point-set-topology

[tranquil cosmos]

Inaccessably many inaccessibles, I'd hope

Worldly, inaccessible and Reinhardt are the only large cardinals that I've been liking so far

[cedar pebble]

Vopenka is nice it makes category theory much nicer

But it’s also one of the most absurdly strong large cardinal axioms and I’ve never needed it

Idk more than enough crazy category theory to do these days just with aleph_1

[tender halo]

aleph 1 is probably the best cardinal tbh

[unreal stratus]

Fake axiom

[waxen oxide]

Can anyone explain this?

[quick crane]

the rest of the connected components of $G$ are contained in the complement of ${z : |z| > R}$

[gentle osprey]

L

[waxen oxide]

but why is there some R such that {z : |z| > R} \subset G

[whole sapphire]

(1) the continuous image of a compact space is compact.
(2) every compact subset of R2 is closed and bounded.

[keen gust]

im confused how to prove this

I just need to show comparability, non reflexivity , and transitivity right?

But for comparability , how do i show it ? the definition itself is a bit dodgy

[thorny agate]

what is the definition of an order relation?

it's a collection of a few axioms you have to check, what are the definitions of those axioms?

[thorny agate]

ok so let (a1, b1) != (a2, b2) be in A x B

what do you have to show for comparability?

[keen gust]

that either (a1, b1) < (a2, b2) or (a2, b2) < (a1, b1)

[thorny agate]

Ok, so can you apply the definition of < on A x B to show that? Note that <A and <B are both order relations

this really is just definition chasing

[keen gust]

so can you apply the definition of < on A x B to show that?
thats where i am confused

[thorny agate]

what does it mean for (a1, b1) < (a2, b2)

[thorny agate]

one thing that would help is if you could elaborate on what you mean by "a bit dodgy"

cause idk what that means

[keen gust]

i mean its a bit weird

and kinda sketchy

if a1 <A a2 or if a1 = a2 and b1 <B b2

this condition i am not understanding

[thorny agate]

what don't you understand?

I mean lets consider an example

we can take our normal ordering < on the integers Z

so then we get an ordering on Z x Z

using the definition, can you tell me if (1, 2) < (3, 4) and why? What about if (1, 2) < (1, 3)?

[keen gust]

1 < 3 so (1, 2) < (3, 4)

for the next one, 1 = 1 so since 2 < 3, (1, 2) < (1, 3)

is that right

[thorny agate]

yes!

[thorny agate]

Ok so it seems you understand the definition, so I'm still not sure what you mean by this

[keen gust]

ok for (1) comparability

Let (a1, b1) != (a2, b2) in A x B.

Case 1: a1 != a2
If a1 != a2 then by definition , we have either a1 <A a2 or a2 <A a1

Case 2: a1 = a2
if a1 = a2 then by definition , we either b1 <B b2 or b2 <B b1

combining both cases, we have either (a1, b1) < (a2, b2) or (a2, b2) < (a1, b1), satisfying comparability.

Is that right

[thorny agate]

yup!

perfect

So now try to do similar definition chasing to get the other conditions

[keen gust]

Thanks!

How should i go about proving this now

I only got this definition to work with

[haughty jungle]

hint try and find a set to use the greatest lower bound property on finding a least upper bound

Or the other way around

But find a way to apply one property while finding the other thing

[keen gust]

but howw

[haughty jungle]

I think the general method is like "consider the set of all upper bounds" and proceed from there

[prime elbow]

If f:[0,1] -> [a,b] surjective continuous mapping with usual topology on R, is it necessary that f needs to be open mapping?

I know this is not true in general

But I am trying to construct a counterexample

[tender halo]

no just go back and forth in the middle a little

[prime elbow]

I try basic functions

[plush folio]

My idea was to ||map [0, 1] to S^1||, then ||rotate by 90 degrees||, then||project down||. I think this works

[cerulean oriole]

Any not-(strictly monotone) continuous function will do.

[plush folio]

Are you sure any such function work? Like, isn't x |-> x^2 from [-1, 1] to [0, 1] an open map?

[tender halo]

it is yea

non monotonicity needs to happen in the middle

[cerulean oriole]

Ah. Open in the closed interval codmain.

[heady skiff]

Let X be a topological space. If A is an open subset of Y and Y is a subset of X, is it true that A is an open subset of the interior of Y?

[tender halo]

A might not a be a subset of the interior of Y though

A \cap int Y is an open subset of the interior of Y

just consider int Y as a subspace of Y

[gritty widget]

How hard should topology textbook(James Munkres) exercises be(i think i can with some little difficulty do 50 percent of them)? Asking this because I'm doing self study and I'm wondering if i missed something important....

[plush folio]

50% is great, that's better than most I think

[gritty widget]

It's a rough estimation... but tbf i wonder how much is due to the fact that i basically am immediately scared away from the ones that don't look obvious

[plush folio]

yeah, sometimes you come across a problem you have no idea how to start. If you have a solutions manual, a good idea is to just look at just the first couple of sentences, and see if you can finish the proof from there

or you can come here to ask for hints

it's completely normal to be stuck on problems. If you're not struggling, the problems are probably too easy

[obsidian aurora]

A thought of mine for this , idk if it's right or not

I'm new at latex so it might be messy

[haughty jungle]

make sure to close all ur dollar signs

There's a lot of text in math environments that should just be plain

[prime elbow]

$f:[0,1] \rightarrow [0,1]$ such that $f(x) = 2x$ when $ 0 \leq x \leq \frac{1}{2}$ and $g(x) = \frac{3-2x}{2}$ when $\frac{1}{2} \leq x \leq 1$.

\vspace{0.5cm}
\Take open set $U = (1/2,1]$ it gives $f(U) = [1/2,1)$ which is not open in $[0,1]$

[gentle osprey]

Notknow🙇
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[prime elbow]

Is it correct?

[plush folio]

yep, looks good 👍

[crimson smelt]

Is my proof for this correct Let $X\neq {0}$ be a vector space. Show that the discrete metric on X is not induced by a norm.
Proof: Suppose the discrete metric is induced by a norm $\left\lVert \cdot \right\rVert \implies d(x,y) = \left\lVert x-y \right\rVert$ But then the $\left\lVert 2 \right\rVert =\left\lVert 2-0 \right\rVert= d(2,0) = 1 \text{(by the definition of discrete metric)} \neq 2 = 2 * 1 = 2 * d(1,0) = \left\lVert 2 \right\rVert *\left\lVert 1 \right\rVert = \left\lVert 2 \right\rVert$

[iron bolt]

you have a \implies outside of math mode, that's probably the reason for the error

you can also ask the bot to tell you the error by clicking the second reaction like it says

[gentle osprey]

rabbits_advocate

[crimson smelt]

thanks

does the cat approve reaction mean that the proof is correct?

[young stone]

yep

ok wait, this works for a 1 dimensional vector space

[rancid umbra]

how do you know 2 != 0?

[young stone]

what is "2" ?

[young stone]

since we are talking about discrete metrics I think it's safe to assume the field is R

although op please clarify in case it isn't

[crimson smelt]

the field is not explicitly stated in this question but the definition of norm in my class assumes R or any general field

[rancid umbra]

what about discrete metrics indicates R? you mean norms?

[young stone]

No I meant in a course where you talk about discrete metrics, you would usually talk about real vector spaces but I suppose not

[young stone]

Yeah so did you get my question? Since X is just any vector space, "2" doesn't necessarily make sense as an element of it so you should clarify what you mean

[plush folio]

on the last line you write 2 * d(1,0) = ||2|| * ||1||, but I think you mean 2 * ||1||, since you haven't shown yet that ||2|| = 2

[tender halo]

also I would probably take the discrete metric to be any metric that induces a discrete topology and not just that specific one, makes the question a little more fun

[crimson smelt]

oh thats right so I can specitfy 2 is an element of R

[tender halo]

what is 2

you have a vector space it has a zero and typically no twos

[prime elbow]

So I can map a function from C[0,1] -> R such that f maps to f(1/2), and this continuous function, so F is closed because it is inverse image of closed set {0} and G is open because it is inverse image of open set

Correct?

[tender halo]

yea

[gritty widget]

Bit late, but just consider the complement of a dense set, e.g. R\Q in R. A non-empty open set of R\Q is not an open subset of the (empty) interior of R\Q in R.

[grizzled lagoon]

this might be a silly question, but how do we know that A and B are open sets? (From mse)

[ruby delta]

It’s the definition of a disconnected topo space, there exists nonempty disjoint open sets whose union is the entire space

[grizzled lagoon]

i guess i was given a different definition, this was the definition i have been given

if (X,d) is a metric space and A subset X, then A is disconnected if it can be written as the union of two non-empty separated sets (where separated means closure(B) cap C = closure(C) cap B = phi) i.e. there exists B,C subset X such that B,C nonempty, closure(B) cap C = closure(C) cap B = phi, A = B cup C

theres no mention of B and C being open, is there a way to deduce that from the definition i was given?

[ruby delta]

Oh, you’re using the metric space definition, let’s see

Also using phi to denote the empty set is kinda genius

[ruby delta]

Closure B is closed <=> (Closure B)^C is open
So C = A cap (Closure B)^C is open in A, and should be the open sets you’re looking for

[gritty widget]

Each space is disjoint from the closure of the other, so inf({d(b, c)|for all c in C)}>0 for any particular b in B, and vice versa for C. B is the union of the open balls centered at any point b of B of said distance, same with C, so B, C are open disjoint subsets who’s union is A.

[gritty widget]

Wow that’s a lot simpler lmao

I hope some more people ask some more questions soon cause I really wanna do some point-set topology rn but don’t know what to do

Might just pop open a random page of Munkres and do some exercises

[grizzled lagoon]

why do we know B is the union of such open balls? like this would rely on the assumption that B is open right?

[grizzled lagoon]

sorry im still a bit confused :<

[gritty widget]

Since BnCl(C) is empty, for every point b in B, d(b, C)>0, so we can pick an open ball around b disjoint from C.

Pick one for every b

Since every set is disjoint from C, so is the union

[ruby delta]

If B and C are separated, then the closures of their respective complements cover each other. That tracks?

[grizzled lagoon]

right, but i guess why is B equal to the union? like for instance if B is a closed set

[gritty widget]

Because you’re picking an open ball around each point b in B, which in particular must contain b

Also, in this case B will be both open and closed(in A)

[grizzled lagoon]

i guess that makes sense intuitively im trying to see why that is true symbolically rn

[ruby delta]

Lemma. If U is an open set, and A is any set, then U cap A is open in A

That’s kind of the lemma that my proof hinges upon

[grizzled lagoon]

hmmm okay i agree with that

Oh so C = A cap cl(B)^c is open in A and thus open in X?

[gritty widget]

Unless A is open, open subsets of A need not be open in X

They’re the intersection of an open subset of X with A, though

[gritty widget]

Also, not sure if I was misleading earlier, B and C are open in A, but not necessarily open in X

[grizzled lagoon]

ahhhh okay i think i kinda get it now

thanks! @ruby delta @gritty widget

[gritty widget]

No problem

[hot pivot]

Definition of "Normal" in these notes: given any two closed disjoint subsets E,F of a topological space X, there exist disjoint open supersets U,V of E,F respectively.

[tender halo]

assume they intersect, then there are two B_{\epsilon_x / 2} (x) that intersect, derive a contradiction

[hot pivot]

Since $U$ and $D$ are disjoint, and since $V$ and $C$ are disjoint, the only way $U$ and $V$ can intersect is if the point is neither in $C$ nor in $D$. The same is true for $U'$ and $V'$, since these are subsets of $U$ and $V$. It feels like we're going to invoke the triangle inequality somewhere but I don't see how we can make progress.\
So for some $x\in C:, y\in D$, we have $B_{{\epsilon_x}/2}(x) \cap B_{{\delta_y}/2}(y) \neq \emptyset$. So far this seems plausible.

[gentle osprey]

Chaos22

[tender halo]

it means the distance between x and y is at most (\epsislon + \delta) / 2

that cannot happen

you just need to realize why

[tender halo]

well i guess technically it needs to be strictly less than that which it is

[hot pivot]

Ah I think I get it now. WLOG, assuming epsilon>delta>0,
This would mean that y is contained in the original Be(x) since (epsilon+delta)/2 < epsilon. Which by construction isn't the case. Contradiction achieved.

[tender halo]

yep

[haughty yew]

By the way, another neat way to show that a metric space is normal is, for two disjoint closed sets A and B, to consider the distance functions x |-> d(A, x) and x |-> d(B, x), which are defined as the infimum of the distance from x to points of A, and the same but for points of B. Then show that these distance functions are continuous.
Next, consider the difference, f, of these distance functions, which is also continuous. Then the inverse images f^-1 ((-infty, 0)) and f^-1 ((0, infty)) are the required disjoint open sets to prove normality.

[gritty widget]

I would think just taking the open balls of respective distance around each point in each would be a simpler way to provide explicit open sets, no?

[hot pivot]

Oh, this is wonderful! When I first read the definition of normality, it felt reminiscent of this real-analysis exercise I did from Abbott. The connection felt vague but nothing precise. Thanks for explicitly showing the link here.

I can't recall if I had to invoke completeness to prove this, but I'm assuming it can be proven just by using the properties of a general metric space. Thus, am I correct in saying that every metric space is normal?

[quick crane]

this is true for any metric space. It's written $g(x) = d(x, F)$, and $d(x, F) = 0$ iff $x \in F$.

[gentle osprey]

L

[tender halo]

yea but its less illustrative

with that approach you get the Uryhson function for the two sets for free (take d(x, A) / (d(x, A) + d(x, B)))

[gritty widget]

Ic

[heady skiff]

Why is this true? I know that f^{-1}(W) is open in |K|, but I'm trying to see why it's an open subset of the interior

Moreover how do we even know the preimage lives in the interior

Am I supposed to interpret it this way? $f^{-1}(W)$ is an open subset of $|K|$, so $f^{-1}(W) \cap \sigma$ is open in $\sigma$. Therefore, $\varphi^{-1}\bigl(f^{-1}(W) \cap \sigma)$ is open in $\Delta^m$; as such, $(\Delta^m)^{\circ} \cap \varphi^{-1}\bigl(f^{-1}(W) \cap \sigma)$ is open in $(\Delta^m)^{\circ}$

[gentle osprey]

okeyokay

[west void]

Mathematicians make up any term atp 💀

[gritty widget]

Is there an example of a topological space X such that, given any two points x, y, there is an automorphism f such that f(x)=y, but such that there is two pairs (x, y) and (a, b), assuming x≠y and a≠b, such that there is no homeomorphism f from X to X with f(x)=a and f(y)=b?

All the examples I can think of satisfying the first property satisfy the second

E.g. $\bR^n$, seemingly, for one

[gentle osprey]

NAT Enthusiast

[gritty widget]

Actually no I don’t think R^n is an example?

Let (0, 1) and (2, -1)

There is no homeomorphism sending 0 to 2 and 1 to -1, since every open, continuous function from R to itself is strictly monotone

So nvm

[queen prism]

?

it's a homeomorphism although it doesn't preserve orientation if that's what you're wondering

[gritty widget]

Now I’m really confused cause I saw an MSE post saying a continuous function from R to R is open iff strictly monotone, so 0<1 implies F(0)<F(1). Maybe I’m just misunderstanding what strictly monotone means?

Oh wait no I think that’s strictly increasing lmao

Yeah mb

thanks

[queen prism]

monotone means it's either increasing everywhere or decreasing everywhere

[gritty widget]

Oh that makes more sense

[gritty widget]

Do you know if there’s any example of a space like this then?

[queen prism]

by automorphism I take it you mean a homeomorphism from a space onto itself

which I don't think is something I've heard before but idk

answer to that question is also idk

[gritty widget]

Yeah

[iron bolt]

I don't have a topological example at hand unfortunately, so I don't know if there is one (even though I would not be surprised)

I do have a neat example from Riemannian geometry though

actually nevermind, my example was overcomplicated and for Riemannian manifolds even something like S^1 is an example

for every two points there's an isometry that maps one to the other, but if you pick two pairs of points such that the distance between the first two does not equal the distance between the second two, you can't have an isometry that maps one pair onto the other

[gritty widget]

That’s actually very helpful! Thank you.

The question was for defining a certain kind of structure on objects and I just arbitrarily picked topological spaces to see if it was interesting for them

But this is just as helpful as a topological example would be

[iron bolt]

oh, now I actually have a really simple topological example too

take something like the disjoint union of two copies of R

for every two points there's still a homeomorphism taking one to the other, but if you pick a pair of points from one component and want to take it to a pair where one point is in one component and the other is in the other...

I think the standard way of saying this is that the homeomorphism group acts transitively but not 2-transitively btw

[gritty widget]

Alright, thank you!

Now I just wanna find an example that’s connected lol

[iron bolt]

I have an example, but it's the opposite of pretty, lol

[gritty widget]

https://math.stackexchange.com/questions/174280/further-examples-of-1-transitive-but-not-2-transitive-groups-of-self-homeomorphi

[iron bolt]

hm, that's an interesting example too

what I was thinking of was to take the real line and replace every point with two indistinguishable copies of itself

[gritty widget]

Ideally I’m looking for a space X such that the equivalence classes of XxX under the relation (x, y)~(a, b) iff there’s a homeomorphism sending x to a and y to b are numerous, but (x, x)~(y, y) for all x, y, e.g. it is homogeneous

[gritty widget]

So line with two origins but line with two everything

[iron bolt]

kind of, but with the two always collapsed together

0 and 0' are not topologically indistinguishable in the line with two origins

here they are

[gritty widget]

Oh yeah

[gritty widget]

Is this homeomorphic to the product of R with {0, 1} with the indiscrete topology?

[iron bolt]

hm, I hadn't thought about it like that, but probably

that's nice actually, because you get 1-transitivity for free

just need to show that if self-homeomorphisms of both X and Y act transitively, those on X ⨯ Y do too

the downside of this example is of course that it isn't Hausdorff, not even T0. that's probably why the math.se answer was so much more complicated

[gritty widget]

I might ask in other channels for other examples with other types of objects.

[gritty widget]

All of these seem to have very small equivalence classes under this relation

[iron bolt]

the R ⨯ (indiscrete {0,1}) example has quite large equivalence classes actually

in fact, I think (a,b) can be mapped to (a',b') by a homeomorphism iff the projection to R is either injective on both pairs or non-injective on both

which, uh. means there's only three equivalence classes

[gritty widget]

Sorry, meant “not very many”, lmao

[iron bolt]

ah

[gritty widget]

Preferably I want an example with $\gneq\mathfrak{c}$ classes

[gentle osprey]

NAT Enthusiast

[iron bolt]

I don't have an answer to that, but I do have one more example in another category I'd like to share

the one about Riemannian manifolds I had in mind before I notice the circle worked too

if you take the cylinder R ⨯ S^1, its isometry group acts transitively on it, but not 2-transitively

that's not super interesting, because one counterexample is still just two pairs of points with d(x,y) ≠ d(x',y')

but, even if you restrict yourself to pairs with some fixed distance, it doesn't act transitively on them

[gritty widget]

That’s very interesting, then!

[iron bolt]

furthermore, it doesn't even act 1-transitively on the tangent bundle

so the cylinder is a really simple example of a space that, roughly speaking, looks the same at every point but not the same in every direction

physics has some word for that probably, but I forgot. homogenous and symmetric maybe? I'm not sure

oh right, it's anisotropic

[gritty widget]

Never heard that before lmao

[iron bolt]

I will resume being productive now again, have a nice day

[gritty widget]

You too!

[unreal stratus]

Good word

[balmy nexus]

is there a topology that is the finest topology bar the discrete topology?

im working on an excersize where i am trying to see if totally disconnected implies that the topology is discrete. Im trying to disprove it by creating a topology that is as fine as possible but isn't discrete

well i have a counterexample ready on a finite set but still i got interested in what happens if i try to remove the finite open sets from the discrete topology

[tender halo]

no

[tender halo]

think about the popular disconnected metric spaces you know, one of them is bound to be totally disconnected

[tough shuttle]

hi, could you tell me which is the best general topology course on Youtube pls.

[hot pivot]

Please verify this proof that complete regularity is finitely productive.
https://media.discordapp.net/attachments/690488884747960331/1350169733122621490/478125996885934082.png?ex=67d5c30d&is=67d4718d&hm=c61d320f167c6f3882ae620ea8612dd7092e3a10e09d0e9576f0124f3c8c5fae&=&format=webp&quality=lossless&width=505&height=824

[hot pivot]

One problem I think I see is that the logic to deduce the continuity of h seems faulty since we're not taking the maximum of two functions at the same input. So is there a neat modification?

[tender halo]

you can just consider them as a composition of a projection and a continuous function

also tbh you only need to establish regularity and complete regularity on subbasic sets

so you get multiplicativity of regularity and complete regularity for free because of that

[hot pivot]

Attempt:
we first define h'(x,y) to the ordered pair on the unit square: (f(x),g(y)).
And note that this is continuous since each projection is continuous.
And then compose h' with the maximum function m: R^2 -> R, (which we know is continuous).
So then h = m composed h', which is continuous.

[hot pivot]

i've seen this trick tossed around where to prove a property about the product topology, it suffices to prove it for the basic or even the subbasic sets. Although I don't really understand when we can rely on this trick, or even, why it works.

Hint for the reverse direction please.
Current idea: A and B are separated. So we should try to find a subspace where they are both closed. Thus under that subspace we can leverage normality and somehow come out of the subspace with non-disjoint open sets.
I tried the subspace of A∪B, but that doesn't seem useful. A and B are both clopen here so the open sets we end up getting are just A and B again.

[tender halo]

i mean you need to prove it separately for each property

here it happens to work

the reason it gives you free proofs about multiplicativity is because its usually really easy to prove stuff about the canonical subbasic sets of the product topology

[main jay]

HELP I’m trying to understand the concept of the closure of a set in topology. What does it mean for a set A to have a closure, and how is it relevant to the idea of closed sets in a topological space???? 😭

Will it include the endpoints of 0 and 1?? 😀

[gritty widget]

The closure is the smallest closed set containing the original set

Equivalently, it’s the set of all points of closure of that set. Informally, these are points which can be approximated to arbitrary degree without leaving the original set.

For example, the closure of (0, 1) is [0, 1]

Because you can approximate 0 within any arbitrarily small positive degree in (0, 1)

Same with 1

[main jay]

How does that make me get any closer to the original set tho??

[main jay]

So we can basically approximate as close as we want ?

[gritty widget]

Not sure what you mean

[gritty widget]

Yeah, any point in the closure can be approximated to arbitrary degree by points of the original set

[main jay]

How does the closure give a fuller view of the set's behaviour ?

[gritty widget]

It tells you which points are arbitrary close to the set

That’s about it

[main jay]

Thank you thank you thank you thank you thank you 🙏 my ahh was so cooked

[main jay]

Literally saved me

[gritty widget]

Np! Glad I could help.

[gritty widget]

Let X be the real line with the cocountable topology

Then, any cocountable set is dense, say, $\bR\setminus{1}$, but there is no sequence of elements of $X$ converging to $1$, because the convergent sequences here are eventually constant.

[gentle osprey]

NAT Enthusiast

[spare light]

holy shit

[gritty widget]

I can’t think of any example atm except where all convergent sequences are eventually constant

[prime elbow]

Discrete metric space

[tender halo]

stone cech compactification of N

certain subspaces of function spaces also have this property

[tranquil cosmos]

Polynomials?

Nah nvm

[haughty jungle]

is it true for cont functions f(a+b) = f(a) + f(b)

sry if this is a rly dumb question

[queen prism]

can you think of any counterexamples to your claim

for example on R

[haughty jungle]

ok yes

x^2

im trying to prove that the closed set definition of continuity is equivalent and in munkres it has in the proof (page 104) f^-1(B) = f^-1(Y) - f^-1(V) where B = Y-V and im not srue how he did that

[queen prism]

do you know what (set) minus (set) means

[tranquil cosmos]

Same thing as X\Y

[haughty jungle]

it's everything that's in b but not a

[queen prism]

ok

what does continuity of additive functions have to do with this

[haughty jungle]

💀 im acc fucking stupid nvm

[queen prism]

many topological spaces don't have an additive structure btw

[gritty widget]

A discrete metric space is definitely sequential lol, not sure what you mean

[iron bolt]

what even was the original question? I don't see it

[tender halo]

it is sequential but also only trivial sequences converge

[tranquil cosmos]

How do you define sequence convergence in top spaces

A sequence converges iff it hits a cofinal sequence in a neighborhood filter?

[tender halo]

thats an interesting way to put it but yeah

im not sure what you mean by a cofinal sequence in a nbhd filter though, it might not have one

the traditional definition is something like "for every [open] nbhd the sequence eventually lies inside the nbhd"

[queen prism]

nets

[tranquil cosmos]

Ahh

[silver ridge]

Can someone help with the details here? I think I've got the idea but I'm struggling to make it fully rigorous.

I'm pretty sure that hausdorffness fails at the two origins. By definition of the quotient topology, given the projection q: X -> X/~, U in X/~ is open if q^-1(U) is open in X. In X, notice that something like ((2,3) x {0} U (2,3) x {1}) is open with respect to the subspace topology, X \subset R^2. The previous set also happens to be q^-1(U), where U = (2,3)/~ (the equivalence classes [(x,0)], with x in (2,3)).

So, now that we kind of see how open sets look like in X/~, consider a neighbourhood V of [(0,0)] of the form (-a, a)/~. Then q^-1(V) contains points arbitrarily close to (0,1) but not (0,1) itself, (e,1) for all 0 < e < a. My suspicion is that q^-1(V) isn't open, but I'm struggling to show this.

As always, appreciate any help.

[gritty widget]

It was in #real-complex-analysis

[gritty widget]

Let 0 be the origin from the first line and 0’ the origin of the second

A neighborhood of 0’ not containing 0 in the new space is just a neighborhood of 0’ in the line, so it has interior say (-a, a) for a>0. Same thing with 0, so a neighborhood of 0 not containing 0’ is essentially a Neighborhood of 0 in the original line, so it has interior say (-b, b) for b>0.

Assume wlog a>b

Then, the projection maps b/2 in both lines to the same point in the line with two origins, which is hende in both neighborhoods.

[prime elbow]

I understand this but how do I visualise this, I can visualise for B[0,1] in R but for R^2 it is a circle so collapse the all points x^2 + y^2 = 1, how does this give me a sphere in R^3?

[prime elbow]

How it collapses the boundary points of B[0,1] in R^2

[hidden crag]

take a sheet of paper, cut out a disk and try

according to my experience this won't exactly look like what you want but it might give you an idea

[prime elbow]

okay thank you

in a topology, if f: X -> Y and C \subset X called saturated when C = f^-1(f(C)), right?

So if p : X -> Y is a quotient map then i can show that p is continuous and saturated open sets maps to open sets

but how do i show converse?

let p: X-> Y is surjective mapping such that it is continuous and saturated open sets maps to open sets then a subset of U is open in Y iff p^-1(U) is open.

one direction is true by continuity of p.
Now i have to show if p^-1(U) is open then U is open, it is easy if p^-1(U) is saturated , how can i modify it?

[haughty jungle]

Anyone have an ordered field that's a linear continuum other than the reals

[molten whale]

are the rationals?

[young stone]

isnt every such field isomorphic to R as an ordered field (hence homeomorphic under the order top)

in fact you don't even need the second condition, any ordered field with lub property is isomorphic to R

[haughty jungle]

Doesn't have least upper bound

[haughty jungle]

Idk that's what I'm wondering but that sounds believable

Is there an easy proof

[young stone]

they both have copies of Q and you can use dedekind cuts

https://mathoverflow.net/questions/362991/who-first-characterized-the-real-numbers-as-the-unique-complete-ordered-field

this question describes the proof

[haughty jungle]

also i think linear continuum and lub property are the same thing for ordered fields

the x < y -> there exists z s.t. x<z<y property can be proven as (x+y)/2

what does complete mean in this context?

[tender halo]

no nontrivial dedekind cuts

i.e. no gaps

[haughty jungle]

Ty

[desert vortex]

should I learn about nets?

what is your review

[quick crane]

they are useful for functional analysis

[queen prism]

I think the modern perspective is that you should know both nets and filters as a researcher

but I'm not a researcher so that could be bs

nets are closer to the idea of a sequence I think

[tranquil cosmos]

An algebraic topology postdoc told me "point-set topology is dead"

[desert vortex]

what book to learn about nets?

[gritty widget]

This seems to be, in some sense, true, afaik

Most people doing topology these days aren’t really doing pointset

But there’s definitely still some

[tender halo]

it is just a subfield of set theory at this point really

and as such its like ok

[tranquil cosmos]

Lol

I once read (and agreed tbh) that the compactness of an infinite number of compact spaces shouldn't be intuitive because, well, infinite number of terms

What are some philosophical points for/against tychonov

[gritty widget]

Do you mean an infinite product of compact spaces?

[tranquil cosmos]

Yes yes

[gritty widget]

I feel like the most obvious one is “every product of compact spaces that we can decide in ZFC whether it’s compact or not is always compact”, feels pretty intuitive.

[tranquil cosmos]

Any infinite such products we can decide?

Cause if we can only decide for finite products that's pretty moot

[gritty widget]

If Tychonoff’s theorem is false, that means there’s some very weird product of compact spaces, which, unlike the ones we can compute and decide compactness of, simply isn’t compact.

[queen prism]

it's not that weird when you remember that an "infinite product" of topological spaces still contains some level of finiteness to it

the basis elements of this topology are the products of (i) the entire space X_i, in which case it doesn't matter, and (ii) a finite number of nontrivial open subsets U_j c X_j

[gritty widget]

$2^\mathbb{N}$ is compact, with or without Tychonoff’s theorem

[gentle osprey]

NAT Enthusiast

[gritty widget]

It’s homeomorphic to the cantor set

[tranquil cosmos]

Interesting

[gritty widget]

The axiom of choice is also “obviously true” so 🗣️

[desert vortex]

facts

[tranquil cosmos]

Lel

[gritty widget]

Topology without AC is actually pretty bad anyhow

[tranquil cosmos]

I need to work on my set theory where AC doesn't imply Zorn and WOT 👀

[gritty widget]

Second countability need not be hereditary

[gritty widget]

WOT?

[tranquil cosmos]

Well ordering thm

[gritty widget]

What’s wrong with that

[desert vortex]

its obviously false

[tranquil cosmos]

Lollll

[gritty widget]

I think “every set bijects with some ordinal” is pretty intuitive, tbf.

[tender halo]

i mean thats not particularly interseting, it's true for any spaces on (uniformly) well orderable sets

2 is somewhat easy to well order

[tranquil cosmos]

Oh yeah? Which ordinal does R biject to?

[gritty widget]

Yeah idk how interesting it is

But they asked for an example

[tender halo]

aleph 1 of course

[gritty widget]

$|\mathbb{R}|$

GG

[gentle osprey]

NAT Enthusiast

[desert vortex]

R bijects to the R with a well ordering

pretty basic

[gritty widget]

R bijects to whatever I want it to for the purposes of some example

[tranquil cosmos]

Set theory is really held together by toothpicks and spit

[gritty widget]

E.g. Assuming $|\mathbb{R}|\geq\omega_\omega$, the cocountable topology on $R$ has some properties it wouldn’t have if $|\mathbb{R}|=\aleph_1$

[gentle osprey]

NAT Enthusiast

[gritty widget]

I don’t see how you got this conclusion at all 💀

[tender halo]

i still want to learn the mysteries of aleph 4 and Shelah's pp

I was told by someone that something happens in the order topology of aleph 4

and im really curious what and why but damn you need to wade through a lot of stuff to get to it

[gritty widget]

Neat

[tender halo]

shoutout to the name of this paragraph in shelahs book

[gritty widget]

It’s always 4 fr.

R^4, 4-manifolds, Aleph 4..

[tranquil cosmos]

Wait what's with aleph 4

[tender halo]

if 2^\omega is less than \aleph_\omega, then its just

why 4? i dunno i havent read it

[tranquil cosmos]

Oh that's not aleph 4

Tsk

Also isn't woodin obsessed with making |R| = aleph 2

[solemn iris]

Hiya! I'm still in the early parts of my topology course so we're still discussing like continuous functions, types of topologies, and closures etc. There's this passage in another one of my courses, differential geometry (the one highlighted in blue).

I'm not looking for an explanation or anything, just curious. What topic does this correlate to? Any particular chapter (or subchapter) in Munkres' topology?

[sturdy wasp]

I would probably used local homology to show that, the idea is that removing any point on a smooth surface should leave you locally to something that looks like a puncted disk (in particular still connected). However here removing (0) leave you with two connected components thus S cannot be smooth. If you use path connectness you can probably show directly that a smooth surface without a point is sill path connectness by writing path explicitly.

[haughty yew]

local homology isn’t necessary, just consider that double cone with the central point removed, then show that the resulting thingy is disconnected

which cannot happen for a 2-dimensional locally euclidean surface

[tranquil cosmos]

Tbt when I thought that was an immersion of the cylinder

[young stone]

Do we get any interesting theorems if we take this?

[tranquil cosmos]

I think he just found some "natural" principles that lead to this result

Fwiw it's also consistent that |P(aleph1)| = aleph2 in that case

[young stone]

that's kinda cursed

damn I guess I am a continuum hypothesis believer that has no reason to feel cursed

[tranquil cosmos]

Well it doesn't need to be the case

https://en.m.wikipedia.org/wiki/Easton's_theorem

[pulsar lynx]

I'm trying to show that the definition of the quotient topology is forced by the universal property

That is, if X/~ is endowed with a topology 𝜏 such that (X/~, 𝜏) satisfies the UP of the quotient, then 𝜏 is actually the ordinary topology on X/~

[gritty widget]

This isn’t true, though(unless you include the specific quotient map)

Any non-trivial bijection from X/~ to itself(as a set) gives it a distinct topology which is homeomorphic to the previous one(by the homeomorphism)

Universal property determines the quotient space up to isomorphism

If you do mean with the projection p:X to X/~ this is pretty straightforward, though, since it’s a quotient map

[pulsar lynx]

oh interesting

I'm trying to justify the definition "U open in X/~ if and only if π^-1(U) open in X"

specifically, the <= direction of this equivalence

I want to appeal to a category-theory type result, that this topology is somehow forced by other considerations

but I'm not sure what the formalization would be

[gritty widget]

Oh so you’re trying to prove it is a quotient map from the UP then

[pulsar lynx]

wdym by a quotient map

[gritty widget]

A quotient map is just a map such that a set is open iff it’s preimage is

Afaik usually defined to be surjective too

[pulsar lynx]

oh

but that's exactly what I'm trying to motivate

[gritty widget]

Yeah

[pulsar lynx]

a set is open iff its preimage is open

why is this the "finest topology such that projection is continuous", and why would such a property matter

[gritty widget]

It follows pretty readily from “the finest topology making the projection continuous”

[pulsar lynx]

I would like to say, if f: X --> Y is a continuous map such that x ~ x' implies f(x) = f(x'), then f factors uniquely through the projection X --> X/~, and the topology on X/~ is forced in this UP

yes why is that the definition

is what I mean

like, it wasn't made out of thin air

I'm sure you could argue you want "the most open sets" from the source X

but I also want to see this from a categorical persective, if that's possible

[gritty widget]

I’m a bit confused, are you trying to show the quotient topology satisfies the UP? Generally, trying to do the inverse isn’t very helpful imo, the UP makes no reference to the topology on it or whatever, and UP’s only defines up to(unique) homeomorphism in general

[pulsar lynx]

oh I guess that's true

[gritty widget]

Going from the UP to the quotient topology is very hard, since the former is incredibly abstract

Going from the quotient to the UP is far easier

[pulsar lynx]

I'm just trying to motivate the topology we define on X/~, that is, "the finest topology such that projection is continuous"

like why do we choose this definition

(also, I see what you mean, yeah. what you said about bijections inducing homeomorphisms is a big issue)

[gritty widget]

I guess we don’t want the quotient to have any “missing” open sets, so to speak? So naturally making it the finest possible seems to fix that issue

[pulsar lynx]

I've been doing category theory recently, so I was wondering if the UP of the quotient, applied to the category of topological spaces, somehow forces the topology we pick on X/~. But it seems like it doesn't actually do this, since homeomorphisms don't see individual topologies.

[pulsar lynx]

yes, someone told me the indiscrete topology on X/~ certainly satisfies pi: X --> X/~ being continuous, but obviously this is a meaningless topology

[gritty widget]

Lots of constructions are like this, e.g. “the freest/most restricted object restricted enough/free enough to still do X”, but I’ve always really just thought that was natural, intuitively

[pulsar lynx]

so we want "the most open sets"

[pulsar lynx]

ah I see what you mean

[gritty widget]

Yeah, we want to impose the least restrictions on it, but we don’t want it to be too free(otherwise it’s indiscrete)

[pulsar lynx]

maybe I don't have enough practice -- why can't we pick the discrete topology on X/~? Isn't that the finest topology possible, and it would mean nothing?

[gritty widget]

Dually for, say; subspace

[gritty widget]

We want the projection map to be continuous, “just barely”

[pulsar lynx]

oh my friend also mentioned if you let ~ be the trivial equivalence relation where a ~ b iff a = b, then you want the map X --> X/~ to be the identity map essentially

so that's another consideration

[gritty widget]

Subspace topology has the least open sets making inclusion continuous, for example

[gritty widget]

Yeah

[pulsar lynx]

I think the bigger issue is that if X doesn't have the discrete topology, then a map X --> Y where Y has discrete topology is never continuous
edit: not sure if this is true actually, but in many cases it is

[gritty widget]

Lots of pointset topology stuff can feel unmotivated icl

[gritty widget]

Well this isn’t in general true(any constant map is continuous, and a surjective one may be, if the connected components of X are more numerous than the points of Y)

[pulsar lynx]

yes

right

[gritty widget]

But for equivalence relation projections typically yeah

[pulsar lynx]

hm ok

[gritty widget]

If pi:X to X/~ is continuous when X/~ is discrete, X/~, well, is discrete(since it’s the finest one), so you won’t find very many examples of quotients being discrete

[pulsar lynx]

also, why is it true that "the finest topology such that π is continuous" is exactly the topology "U open in X/~ iff π^-1(U) open in X"

[gritty widget]

Add some new open set to X/~

It’s preimage is not open

Because, if it was, the image would already be open

Breaking continuity

Hence that’s the finest topology making it continuous

Other direction is similar

[gritty widget]

Also, just thought I’d add, given any set function f:X to Y, if f is an injection, and Y is a topological space, we can give X the coarsest topology making f continuous(this is just the subspace topology)

If f is surjective, and, instead, X is a space, we can give Y the finest topology making it continuous, which you already know about

Subspaces and quotient spaces are dual like that

[pulsar lynx]

oh wow that's nice

ok that's quite cool

[iron bolt]

depends on the exact formulation of the UP. for universal properties in general you're right, but this one is typically stated specifically for X/~ with the quotient projection, and then it determines the topology uniquely

[iron bolt]

if you're curious, the exact condition for maps to discrete spaces to be continuous is that they're locally constant - you can probably guess from the name what this means, and verify that it is indeed equivalent

for quotient projections this happens iff all equivalences classes are open

[gritty widget]

Ah, ic

[iron bolt]

this all works without any restrictions on f btw. they're then called the induced/coinduced or pullback/pushforward topologies

mostly unrelated to pullbacks and pushforwards in category theory though

actually, that was a lie

you can understand them as literal pullbacks and pushforwards. if f : X -> Y is any function between sets and Y carries a topology T, the induced topology on X is the unique topology T' such that (X, T') with the obvious projections becomes the pullback of (X, indiscrete) -> (Y, indiscrete) <- (Y, T)

since pullbacks are unique up to isomorphism, that determines the induced topology uniquely

dually, the same holds for the coinduced topology if you replace the pullback with a pushforward and the indiscrete with the discrete topology

this is handy because it for example tells you that colimit-preserving functors from Top to itself preserve quotient spaces

I wouldn't generally advise viewing the induced/coinduced topologies like that though. It's a bit backwards, in that they actually come first and are how you construct limits and colimits in Top

namely, you can show that for any diagram of topological spaces, taking its limit in Set and equipping it with the induced topology yields a (and thus the) limit in Top, and of course the same dually for colimits and the coinduced topology

the same works in other categories with induced / coinduced structures, so it's quite powerful. for example measurable spaces, diffeological spaces, bornological spaces, uniform spaces etc.

I think all you need is a concrete category C (with forgetful functor F : C -> Set) with the property that for every set X, the poset F^-1(id_X) of C-structures on X forms a complete lattice, and that for maps f : X -> Y the property of f being a morphism as a function of structures on X and Y is closed under infima in the first argument and suprema in the second

[pulsar lynx]

ok so let me be very specific. Is this true:

Let X be a topological space. Let 𝜏 be a topology on the quotient set X/~, and let π: X --> (X/~, 𝜏) be a continuous map such that x ~ x' implies π(x) = π(x'). Suppose that (X/~, 𝜏) satisfies the following property: for any continuous map f: X --> Y such that x ~ x' implies f(x) = f(x'), then there is a unique continuous map g: (X/~, 𝜏) --> Y such that f = gπ.

Then 𝜏 is the ordinary quotient topology on X/~.

[iron bolt]

then you get induced/coinduced structures, substructures, quotients, limits, colimits, all that nice stuff

[pulsar lynx]

is this because the forgetful functor reflects colimits or something, and Set has all colimits?

[iron bolt]

you can verify that it is a limiting cone directly

[pulsar lynx]

yes

but the general idea is that Top is a category of set-like objects, so colimits are explicitly constructed like colimits in Set with the added structure

i'm still a bit fuzzy on the details of this

[iron bolt]

yep, exactly

[pulsar lynx]

like how we can say the p-adics are an inverse limits, so they are isomorphic to the direct product of Z/pZ satisfying the ring homs

this seems very similar to the explicit construction of the inverse limit in Set

but isn't there a general way to understand this sort of thing, like the forgetful functor preserving/reflecting colimits, so they should look the same

[iron bolt]

it preserves colimits because it has a right-adjoint, the indiscrete topology functor

so that tells you that if a colimit exists, it has to look like the colimit in Set equipped with some topology

I don't know if you can show abstractly that one exists though

[pulsar lynx]

ah

so the answer here is adjoint stuff

hm i have more to read then haha

[iron bolt]

I think if you're looking to understand categories like the ones listed here well, the answer is actually this

you get e.g. the existence of discrete / indiscrete structures and hence adjoints to the forgetful functor from it too

[pulsar lynx]

is this related at all to filtered colimits? (sorry this is something my prof keeps putting on hw lol)

[iron bolt]

I don't see a direct connection, no

I mean, filtered colimits are a special case of colimits, so this allows you to construct them too

but they're not largely important at least here

[pulsar lynx]

ah

ok ok

thank you

[iron bolt]

np ^^

it's something that has been on my mind for a while anyways, so I'm happy to share

just two more things:

[iron bolt]

1. I think if you replace Set with some other complete and cocomplete category, iirc all of this still works too - though I don't quite remember if you might need some more mild conditions

so this should get you limits and colimits of e.g. topological groups or topological modules too, if that's something you're interested in

and 2) you can also find this as "topological concrete category" on nlab, though it is written down quite a bit more abstractly there

oh, also another cool theorem you get for free: if a subcategory of a topological concrete caregory is closed under products (coproducts) and subspaces (quotients), it is automatically reflective (coreflective), with again a nice formula for limits and colimits

you can use that for example in Top to show that the subcategories of discrete spaces, locally connected spaces, sequential spaces, locally path-connected spaces, compactly generated spaces and delta-generated spaces are coreflective

or that categories like that of Hausdorff spaces are reflective

[gritty widget]

is this john m lee

[tranquil cosmos]

Is the universal cover of $\bC^*$ a group?

[gentle osprey]

PKThoron

[tender halo]

where can i read about that?

[desert vortex]

the universal cover of a lie group is always a lie group, you can lift the multiplication map

u get lift of U x U -> G x G -> G, to a map UxU -> U, for U universal cover

[tranquil cosmos]

Actually, I guess I have to ask this embarrassing question

[desert vortex]

also this group structure in this case on U = complex plane is just addition, and the map C -> C* is the exponential map

[tranquil cosmos]

Yeah was gonna ask lol

Is it just C

The silly reason why I asked is cause I wondered if you can write C multiplicatively

(C,+) that is

[desert vortex]

what do you mean?

[tranquil cosmos]

As (R,+) is the same as (R>0,×)

[desert vortex]

yeah thats right, isomorphism is exponential map again

and inverse is logarithm

[tranquil cosmos]

Of course

[desert vortex]

and you get universal cover of C* = S^1 x (R>0,×) is the (universal cover of S^1)x (universal cover of (R>0,×))

[tranquil cosmos]

I was just asking if you can write (C,+) multiplicatively

[desert vortex]

what does that mean?

[tranquil cosmos]

Wait a second

Duh

S^1 × (R>0,×)

That's all I needed lol

[desert vortex]

yeah thats your C* 😄

[tranquil cosmos]

I'm stupid yet again

[desert vortex]

everything is clear?

if not put on your thinkin cap 🧠

[tranquil cosmos]

Yeah I didn't want the multiplicative group of C* written multiplicatively lol

But the additive group of C

C = uc of S1 × R>0 = R × R>0 = R>0 × R>0

Which is a bit contrived but hey

The cover map is: (x,y) maps to y*e^(i ln(x))

[desert vortex]

C = (R,+) x (R,+)

[tranquil cosmos]

Man I'm dull rn

Right, it's that easy

[desert vortex]

💡

its hard until its easy my friend

[iron bolt]

https://wildtopology.com/tag/coreflective-hull/

sequential spaces aren't mentioned there, but they're the coreflective hull of the one-point compactification of the natural numbers, for example

or of the cantor set or of all metric spaces or of all first-countable spaces, equivalently

meanwhile delta-generated spaces are the coreflective hull of either the standard simplices, the spaces R^n, just R, just the unit interval, all manifolds, or all CW complexes

and compactly generated spaces are the coreflective hull of compact Hausdorff spaces

seeing the various inclusions here isn't too hard because the coreflective hull is a genuine hull operator - so if J and J' are two classes of spaces such that J is contained in the coreflective hull of J', then the hull of J is contained in the hull of J' as well

so for example the hulls of N* and all metric spaces agree because N* is metrisable and all metric spaces are sequential

the hulls of the Cantor set C and N* agree because N* is a quotient of C and C is metrisable, hence sequential

discrete spaces are the coreflective hull of the one-point space, because that space is discrete and every discrete spaces is a disjoint union of copies of it

I wonder what the coreflective hull of the Sierpinski two-point space is

[plush folio]

In a topological space X, a finite set is discrete if X is T1 and a discrete set is finite if X is compact, right?

[iron bolt]

no to the latter

discrete compact spaces are finite, yes

but why would the discrete subset of the compact space have to be compact?

you need some extra condition on how it sits in that space

[plush folio]

Ah, I see I'm trying to think of a counter example... { 1/n | n \in N } in [0, 1]?

[iron bolt]

looks good, yeah

notice that the set has an accumulation point that lies not inside the set, so it's discrete anyways

[queen prism]

just a notation question but what is this product

[iron bolt]

if you ask your subset to be discrete in the sense that every point in X has a neighbourhood containing only finitely many points of the set, that can't happen, and it becomes easy to show from compactness thst the set has to be finite

alternatively, closed discrete subsets of compact spaces are obviously finite

[iron bolt]

pullback / fiber product

[queen prism]

ono I don't know this term

ok thanks

[desert vortex]

congratulations its your first encounter with category theory

[queen prism]

nonono not my first but I am bad at categories

[desert vortex]

thats alright its just nonsense anyway

[iron bolt]

C(K,U) and C(X,Y) both admit maps to C(K,Y). it's the pullback of those two maps

they're just hidden from the notation

[queen prism]

true

[plush folio]

yep, I noticed that if we consider {0} U { 1 / n | n \in N } in [-1, 1] then it's closed and therefore compact, but no longer discrete. So everything seems to make sense

[gritty widget]

Well, they must all be locally connected locally path-connected spaces, and they’re T1 iff all the quotient maps collapse each Sierpinski space to a point, since every map from the Sierpinski space to a T1 space is constant(and this characteristizes T1 spaces). Also, every coproduct of extremally disconnected spaces is extremally disconnected(I think..?), so every such space is extremally disconnected, because the only quotients of the sierpinski space are itself and the point, and they are all locally path connected and locally connected, and the only T1 ones are if the quotient maps collapse each Sierpinski space to a point, which would just be a discrete space

So far the necessary properties would be:

Alexandrov

Extremally disconnected

No clue about sufficient properties(aside from trivial ones like “discrete”)

[gritty widget]

Also they must be finitely generated(in the sense of the Alexandrov topology), it’s preserved under quotients and every coproduct is Alexandrov because the smallest neighborhood around a point in a coproduct is just the Sierpinski space it belongs to(if it’s the closed point), or the point itself if it’s the open one

Oh most of these properties are implied by this one anyway lol

[iron bolt]

yep, Alexandrov-discreteness definitely seems the most relevant here

[gritty widget]

The only property I mentioned earlier that isn’t implied by that is extremally disconnectedness

[iron bolt]

I wonder if it's called finitely generated because it's the coreflective hull of all finite topological spaces?

[gritty widget]

It seems reasonable enough to conjecture the coreflective hull is the extremally disconnected alexandrov spaces, I gtg to work soon so I’ll try working out some more properties and/or a (dis)proof later

[gritty widget]

Yeah, and it’s coherent with the finite subspaces

[iron bolt]

it's also the only one I'm not yet fully convinced of, lol

[gritty widget]

A set is open in it iff the intersection with all finite subsets is open or something

[iron bolt]

yeah, that would mean it's the coreflective hull

[gritty widget]

Yeah

[iron bolt]

nice

[gritty widget]

Yeah now that I think about it, I was only considering the open sets as the disjoint pieces of the union. I have no clue if that’s even actually true lmao

[iron bolt]

ok, I think all finite topological spaces actually lie in the coreflective hull of the Sierpinski two-point space

which is nice because it means that that hull is just all Alexandrov-discrete spaces

fun corollary: all Alexandrov-discrete spaces are delta-generated and in particular sequential & locally path-connected... simply because the Sierpinski two-point space is a quotient of the unit interval

every finite topological space is a quotient of a CW complex

[unreal stratus]

Oh lol

[unreal stratus]

This amusing to me because "finite space" has a different meaning in homotopy theory

(Which also has smth to do w cw complexes)

[iron bolt]

huh

I can't find it, what does it mean there?

[unreal stratus]

Spaces weakly homotopy equivalent to a finite cw complex

(I guess some might omit the weakly)

(You can also make this definition like internal to the homotopy theory of spaces since cw complexes can)

[gritty widget]

Yeah

That makes sense ig

[gritty widget]

Yeah, a space is extremally disconnected iff every point has a minimal open set, so coproducts of the sierpinski space are extremally disconnected

[iron bolt]

"every point has a minimal open neighbourhood" is just Alexandov-discreteness, no?

extremally disconnected is when the closure of every open set is open

that's preserved under coproducts too, but idk about quotients

[gritty widget]

Wait shoot

Let me recheck

Yeah no you’re right

It’s not

[iron bolt]

not preserved under quotients?

[gritty widget]

Yeah

[iron bolt]

yup. so no coreflective subcategory then

[gritty widget]

https://math.stackexchange.com/questions/1991411/quotients-of-extremally-disconnected-compact-hausdorff-spaces

[gritty widget]

So the coreflective hull for the Sierpinski space is just the alexandrov-discrete spaces right?

[iron bolt]

yep

[gritty widget]

Yknow

Is the coreflective hulk of the subcategory of Euclidean spaces just the category of manifolds(not necessarily Hausdorff or second countable, like, just locally Euclidean spaces)?

[iron bolt]

it's this, actually

[gritty widget]

Ah

That makes sense since how CW fomplexes are defined

[iron bolt]

yep

they're colimits, and coreflective hulls are closed under colimits by construction

[unreal stratus]

Delta generated sets are cute

[iron bolt]

is that what you call them to distinguish them from higher categorical variants?

[unreal stratus]

Wdym

[iron bolt]

delta-generated sets instead of spaces

[unreal stratus]

Oh sorry mistake lol I meant spaces

This was basically cause I had delta sets in mind too lol

[iron bolt]

lol. would've sort of made sense though

[gritty widget]

Is the reflective hull just products and subsets

[iron bolt]

I hear some people call smooth spaces smooth sets to emphasise that they don't have much homotopy going on

[unreal stratus]

I guess really like in higher category theory all of these guys are the same lol

[gritty widget]

[unreal stratus]

Just "different models"

But I think delta-generated spaces seem rly cool and maybe underrated lol

[gritty widget]

I guess that’s a lot less interesting

[unreal stratus]

Like I wonder if they would be more popular if they had been discovered/studied 20 years earlier or smth lol

Like rather than cgwh spaces, say

[iron bolt]

they do seem pretty popular among the nlab crowd at least

[unreal stratus]

Yeah fair

[iron bolt]

less so outside of that though 😔

[iron bolt]

honestly? I don't know

I would've expected it to work dual to coreflective hulls, but that would mean that the reflector just changes the topology of each space without changing the underlying set

but as far as I can see, the reflector of Hausdorff spaces for example actually involves taking a quotient?

I'm not sure how to make sense of that. one of those two can't be right

[gritty widget]

Huh that’s unintuitive

Hmmm

[iron bolt]

in that specific context it's intuitive enough, you just quotient out point that have no disjoint neighbourhoods

[gritty widget]

What are some good pointset properties that are perserved under colimits? 🤔

[iron bolt]

but abstractly it's not what I would have expected

[gritty widget]

Connectedness/Compactness ofc

[iron bolt]

nope, lol

[gritty widget]

Isn’t Local Compactness not preserved in general

[gritty widget]

Oh yeah lmao

Disjoint union isn’t connected

[iron bolt]

and not compact

[gritty widget]

And infinite union of compact spaces need not be compact

Shoot

Local connectedness is one right

[iron bolt]

connected spaces are preserved under connected colimits iirc

[gritty widget]

Discreteness is a pretty trivial one haha

[iron bolt]

discrete, almost discrete, Alexandrov discrete, delta-generated, sequential, compactly generated, locally connected and locally path-connected are the ones I remember

oh, there's also one really trivial one, lol

emptyness

[gritty widget]

True lmao

hmmm what else

Man I love http://topology.pi-base.org for this stuff fr

[iron bolt]

right, "True" is another one

totally what you meant there

[gritty widget]

Is first countability preserved by quotients?

[iron bolt]

nope

quotients of first-countable spaces are actually just the sequential spaces again

[gritty widget]

Ah

Darn

[iron bolt]

same for metric spaces

I think it's kind of cool that so many of these hulls agree

[gritty widget]

Is there any non-trivial one whose hull is just Top

[iron bolt]

probably lots, but I don't have a good example

"nonempty" is one, but I guess that's still trivial

[gritty widget]

Yeah I mean one that isn’t basically just “almost every space”

Is every space a quotient of a Hausdorff space?

[iron bolt]

honestly, no idea

[civic verge]

Looks interesting

[balmy briar]

Could someone help me with c). If I fix some x \in X, then there exists some U_x which intersects the family for finite sets A_1,A_2,...,A_n. idk what to do from here

I think also $U_x = \bigcup_\alpha ( U_x \cap A_\alpha)$

[gentle osprey]

XDStar

[iron bolt]

do you know that if f is continuous on each U_x, it is continuous on all of X?

[balmy briar]

why is that

[iron bolt]

another lemma. it essentially reduces c) to a), because then you only need to show continuity on each U_x

this "continuous on each set of the covering => continuous on X" works both for locally finite closed covers and arbitrary open ones

I'm a bit surprised that the exercise doesn't include that, but it also shouldn't be too hard to show without it that f being continuous on every U_x is enough

[balmy briar]

this lemma?

[paper wedge]

I think munkres proves this

[iron bolt]

nope, that's just one of the equivalent definitions of continuity

[paper wedge]

in this section

iirc

and then gives the closed version (finite one) as an exercise

[iron bolt]

I mean "if U_i is an open cover of X and f : X → Y is continuous on each U_i, it is also continuous on X"

[paper wedge]

yes im almost sure munkres proves htis

[paper wedge]

anyways, try to write out the condition u need to have f continuous

u should have some sort of intersection/union. u wouuldn't be able to conclude since u are going to be dealing with a non-finite collection of closed sets

now if u go back a couple sections u see that u cna prove that for a locally finite collection of closed sets, they respect the closure under intersection iirc

u should be able to use this to finish up the proof

[paper wedge]

@balmy briar do u see how u prove this

[balmy briar]

im not sure

[paper wedge]

if you have a covering of your space

and you have some open set B, try to write f^-1(B) as some union of f^-1|U_i some-how

this would imply its open and u would be done

now this doesn't work with an infinite collection of closed sets cuz the union of those isn't closed

so then u might say to urself "well this locally finite condition must make this happen"

and indeed a union of a locally finite collection of closed sets is closed

if u prove that u would be odne

[paper wedge]

after seeing how to prove this

[iron bolt]

small nitpick: (f|U_i)^-1

[paper wedge]

yes

sorry

[iron bolt]

all good, just wanted to rule out one more possible source of confusion ^^

[paper wedge]

yeah ur right i should learn latex

[iron bolt]

the basic syntax is quite simple actually. basically just write your equation normally and use curly braces to put stuff in parentheses if necessary

[paper wedge]

i can do stuff in over-leaf where like one can see the output in the code by hovering around it

b4 compiling it

but doing it raw like this is just impossible for me

[iron bolt]

like, $f^{-1}|_{U_i}(B)$ already works

[paper wedge]

yeah

[iron bolt]

remembering all the commands for extra stuff is the more tricky part

[gentle osprey]

undefined

[iron bolt]

wait, now I've put the ^-1 in the wrong place too 😆

oh well

[paper wedge]

haha

[heady skiff]

Is there a reason why invariance of domain is not stated as a blackbox theorem in most point set texts (besides the proof involving algebraic topology)? It seems as if it's very powerful and would be nice to have when working in a point set context

[paper wedge]

ig to do this u would need like

a very big part of ur theory depending on it

which probably isn't the case with point-set

maybe when talking about manifolds then u might say that cuz without it u cant even say that the dimensino ofa manifold is well-defined

and so u would just say well okay this is true but we won't give the proof now

[heady skiff]

Yeah I guess that makes sense

[unreal stratus]

Hm yeah I feel like uh

If you want to work with smooth manifolds you don't need invariance of domain (at least initially) as it is clear for smooth maps etc

And if you want to work seriously with topological manifolds you should probably know some algebraic topology

to the point where invariance of domain is smth you'd know anyway

[paper wedge]

yeah ig ur right

other-wise ur not doing manifolds, ur really doing multi-variable calculus and rank theorem

[unreal stratus]

And ye I agree with this

[paper wedge]

without AT i mean

[unreal stratus]

Oh I mean you can do some smooth mfds stuff beyond that lol tbf

[paper wedge]

if u thikn about it however thats how most courses in diff topology begin

[unreal stratus]

But I mean like doing pure topological manifolds

[paper wedge]

things like diff forms and de-rham?

yeah like lee's textbook does not assume much AT if any

[unreal stratus]

Yeah exactly

[paper wedge]

and yet is a standard ig

but i think ur talking about the more serious things like

hirsch differential topology or smth

which then yeha

but its kinda weird tho cuas

its hard to have motivation for AT without having an exposure to manifolds ig

cuz AT is just a bunch of cool tools at the end. like imagine learning AT untill lke spectral sequences and characteristic classes but haven't dealt with manifolds at all

[civic verge]

Steven H. Weintraub (Auth.) - Differential Forms. Theory and Practice

[paper wedge]

i think it would be dry for someone who wants to do low-dim stuff for examplep idk

but yeah at the same time, alot of the things in manifold courses can be done much more cleanily with AT. things like orientation/degree stuff.

[unreal stratus]

Well this is smth people should not allow to happen aha

[paper wedge]

yeah manifolds are not that bad to define and alot of cool manifold theory comes from bundles which can be taught in contexts of AT like milnor and stasheff ig

[balmy briar]

okay i got it thank you guys

[plush folio]

Can I say that q^-1(U) is evenly covered iff every component of q^-1(U) is homeomorphic to U (based on the definition given here)?

[plush folio]

Hmm, I think we need every component of q^-1(U) to not just be homeomorphic to U, but be mapped homeomorphically by q to U

Which I guess is why the exponential map from E = (0, 2) to S¹ isn't a covering map. If we take a small neighbourhood U around 1 in S¹, then every component of f^-1(U) is homeomorphic to U, but it's not mapped homeomorphically to U by f

[crimson smelt]

Is my proof correct?
Let Y be a subspace of a topological space X. Prove that Y (equipped with the subspace topology) is Hausdorff if X is Hausdorff.

Proof by Contradiction
Suppose X is Hausdorff and Y is not Hausdorff. Then for all distinct x,y in X, there exists $U_x, U_y$ in $T_x$ such that $x \in U_x$ and $y \in U_y$ and $U_x \cap U_y = \emptyset$.

Similarly There exist distinct a,b in Y such that $ \exists U_a, U_b \in T_y$ such that $a \in U_a$ and $b \in U_b$ and $U_a \cap U_b \neq \emptyset$
Now $\forall d, e \in X, U_d \cap U_e = \emptyset$ in X but also $ U_d \cap Y U_e \neq \emptyset$ in Y which is a contradiction

[gentle osprey]

rabbits_advocate

[ruby delta]

your proof seems really roundabout - there is no need for contradiction here, since open sets in X when intersected with Y is open in Y

so disjoint neighbourhoods in X can be made into disjoint neighbourhoods in Y

[waxen oxide]

Let $X$ and $Y$ be sets with some topologies, $f: X \rightarrow Y$. if $U \subset Y$ is open, such that $B \subset U$ is a basis element for the topology of $Y$, and you show that $f^{-1}(B)$ is open, have you shown $f$ is continuous?

[gentle osprey]

Luke

[tender halo]

if you are asking whether just proving that preimages of basic open sets are open is enough for continuity then the answer is yes

[rancid umbra]

if you can do this for each open U and for each B contained in U, then yes

this is because U is the union of basic open sets in Y

[young stone]

in U*

[rancid umbra]

sure, was specifying that we were working in the codomain tho

[median sand]

If X is regular (points can be separated from closed sets), then the standard proof of Urysohn's lemma goes through verbatim to prove that for any closed A and x\notin A there exists f:X->[0,1] with f|A=0 and fx=1, right?

[tender halo]

no, that property is called complete regularity and its strictly stronger

[tender halo]

its fine lol we all make stupid mistakes

[keen gust]

What is this example on lol?

Should i just simply say

(a, b) <-> (b, a)

just like in the book

confused

[tranquil cosmos]

yeah

[lilac musk]

Yes, this map is indeed a bijection (in either direction)

[keen gust]

alright thanks i just wrote (a, b) <-> (b, a) as the answer lol

For this i wrote

(a_1, ..., a_n-1, a_n) <-> ( (a_1, ..., a_n-1) , a_n )

Is that right

[unreal stratus]

is this munkres chap 1

[keen gust]

ye

[heady skiff]

Yeah

[queen prism]

prob the intention was to also ask for statements showing that this is a function and it's bijective

but yea there's not much to say here

[gritty widget]

What are some uses of the Baire category theorem(for locally compact regular spaces)?

[tender halo]

(BCT for locally compact spaces and BCT for complete spaces are actually the same theorem)

[gritty widget]

I’m assuming you mean that like their proofs are similar/nearly identical or whatever

Regardless, are their any notable usages of the fact locally compact regular spaces are Baire?

[tender halo]

you can prove BCT for Cech-complete spaces, and complete spaces and locally compact spaces are both Cech-complete

[gritty widget]

Off of a quick Pi-base search, the irrationals are Cech-complete and not locally compact. What definition of locally compact are you using?

Oh you’re probably assuming T2.

BCT applies to all locally compact regular spaces, not just locally compact T_3 ones.

[tender halo]

the irrationals are completely metrizable

[gritty widget]

Oh I got that backwards lmao mb

[gritty widget]

Regardless, this is still true

[tender halo]

yeah, anyway idk about your actual question sorgy

[gritty widget]

Aight, no problem. Also, thanks! I will definitely look into BCT for Cech complete spaces.

[vernal gate]

being in a topological space is a necesary condition for convergence, right?

[lavish bluff]

yes

u cant even define convergence w/o a topology

[vernal gate]

I see, thanks .

[gritty widget]

Why are continua relevant enough to be defined as their own thing?

Like

what’s so special about compactness+connectedness+Hausdorffness

[iron bolt]

convergence is about filters really

neighbourhood filters are the main example - to say that a function f converges to y at x is to say that the preimage of every neighbourhood of y is a neighbourhood of x, i.e. that the image of the neighbourhood filter at x contains the neighbourhood filter at y

but even in basic real analysis you already encounter other filters, just without calling them such. convergence of a sequence in a space X to x for example means precisely that the neighbourhood filter at x is contained in the image of the "filter at infinity" on N

or for example that a real function f : R → R tends to +∞ as x tends to -∞, that it tends to y as x tends to 0 from below etc. are all statements about certain filters

[iron bolt]

compact Hausdorff spaces have the property that continuous maps between them are automatically closed, so in particular continuous maps in them are homeomorphisms iff they are bijections - or phrased categorically, the forgetful functor CompHaus → Set reflects isomorphisms while the Top → Set one does not

for connectedness idk, I guess that's also just a nice property for a space to have? 💀

not really a continuum otherwise

[queen prism]

hausdorff spaces became more special to me when I found out that hausdorff = every convergent net/filter has a unique limit

[gritty widget]

Ig that’s neat

[iron bolt]

is that an actual equivalence for nets?

for sequences it's just an implication

[queen prism]

yea it is, see this problem from folland (pg 127) for example

might also be that a definition somewhere was cooked up to make this happen but either way I really like it

[storm night]

Suppose τ₀ ⊆ ℙ(A) and τ₁ ⊆ ℙ(B) are topologies. Is the following a correct definition for the product space?
{S ∈ ℙ(A × B) | (∀ a ∈ A, {b ∈ B | (a, b) ∈ S} ∈ τ₁) ∧ ∀ b ∈ B, {a ∈ A | (a , b) ∈ S} ∈ τ₀}

Similarly, here's how I defined the disjoint union of two topologies.
{S ∈ ℙ(A ⊔ B) | {a ∈ A | inl(a) ∈ S} ∈ τ₀ ∧ {b ∈ B | inr(a) ∈ S} ∈ τ₁}
Where inl : A → A ⊔ B and inr : B → A ⊔ B are constructors for the disjoint union of sets.

[iron bolt]

inl and inr? the only place I've seen that name for them before is lean, are you trying to formalise something?

the disjoint union one is correct, the product one isn't

[storm night]

I already did. https://github.com/jhimes2/Math

[iron bolt]

ah, agda

that's one I've been thinking about trying for a while now, but so far I haven't found the time

[storm night]

Look in the classical directory, where you'd find topology.agda

I proved they are both topologies, but I'm not sure if it's the right topology.

[iron bolt]

the disjoint union one is the correct topology, the product one isn't

it's finer than the product topology

consider ||in R^2 the complement of {(x,y) | x = y } \ {(0,0)} ||

it's open under your topology but not open at the origin under the product topology

(by which I mean, the origin lies in the set but not in its interior)

[storm night]

I'm making sure I'm on the right track. The complement of the set to consider is {(x,y) | x≠y}∪{(0,0)}. You're saying that ∀ a, {b | (a,b)∈({(x,y) | x≠y}∪{(0,0)})} ∈ Γ where Γ ⊆ ℙ(A) is the topology for ℝ. I'm currently figuring out if the following statements hold:
{b | (5,b)∈({(x,y) | x≠y}∪{(0,0)})} ∈ Γ
{b | (0,b)∈({(x,y) | x≠y}∪{(0,0)})} ∈ Γ

[iron bolt]

that first set is R \ {5}, while that second set is all of R

both are open

you could even do something more dramatic like take the complement of {(x,x) | x ∈ Q}

the point really is, the set is open no matter what you remove from the diagonal, because the diagonal intersects every vertical or horizontal line only at one point

so your definition of openness can't detect any information about the subset of the diagonal that you remove

[storm night]

Second attempt. Is this the definition of a product space?
Suppose τ₀ ⊆ ℙ(A) and τ₁ ⊆ ℙ(B) are topologies.

  {S ∈ ℙ(A×B) | ∀(a,b) ∈ S. ∃ X ∈ τ₀. a ∈ X
                          ∧ ∃ Y ∈ τ₁. b ∈ Y
                          ∧ ∀ x ∈ X. ∀ y ∈ Y. (x,y) ∈ S}

[median sand]

If x_n is a sequence in a compact space X with all elements distinct that has a unique limit point x\neq x_n, then necessarily x_n->x, right?

[tender halo]

i dont think so

[median sand]

How so? Fix a neighbourhood U of x, because each x_n is not a limit point of {x_n} there is a neighbourhood U_n separating it from all other x_m. This gives you a cover {U_n}\cup{U} of the compact closure{x_n}={x_n}\cup{x}. The finite subcover cannot be U_n1,...,U_nk by construction, so it must be U_n1,...,U_nk,U and U contains all x_n with n>n_k.

[tender halo]

ah hm

im tripping

yea ur right

[rancid umbra]

yes, it is a correct description of the box topology.

a more concise way to describe this set would be
{S ∈ ℙ(A x B) : ∀ (a,b) ∈ S. ∃ X x Y ∈ 𝜏_0 x 𝜏_1. (a,b) ∈ X x Y ⊆ S}

an easy way to see that it is correct is that the basis 𝜏_0 x 𝜏_1 generates both topologies

[storm night]

What is an example of two compact topologies that are not compact under the box topology?

Would [0,1]² be compact if constructed using the box topology?

That's not what I'm asking.

Ah, I misread that as a finite topology.

[storm night]

What is an example of a set that is not open in ℝ^∞ using the product topology but open using the box topology? I like to represent countably infinite real vectors as functions ℕ → ℝ.

[storm night]

@short path Interesting. It makes more intuitive sense for me to think of an infinite dimensional "open" cube as open.

[iron bolt]

{0,1}^∞ is another interesting example of an infinite product space

it's discrete with the box topology, but with the product topology it's homeomorphic to the Cantor set

I don't think I know of a nice intuitive reason for why arbitrary products of open sets don't need to be open - it just happens to be the case that that's the topology on products with the right universal property

continuity of the projections forces cylinders to be open, and the fact that only finite intersections of open sets need to be open too means that you get only openness of finite products of open sets from those, not infinite ones

[gritty widget]

Are there any equivalent characterizations for a space to be a product of non-trivial(neither empty nor a singleton) discrete spaces?

Obviously it’d be totally separated, that’s one necessary condition

[iron bolt]

by product you probably mean an indexed/infinite product, right?

because binary products of discrete spaces are discrete

so a space is a product of two discrete spaces iff it is discrete and its cardinality can be nontrivially factored

for arbitrary products, you get some subclass of the class of all Stone spaces

nevermind about that. arbitrary products of finite discrete spaces are Stone spaces, but as soon as one of the factors becomes infinite the space isn't compact anymore

[opaque scroll]

I guess a pretty easy reason is that you want the diagonal map to be continuous.

If you consider for example
(-1, 1)x(-1/2, 1/2)x(-1/3, 1/3)x...
then the preimage of this by the diagonal map is {0}

[unreal stratus]

It's also the smallest topology such that the projections are continuous, which means defining maps into it can be done componentwise (with the diagonal map being one example)

[dark sapphire]

Hi is there someone who is interested in set theory and have had experience in writing research papers? I want to write my first research paper and I need a coauthor (the topic and content is kinda basic so no need for advanced expertise)

[dry seal]

Feels very basic but why are there seemingly two different definitions for neighborhoods? In munkres I'm seeing "U is an open set containing x" to mean "U is a neighborhood of x" but elsewhere I'm also seeing "A subset U of a topological space X is called a neighborhood of a point x if there exists an open set O such that O is a subset of U and contains x."

What's the deal?

[tender halo]

just another inconsistency in definitions, doesnt have any particular reason for it as far as im aware

[unreal stratus]

There are indeed two definitions, but it very rarely matters

Because usually when you consider neighbourhoods the idea is they are "small" and you can shrink them at your convenience etc

(I would say the first definition is more common though)

[tender halo]

sometimes it can make a statement or a proof easier

[dry seal]

Well that's fair enough ig

At least for the definition of first countability, should I be concerned about which is which or are they the same?

I guess the question is really now about the definition of basis

[tender halo]

bases are open

[dry seal]

Ah that's right

They're equivalent anyways

[tender halo]

technically having first countability is equivalent to having a countable amount of nbhds of a point that such that at least one lies inside any open set containig that points

but that would be a strange way to think about first countability

[dry seal]

Still kinda annoying how there are two different definitions for this sort of thing that's supposed to be so basic to the overall theory

[tender halo]

wait till you hear about natural numbers

[dry seal]

I mean there's rarely confusion there in my experience

[tranquil cosmos]

Are there even fc spaces that aren't sc that show up in nature

Sounds like a case for the long line again

[dry seal]

Authors are usually careful with saying positive integers etc

Ok but wait we're defining convergent sequences in general topological spaces using neighborhoods

[tender halo]

rarely confusion here too

[dry seal]

And at least in munkres it doesn't say open neighborhood

[tender halo]

doesnt matter which one you use, convergence is the same

[tender halo]

discrete spaces

[tranquil cosmos]

o.

[dry seal]

How? In one definition you're taking just subsets of X but in the other they each need to be open. Is it because you can just take the required open subset of each nbd in that definition?

[tender halo]

river metric too

[tender halo]

yep

[dry seal]

Wouldn't there be the possibility of there being some overlap somewhere?

[tender halo]

if your sequence is eventually in every nbhd, then its obviously eventually in every open nbhd, and vice versa

[dry seal]

I can see the first direction but not the other

I don't see how being in every open nbd implies being in every nbd

[tender halo]

because every nbd has an open nbd inside

and you are eventually inside that

[dry seal]

Ah ok it makes more sense now

Thanks

My intuition for point-set really needs to improve

[dry seal]

So I know that countable compactness is equivalent to sequential compactness in first countable spaces but is there some analogous condition for countably compact to compact?

Other than being metrizable ofc

[faint pewter]

Hello all!

I need a little help with a problem

I have a topology problem on the topic of convergence and the subtopic of nets.

In that problem I have already shown that the proposed relation is a pre-order and I have also shown that the pair is a directed set, without any problem, it was easy.

But I would like help to show that certain parametric expressions satisfy the previously defined relationship.

I attach an image with the necessary definitions and the exercise.

[tender halo]

in second countable spaces they are the same

or like in lindelof spaces in general

[faint pewter]

I really appreciate your attention and willingness to help.

[tender halo]

countable spaces also count

[dry seal]

Yeah Lindelöf was the obvious one

[faint pewter]

Reference: Tej Bahadur Singh - Introduction to Topology, 2019 (Springer), section 4.2: Nets

[dry seal]

What about Tychonoff? Seems "nice" enough and I can't seem to find counterexamples

I am waiting on my counterexample book to come in lol

[tender halo]

thats definitely not close to enough

[dry seal]

Ah but that wouldn't make sense yeah

Because then sequentially compact implies compact

In non metric spaces

[tender halo]

you can't do better than lindelof though

if your space turns out to be compact, then it will be lindelof

[dry seal]

That does make sense but it's a bit disappointing since it's so "obvious" lol

[devout karma]

This should be a simple intro question but I am stupid

(a) I've shown. Assume T is the standard product topology, P is some other topology satisfying the UP, X is the cartesian product of X_1...X_k

We know $\pi_k : (X,T) \twoheadrightarrow X_k$ is continuous through verification, Thus $\mathrm{id} : (X,T) \rightarrow (X,P)$ is continuous by the Universal Property on $T$, equivalently meaning that $P$ is finer than $T$

[gentle osprey]

Mizalign (Study acc)

[devout karma]

but we don't know that $\pi_k : (X,P) \twoheadrightarrow X_k$ is continuous a priori

[gentle osprey]

Mizalign (Study acc)

[iron bolt]

the projections are the component functions of the identity

[gritty widget]

discrete spaces

[devout karma]

I know that. But P some topology on the cartesian product. We don't know a priori that the projections of the constituents of P are open in the respective topologies

T has been explicitly described via the basis of boxes, so we can show explicitly that the projections are open continuous maps

[iron bolt]

idk what you mean by constituents, but if P satisfies the universal property, the projections to the factors are continuous because they're the components of the identity function, which is continuous

[rancid umbra]

the factor spaces of X have their original topologies.

if (X, P) satisfies the universal property, then it is equipped with continuous projections

[devout karma]

how do we show that the projections are continuous off the UP

i.e that the projections of the open sets of this other topology satisfying the UP are in the respective topologies

we know it's true for the constructed box topology

[iron bolt]

that's not what continuity of the projections means

[devout karma]

if we have a statement like "the coursest topology satisfying..." then

[iron bolt]

a function is continuous if all preimages of open sets are open

that's not equivalent to all images of open sets being open

[devout karma]

preimages of open sets in X_i under the projection are open in the new topology P, i.e the "tubes" are in P

which once again

we need to show that they are off the UP

we know that it's true for T, the topology we constructed off of boxes / products of open sets

[iron bolt]

just think about what the universal property means in the case F = id

[devout karma]

that's how I got the one direction.

that P is finer than T