanks, that was indeed much easier. I didn’t even have to use the Hausdorff property or require that H is closed, so amusingly it also applies to algebraic groups (vacuously of course because Zariski topologies are always compact)

[uncut minnow]

Glad! And indeed it does!

[viral wing]

Try picking up Milnor’s short lecture notes, they’re published by Dover and very cheap. They should give you a feel for the subject, even if you won’t necessarily be able to follow some arguments

[viral wing]

BTW Chevalley’s textbook on Lie groups is a surprising mix of a very modern, essentially sheaf theoretic treatment of manifolds, and a very messy old style treatment of topological groups

I wonder how it came to be this way

[uncut minnow]

Don't know if chevally had some fight with topological groups lol

[upbeat scarab]

I've always thought that book was good. I'd like to buy it and read it.
Thank you for your reply

[viral wing]

Holy cow, Chevalley's book includes the pre-categorical formulation of Tannaka-Krein duality!

[main pulsar]

They don't seem to be published by Dover

If you're talking about Topology from the Differentiable Viewpoint

[viral wing]

Yep, sorry, it's by Princeton

[sweet arrow]

is this proof correct for part (a)

[warm kettle]

What do you mean by an interior point? Did you mean a boundary point?

[sweet arrow]

wait i think i meant limit point

[hidden abyss]

Also you should probably prove that boundary points of A_i are boundary points of B and vice versa

[sweet arrow]

E' denotes the set of all limit points of E

[warm kettle]

I see. The first inclusion looks correct (but could be slightly more rigorous: why a limit point of some A_i is a limit point of B_n?). For the other inclusion, you should elaborate on why every limit point of B_n is a limit point of some A_i

Hint: ||Pigeonhole principle might be useful here||

[sweet arrow]

to show that you can just unpack the definition right? hold on i can write something

something like this

[warm kettle]

Yes, this is correct. You could also take a sequence of elements of A_i tending to x and say that those elements also belong to B_n, so the sequence is there too and so x must also be in the closure of B_n

[sweet arrow]

and for the other inclusion u have a limit point p of $B_n$ so $q \in \cup A_i$ then q is in some $A_i$ and thus p is a limit point for that $A_i$

[gentle osprey]

heavenly_lamb_61155

[warm kettle]

What is q?

[sweet arrow]

a point in the neighborhood of p

not equal to p

just a sketch

[warm kettle]

I see. But then it depends on the neighbourhood, and for different neighbourhoods, it might intersect different A_i's

[sweet arrow]

it doesnt matter tho right? we just wanna know if its an interior point of some A_i (maybe multiple if they intersect)

[warm kettle]

It's very important. That's the core of what needs to be proven

Or actually not

[sweet arrow]

im confused

[warm kettle]

If that were it, then it would also hold for a countable sum, right?

Beause we don't use the finiteness of n anywhere

But you showed that there's a counterexample for a countable sum

[sweet arrow]

that was going to be my next question xd

[warm kettle]

Let $x \in \overline{B_n}$. It means that every neighbourhood of $x$ has a non-empty intersection with $B_n$, right? If $x \in \bigcup_{i = 1}^n \overline{A_i}$, then $x \in \overline{A_i}$ for \textbf{some} $A_i$

[gentle osprey]

thingoln

[sweet arrow]

yes

$\overline{B_n} = B_n \cup B_n'$

[gentle osprey]

heavenly_lamb_61155

[sweet arrow]

but then why would this work but the other way around it wouldnt

[warm kettle]

Because if every neighbourhood of a point intersects a specific Ai, then every neighbourhood of that point also intersects Bn (because Ai is a subset of Bn)

But here you divide Bn into smaller sets

[sweet arrow]

ohh ok so in the first one we have a specific Ai then we show any neighborhood of Bn intersects that specific one

[warm kettle]

Yes

I think the easiest way to prove the other inclusion is to do a proof by contradiction

And ||use some property of open sets||

[sweet arrow]

so if I have something like $q \in \cup_{i=1}^{\infty} \overline{A_i}$ I can still say that $q \in \overline{A_i}$ for some $i$? even if its infinity?

[warm kettle]

No

This doesn't hold even for a finite number of sets, right?

Since the closure of Ai might be a larger set than Ai

[gentle osprey]

heavenly_lamb_61155

[warm kettle]

Oh, yes. You can

[sweet arrow]

typo

[sweet arrow]

hmm ok so since we can do this, I see how the second inclusion leads to a contradiction since it would hold for infinity also but we showed an example of a proper inclusion, but I still dont see whats wrong with the actual reasoning

[warm kettle]

In your reasoning, you say that every neighbourhood will have a non-empty intersection with \emph{some} $A_i$. However, you need to prove that every neighbourhood of a point will have a non-empty intersection with \textbf{one specific} $A_i$

[gentle osprey]

thingoln

[warm kettle]

Because of this

Try to imagine a situation like this: you have a point p and n neighbourhoods of it: N1, N2, ..., Nn. Neighbourhood Ni intersects Ai but no other set from {A1, ..., An}

Then you immediately know that that point cannot be in the sum of closures of Ai's, but it still has a chance to be in the closure of Bn

[sweet arrow]

thats quite hard to imagine

should we consider the example from earlier

Let $A_i = \left{1/i \right}$. Then $\cup_{i=1}^{\infty} \overline{A_i} = \left{1/i \colon i \in \mathbb{N} \right}$ but $\overline{B_\infty} = \left{1/i \colon i \in \mathbb{N} \right} \cup {0}$

[gentle osprey]

heavenly_lamb_61155

[sweet arrow]

so we have a point $p \in X$ right in an metric space X

[warm kettle]

I'm not sure if this shows us something useful in the finite case

[gentle osprey]

heavenly_lamb_61155

[warm kettle]

Maybe let's start with the case of a metric space

If x is in the closure of B_n, then there exists a sequence of elements of B_n that converges to x, right?

[sweet arrow]

i dont see why thats true

[warm kettle]

Oh, I see. It's a characterisation of the closure in metric spaces

Let's forget it then

[sweet arrow]

are we considering an arbitrary subset B_n or the B_n we have from the problem

[warm kettle]

From the problem

[sweet arrow]

I am beginning to learn real analysis for context

[warm kettle]

Try to prove it by contradiction. I'll start: let $x \in \overline{B_n}$. We want to prove that there exists an index $i \in { 1, \ldots, n }$ such that $x \in \overline{A_i}$. Assume, to the contrary, that for each $j \in { 1, \ldots, n }$ there exists a neighbourhood $N_j$ of $x$ such that $N_j \cap A_j = \varnothing$. We can assume, without loss of generality, that the sets $N_1, \ldots, N_n$ are open sets. [...]

[gentle osprey]

thingoln

[prime elbow]

I think it helps you if you prove that cl A u B = cl A u cl B

Then you can use induction

So cl A u cl B \subset cl ( A u B) is easy

Now take the contrapositive to show the converse part say x \not \in cl A u cl B then show x \not in cl(A u B).

[sweet arrow]

is this good

[sweet arrow]

also for part (b), the reasoning is almost the same right? the only part we have to change in the first inclusion is change (1 <= i <= n) to (i >= 1)

[warm kettle]

Yes

[warm kettle]

Looks good!

[sweet arrow]

This might be the most formal thing I have written xD

I was thinking for a couple of hours how to come up with it

[warm kettle]

The point where the finiteness of n matters here is that you can always choose a positive radius among r1, r2, ..., rn. If you had an infinite number of balls, the infimum could be equal to 0, which is why the countable case (b) doesn't hold

When you start studying general topology, you'll see that the corresponding proof for an arbitrary topological space will be pretty much the same

[sweet arrow]

yes I see

[warm kettle]

That's great. It always takes a loooot of time at the beginning, but it'll be better with time

🎉

[sweet arrow]

thank you for the help !

[warm kettle]

no worries, I'm glad it was helpful

[uneven bronze]

In some introductory text to topology, I read:

However, metric spaces are not sufficiently general to describe even some very classical modes of convergence, for example, pointwise convergence of functions on R.

Sorry if this is too basic, but what is it about pointwise convergence of functions on R that requires topology?

[tender halo]

C_p(X) (the space of real valued continuous functions on X) is too coarse to be metrizable

in particular, unless X is particularly pathological, C_p(X) is not first-countable at all, at any point

[uncut minnow]

Any convergence requires a topology. When we talk of metric spaces, we consider convergence under the topology induced by the metric. And as bussy beaver mentions, the space of real valued continuous functions (on good enough spaces) has a natural topology which is coarse enough to not be metrizable.

[heady skiff]

In a compact Hausdorff space can any two distinct points be separated by disjoint closed neighborhoods

[plush folio]

{x} and {y} are closed neighborhoods of x and y respectively

[unreal stratus]

Well neighbourhood of x almost always means it contains an open set containing x

[unreal stratus]

But yes like this should basically be true by picking disjoint open neighbourhoods, shrinking them and then taking closures (where the shrinking is so they remain disjoint afrer taking closures)

[plush folio]

Ah, good point 👍

[heady skiff]

Oh right duh

[heady skiff]

Mm I see

Thank you both

[heady skiff]

Can I have a hint in seeing that the U_f form a basis for the topology on X. Here, X is compact Hausdorff.

For example let $O$ be an open set of $X$. Then why must we have $O = \bigcup_{\alpha \in A} U_{f_{\alpha}}$? If $x \in O$, then we can just choose some constant function $f_x = 1$, and $f_x(x) = 1 \neq 0$. Thus $x \in U_{f_x}$. However, the other direction is not really clear to me

[gentle osprey]

okeyokay

[gaunt linden]

Do you know that a compact Hausdorff space is completely regular?

For each x in O you'll want a function such that f(x)=1 and f is 0 outside O.

[uneven bronze]

Consider the proposition that states that if $\mathcal E\subset\mathcal P(X)$, in order for $\mathcal E$ to be a base for a topology on $X$ it is necessary and sufficient for the following two conditions to be satisfied: \
a) each $x\in X$ is contained in some $V\in\mathcal E$;\
b) if $U,V\in\mathcal E$ and $x\in U\cap V$, there exists $W\in\mathcal E$ with $x\in W\subset (U\cap V)$.\

Now I'm reading about that if $\mathcal E\subset\mathcal P(X)$, the topology $\mathcal T(\mathcal E)$ generated by $\mathcal E$ consists of $\varnothing, X$ and all unions of finite intersections of members of $\mathcal E$. \

In the proof, the author claims that the family of finite intersections of sets in $\mathcal E$, together with $X$, satisfies the conditions of the proposition. Suppose $\mathcal E$ is a proper subset of the power set. Couldn't there be some $x\in X$ that is not contained in some finite intersections of $\mathcal E$?

[gentle osprey]

psie

[ruby delta]

we are defining X to be the topology generated by E

so, by definition, it can't contain any point outside of E

we're just saying that if a topology T is generated by a set E, then we can refine E a little bit more to make it into a basis for T

[tender halo]

also note that an empty intersection is the whole space

[ruby delta]

also note that you can intersect a set with itself once, so the set of all finite intersections contains E at least

[tender halo]

people usually define subbasis to be a cover but thats not strictly needed

basis needs to be a cover though

in the end its a question of niceness really

you can change definitions around in the trivial cases

nobody will mind

[uneven bronze]

right, but isn't this what one needs to prove? We are given X and an arbitrary subset of the power set P(X) and nothing more. I don't see why if say X = [0,1] and we pick [0,1/2] and [1/4,3/4] as our collection of subsets, how x = 0.9 would be contained in some subset of the finite intersections of [0,1/2], [1/4,3/4] and X.

[ruby delta]

you're confusing X and T(E)

X is some random set, and we're creating a collection of subsets E, and looking at the topology generated by E, which may or may not be X

it's exactly like how if we start with some arbitrary vector space V, and pick a random collection of vectors C in V, then we can refine C into a basis for span(C)

we dont know if span(C) is actually equal to V or not, and we don't really care

it's not a coincidence that they're both called a basis, by the way, it's exactly because they're essentially the same idea

[uneven bronze]

Hmm, ok. 😔 I still don't quite understand what I'm misunderstanding. I feel like my example with X = [0,1] and E = {[0,1/2],[1/4,3/4]} is a counterexample to finding a set in the finite intersections of E and X such that it contains x = 0.9

[ruby delta]

because T(E) is not X

T(E) is the discrete topology {∅, [0,1/2], [1/4,3/4], [0,1/2]u[1/4,3/4]}

[tender halo]

thats not discrete?

[ruby delta]

oh they have an intersection let me edit that

[tender halo]

its Kolmogorov quotient is discrete

[uneven bronze]

T(E) should be the union of the sets in {∅, [0,1/2], [1/4,3/4], [0,1/2]n[1/4,3/4], [0,1]}, no?

[ruby delta]

why would it contain [0,1], it doesn't

the biggest element is just [0,1/2]u[1/4,3/4]

and that's the (union of the) entire topology

[ruby delta]

otherwise I can embed [0,1] into basically any subset of R

so that would be the same as requiring that T(E) contain [0,1], [0,2], [0,3], [0,4] and so on

where does it end?

[uneven bronze]

This is what I'm reading, word by word:\

Prop. If $\mathcal E\subset \mathcal P(X)$, the topology $\mathcal T(\mathcal E)$ generated by $\mathcal E$ consists of $\varnothing, X$ and all unions of finite intersections of members of $\mathcal E$.\

You see, they say $\mathcal T(\mathcal E)$ consists, among other sets, of $X$. That is why I thought $[0,1]$ is in $\mathcal T(\mathcal E)$.

[gentle osprey]

psie

[ruby delta]

am I crazy? let me look this up

[uneven bronze]

I'll post a screenshot in a second or two

I have been stuck understanding how the family of finite intersections of sets in curly E in Prop. 4.4, together with X, satisfy the conditions in Prop. 4.3. I'd be extremely grateful for some concrete examples of sets V and W as in (a) and (b) of Prop. 4.3.

[ruby delta]

together with X

is the key

they're definiting it differently

[gaunt linden]

That doesn't satsify condition (a) in your definition.

Oh, your original post was using curly E for two different purposes.

[uneven bronze]

Ok. 👍 Yes, it was all a bit confusing, but I think I've made some progress. 😅 I'm still stuck on the last sentence of the proof of Prop. 4.4 though. Do you by any chance know why the topology is "obviously" contained in T(E) and why they hence equal?

I keep thinking, is E closed under finite intersections? I have a hard time accepting the statements if it isn't.

[gaunt linden]

No, but T(E) is closed under finite intersections because it is (by definition) a topology.

I suppose T(E) has been defined as something like the coarsest topology that has E as a subset, or at least a definition that lets you know that T(E) is some topology that has E as a subset.

[uneven bronze]

yes, indeed

[gaunt linden]

From there you can conclude that T(E) contains every finite intersection of elements of E.

And since it contains all those, it also contains all unions of those finite intersections (again by virtue of being a topology).

And of course T(E) contains Ø and X, since it is a topology.

The text of the proof seems to waffle a bit on whether it wants to treat X specially or just say it is an intersection of no sets (which makes it covered by the "finite intersections" part). Ah no, X is explicitly one of "such sets".

[uneven bronze]

Yeah, I feel there's a bit of an uncertainty in the proof, but the statement of the proposition makes it clear. Anyway, so the other topology of the proof of Prop. 4.4, the one that is proved to be equal to T(E) ... why, if it is contained in T(E), is it equal to T(E)? It's not clear to me that this other topology contains E, it just contains finite intersections of E.

[gaunt linden]

Because the first part of the proof implies that T(E) is contained in the unnamed prop 4.4 thing.

The unions-of-finite-intersections is argued to be a topology, and it obviously has E as a subset, so by definition of T(E), unions-of-finite-intersections has T(E) as a subset.

[uneven bronze]

I will make a confession. I still don't quite understand why the unions-of-finite-intersections topology has E as a subset. That topology has as a base finite intersections of E. Why it would contain all of E eludes me.

[gaunt linden]

Let A be an element of E. Then A is the intersection of just itself, and 1 is for sure a finite number of sets to take the intersection of. Next, A is the union of just itself too.

[uneven bronze]

Ok, thanks. 👍

[uneven bronze]

$T_4$: $X$ is a $T_1$ space, and for any disjoint closed sets $A,B$ in $X$ there are disjoint open sets $U,V$ with $A\subset U$ and $B\subset V$.\
$T_3$: $X$ is a $T_1$ space, and for any closed set $A\subset X$ and any $x\in A^c$, there are disjoint open sets $U,V$ with $x\in U$ and $A\subset V$.\

I'm struggling showing $T_4$ space is a $T_3$ space given the fact that $X$ is $T_1$ iff ${x}$ is closed for every $x\in X$. I choose $A={x}$ and $B$ to be any closed set that does not contain $x$ in the definition of $T_4$. Then these are disjoint closed sets in $X$, and so there are disjoint open sets $U,V$ such that $A\subset U$ and $B\subset V$. Now how does this show it is a $T_3$ space?

[gentle osprey]

psie

[prime elbow]

Since X is T1 space so {x} is closed set

[prime elbow]

Yes, take B = {x}

So A and B are disjoint, so by T4, there exists disjoint open sets U and V such that A\subset U and x \in V, which is exactly T3

[uneven bronze]

What is A now? 😅

[prime elbow]

A is given in your definition of T3

[uneven bronze]

Ok, so any closed set that does not contain x.

[prime elbow]

You have to show for any closed set A and x \in A^c, you have disjoint open sets U and V such that x \in U and A \in V.

[prime elbow]

Yes

So by using definition of T4, I am taking B = {x}, which is closed by T1, so we will get disjoint open sets U and V such that x\in U and A \subset V.

[uneven bronze]

I'm reading about the Zariski topology without going too much into the details. The author lets $k(X_1,\ldots,X_n)$ be the ring of polynomials in $n$ variables over the field $k$. Then also says that there is a one-to-one correspondence between the polynomials $P$ and polynomial functions $p$ if $k$ is infinite. The collection of all sets $p^{-1}({0})$ in $k^n$, as $p$ ranges over all polynomial maps is closed under finite unions. This is all clear. Then, however, he claims the collection of all sets of the form $\bigcap_{\alpha\in A}p_{\alpha}^{-1}({0})$ is the collection of closed sets for a topology on $k^n$. I don't follow this last sentence. Why are the closed sets of that form?

[gentle osprey]

psie

[rancid umbra]

are you asking why the closed sets satisfy the axioms for a topology?

or are you asking why choose those specific sets to be the closed sets?

[uneven bronze]

This one. Why are the closed sets given by that intersection?

Moreover, why do the collection of sets given by that intersection form a topology on k^n?

[tender halo]

its the topology generated by the closed-set base of zeroes of polynomials

the smallest topology that contains roots as closed sets

[rancid umbra]

you can also just check directly

that they are closed under finite unions and arbitrary intersections

[uneven bronze]

right, ok, this sounds familiar. How do you know the collection of sets that are roots to a given polynomial form a base?

[tender halo]

they are closed under unions

that implies that they comprise a closed set base

it is the same as their complements being closed under intersection

so their complements make a regular base

[uneven bronze]

ok, thanks. And why are the sets which are roots to polynomials closed?

[tender halo]

not sure how to answer that

we take the smallest topology in which they are closed

thats why they turn out to be closed

if you are asking for a moral reason we study this topology then i dont know

[rancid umbra]

the sets which are roots of polynomials?

[uneven bronze]

yes

[rancid umbra]

just preimages of zero?

[uneven bronze]

ah true, so you think it's a continuity thing

[rancid umbra]

no

how are you defining sets which are roots of polynomials

[uneven bronze]

well, if P is a polynomial, we take the associated polynomial function p and then simply define p^-1 ({0})

[rancid umbra]

this is just the intersection ∩ p^{-1}(0) over an indexing set with one index

so by def. it’s closed

[uneven bronze]

I don't understand I think. How do you know p^{-1}(0) is closed?

[rancid umbra]

because we declared it so

just like in the topology on the real line, we declare that the open sets are those which are unions of open intervals

in general you can take any collection of subsets and form the smallest topology in which those sets are open (or closed)

[uneven bronze]

hmm, but I only said that the collection of sets of the form p^{-1}(0) is closed under finite unions, not that p^{-1}(0) is closed itself

[tender halo]

you took the topology in which intersections of p_i^{-1}(0) are closed

naturally every individual set of roots is closed

[heady skiff]

I did not know that, but that's helpful lol, thanks

[heady skiff]

I tried looking up a proof of compact Hausdorff => completely regular, but I couldn't find anything, is the approach to show that compact Hausdorff => normal => completely regular?

[plush folio]

I haven't read the preceding discussion closely, but are you mixing up a set being topologically closed with a set being closed under an operation? These are very different things

[tender halo]

yea thats easier to prove

[uneven bronze]

Ok. I think I get some parts of what has been said. Correct me if I'm wrong, but we have a collection of sets C such that they are closed under finite unions. This collection contains the whole set k^n = X, since we can simply pick p=0.

Now let C^c be the set consisting of all complements of sets in C, together with the whole set X. This is closed under finite intersections and it satisfies the necessary and sufficient conditions of a base, namely that (a) each x in X is contained in some V in C^c ( we can take as V simply X) and (b) if U,V in C^c and x in U\cap V, there exists W in C^c with x in W \subset U\cap V (since U\cap V is in C^c, this is immediate).

So the topology with base C^c consists of unions of members of C^c, or by deMorgan, the closed sets are intersections of members of C.

[gentle osprey]

c squared

c squared

c squared

[uneven bronze]

thanks, I read through all of it

[rancid umbra]

typo in the last paragraph. intersections of unions of finite subsets of S

[gaunt linden]

That's the path I can google up too. The reason why I stated just the conclusion you need is merely that I don't myself feel confident to state whether it's the standard path or not ...

[uneven bronze]

Just want to check a basic claim. Is every collection of sets that is closed under finite intersections, contains the whole set X and the empty set, a base for a topology of X? According to my reasoning, I believe yes.

[rancid umbra]

yes. you can even apply some of the same free construction reasoning to this

btw, it doesn’t need to contain the whole set or the empty set. just needs to cover X

[uneven bronze]

yes, right

[rancid umbra]

the empty set comes for free as the empty intersection

[uneven bronze]

ok, didn't know. So the whole set can be obtained as a finite intersection of empty sets?

[rancid umbra]

no, the whole set can be obtained by the union of all of them. the empty set can be obtained by the intersection of none of them

zero is finite, after all

a finite intersection of empty sets is still empty

[uneven bronze]

ah ok, so what you meant was an intersection where the index set is empty, correct? 🙂

[unreal stratus]

Eh usually the empty intersection is undefined or is meant to be the whole set by convention

[rancid umbra]

oh shoot u right

[unreal stratus]

(Informally because it's the set of all x such that for all things in the empty set blah blah, so the condition is empty)

[rancid umbra]

it needs to contain the empty set then

[unreal stratus]

I don't think it does cause usually you just want every open to be a union of base elements and the empty set is the empty union

But this is just like technicalities at this point aha

[rancid umbra]

like, for a basis, you want it to cover the whole space and satisfy the downward closed condition tho. if two basis elements are disjoint, you want the empty set as a basis element, right?

[unreal stratus]

Wdym by downward closed

[viral wing]

I need a double-check: are locally connected spaces coreflective in Top?

[rancid umbra]

like if i take two basis elements B1 and B2, there is a B3 contained in their intersection

[unreal stratus]

Oh sure then ye

I guess I just operate with "every open is a union of basis elements" and this should be equivalent except for the empty set stuff

Well actually they assumed here that it is closed under finite intersections

So you wouldn't encounter such a problem

[rancid umbra]

true. okay. i guess i noticed that you can get the empty set and the whole set for free but for the wrong reasons lol

[uneven bronze]

I think you had the right idea except a typo. To summarize...

Every collection of sets that is closed under finite intersections and whose sets cover X is a base for a topology on X. The union of all elements should be the whole set X and the empty union should be the empty set.

In the same vein, a collection of sets that is closed under finite unions whose sets intersected is the empty set is a base for closed sets. The whole set is the empty intersection.

[barren grail]

I'm not sure what channel to ask this in, but I'm trying to prove that every countable ordinal is order isomorphic to a subset of Q which is closed in R. The hint in the textbook says to use transfinite recursion, but I can't figure out what to do at limit stages.

I've been trying find a way to break up my ordinal into "chunks" ordered in type omega then use the inductive hypothesis as a way to embed each chunk into the unit interval, then use (n, n+1) for the n-th "chunk", but I'm having trouble making this formal. I have a feeling it's because for limit ordinals, the corresponding subset will need to be unbounded, so I have to be careful about squeezing them into the unit interval in a way that keeps the whole thing closed. Does anyone have any further hints or tips?

[austere flare]

Seems like a better fit for #foundations

[uneven bronze]

Consider the Zariski topology. I'm trying to show if the field $k$ is infinite, then any two nonempty open sets $U,V$ have nonempty intersection. They have associated with them two families of nonzero polynomials $f_1,\ldots,f_m;g_1,\ldots,g_l\in k[x_1,...,x_n]$ such that $$f_i(a)=0,\forall a\not\in U,\forall i\qquad g_j(b)=0,\forall b\not\in V,\forall j.$$Now just so I understand this correctly, $U^c, V^c$ are finite sets, right? So not all $f_i$ vanish for all $a\in U^c$, correct? Instead I would say $f_i$ vanishes for some $a\in U^c$.

[gentle osprey]

psie

[gaunt linden]

Now just so I understand this correctly, $U^c, V^c$ are finite sets, right?
Not generally. For example if U if { (x,y) | x != 0 }, then U^c is the entire y-axis, which is infinite.

[uneven bronze]

Ah ok, so there are closed sets that are not finite. I totally forgot.

Though every finite set is closed in the Zariski topology.

[gaunt linden]

As long as we're talking about k^n, yes.

You can assume without loss of generality that m = l = 1. (Can you see why?)

[uneven bronze]

Hmm, not really. Why can we do that?

[gaunt linden]

Suppose you want to show that $$(U_1 \cup U_2 \cup \cdots \cup U_m) \cap (V_1 \cup V_2 \cup \cdots \cup V_l)$$ is nonempty. Now if only you had a way to show that $$U_1 \cap V_1$$ is nonempty, then ...

[gentle osprey]

Troposphere

[gaunt linden]

(where each of the U_i and V_j is the nonzeroset of a single polynomial)

[uneven bronze]

Ok, I think I understand. 👍 And just to be certain, these U_i and V_j might be infinite?

[gaunt linden]

Yes. (In fact they have to be).

[uneven bronze]

but here, f_2(x) doesn't equal 0 for x in say U_1, or? I'm just trying to check my understanding of the problem. So f_i(x) does not equal zero for all x in U, for all i. Rather, for each i, f_i(x)=0 for some x in U.

I'm not saying you wrote something incorrect.

[gaunt linden]

That seems to be a very roundabout way of thinking of it.
I'd rather say U1 = { x | f1(x) != 0 }, U2 = { x | f2(x) != 0 }, ...

[uneven bronze]

yes, very roundabout way indeed 😅

[gaunt linden]

And phrasing things as for each i (something holds) for some x is in itself confusing -- having quantifiers both before and after the statement obscures whether x is allowed to depend on i or not.

[uneven bronze]

yes, I agree

[uneven bronze]

a final question maybe; if U is open and U^c is infinite, how do we know there are only finitely many polynomials associated with U (or U^c)?

[gaunt linden]

We don't, in general. (It turns out that the Zariski topology is compact, so we only need finitely many), but you don't actually need to care about how many for this proof.

[uneven bronze]

Hmm, I wonder then where one uses the assumption that k is infinite...

[gaunt linden]

The solution I can see contains a step that says "a nonzero polynomial in one variable over a field has only finitely many zeroes, so there must be some values left over where this one is nonzero".

[uneven bronze]

that's a very puzzling hint 🥲 but thanks, the problem is a bit clearer

[unreal stratus]

What are these U,V open sets of

[uneven bronze]

ah yes, I was sloppy. They are open sets of k^n, where k is the field.

[gaunt linden]

Zariski topology on k^n.

[unreal stratus]

Sure

[gaunt linden]

It's also possible to get through without getting into that level of detail if you have a theorem such as "if K is infinite, then the only polynomial in K[x,y,...,z] that evaluates to 0 at all points is the zero polynomial".

[tribal palm]

compact unless you’re french 😌

or over a finite field

wait no over a finite field is also compact but in the stronger but in the french sense

ok i cant sensentences and should sleep bye

[alpine hound]

If X is a topological compact space, and ~ is an equivalence relation on X, does it mean that X/~ is also a compact space with the quotient topology?

(I specifically want to know about S^n and RP^n -- which are Hausdorff)

I know that S^n is Hausdorff and compact. and RP^n is obv Hausdorff but why is it compact?

[rancid umbra]

yes

continuous maps preserve compactness

[alpine hound]

explanation?

[alpine hound]

oh fck

thx

[uneven bronze]

Here's my attempt at a solution. Let $X=k^n$. Recall we are trying to prove that if $U,V$ are nonempty open sets, then $U\cap V\neq \varnothing$. If $U$ or $V$ are equal to $X$, then the result is immediate. So suppose they are both proper subsets and for contradiction, that $X=U^c\cup V^c$. Since $U^c$ is closed, it is the (possibly arbitrary) intersection of zero sets of polynomials. So we know there is some nonconstant polynomial $f$ vanishing on $U^c$. Likewise, there's some nonconstant $g$ vanishing on $V^c$. Now $h=fg$ is also nonconstant and vanishes on $U^c\cup V^c=X$.

[gentle osprey]

psie

psie

[uneven bronze]

The only thing I'm a bit unsure about is whether or not f and g vanish on all of U^c and V^c respectively?

I'm also a bit unsure if the product h really is nonconstant or not.

[gaunt linden]

You don't really need the "for contradiction assume that X = U^c cup V^c".
Just go on to derive that h is nonzero and vanishes on U^c cup V^c.
Then prove in general (e.g. as in your second image) that a nonzero polynomial must have a point where it doesn't vanish; that point cannot be in U^c cup V^c, so it is in U cap V, q.e.d.

[gaunt linden]

The product of nonzero polynomials over an integral domain is nonzero.

(It would be okay if h (or f or g) were nonzero constant; that just means that the respective open set is all of X, which is a happy conclusion here).

[gaunt linden]

U is the union of a bunch of basic open sets. If you ignore all but one of them, the U gets smaller, so its complement gets larger. Thus you're looking at an f that vanishes on the now larger complement -- in particular it also vanishes on the complement on the original U.

[uneven bronze]

Makes sense.

[gaunt linden]

You'll need to appeal to induction to conclude that (a2,...,an) exist -- the claim that a nonzero polynomial in several variables has a point where it doesn't vanish is what you're proving here.

[uneven bronze]

Hmm, couldn't we just say that if (a2,...,an) didn't exist, or for that matter any (n-1)-tuple, then h would be a constant or the zero polynomial since it wouldn't have a nonzero term with a nonzero power?

[gaunt linden]

That is true, but it is the same thing you're proving about h.

FWIW, my approach that would avoid the induction would be something like:

Because U is nonempty, there is some fi and p such that fi(p) != 0.
Because V is nonempty, there is some gj and q such that gj(q) != 0.
Now define s1(t) = fi((1-t)p+tq) and s2(x) = gj((1-t)p+tq).
Each of s1 and s2 is nonzero because s1(0) = fi(p) != 0 and s2(1) = gj(q) != 0.
Since they are single-variable polynomials, s1 has only finitely many roots and s2 also has only finitely many roots, so there must be a t0 that is a root in neither of them.
Then both fi and gj are nonzero at (1-t0)p+t0q, so that point is in U cap V.

But yours (with the induction) is more likely to be the intended solution.

[little vine]

is every set compact relative to itself?

[tender halo]

by set do you mean space?

compactness is not a relative property

[little vine]

Yes

[tender halo]

then no, not all spaces are compact

[iron bolt]

might be a good exercise to find out when exactly a discrete space is compact

[prime elbow]

|| when underlying set is finite ||

[uneven bronze]

Consider the following exercise:\

Exercise: Every metric space is normal. (If $A,B$ are closed sets in the metric space $(X,\rho)$, consider the sets of points $x$ where $\rho(x,A)<\rho(x,B)$ or $\rho(x,A)>\rho(x,B)$.)\

I have a couple of questions about this exercise. First, $A,B$ are disjoint closed sets, right? Second, in the proof of this, we consider the sets ${x\in X: \rho(x,A)<\rho(x,B)}$ and ${x\in X: \rho(x,A)>\rho(x,B)}$, which contain $A,B$ respectively (here $\rho$ is the metric on the metric space $X$). Why does e.g. ${x\in X: \rho(x,A)<\rho(x,B)}$ contain $A$?

[gentle osprey]

psie

[solemn iris]

what are they doing with the b-hat?

aren't you supposed to show bijective

[hidden abyss]

Yes, A and B are assumed to be disjoint.

For your second question: suppose x ∈ A. We want to prove ρ(x,B)>ρ(x,A)=0. Since A and B are disjoint, x is not in B. Since B is closed, there is some epsilon-ball U(x, ε) which contains x but doesn't intersect B. This means that ρ(x,B)≥ε

[uneven bronze]

Thank you, makes sense. 🙂

[hidden abyss]

You can show bijectivity by giving an explicit inverse map

[solemn iris]

oh i see so that's what they're doing. the notation's a bit new for me T.T thanks!

[rotund halo]

I'm trying to understand this proof that a covering map with finite fibers is closed. Take $p:C\to X$ to be a covering map with finite fibers. Fix $A\subseteq C$ closed. Let $x\in C-p[A]$. By our finite fiber assumption we can write $p^{-1}[{x}]={x_1 , ... , x_n }$. As $p$ is a covering map for each $i$ we can take a nbhd $U_i$ of $x_i$ so that $U_i \cap A = \emptyset$ and $p$ restricted to $U_i$ is a homeomorphism onto $p[U_i ]$. We can also take our $U_i$ disjoint. Now, the claim is $U=\cap_{i=1}^n p[U_i ]$ is a neighborhood of $x$ disjoint from $p[A]$. Everything so far to me is kosher except I don't see why disjointness holds?

[gentle osprey]

DootDooter

[rotund halo]

If I take $y\in U \cap p[A]$ for example, then I can let $y=p(a)$ for some $a\in A$ and then by assumption I have $a\in p^{-1}[{y}]={b_1 , ..., b_k}$. I also get $y=p(u_i)$ for some $u_i\in U_i$. So, then I get ${u_1,...,u_n}\subseteq{b_1,...,b_k}$.

[gentle osprey]

DootDooter

[rotund halo]

And a has to be one of the b_i from earlier too.

But I don't see how this leads to an issue?

[hidden abyss]

Each of the u_is are distinct and different from a right? So we would have n+1 distinct elements of {b1,....,bn}

[rotund halo]

Yes I believe so. It contains at least all the u_i and a as well. I used k for the b_i since we don't know how many elements lie in that set without more info.

[fringe thorn]

didn't you already write that the number of elements in each fiber is n, though?

By our finite fiber assumption we can write p^{-1}[{x}]={x_1 , ... , x_n}

y is also a singleton, so why would its preimage have a different number of elements?

[rotund halo]

No, I wrote that for the fiber of x.

There is a later exercise that I think proves that the fibers have to all have the same number of elements?

I do not know if invoking that leads to circularity though?

[fringe thorn]

ahh, I see

[rotund halo]

The proof is done if it's not circular (provided I prove that part).

[fringe thorn]

I'm super rusty with covering maps, so I'll let somebody else answer you

[rotund halo]

Ah I think I would need connectedness in the codomain for this actually.

[hidden abyss]

Oh right, I'm too used to assuming locally+globally path connected

[tribal palm]

https://arxiv.org/abs/math/0609098

“the complete feather” constructed here is really funny

an example of a homogenous non-Hausdorff 1-manifold

[uneven bronze]

Is it correct that if in the definition of a normal and regular space they are assumed to be T1 spaces, that not regular does imply not normal? I'm working this exercise to show that a topology on R is Hausdorff but not regular, yet contains the Euclidean topology, i.e. a normal topology, and the author says "This exercise shows that a topology stronger than a normal topology need not be normal or even regular."

[deft crow]

I have a conjecture that I can't find any writings on so it might just be straight up false, but I think it's true, at least intuitively. Suppose $E$ is a finite dimension normed vector space over $\mathbb{R}$, $K$ be a compact convex subset of $E$ and $\Omega\subset K$ such that $\overline{\Omega\cap\partial K}=\partial K$. Then the convex hull of $\Omega$ is $K$. Any ideas on how to show this? I've quite easily shown that the convex hull of $\partial K$ is $K$, but that's not anywhere near the result I'm looking to show.

[tender halo]

yes, normal T1 spaces are regular

[gentle osprey]

Dealersgrip

[tender halo]

T1-T6 properties have strictly increasing strenth

[uneven bronze]

oh wow, there is even T5 and T6, my book only considers up to T4

[alpine nest]

There's also T2.5 and T3.5

https://en.wikipedia.org/wiki/Separation_axiom#Main_definitions

I quite like point set topology, but it's sometimes hard to defend.

[ruby delta]

absolutely insane, why would anyone want to investigate the minutia of how to separate two points in a space. they mustve been really bored back then

[tender halo]

i mean its pretty motivated i'd say

T2.5 is meh

never seen it out in the wild

regularity is the worst one

its almost useless, all nice theorems use tychonoffness

[tender halo]

i think its a bit misleading to call them "minutia of how to separate points", they capture much more interesting properties of the space
T0 spaces are ones where there are no duplicated points
T1 spaces are spaces where intersection of all open set containing x is x
T2 is the same as T1 but with closures of open sets
T3 is space having a base of regular open sets
T3.5 is the space being fully described by the continuous function on itself/having a base of cozero sets
T4 is "any pair of disjoint closed sets has a pair disjoint zero sets respectively containing them"
T5 idk i dont have anything for that one tbh its more of an oddity i guess
T6 is every open set being cozero/closed set being zero

also normality is in a certain sense a finiteness/compactness condition, about whether the closed sets in the space can be differentiated in finite time

[alpine nest]

Yeah, also I think the motivation wasn't so much "in what various manners can points be separated", but rather "If my space was T4, I could do <thing>, but this space is only T3. On the other hand it does have some extra properties, so maybe with those I can do <thing>"

And if this kind of "T3 with some extra properties" were used often enough, it eventually merited a name of its own.

[uneven bronze]

Let $\mathcal E={(a,b]:-\infty<a<b<\infty}$. I want to show this is a base for a topology on $\mathbb R$. Clearly $\mathbb{R}=\bigcup_{n=1}^\infty(-n,n]$. Let $a,b,c,d\in\mathbb R$ and suppose that $a<b$ and $c<d$. Then $(a,b]\cap(c,d]=\varnothing$ or $\max{a,c}<\min{b,d}$, in which case $(a,b]\cap(c,d]=(\max{a,c},\min{b,d}]$. \

Now, what I've shown doesn't show yet that $\mathcal{E}$ is a base for a topology, right? If it only contained the empty set, then it would be a $\pi$-system that covers $\mathbb R$ and hence a base for a topology, but it doesn't contain the empty set, right? What else do I have to show?

[gentle osprey]

psie

[tender halo]

no, you've shown that it is a base for a topology

any pi-system that covers X is a base

[uneven bronze]

indeed, but E does not contain the empty set, unless I'm overlooking something obvious?

[tender halo]

empty union of members of a base is an empty set

ur getting really hung up on the whole "does it contain the whole set/the empty set" thing

thats minutia

you just add it if it doesnt have it

its not that big of a deal

[uneven bronze]

Ok, yes. E is not a pi-system, but if we add the empty set, it is. And we are all good.

[uneven bronze]

although your statement is true, we could weaken it a tiny bit by saying any collection that is closed under finite intersections that are nonempty is a base for a topology, or? As you said, the empty set comes for free as the empty union eventually. Such a collection would not be a pi-system, yet be a base nevertheless.

And covering the whole set, of course.

[tender halo]

hm? being closed under finite intersection is what being a pi system is, no?

[uneven bronze]

yes, but E above is not closed under finite intersections yet a base for a topology. It is closed under finite intersections that are nonempty only.

[tender halo]

ah i missed the nonempty part

i mean yea

that works

[solemn iris]

ik it's an illustration but i'm not getting the idea why the pancakes are of the same size as U

[ruby delta]

because V_a is homeomorphic to U

not strictly the "same size" per say, but the "same shape"

[solemn iris]

oh i get it, but i kinda imagined it as U being chopped into a few pieces and the restriction of each of these pieces is sent to U. therefore the pancakes are like little chopped pieces of U.

[ruby delta]

if you want a concrete example to work with, you can consider the map from R to S^1 via x -> e^{2pi i * x}

then you're cutting R into intervals [n, n+1) and they spiral onto the circle

[solemn iris]

oo okay i'll try doing that

thanks hchan

[ornate blaze]

in R², can i have a bijection between an open ball and a closed ball? sorry if this isn't the right channel for this question.

[untold lily]

continuous bijection? or just a bijection?

[ornate blaze]

continuous

[austere flare]

No; the preimage of a closed set under a continuous function has to be closed

[untold lily]

a closed ball is compact, an open ball is hausdorff

any continuous bijection from a compact space to a hausdorff space must be a homeomorphism, so I know you can't do this from closed to open

[ornate blaze]

thank you guys, that helps a lot

[untold lily]

wait I'm confused, how does this help?

[austere flare]

That shows that you can't have a cts bijection from open to closed

[untold lily]

why not

open ball is still closed in its subspace topology?

[austere flare]

Ah, good point

[untold lily]

https://math.stackexchange.com/questions/42321/continuous-bijection-from-open-n-ball-to-closed-n-ball

so open to closed also seems to be no, but uses brouwer's theorem apparently (so far from trivial)

[unreal stratus]

Yeah invariance of domain is the natural way to me

[ornate blaze]

i actually just wanted to prove that ${(x,y,z)\in\mathbb{R}^3;z=0$ and $x^2+y^2\leq 1}$ is not a regular surface. i'm sure this proves but idk if there's an easier way

[gentle osprey]

jango

[tardy carbon]

It's more useful to prove that it's a regular surface with boundary, then prove that a regular surface with boundary is a regular surface if and only if it's boundary is empty

but the method of proof would be similar

[plush folio]

I don't quite see how you prove this using the fact that there is no continuous bijection from an open ball to a closed ball. Can you show the whole proof?

[ornate blaze]

i'll try that, thanks

[ornate blaze]

i still haven't written it but, like, i was trying to show that there's no homeomorphic function between an open set and a neighbourhood of a boundary point of the closed ball

come to think of it, i guess the intersection between the closed ball and a neighbourhood of a boundary point would neither be closed or open so i gotta consider that

[rancid umbra]

did you ever figure this out?

[warm hedge]

whats the difference between standard borel space and a polish space ?

[uneven bronze]

If $X$ is a linearly ordered set, the topology $\mathcal T$ generated by the sets ${x:x<a}$ and ${x:x>a}$ $(a\in X)$ is called the order topology.\

Exercise: If $a,b\in X$ and $a<b$, there exist $U,V\in\mathcal T$ with $a\in U,b\in V$, and $x<y$ for all $x\in U$ and $y\in V$. The order topology is the weakest topology with this property. \

Attempt: obviously $a\in{x:x<b}=U$ and $b\in {x:x>a}=V$. Now how do I show that $x<y$ for all $x\in U$ and $y\in V$? Moreover, what do I have to do in order to show this is the weakest topology with this property?

[gentle osprey]

psie

[hidden abyss]

Your definition of U and V doesn't work: if c lies between a and b, then c will be an element of both U and V

And U and V need to be disjoint in order for the x<y property to hold

[hidden abyss]

Maybe you could use c to construct U and V (if such a c exists)

To show that it's the weakest such topology, you need to show that all sets of the form {x: x<a} and {x: x>a} are open in any topology with the property

[uneven bronze]

Ok, thank you. If there is a c inbetween a and b, then we have b in {x: x>c} and a in {x: x<c} and the x<y property holds, right?

[uneven bronze]

regarding the second part of the exercise, we want to show for any $\tau$ that satisfies the property above, that $\tau\supset\mathcal T$. Fix $a\in X$. For each $x<a$ there are $U_x,V_x\in\tau$ such that $x\in U_x$, $a\in V_x$, and $y<z$ whenever $y\in U_x$ and $z\in V_x$. In particular, $y<a$ for every $y\in U_x$. Now the union of all the $U_x$ is ${x:x<a}$, right? What about the union of all the $V_x$; do they have any significance?

[gentle osprey]

psie

[hidden abyss]

Yeah that works. The union of V_xs isnt relevant, you just need them for y<a

[rotund halo]

No. I've still got no idea about disjointness bit.

[warm hedge]

whats the deference bewteen a polish space and a standard borel space ?

like saying (X,B) is a standard borel space means we dont care about which topology we have in X but we just care that we can make B a borel s-algebra for a polish topology on X ? So for example R with the usual borel family is a standard borel space but it could have the discrete topology (which isnt polish )?

[deft crow]

ANy ideas...?

[tardy carbon]

this is false. Let $K$ be the unit square in $\mathbb{R}^2$, and $\Omega$ be the boundary of $K$ with the four corners removed.

[gentle osprey]

L

[deft crow]

terribly sad

[worn mortar]

A standard Borel space is a measurable space isomorphic to a Borel subset of a Polish space. In particular, a metrizable space $X$ is standard Borel if $(X, \mathcal{B}_X)$ is standard Borel.

[gentle osprey]

calebuic魏凯布

[arctic mural]

In step 2, is the set $V_x$ the product of the two balls of radius $\varepsilon /3$ with centers $g(x)$ and $g(x_0)$, respectively, times many $Y$ set copies to fill $Y^X$ right?

[gentle osprey]

pingolfingo

[arctic mural]

btw book is Munkres

[uneven bronze]

The author of my book says that a base for the product topology is given by sets of the form $\bigcap_1^n\pi_{\alpha_j}^{-1}(U_{\alpha_j})$ where $n\in\mathbb N$ and $U_{\alpha_j}$ is open in $X_{\alpha_j}$ for $1\leq j\leq n$. Then goes on to say that these sets can also be written as $\prod_{\alpha\in A}U_\alpha$ where $U_\alpha=X_\alpha$ if $\alpha\neq\alpha_1,\ldots,\alpha_n$. Why can the sets be written in two ways?

[gentle osprey]

psie

[tender halo]

they are uhh equal

if you compute what the first thing gives you get the second thing

[uneven bronze]

hmm, but what is $U_\alpha$ if $\alpha=\alpha_1$ say?

[gentle osprey]

psie

[uneven bronze]

In the product, that is.

[tender halo]

some open subset of X_\alpha_1

[uneven bronze]

In my book, it says that if $X$ is any set and ${f_\alpha:X\to Y_\alpha}{\alpha\in A}$ is a family of maps from $X$ into some topological space $Y\alpha$, there is a unique weakest topology $\mathcal T$ on $X$ that makes all the $f_\alpha$ continuous. How do I show this topology is unique?

[gentle osprey]

psie

psie

[ruby delta]

well since you know what that topology looks like

you can show that the topology embeds into any other such topology

[uneven bronze]

yeah, perhaps one could simply say, because it is the weakest such topology, it is unique

[ruby delta]

sure, if you want to

[unreal stratus]

well it's the fact that like intersection of topologies is a topology ig

as in, there is indeed "a" weakest rather than idk many minimal ones

[rancid umbra]

what do you think of this:

let $p : C \to X$ be a covering map with finite fibers, $A \subseteq C$ closed.
let $W$ be an evenly covered neighborhood of some $x_0 \in X - f(A)$, say $p^{-1}(W) = \bigcup_{i = 1}^n U_i$. Since $C$ is closed, we can safely replace each $U_i$ with $U_i \cap (C - A)$ to ensure that $U_i \cap A = \emptyset$.$\newline$

Put $U = \bigcap_{i = 1}^n p(U_i)$ and note that $U \subseteq W$. Set $V_i = U_i \cap p^{-1}(U)$. Then the $V_i$'s are pair-wise disjoint and
$$\bigcup_{i = 1}^nV_i = \bigcup_{i = 1}^n(U_i \cap p^{-1}(U)) = p^{-1}(U) \cap \bigcup_{i = 1}^n U_i = p^{-1}(U) \cap p^{-1}(W) = p^{-1}(U),$$

Now $p^{-1}(U) \cap A =\emptyset$ since each $V_i$ is disjoint from $A$. For $x \in U$, we have
$p^{-1}(x) = {x_1,\dots,x_n}$ with $x_i \in V_i$ for each $i$, and so $p(A)$ and $U$ must be disjoint. Thus, $U$ is an open neighborhood of $x_0$ contained in $X - p(A)$.

[gentle osprey]

c squared

[cunning silo]

well man

my honest opinion:

it is a clear and rigorous demonstration

[rancid umbra]

meant to say, "since A is closed, we can safely replace each ..."

[rancid umbra]

$\pi_{\alpha_0}^{-1}(U) = {x \in X : \pi_{\alpha_0}(x) \in U} = \prod_{\alpha}U_{\alpha}$ where $U_{\alpha} = X_{\alpha}$ if $\alpha \neq \alpha_0$

[gentle osprey]

c squared

[uneven bronze]

👍

[rotund halo]

That makes sense. Thanks for the help!

[tardy carbon]

if you have two topologies, each of which is weaker than the other, than they are the same. So you only need to show that there is a weakest such topology. That's obvious as you just take the smallest topology containing all $f_{\alpha}^{-1}(O)$ for $\alpha \in A$ and $O$ open in $Y_{\alpha}$.

[gentle osprey]

L

[rancid umbra]

the smallest being the intersection of all topologies containing said sets

in general, the smallest topology such that some property holds is the intersection of all topologies having that property

(being informal here)

[untold lily]

are you being informal though?

I guess depends on the property, because some won't survive under intersection

[rancid umbra]

thats what im getting at

i don't know the exact characterizing properties that make this work, but i have found that most of the time, it does work

[amber musk]

Let $\mathcal{T}$ be a topology and $V$ a neighborhood of $x$, then i have to show that $V \subset W \Rightarrow W \in \mathcal{V}(x)$.

As $V \in \mathcal{V}(x), \ \exists U \in \mathcal{T} ; \ x \in U, \ U \subset V$.
Now $V \subset W \Rightarrow \exists U \in \mathcal{T} ; \ x \in U , U \subset V \subset W$ so $W \in \mathcal{V}(x)$.

Are we allowed to use the same U for W?

[gentle osprey]

tm

[ruby delta]

What is V(x) here?

[amber musk]

the set of neighborhoods of x

[ruby delta]

Why isn’t it immediately true

[tardy carbon]

While correct, the intersection characterization doesn't really help with proofs

[amber musk]

yh it’s trivial but i’m doing every proofs in the book i’m reading soo

[ruby delta]

Also, why can you assume that such a U necessarily exists?

[amber musk]

because V is a neighborhood of x

[ruby delta]

Oh, that definition ok

Sure this works then and yes in this case it’s not 100% trivial

[amber musk]

my definition of a neighborhood is $V \in \mathcal{V}(x)$ if $\exists U \in \mathcal{T} ; \ x \in U, \ U \subset V$

[gentle osprey]

tm

[amber musk]

V contains an open set which contains the point x

[ruby delta]

Yeah I know and your proof is fine

[amber musk]

alr thanks

[rancid umbra]

i feel like that depends on the statement you are trying to prove and the nature of the proof haha

[uneven bronze]

Is there a simple, layman example of a regular space $X$ (regular implies Hausdorff in my book) such that all functions in $C(X)$ are constants, where $C(X)$ are the (real- or complex-valued) continuous functions with domain $X$?

[gentle osprey]

psie

[untold lily]

I'm confused about one thing

If X has the weak topology induced by maps f_n : X -> Y_n (countable) and each Y_n is second countable, is it not true that X is second countable as well?

[tardy carbon]

$X = {0}$

[gentle osprey]

L

[uneven bronze]

But is this T1? Is it Hausdorff? Recall a set is T1 iff {x} is closed for every x in X.

[untold lily]

entire space is closed

it's hausdorff because there are no 2 distinct pts to separate

[uneven bronze]

ok, makes sense, so it is vacuously regular. Cool, thank you!

[tardy carbon]

it's a metric space

[untold lily]

to expand on this, here's a book making this argument essentially (maybe this should have gone to #advanced-analysis but here should be ok)

But I know that Frechet spaces have weak topologies induced by countably many maps to R, and yet they are not necessarily second countable, far from it

[tender halo]

i dont think there are any sensible nontrivial examples sadly

[uneven bronze]

ok, yeah, I should have added that I'm interested in spaces that do not have the trivial topology. But a single point set has the trivial topology if I'm not mistaken.

[tardy carbon]

If $X$ is a metric space, it's easy to show that if all functions in $C(X)$ are constant, then $X$ is either empty or a singleton.

[gentle osprey]

L

[uneven bronze]

Ok.

[tender halo]

you only need tychonofness

idk the problem is also that the space needs to be regular but not tychonoff and that is already a little complicated

if you just asked for Hausdorffness, that would be easier, because any connected countable hausdorff space has only constant functions

[unreal stratus]

maybe silly q but how do you do this for general tychonoff spaces?

oh nvm i just misremembered what a tychonoff space is

[tender halo]

lol

[unreal stratus]

i think my pointset topology just gets worse over time

[tender halo]

as does the field as a whole

[alpine nest]

Less and less point

[untold lily]

I think I realized the problem: I don't believe it's technically true that topologies on vector spaces generated by semi norms are initial for spaces, one has to consider the continuity of vector space operations as well

so it is the initial topology for a topological vector space, not initial for an arbitrary space

[uneven bronze]

Consider the proof of Urysohn's lemma. In the proof, one considers a normal space $X$ and any two disjoint closed sets $A,B$. Moreover, from a previous lemma, there are these open sets $U_r$ such that $A\subset U_r\subset B^c$, where $r$ is a dyadic rational number in $(0,1)$ and $U_1=X$. \

For $x\in X$, the author then defines $f\left(x\right)=\inf \left{r:x\in U_r\right}$ and claims clearly that $f(x)=0$ for $x\in A$ and $f(x)=1$ in $x\in B$. I'm wondering, why is $f(x)=1$ for $x\in B$?

[gentle osprey]

psie

[uneven bronze]

It's like we are taking the infimum over the empty set, is that correct?

Oh wait, I'm totally missing that U1 = X. Silly. So if x is in B, then f(x) has to evaluate to 1 since we basically get inf{1}.

[uneven bronze]

I'm confused about a certain part of a proof. Let $f:X\to [0,1]$ (in the proof I'm reading about Urysohn's lemma, this is the continuous function constructed, where $X$ is normal). To show $f$ is continuous, the author shows that $f^{-1}((-\infty,\alpha))$ and $f^{-1}((\beta,\infty))$ are open sets. Then says that $f$ is continuous since the half-open lines are generators of the standard topology on $\mathbb R$.\

Now, since $[0,1]$ is a subspace of $\mathbb R$, don't we have to show that $f^{-1}(V)$ is open for every open set $V$ in $[0,1]$?

[gentle osprey]

psie

[tender halo]

if Z is a subspace of Y, and f is a continuous function from X to Y whose image lies inside of Z, then f is also a continuous function from X to Z

[rancid umbra]

this is the universal property of the subspace topology

[uneven bronze]

Ah ok, so I guess you are taking Y = R and Z = [0,1]. Does this have a name this result?

Ok, the universal property.

[rancid umbra]

yes, of the subspace topology

it’s actually stronger than what bussy beaver stated

f : X —> Z is continuous if and only if i o f : X —> Y is

where i is the inclusion of Z into Y

[uneven bronze]

interesting, is the proof straightforward? 🙂

[rancid umbra]

yes, you should try it. just give Z the subspace topology

[uneven bronze]

ok 👍

[radiant stone]

ah

f

i wrote one wrong

[rancid umbra]

it can also be shown that the subspace topology is the unique topology on Z for which this property holds whenever i is a continuous injection

this is a special case of the more general notion of initial and final topologies

[gentle osprey]

Spec(𝔽₂[x₁,…,xₙ]) Enjoyer

[quick crane]

$V = [0, 1] \cap W$ for some $W$ open in $\mathbb{R}$ and $f^{-1}(V) = f^{-1}(W)$.

[gentle osprey]

L

[unreal stratus]

Lol composition

[uneven bronze]

There's this corollary to Tietze extension theorem:\

Corollary: If $X$ is normal, $A\subset X$ is closed, and $f\in C(A)$, there exists $F\in C(X)$ such that $F|A=f$.\

Proof: Let $g=f/(1+|f|)$. Then $g\in C(A,(-1,1))$, so there exists $G\in C(X,[-1,1])$ with $G |A=g$. And the proof continues. \

My question is; why is there such a $G$? It can't be by Tietze because that theorem says that $G\in C(X,(-1,1))$. Note, I used the open interval $(-1,1)$ instead of the closed $[-1,1]$. Any help is appreciated.

[gentle osprey]

psie

[alpine nest]

If B is a subspace of A, and g: X-> B is continuous, then g can also be understood as a continuous function from X into A

If we define a function as a subset of the cartesian product, then it will be the exact same set of pairs except in a different cartesian product.

f(x) = x^2 is a continuous function from R into [0, infnity), but it is also a continuous function from R into R

[uneven bronze]

makes sense, thanks

[alpine nest]

Note that this also works in the other direction, i.e. if f: X -> A is continuous, and B is a subset of A which contains f(X), then f can also be thought of as a function from X into B, and it will be continuous with respect to the relative topology.

Basically what you do with the codomain outside of f(X) shouldn't matter much

[uneven bronze]

ah ok, good to know

[tender halo]

whats your statement of tietze?

because tietze is about bounded functions, so it should be about functions from X to I, not X to R

[untold lily]

Tietze is more general than that

hm, the corollary above is generally seen as part of tietze

at least it is in willard, and it also seems to be written that way in the wiki

[tender halo]

wiki is wrong for some reason

the proof is for bounded functions

[untold lily]

By "the proof", do you mean Tietze's proof?

it's not uncommon to incorporate an easy corollary into the statement

[tender halo]

no i mean the proof listed on the page for "tietze extention theorem"

[untold lily]

ohhh

right

[tender halo]

i guess its just semantics

to me tietze is the statement "if any two completely separated subsets of A are completly separated in X, then A is C*-embedded in X"

and then there is like an extension that "if A is C*-embedded and completely separated from any zero-sets disjoint from it, then A is C-embedded"

[rotund halo]

I thought about this more today. When you say "Since C is closed, we can safely replace each U_i with U_i cap (C-A)." Why is this something we can do? I see we can intersect the finite union of U_i's by C-A, but how do we know the union will still be the preimage of W if we do this?

[rancid umbra]

i’m just shrinking the U_i’s, which already make up the preimage of W

[rotund halo]

Yeah I see that part. But don't you need the union to still work out to the preimage of W after for this to work?

[rancid umbra]

yea, i do. just realized before you sent lol

how were you making the U_i’s disjoint from A in your attempt?

this should be an easy fix i think

[rotund halo]

I was picking elements of the fiber individually iirc

Actually rereading this I might be being stupid about covering spaces.

Well no okay I'm fine with that I guess. But I think I'm not doing a restriction different from you.

[rancid umbra]

the only part where i use p^-1(W) is for equality with the union of the V_i’s.

do i need it there?

we just need p^-1(U) contained in the union of the U_i’s

now i’m gonna be stuck on it some more lol

i’ll let you know if i get anything

[rotund halo]

I'll keep thinking about it too. It's a goofy problem.

[rancid umbra]

fr

[rotund halo]

Thanks for your help either way

[rancid umbra]

okay, revised version:

Let $W$ be an evenly covered neighborhood of $x_0 \in X - p(A)$ by sheets $U_1,\dots,U_n$. Put $U = \bigcap_{i = 1}^np(U_i \cap (C - A)) \subseteq W$. Note that $x_0 \in U$ since $p^{-1}(x_0) \cap A = \emptyset$. $\newline$

First, observe that $p^{-1}(U) \cap A = \emptyset$: for $x \in p^{-1}(U)$, we have $p^{-1}(p(x)) = {x_1,\dots,x_n}$ with $x_i \in U_i \cap (C - A)$ for each $i$. Since $p$ is injective on $U_i$, then $x_i$ is the unique point in $U_i$ mapping to $p(x)$, and so $x$ can't be in $A$ since it is one of the $x_i$'s. $\newline$

Second, observe that
$$\bigcup_{i = 1}^{n}(U_i \cap (C - A) \cap p^{-1}(U)) = (C - A) \cap p^{-1}(U) \cap \bigcup_{i = 1}^n U_i = p^{-1}(U) \cap p^{-1}(W) = p^{-1}(U).$$ It follows that $U$ is evenly covered by $p$ since $U_i \cap (C - A) \cap p^{-1}(U)= \left(p|_{U_i \cap (C - A)}\right)^{-1}(U)$, but this fact is not necessary for the rest of the proof. $\newline$

Finally, for $x \in U$, we have $p^{-1}(x) = {x_1,\dots,x_i}$ with $x_i \in U_i \cap (C - A) \cap p^{-1}(U)$ for each $i$, so $x$ can't be in $p(A)$. Thus $U$ is an open neighborhood of $x_0$ disjoint from $p(A)$.

[gentle osprey]

c squared

[rancid umbra]

i think this mitigated the issue we had earlier and still does the job

[uneven bronze]

Exercise. If $X$ is a topological space, $A\subset X$ is closed, and $g\in C(A)$ satisfies $g=0$ on $\partial A$, then the extension of $g$ to $X$ defined by $g(x)=0$ for $x\in A^c$ is continuous. \

I'm a bit confused. The way I'm trying to solve this exercise is by cases for $x\in A^\circ$, $x\in\partial A$ and $x\in A^c$. But the definition of continuity at a point is kind of loosely stated in my book. One definition just says "if for every neighborhood $V$ of $f(x)$ there is a neighborhood $U$ of $x$ such that $f(U)\subset V$, then $f$ is continuous at $x$". In all of the three cases I consider in my solution, what would $U$ be a subset of?

[gentle osprey]

psie

[uneven bronze]

I guess just A actually.

I'm not sure treating cases is the right idea, because I feel like that requires some sort of gluing result. But I'm not sure how else to approach the exercise.

[tender halo]

it does need a gluing result

you can glue the function on two closed subsets

and get what you want

namely, on A and cl(X \ A)

[uneven bronze]

hmm, looking through the section on continuity of my text, I can't find any gluing lemma 😔 I'm not sure how the author intended the exercise to be solved

[tender halo]

in this specific case you can just get away with proving that preimage of 0 is closed i think

and prove stuff directly from that

[uneven bronze]

Ok. I have to ponder on the problem a bit.

[quick crane]

The definition you stated is good and you can easily prove the continuity of the extension using that definition

[quick crane]

Say $X, Y$ are topological spaces, and $f : X \to Y$. Then $f$ is continuous at $x \in X$ if for every open set $V$ containing $f(x)$, there is an open set $U$ containing $x$ with $f(U) \subset V$.

[gentle osprey]

L

[quick crane]

it's a generalization of epsilon delta. Roughly, V is the epsilon, and U is the delta

[uneven bronze]

Right. So for example, for $x\in A^\circ$. Let $V$ be an open set containing $g(x)$. Then by continuity of $g\in C(A)$, we can find a an open set $U\subset X$ containing $x$ such that $g(U)\subset V$. Is that everything I have to show in this specific case?

[gentle osprey]

psie

[uneven bronze]

I feel like my approach is a bit strange, since we only know g is continuous on A.

So when I say that we can find an open set U subset X, that's actually not correct I'd say.

[tender halo]

look at a preimage of a closed set under the extension of g

you should easily see that it is closed

[quick crane]

yeah it's not. The definition says you can find a set $U$ open in $A$ such that $g(U) \subset V$

[gentle osprey]

L

[uneven bronze]

Ok, makes sense.

[uneven bronze]

By the way, in this exercise, does anyone know if g is not equal to 0 in the interior of A?

[rancid umbra]

are you asking for an example of such a g?

g could be 0 on the interior of A, it doesn’t have to be

[uneven bronze]

Ok, that complicates a solution significantly if it is.

[rancid umbra]

if g is 0 on the interior of A?

[uneven bronze]

yeah

or just some parts of the interior

[gentle osprey]

tm

[amber musk]

is my proof right ?

[rancid umbra]

why would it complicate things?
if G : X —> R is the extension, then G is continuous since G restricted to A and G restricted to cl(X - A) agree on their common intersection (the boundary of A), and are each continuous as restricted functions

this is the pasting lemma

[uneven bronze]

well, the reasoning I've been following goes like this. It suffices to only consider sets of the form $(a,b)$, since they generate the standard topology on $\mathbb R$ (if the codomain is $\mathbb C$, consider real and imaginary parts separately). The case $0\notin(a,b)$ is quite straightforward. Call the extension $f$. Suppose $0 \in(a, b)$, then
\begin{align*}
f^{-1}(a, b) & =f^{-1}(a, 0) \cup f^{-1}({0}) \cup f^{-1}(0, b) \
& =g^{-1}(a, 0) \cup f^{-1}({0}) \cup g^{-1}(0, b) \
& =g^{-1}(a, 0) \cup \partial A \cup A^c \cup g^{-1}(0, b) \
& =g^{-1}(a, 0) \cup g^{-1}({0}) \cup g^{-1}(0, b) \cup A^c \
& =g^{-1}(a, b) \cup A^c \quad(\ast)
\end{align*}
Since $g$ is continuous, $g^{-1}(a, b)$ is open in $A$, meaning that $g^{-1}(a, b)=U \cap A$ for some $U$ open in $X$. Then
$$(\ast)=(U \cap A) \cup A^c=U \cup A^c$$
which is open in $X$. Thus $f^{-1}(a, b)$ is open in $X$.

[gentle osprey]

psie

[rancid umbra]

yes, it’s correct. it is also nice to prove this with the alternative characterization of hausdorff, that X is hausdorff if and only if {(x,x) : x in X} is closed in X x X with the product topology

[amber musk]

lets gooo i wasn’t rlly sure about max(n_0,n_1)

i’ll check that thanks

[uneven bronze]

here it is assumed the extension not to be 0 in the interior of A, so that f^{-1}({0}) is the union of the boundary of A and A^c

[rancid umbra]

maybe the only thing i would change at the end is the wording. then x_n is in U cap O, but U cap O is empty. contradiction

just makes it more clear

[amber musk]

thanks

[rancid umbra]

right. this will get messy if the preimage of 0 has points in the interior of A

you should try to show that the preimage of any closed set under f is closed instead.

[uneven bronze]

well, eventually one still has to consider if 0 is in the closed set or not

[rancid umbra]

nope!

you have two closed sets at your disposal which union to X, namely A and cl(X - A) and f restricts to a continuous function on each

to start you off: ||f^-1(C) ∩ A = (f|A)^-1(C) = g^-1(C) is closed since g is continuous||

[uneven bronze]

and a similar identity holds for intersecting with cl(X - A) instead of with A, I take it?

[rancid umbra]

yup!

[uneven bronze]

but what about closed sets that overlap the intersection of A and cl(X - A), i.e. contain 0 and possibly nonzero elements?

perhaps I was speaking gibberish

I feel like I have to take a break here, return to this later.

[rancid umbra]

it doesn’t matter anymore:
f^-1(C) = (f^-1(C) ∩ A) ∪ (f^-1(C) ∩ cl(X - A)) is closed for any closed C ⊆ R

btw, this is the exact same proof of the pasting lemma on the wiki page

[uneven bronze]

I will look into it 😉 I have been avoiding it up to now since I can't find it in my book

[rancid umbra]

i get that

although, this is a very standard result that isn’t buried in any heavy machinery or is totally out of scope

[uneven bronze]

yes, just briefly looking at it now, it seems very straightforward. Probably too basic to not bother about (the author of my textbook, that is).

[rancid umbra]

yea, i would prove it once and try to remember and internalize the proof and the statement

the most general form is: if f : X —> Y is a function and (U_i) is an open cover of X or a finite closed cover of X, then f is continuous if and only if f restricted to U_i is continuous for each i.

[rotund halo]

When you prove preimage(U) is disjoint from A how do you know all the x_i lie in U_i cap (C-A)? Is this from preimage(U) being a subset of preimage(W)?

The rest makes sense to me.

Except the last paragraph I have basically the same question.

Feels like that inclusion + the covering for W forces it to work out that way.

[rancid umbra]

yea. preimage(U) subset preimage(W) so for x in preimage(U), p(x) is in U, and so p(x) = p(x_i) for some x_i in U_i ∩ (C - A)

[rotund halo]

Okay yeah that makes sense to me then.

[rancid umbra]

like, you can pick i so that x_i and x lie in the same U_i

then x = x_i due to injectivity on U_i

[rotund halo]

Yeah that part makes sense. My main concern was that since p is surjective but not injective I kept running into shit taking preimages where I'd end up saying things that weren't true due to accidentally assuming the preimage of whatever set I was dealing with was smaller than it was.

This actually avoids that really nicely.

[rancid umbra]

yea, i hate that. i can never remember like, preimage of image is image except when it’s injective or surjective or something

always have to look it up

[rotund halo]

Tfw the function isn't a nice bijection

[gritty widget]

Iirc a locally finite closed cover works

[rancid umbra]

i think if you have Z with the cofinite topology, Sing(Z) is a locally finite closed cover such that the restriction of the inclusion to a single integer is continuous, but the preimage of a bounded open subset of R under the inclusion map is finite, so it’s not open, and the map isn’t continuous.

this example is from the wiki page on the pasting lemma, btw

[gritty widget]

Neat

Wait

That isn’t locally finite, is it? Every neighborhood of an integer contains infinitely many other integers

No neighborhood of an integer will intersect only finitely many integers

The wiki page also says it is true for locally finite closed collections

[rancid umbra]

oh, i misremembered the definition of locally finite, my bad

you right

that is cool, didn't know that

[uneven bronze]

Exercise: If $X$ is a set, $\mathcal{F}$ a collection of real-valued functions on $X$, and $\mathcal{T}$ the weak topology generated by $\mathcal{F}$, then $\mathcal{T}$ is Hausdorff iff for every $x,y\in X$ with $x\neq y$ there exists $f\in\mathcal{F}$ with $f(x)\neq f(y)$. \

Just want to check if my reasoning is correct for the $\impliedby$ direction; we need to use that $\mathbb{R}$ with the usual topology is Hausdorff, right? Because then there are disjoint open $U,V$ such that $f(x)\in U$ and $f(y)\in V$, and by continuity of $f$, the preimages of $U,V$ are disjoint and contain $x,y$ respectively.

[gentle osprey]

psie

[uneven bronze]

I don't know if it is implicit that R has the usual topology all the time, hence my doubt.

[rancid umbra]

yes, R has the usual topology

the usual convention is that, unless explicitly stated otherwise, R has the standard topology

[alpine nest]

Same for R^d for any natural d

[rancid umbra]

this is correct. although, for this direction, you can replace R with any hausdorff space Y

[uneven bronze]

ok, thanks!