point-set-topology

[prime elbow]

In Box topology, R^∞ is closed.
In Product topology, R^∞ is dense, is it correct?

[paper wedge]

yes

[civic verge]

name of book?

[silver ridge]

Lee itm

[elfin fable]

Does anyone have a link where they show that the any of the topological conditions/definitions of a set X being dense in R is equivalent to the usual analysis definition of density in R (for all numbers a,b in R blah blah such that a < x < b)?

[red yoke]

Usual defn = every open interval intersects X = every non-empty open set intersects X

Since open intervals generate the topology

[elfin fable]

That's not the definition I'm refering to, tho. I'm refering to the one that says that for all numbers a, b in R (a < b) there exists a number x in X such that a < x < b, as I've explained in the brackets

[red yoke]

Well open intervals are {a < x < b}

[elfin fable]

Yes and their infinite versions

Okay so if every open interval intersects (a,b) that means that there is always an element inside (a,b)

Hmm okay thanks!

[ruby hinge]

Hi, I've been trying to prove this statement. I've been trying to show that the preimage of $\mathcal{B}_Y$ is a basis for $\mathcal{T}_X$. Is this the way to go?

[gentle osprey]

joel

[ruby hinge]

Is it true that $\bigcup_{B \in \mathcal{B}Y} f^{-1}(B) = f^{-1}(\bigcup{B\in \mathcal{B}_Y} B)$?

[gentle osprey]

joel

[ruby delta]

That’s not necessarily true

For instance the constant function is continuous, but as long as the topology is not trivial than the preimage of a basis is never a basis

[ruby delta]

But then, what’s your definition of continuity?

[ruby hinge]

for any open set, the preimage is open.

like the normal topological definition for continuity

[ruby delta]

So… how does it not follow immediately from the definition

[ruby hinge]

of continuity?

[ruby hinge]

Maybe use this?

[ruby delta]

You don’t even need that, it’s just a straightforward application of the definition

[ruby hinge]

I mean --> follows directly, sure. But what about <--?

[ruby delta]

All you want to show is that U open -> f^-1 U open
Oh, I see yeah you use that, then it follows

[unreal stratus]

Yes

[ruby hinge]

Right?

[unreal stratus]

Basically you just want to write a given open set as a union of basis elements

Now compute the preimage

[ruby hinge]

Like, let U be an open set. then U is a union of basis elements. then union f^{-1}( B) is open

[ruby hinge]

yeah alright 👍

[unreal stratus]

This is not useful because both sides are just X

[unreal stratus]

Exactly

[ruby hinge]

alright, thanks!

[balmy briar]

What does this question mean? are X and X' different sets?

[opaque scroll]

They are the same set, but equipped with different topologies

[balmy briar]

and these are the topological spaces right

[opaque scroll]

Yes, a set together with a topology is a topological space

[balmy briar]

yeah, okay thanks

[unreal stratus]

Yeah this is an unfortunate case where there is a bit of abuse: often given a topological space, which one might formally write as the pair (X, \tau), you write X for the space by slight abuse. But then if you want to consider the same set with two topologies that can be annoying and hence stuff like this

In algebraic geometry for example it is common to say "a scheme X" and then write |X| for the underlying space

[quiet summit]

What does lambda_p^a supposed to mean, is it some kind of function?

[unreal stratus]

Looks like a good typo lol

[opaque scroll]

My guess would be it's supposed to be $\lambda p^a$

[gentle osprey]

jagr2808

[unreal stratus]

Yes

[opaque scroll]

That would make the p-adic topology at least

[quiet summit]

yeah i think so too, thanks

[heady skiff]

Let $n > 1$, and let $e_0, e_1 \in S^n$ and $a_0, a_1 \in S^n$ be distinct points. What is a homeomorphism between $S^n \setminus {e_0, e_1}$ and $S^n \setminus {a_0, a_1}$? I thought of
[
f(x) =
\begin{cases}
x \text{ if } x \notin {a_0, a_1} \
e_0 \text{ if } x = a_0 \
e_1 \text{ if } x = a_1
\end{cases}
]
However continuity is not obvious to me, since the glueing lemma doesn't necessarily apply (${a_0}$ and ${a_1}$ are closed, while $S^n \setminus {a_0, a_1}$ is open)

[gentle osprey]

okeyokay

[young stone]

This map isn't continuous

[heady skiff]

Nvm then

[young stone]

I'm not sure how you would construct an explicit homeomorphism without specifying the points

[mighty hull]

S^n with a point removed is homeomorphic to ℝ^n

[silver spruce]

easiest way to go about this would be stereographic projection i think

[mighty hull]

Yes

[heady skiff]

oh yeah I always forget about stereographic projection lol

[prime elbow]

If J is infinite, we can take an basis element B(x,eps) in uniform topology but this is not open in product topology.

And for box topology I think we can find the open set which is not bounded such that it cannot be open in uniform topology

[unreal stratus]

Honestly this is sort of annoying to write down imo. You can rotate stuff to assume e_0 = a_0, and then you reduce to the assertion between removing two points in R^n, but then you can just translate to get a homeomorphism

Then you can stitch all of those together

More generally, for any manifold M (in dim >= 2 i think?) and any finite ordered lists X, Y of points (of the same cardinality), you can find a homeo from M to itself taking X to Y

[heady skiff]

Okay okay thank you

At what point can one just blackbox these explicit homeomorphisms and rely on visual justification

(since I'm working through an alg top textbook)

[glossy wagon]

Hello

With a manifold, can we always reconstruct a total continuous map from the space into R^n I guess using just translations and scalings and some other "movements" xd

By movements I don't mean just movements

[unreal stratus]

Well it's not entirely obvious you can get an embedding into R^n

[glossy wagon]

Yeah I mean a general continous map just patching it up with the open balls or something, need not be injective

It seems like so to me but in this book they're defined not as being "locally homeomorphic to R^n" but "each point having a neighborhood homeomorphic to (an open subset of) R^n" so I wasn't sure if these 2 basically came out to be the same

[unreal stratus]

That's the same thing as any open subset of R^n contains an open subset homeomorphic to R^n

[glossy wagon]

I mean the part where the second definition says nothing about the existence of a continuous function on the whole space that facilitates this local homeomirphism

[unreal stratus]

Well the same is true of the first

[unreal stratus]

Idk what you mean by this

[glossy wagon]

Like the first part says there's a function X -> R^N that is a local homeomorphism

Second says that each point x of X has some open neighborhood U_x that has its own homeomorphism f_x : U_x - > R^n

So can we always combine the f_xs to get an f?

[unreal stratus]

That isn't what people usually mean when they say locally homeomorphic to R^n

[unreal stratus]

It means this

[glossy wagon]

But from the first you get the second right?

[unreal stratus]

Sure

[glossy wagon]

So can you go backwards

With some choice maybe

[unreal stratus]

Well there's the Whitney embedding theorem etc

(Really that is stronger than what you need but yeah)

Any manifold can be embedded into R^N for some large N

Well maybe actually no if you want the same n then that probably shouldn't be possible

[glossy wagon]

It doesn't have to be injective

[unreal stratus]

No I mean like

There does not exist a local homeomorphism S^1 -> R for example

[glossy wagon]

Ah truu

[unreal stratus]

(And similarly for any closed manifold)

[glossy wagon]

I see thank you

[plush folio]

The Whitney embedding theorem is only for smooth manifolds, right?

[plush folio]

I see, thanks

[unreal stratus]

sure ye

i guess maybe this is bad but when people say manifold it seems safe to assume they mean smooth

but then again this is #point-set-topology...

[lilac musk]

How does one show that the projection map X × Y → X is a closed map when Y is compact?

[lilac musk]

Thanks

I get it now

[prime elbow]

#1332828732578074754

[tranquil cosmos]

Oh I know this, this is really cool

[prime elbow]

Yes

[prime elbow]

let X be non-empty set, and A \notempty be subset of X, then T = { U \subset of X | A \subset U } u {empty set }, makes topology, A will be dense in this topology right?

[plush folio]

Yes, all non-empty open sets contain A, so the only closed set containing A is X

[plush folio]

Just need a few sanity checks here: X being path connected is equivalent to saying that any two constant maps on X are homotopic, right? And homotopy is preserved by pre- and postcomposition, ie. if f ~ g, and h is a map with appropriate domain/codomain then f o h ~ g o h and h o f ~ h o g, correct?

[granite slate]

stupid question

Let (X, d) be a metric space and A is a non-empty subset. Define $f_A(x) = inf_{y \in A} d(x, y)$. Prove that $x \in \bar{A} \Rightarrow f_A(x) = 0$.

Surely there's some added condition about connectedness missing from this question? If A = [0, 2] $\cup {3}$ for instance then 3 is in the closure and f(3) isnt 0?

[gentle osprey]

Dompa

[prime elbow]

f(3) = 0

See 3 in A, f(A) = 0

[granite slate]

oh right how silly of me i thought y and x had to be distinct

[drifting ravine]

Nope, if $x\in \bar{A}$, then there is a succession of points of $A$ which accumulates on $x$; this realizes $0$ as the infimum

[gentle osprey]

Enrico

[drifting ravine]

No need to assume connectedness of any kind

[prime elbow]

Any hint to find a uncountable collection of pairwise disjoint open sets in lower limit topology on R?

[hoary ginkgo]

Isn't it impossible? 🤔

[rancid umbra]

if you could do that, you could prove that Q is uncountable

[haughty jungle]

sorry if this question is really stupid but can someone help me see real quick why [a, b] isn't open in the standard topology

can't it be represented as the union of (a + 1/n, b - 1/n) as n approaches infinity?

infinite union of open sets so open i thought

but it doesnt seem like closed intervals are open

but if u make it infinite doesnt it include them cause they r the limits

[paper wedge]

i think if you switch the minus and +

and take intersection

then it would be [a,b]

but that doesn't mean it's open either

[haughty jungle]

why is it not open if it can be writen as an infinite union of open sets

[paper wedge]

it can't

it would be an infinite intersection of open sets which need not be open

[haughty jungle]

holup lemme consider

n approaches infinity

[paper wedge]

a isn't in (a,b)

[haughty jungle]

wait nevermind

i think im dumb

yea 💀

ok ty

like you can take the union as much as you want but you never get the endpoints

i js got sloppy with the definitions

this is rly helpful now ig i see why infinite intersection isn't allowed

🙏

[gritty widget]

Yes to both, a path is literally a homotopy between points(when considering {a}x[0, 1] as [0, 1])

Actually I don’t know the latter off the top of my head, thought you meant path homotopy concatenation.

I assume it’s true but I don’t have a proof

Let H be a homotopy f~g, let h•f and h•g be defined

Let J(x, t)=H(h(x), t)

Shouldn’t be difficult to prove this is continuous, no?

[plush folio]

Yeah, that was the proof I had in mind, but wasn't 100% sure it was correct. Thanks

[haughty jungle]

i lowk have beef with the pin that says munkres is popsci for nerds cause it's getting rly hard for me 😭

[plush folio]

My lecturer told me something crazy, that every algebraic invariant on top spaces factors through the homotopy category. I only know the fundamental group right now, but it seems like a pretty powerful statement, looking forward to learning more alg top

[plush folio]

I also found Munkres kinda hard, I recommend checking out Lee's introduction to topological manifolds if you want something different

[haughty jungle]

I mean I'm going to work through all the point set stuff of munkres

after that i hear rotman + hatcher is a great alg top combination

what does lee cover

[plush folio]

Lee covers almost the same as Munkres, but a bit more focused on manifolds

I prefer the exposition in Lee, and I don't think there's anything very important that isn't covered in Lee

[haughty jungle]

hm ok

the exposition in munkres is like fine it's just kinda hard to like juggle all the new definitions and objects in play yk

you prolly need a couple times through this stuff to actually understand it well

[plush folio]

Yeah, it didn't really click for me until well after I finished my first topology course

There's just a handful of concepts that are really important, which you'll get very used to eventually, like connectedness, compactness, the separation axioms etc. But it feels like a lot to begin with

Oh, one thing Lee does that I really miss in Munkres: he highlights the characteristic property of the subspace topology, product topology and quotient topology. Munkres mentions maybe only the latter in passing. I feel like these are pretty important to know

Another good source is chapter 1 here: https://topology.mitpress.mit.edu/
They show various equivalent ways of defining the product, subspace and quotient topologies

[gritty widget]

munkres is goated fr

[gritty widget]

I’m not entirely sure what this means

Most algebraic invariants are only up to homotopy equivalence if that’s what that means

[cerulean oriole]

It most likely means the slightly stronger statement that homotopic maps induce the same homomorphism of the algebraic invariant.

[plush folio]

I'm not sure either, but he drew a diagram, I think a functor from Top to an algebraic category like Grp or something, and it factored through hTop. I guess the important takeaway is what you say, that they are defined up to homotopy equivalence

[cerulean oriole]

Yep, that amounts to what I said.

[plush folio]

And he mentioned something about homotopy in other categories like homological algebra, but it went slightly over my head. But I was intrigued

[unreal stratus]

Most also factor through weak equivalences, which is fun

[gritty widget]

This is usually true

Not always though

The fundamental groupoid functor maps homotopy equivalent topological spaces to homotopy equivalent groupoids(not necessarily isomorphic)

It preserve homotopics maps(e.g. $f\cong{g}$ implies $\pi(f)$ is homotopic to $\pi(g)$) but homotopic maps aren’t always mapped to the same one

[gentle osprey]

NAT Enthusiast

[gritty widget]

I recommend T&G by Ronald Brown for AT

[prime elbow]

I mean I have to show lower limit topology is not second countable

It is a separable space so it is not possible

[rancid umbra]

the sorgenfrey line is not second countable, yes, but the reason is not that there is an uncountable collection of disjoint basis elements

if there were such a collection, then for each of the disjoint basis elements, you could choose a unique rational number within it, which would be an injection from an uncountable set into Q

the reason that it is not second countable is because for any basis of the sorgenfrey line, you can find an injection from R into the basis

[prime elbow]

Yes

So if I let B be the basis of lower limit topology on R.

Then for each x there is B_x \ [x, x+1)

Let x and y such that x < y, then x not in B_y, thus B_x ≠ B_y.

Then we map x to B_x, since it is injective hence B is uncountable

Correct?

[rancid umbra]

do you mean B_x = [x, x + 1)?

i am confused by the second line in your proof attempt

[prime elbow]

[x, x+1) is open set so there exists B_x \subset [x, x+1)

[rancid umbra]

right

you need to specify that x is in B_x tho

for this to work

[prime elbow]

Yes

I forgot to write

[rancid umbra]

all good

but yea, that is why it is not second countable

[prime elbow]

Second countable implies separable but in non-metrizable space separable doesn't implies second countable

[rancid umbra]

right, since Q is still dense in the sorgenfrey line

[prime elbow]

Yes

Thanks ❤️

[signal kite]

does tychonoff say that countable product of compact sets is compact in the countable product metric space

?

<@&286206848099549185>

[cloud kindle]

Could anyone point me to a textbook or a paper which defines a topology via neighbourhoods rather than open sets, I know how they’re equivalent I’m just looking for a reference

[prime elbow]

I think kelley

I am not sure

[prime elbow]

I know R with co-countable topology is not separable, can I find the uncountable collection of pairwise disjoint open sets?

But there are no disjoint open sets

[prime elbow]

If I define ray topology on R by the basis element [a, ∞) then it is not second countable, because the least basis set is the collection of {[x, ∞) | x in R }, right?

If [x, ∞) is not in basis then there is no basis element x \in B_x \subset [x, ∞)

[drifting ravine]

Arbitrary products

[drifting ravine]

Anything on topological vector spaces and topological groups is done via neighborhoods of the origin. Look that up

[drifting ravine]

Also, Tikhonov's theorem does not deal with metric spaces, but with topological spaces. The theorem states that any product of compact topological spaces is compact. Beware that compactness does not depend on the topology of any ambient space, but is an intrinsic property of your given space.

i.e., it doesn't make sense to ask "Is A compact in B"; if A is compact, it will be compact no matter what you embed it into

[prime elbow]

how to show that lower limit topology is first countable?

[gritty widget]

Every neighborhood of x contains [x, r) for some rational number r(of which there are countably many)

[prime elbow]

i got it, thank you

[prime elbow]

is it correct?

[gritty widget]

I’m not sure what you mean by “least basis set”

[prime elbow]

i mean let B' be any other basis set then B \subset B'

[gritty widget]

Just assume you have a countable set of neighborhoods of every point in the space and produce some neighborhood which can’t be on the list

If [x, r) is a union of neighborhoods, one of the elements of the union must be of the form [x, q) for some q<r

[prime elbow]

but it is ray topology

so every open set is of the form [a, \infty)

[gritty widget]

I’m not sure off the top of my head how you’d prove that without making it more complicated than just proving there’s no countable base directly tbh

[gritty widget]

Oh that’s my bad lmao.

Idk why, but I still thought you meant the sorgenfrey line.

Ignore what I said then

[prime elbow]

is N with co-finite topology, second countable?

[warm kettle]

Yes. You could consider the family of all closed sets in that topology. There's a bijection between that family and the topology. Note that that family is the family of all finite subsets of N, and that is countable

[prime elbow]

it is first countable right?

[warm kettle]

So the whole topology is countable

[prime elbow]

yes

[warm kettle]

And in particular, you have a countable base

[prime elbow]

i got it, thanks

yeah basically we can write every open sets as N\S, where S is a finite subset of N, since collection of S's are countable so our topology is countable

[prime elbow]

I made this table, is it correct?

[cerulean oriole]

I don't know/remember what rows 5-7 and the last column mean. In the rest table, the only mistake I see is the separability of the discrete topology on ℝ.

[peak dove]

If we have the product topology on $(\mathbb{R}, \tau) \times (\mathbb{R}, \tau)$ where $\tau$ is the lower limit topology, then would the subspace topology for a (non-vertical) line, $L$, in 2-space, $\tau_L$, be comprised of unions of segments of $L$ where each end can be open or closed (in any of the four configurations)?

[gentle osprey]

JJCUBER

[prime elbow]

Yes it is not, last column ccc, if we take any collection of pairwise open disjoint sets and this collection must be countable.

T_ray topology on R generated by basis element [x, ∞).

Sorgenfrey line, a topology on R generated by basis element [a,b).

[prime elbow]

if there is some mistake please tell me

[ruby delta]

I think you might be misinterpreting the question a bit

[prime elbow]

how?

[ruby delta]

your definition of d is defined on X x X x X x X, because it's now taking in two distinct points in XxX

not X x X

it's just asking to show the metric, defined on X, is continuous

[prime elbow]

i don't get it

[ruby delta]

and the metric on X is a function from XxX to R

[prime elbow]

yes

so we need a topological structure on X\times X which is product topology

[ruby delta]

so, d(x,y) -> R takes in two inputs from X, and gives one output in R

[prime elbow]

so i showed that it induce by \overline d

[ruby delta]

but your d(x,y) = max{d(x1,y1) d(x2,y2)} takes in four inputs from X and gives one output in R

it's a function from X^4 to R, not X^2 to R

[prime elbow]

that is \overline d not d

[ruby delta]

point is, the question is asking you to show that d is continuous on X

not that the induced d is continuous in the product metric space XxX

[prime elbow]

d is continuous on X \times X

[ruby delta]

it is, but that's not what the question wants

[prime elbow]

please go through whole proof, i think there is misunderstanding

[ruby delta]

ok

oh, you create a new d bar, ok

yeah that looks right

[prime elbow]

okay thank you

[prime elbow]

@ruby delta do you know other ways to prove

I know one sequencial proof

[prime elbow]

@opaque scroll this one

[plush folio]

Lol, I like how you went to #groups-rings-fields to ask him first, as if he has to physically walk from that channel to here

[prime elbow]

he is usually active on group ring channel so i asked him on that channel

[spark shale]

Is this proof ok? It seems fine to me but I'm not entirely sure.\
Let $X$ and $Y$ be topological spaces. Let $f:X\to Y.$ Let ${ A_\alpha }$ be a collection of closed subsets of $X$ such that $\bigcup A_\alpha=X,$ and for each $x\in X$, there exists a neighborhood of $x$ which intersects finitely many elements of ${ A_\alpha }$ (${A_\alpha}$ is locally finite). Show that if the restriction $f|A_\alpha$ is continuous for each $\alpha$, $f$ is continuous.
\begin{proof}
To show this, we will construct a collection of open subsets of $X,$ ${U_\alpha},$ such that $\bigcup U_\alpha=X,$ and $f|U_\alpha$ is continuous for each $\alpha.$ We define the collection ${ V_x }{x\in X},$ where $V_x$ is a neighborhood of $x\in X$ such that $V_x$ intersects finitely many elements of ${A\alpha }.$ Clearly $f|V_x$ is continuous because $V_x$ is a subset of the union of the elements of ${ A_\alpha }$ which $V_x$ intersects*. Then, as desired, we have that $f|V_x$ is continuous for each $x\in X,$ and of course $\bigcup V_x=X,$ hence $f$ is continuous.
\end{proof}
*I proved in a previous exercise that if the collection ${A_\alpha}$ of closed subsets of $X$ were finite, $\bigcup A_\alpha=X,$ and $f|A_\alpha$ were continuous for all $\alpha,$ $f$ is continuous.
\end{document}

[gentle osprey]

Maxwell

[prime elbow]

Yeah but you have to show where you used the finite property here?

[spark shale]

Do you mean the locally finite property? I used it to construct the collection of neighborhoods which only intersect finitely many elements of {A_alpha}.
I suppose I should also probably reiterate that fact when I say that f|Vx is continuous and say "f|Vx is continuous because Vx is a subset of the union of the finitely many elements of {A_alpha}"?
is that what you meant?

[prime elbow]

yeah so here you used pasting lemma, right?

[spark shale]

In this proof I used the "local formulation of continuity" as munkres calls it:
The map f: X -> Y is continuous if X can be written as the union of open sets U_alpha such that f|U_alpha is continuous for each alpha

should I explicitly state that as well?
probably

[prime elbow]

but how did you show that f|V_x is continuous?

if you want then i can attach solution, your solution is good but i think you have to use pasting lemma to show f|V_x is continuous

[spark shale]

ohh no you're right I see what you mean now

[spark shale]

i'll take another solution

[spark shale]

I got the solution to the other exercise confused with the pasting lemma and was referencing that in my proof

[spark shale]

That also makes sense
thank you! :D

[prime elbow]

Any hint for counterexample, x is accumulation point of sequence but x is not a limit point of sequence

[spark shale]

I'm not sure but it might have something to do with the fact that it only requires neighborhoods of the accumulation points to have infinitely many x_n whereas I think the limit of a sequence requires all but a finite number of points to be in neighborhoods of the limit points?

[prime elbow]

actually no

i mean your limit defination is correct, but i want limit point of sequence not limit

[prime elbow]

2 =>1, let C be closed set in T1 then i have to show C is also closed in T2, since we proved that C is closed iff all limit point of convergent net is in C.

let w be a convergent net in (C,T2 ) then by 2, it is also convergent in T1 and C is closed in T1 so limit is in C imply C is closed in T2.

similarly for T2 to T1, so T1 = T2

is it correct?

[solemn iris]

Can I get some help on understanding what this means?

[prime elbow]

Do you understand how the order topology works?

Let's say {2} is open in Y with order topology so there must be basis element B such that 2 in B \subset {2}.

Now how does the basis elements look like?

[solemn iris]

hmm since 2 is a largest element then i suppose we want some interval (a, 2]?

oh, because (1.5,2] is not contained in Y, then it’s not in any basis element?

[prime elbow]

Not (1.5, 2], here your basis element for 2 is (a, 2] where a in Y

so that means a in [0, 1)

So we can always find b such that a< b <1 <2 so b in (a, 2] and b in Y

[solemn iris]

oh okay

I'm still not exactly sure what Example 2 is trying to say here

[prime elbow]

{2} is open in subspace topology on Y but {2} is not open in order topology on Y, so topology induced by order is different as topology induced by subspace topology on Y

[prime elbow]

I proved 2 => 1 by showing f(clA) \subset cl f(A).

But if I have fix x in X1 such that if w converge to x then f(w) converge to f(x), then how do I prove that ?

[prime elbow]

I learnt about net and filter and I was expecting something big tool for my point set topology but I don't know how it helps me

[tender halo]

filters are much more useful than nets really

at least most of the time

but its less of a tool like a theorem is, it's more of a lens to look at topological spaces and to state theorems

say, you can state compactness as "every z-ultrafilter converges"

if you are interested in how to use filters to investigate topological spaces, i would recommend gillmans book "rings of continuous functions"

[prime elbow]

Okay

[woeful coral]

Are they not equivalent within ZFC

[tender halo]

depends on what you mean by equivalent

for one, subnets and finer/coarser filters are not equivalent

they are equivalent in that you can present a bijection between the two that respects convergence

uhh i guess not a bijection, just a correspondence

to every point of a (Tychonoff) space (or to every point of its Stone-Cech compactification) corresponds a natural ultrafilter, which I dont know how i would express with nets for example

[novel compass]

is there a largest type of convergence that allows the space of continuous functions C(X, Y) (viewed using product topology as Y^|X|) to be closed under it? Obviously pointwise convergence does not work, compact convergence does but I wonder if that notion of convergence gives us the largest class of sequence continuous functions are closed under limits of, or if we can still do better.

[ruby delta]

uniform convergence should be the necessary & sufficient condition

[novel compass]

Nah, compact convergence is strictly weaker than uniform convergence yet there are functions with compact convergence that don’t have uniform convergence (as in continuous functions converging to continuous functions)

I think I found my answer though

https://math.stackexchange.com/questions/2294852/the-equivalence-of-alexandroff-convergence-and-continuity

[ruby delta]

cool, maybe in topo spaces it's different but I'm pretty sure it's true in metric spaces

let me check

[civic verge]

name of book ?

[spare light]

hahaha, good job i was thinking of pinning down the arguments too

[prime elbow]

https://www.math.toronto.edu/ivan/mat327/?resources

[wraith edge]

Can anyone help me with this problem?

[prime elbow]

Think about E_1 and E_2 such that there is no point common but there are points (e_n, f_n) in E_1 \times E_2 such that d(e_n, f_n ) = 1/n

Think about in R

[wraith edge]

Ahhh. I have a thought now. If I set $E_1={\frac 1{2n}:n\in\mathbb{N}},E_2={\frac 1{2m+1}:m\in\mathbb{N}}$, since for $x_n=\frac 1{2n}\in E_1,y_n=\frac 1{2n+1}\in E_2$ we have $d(x_n,y_n)=\frac{1}{2n(2n+1)}$. Taking $n$ to infinity, we'll have $d(E_1,E_2)=0$

[gentle osprey]

Owen

[prime elbow]

But E_1 is not closed and E_2 is not closed

[wraith edge]

Oh right..

What about $E_1=\bigcup_{n=1}^\infty [\frac 1{4n+1},\frac1{4n+2}],E_2=\bigcup_{n=1}^\infty[\frac 1{4n+3},\frac1{4n+4}]$, so that the distance of it, $x_n-y_n=\frac1{4n+2}-\frac1{4n+3}=\frac1{(4n+2)(4n+3)}$ gradually approaches 0 when taking $n\to\infty$. I think this will work.

[gentle osprey]

Owen

[wraith edge]

Oh I thought countable union of closed set is closed, but only finite cases work after a search.

[ruby delta]

it doesn't sound possible....? I see where my argument fails

doesn't ||the inf of the distance being 0 imply that there exists limit points of E_1 and E_2 that have distance 0 from each other||

unless you're thinking of ||the two closed sets being the entire space itself and the empty set, but I'm not sure the distance is even defined in that case||

[ruby delta]

tell me why this isn't true see above, I know

[prime elbow]

|| take E_1 = {n | n ≥ 3 }and E_2 = { n + 1/n | n ≥ 3 } ||

[tender halo]

its true if the space is compact

take idk a hyperbola and a vertical line and x = 0

[cloud kindle]

I asked for this like a week ago but I’ve come up short, anyone know of a reference which defines a topology in terms of neighbourhoods rather than open sets?

[prime elbow]

so kelley doesn't?

[cloud kindle]

Not from what I could see, but I also couldn’t find the full book (my university library had a copy but I’m away from home for a few days)

[sonic crane]

I think Janich gives that definition

[tender halo]

engelking has one

[red yoke]

#advanced-lounge message

I don't know if there are any advantages to this defn though, other than serving as motivation

[wicked pollen]

The standard proof I've seen that arbitrary products of connected spaces are connected relies on the axiom of choice to pick a "starting vector", then look at finite perturbations of it, and show these are dense.

Is there a way without the axiom of choice?

[unreal stratus]

Well I guess lol like

If your convention is that empty spaces are connected, then you just condition on whether it is empty or not

If your convention is that empty spaces aren't connected, then this can fail in the absence of axiom of choice

[wicked pollen]

Oh, that is true I suppose.

[rancid umbra]

the absence of it, right?

[unreal stratus]

Of course, thanks

[wicked pollen]

But then wouldn't the same thing work with Tychonoff? If you take the empty set to be compact

And Tychonoff is usually said to be equivalent to the axiom of choice

[rancid umbra]

for the analogy to work, the empty space would need to be considered not compact.

you would have the same conditioning (with the convention that the empty space is compact), but in the absence of choice, there are infinite products which are empty, and so tychonoff would fail (with the convention that the empty space is not compact)

[wicked pollen]

Why shouldn't the empty space be compact?

Any open cover (there is only one, {{}}) has a finite subcover (itself)

[rancid umbra]

this would just be a convention.

the empty set is vacuously connected (it cannot be disconnected) but some make the convention to say that it is not connected

[wicked pollen]

Right, but if the empty set is compact, Tychonoff would not necessarily imply AoC right?

[rancid umbra]

tychonoff already implies the axiom of choice with this convention

[wicked pollen]

I should probably just check the proof of that implication

[rancid umbra]

if you assert that the empty set is not compact and try to prove the axiom of choice from the tychonoff theorem, you have to weaken the statement of the tychonoff theorem slightly

instead of any product of compact spaces being compact, it must be weakened to the statement that the product of any non-empty collection of compacts spaces is compact

both statements of the tychonoff theorem imply the axiom of choice, but i believe the axiom of choice only implies the weaker one.

[cerulean oriole]

Presumably to show this, one takes a family of non-empty sets and constructs related but different compact spaces, whose product might be obviously non-empty but non-obviously compact, and it's the compactness that leads to the original Cartesian product being non-empty.

This is speculation but I imagine given X you construct a compact space with an filter whose cluster points are exactly X.

[drifting ravine]

Their closures have 0 in common

[cloud kindle]

Yeah I mean I know how to define it that way, I’m just looking to cite it for something I’m writing, motivation is the exact thing I want. It’s a project about homological algebra that needs to be vaguely self contained so I’m including enough point set for that to be legally true

It’s just that after I define a topological space I say something along the lines of “there is an equivalent axiomatisation which makes this notion of “closeness” more apparent” so I’m just looking to throw a reference to the neighbourhood definition

I just give the motivation of like it generalises metric spaces, we no longer have a notion of distance, but instead one of “closeness” type thing

[wraith edge]

https://math.stackexchange.com/questions/420122/find-two-closed-subsets-or-real-numbers-such-that-da-b-0-but-a-cap-b-varno

[prime elbow]

i think i gave same answer

[tranquil cosmos]

That's cool

[hot pivot]

{\bf Exercise.} Construct a topological space $(X, \mathcal{T} )$ which is countable (i.e.\ $X$ is countable) but not first
countable.\
{\bf Attempt.} As a guess, I {\sl think} if we took the basis of the Sorgenfrey line, and restricted it to $\mathbf{Q}$, that would generate a topology with the required properties. So let's try proving that. Firstly, I think it's safe to refer to each basic open set as $[a,b) \cap \mathbf{Q}$, where $a,b\in\mathbf{R}$. This is because the subset of $\mathbf{Q}\subset\mathbf{R}$ is ``the same'' as $\mathbf{Q}$, so in theory, describing this basis should be possible without having to define $\mathbf{R}$; it would just be very cumbersome to do. So we proceed: The above is certainly a basis of $\mathbf{Q}$, since for one, they cover $\mathbf{Q}$, and two, for $x\in [a_1,b_1)\cap[a_2,b_2)\cap\mathbf{Q}$, well, if not for the $\cap\mathbf{Q}$ at the end, the Sorgenfrey line provides us with some $[a_3,b_3)$ that contains $x$ that is a subset of $[a_1,b_1)\cap[a_2,b_2)$. But $x$ is rational, so $[a_3,b_3)\cap\mathbf{Q}$ will do the trick, proving that this is a basis on $\mathbf{Q}$.
Thus this generates a topology on $\mathbf{Q}$. I claim this to not be first countable.
There's a preliminary result that for a countable set, first countability and second countability are equivalent notions (of which I omit the proof).
So for the sake of contradiction, assume this topology is second countable,
Here's where I'm stuck; I don't know how to arrive at a contradiction, although intuitively I feel that this will lead to showing that a countable basis exists for $\mathbf{R}_{\rm Sorgenfrey}$, but I don't see how to make that connection.
Hints?

[gentle osprey]

Chaos22

[rancid umbra]

the sorgenfrey line is first countable

[hot pivot]

Hmm. Does that mean its restriction to Q is also first countable?

[rancid umbra]

yes

you would take the neighborhood basis of a rational in R to the neighborhood basis of that rational in Q by intersecting each neighborhood with Q

[hot pivot]

So that entire attempt was all for naught?

[prime elbow]

Can I ask the book?

It seems interesting construction

[hot pivot]

The Big List, problem 5.14.

[prime elbow]

Oh you are doing big list

Good

Then I think I done it

No, three star 🙂

Have you done 5.8?

[rancid umbra]

something like the long rationals inside of the long line should work

oh wait

that’s not countable

[hot pivot]

5.8 is the only other one I didn't do within this section oops,
with the air of "i'll do this later when I cover more topological properties"

[prime elbow]

I made the table, you can verify if you want

[hot pivot]

i'd appreciate that

[prime elbow]

|| it has one typo ||

[hot pivot]

Hmm, $\mathbf{R},\mathcal{T}_{\rm discrete}$ shouldn't be separable, right? Since for any countable set $K\subset\mathbf{R}$, the complement of $K$ is open but doesn't intersect $K$, so $K$ cannot be dense in $\mathbf{R}$. Is that right?

[gentle osprey]

Chaos22

[prime elbow]

Yes that was a typo

And also in discrete space only X is dense set, so when X is countable then X with discrete topology is separable

[rancid umbra]

for this exercise, you want a countable space with no countable basis. so intuitively, you want to maximize the number of open sets

[prime elbow]

https://www.physicsforums.com/threads/countable-but-not-second-countable-topological-space.187934/

[prime elbow]

I see

Set is countable so first countable implies second countable

[hot pivot]

Checking for my own understanding, $\mathbf{R},\mathcal{T}_{\rm ray}$ is only ccc because it's a vacuous statement right? Since no two non-empty sets can be disjoint.

[gentle osprey]

Chaos22

[prime elbow]

Yes

[hot pivot]

Same deal with the particular point.

[prime elbow]

Same with R co-countable topology

Co-finite topology

[spare light]

https://discord.com/channels/268882317391429632/1332828732578074754
haha

[prime elbow]

any hint?

[rancid umbra]

one example has X = R

[prime elbow]

i tried with R by usual topology, discrete topology, particular point topology

[rancid umbra]

R with the usual topology works

to find Y, think of spaces Y that contain a copy of R which are also contained in R

[prime elbow]

what if i take Y = R?

[rancid umbra]

that’s a good start, but then X and Y are homeomorphic

[prime elbow]

what if i take different topology

[rancid umbra]

hmm. R has some self-similarity that you can exploit

[prime elbow]

can i take Y = R{x}

R\{x}

[rancid umbra]

yes

where x is in R

[prime elbow]

yes

but what will be the function then?

[rancid umbra]

just start thinking of some maps between the two spaces, to start

and you need two different functions

[prime elbow]

yes

[rancid umbra]

one which embeds R \ {0} in R and one which embeds R in R \ {0} (choosing x = 0 for concreteness)

[prime elbow]

i take Y = R \ {1}, then map R to R \ {1} by x-> x/ (x-1)

my bad

it is not defined at 1

[rancid umbra]

ye. there is no way to continuously extend it either

because of the vertical asymptote at x = 1

the easier of the two maps is taking R \ {1} into R

[prime elbow]

i don't know, i know there is bijection between them but thats not point here

R to R \ {0} injective and also continuous

[rancid umbra]

an embedding isn’t a homeomorphism, its a homeomorphism onto its image

[prime elbow]

Yes that's why it is not a point here

[rancid umbra]

im confused by what you mean

[prime elbow]

Nothing ignore

[prime elbow]

But can you give me more hints on the construction of function here?

[rancid umbra]

R \ {1} is a subset of R

is my first hint

it literally sits inside of R without modifying the set at all

[prime elbow]

Okay

[wraith edge]

Yep you did, thanks!

[prime elbow]

(0,p) \cup (p,1) , 0 < p < 1 with subspace topology of R - usual topology, is connected, right?

[rancid umbra]

no

[prime elbow]

Got it

(0,p) will be clopen in this topology

I have to show [0,1) and (0,1) is not homeomorphic, I know how to show but I am thinking to do by cut-points

formal proof: if they are homeomorphic function f then (0,1) is homeomorphic to (0,1) \ {f(0)}.

But one is connected and the second one is not

[prime elbow]

In this question, I think we have to use cut-points ideas only

[hot pivot]

{\bf Exercise.} Find a function $f: \mathbf{R}\to\mathbf{R}$, that is continuous if both sets have the usual topology, but discontinuous if both sets have the Sorgenfrey line or both sets are the ray topology.
{\bf Attempt.} Define $f:\mathbf{R}\to\mathbf{R}$ by $f(x):=e^{-x^2}.$ This is continuous for $\mathbf{R}{\rm usual}\to\mathbf{R}{\rm usual}$, since the composition of two functions (polynomial and $\exp$) are both continuous. For $\mathbf{R}{\rm ray}\to\mathbf{R}{\rm ray}$, $\mathbf{R}$ is an open subset of the co-domain, but the pre-image is $(0,1]$, which is bounded, and hence, not open in $\mathbf{R}_{\rm ray}$. For the Sorgenfrey line, it's continuous almost everywhere I believe, except at $x=0$. This is because for the open set $[1,\infty)$, the pre-image is ${0}$, which is not open in the Sorgenfrey line.

[gentle osprey]

Chaos22

[hot pivot]

Is my understanding correct that this is the only weird way where functions between Sorgenfrey lines are discontinuous where it otherwise would be continuous in our usual topology? i.e., specifically when the function has a local maximum, since, we know monotone increasing sequences in the Sorgenfrey line do not converge, even if they are bounded.

So, |x| would be continuous, but -|x| would not be.

[prime elbow]

I did this in today morning

I take the function || f(x) = 1-x, so in lower limit topology inverse image of [0,1) is (0,1] which is not open in lower limit topology||

And I didn't try for ray topology because the question asked us for ray or lower limit topology

[prime elbow]

Are you sure the pre-image is (0,1] ?

[hot pivot]

Oops, the pre-image is R. Ah.. I guess I shouldn't take R as the open set in the codomain, maybe something like (1/2,infty), then the pre-image will only be the numbers from (-sqrt(ln2),sqrt(ln2)), which is bounded and hence not open in RayTopology.

[hot pivot]

{\bf Proposition.} Let $f:\mathbf{R}{\rm co-countable}\to\mathbf{R}{\rm usual}$ be continuous. Then $f$ is constant. \
{\bf Warning.} This proposition may even be false! \

{\it Proof.} If $f$ is constant and always equals some $k\in\mathbf{R}$, then for any open set containing $k$, the pre-image is $\mathbf{R}$, which is open. Any open set that doesn't contain $k$, is the empty set $\emptyset$, which is also open. Now for the sake of contradiction, assume $f$ isn't constant. Say $f$ takes on two distinct values, $a$ and $b$. But now continuous functions also preserve closed sets in their pre-images. So the pre-image of ${a}$ and ${ b}$ must each be countable (neither can be $\mathbf{R}$ otherwise we'd arrive back to $f$ being constant). But then the union of these two countable sets should be the entire domain, $\mathbf{R}$, which is uncountable, and hence we have arrived at a contradiction. This similar logic can show that $f$ cannot take at most countably many values, since the union of countable sets must be another countable set. Thus we are forced to conclude that if $f$ is not constant, it must attain uncountably many distinct values.

But here's where I cannot continue the argument, because now, this seems feasible. However, with the restriction that $f$ was bounded, I can make progress:

If $f$ is bounded (and continuous), we can show it cannot have uncountable range. Suppose, for the sake of contradiction, {\it that it did:/} Then there is some interval $[-M,M]$ that contains the range of $f$. Bisect this into two halves. At least one of them must contain uncountably many points of ${\rm range}: f$. If its pre-image is countable, then it would be a contradiction, because you cannot map countably many points onto uncountably many points. Therefore the pre-image must be $\mathbf{R}$ itself. But then we can keep bisecting the intervals like this, ending up with a nested sequence of intervals, all containing uncountably many points of the range, and all whose pre-image is $\mathbf{R}$. Thus if we take the intersection of these intervals, we'll get a singleton set, whose pre-image must still be $\mathbf{R}$. This means the range is a singleton, contradicting the fact that the range of $f$ is uncountable.
Thus, combined with the result above, $f$ can only be constant.

How do I make progress in the case where $f$ is possibly unbounded?

I should clarify, this is a proposition that I don't know to be true or false. Specifically, the question asked me to characterize the kinds of continuous functions from the co-countable topology on R to the usual topology on R. It is my conjecture that it's only the constant functions (but I cannot prove this).

[gentle osprey]

Chaos22

[prime elbow]

You are on right track, but use the Hausdorff property of R with usual topology

[hot pivot]

To extend the above argument to the case where f can be unbounded, if it was possible to find a bounded interval that intersected the range at uncountably many points, then we would be done, since then the rest of the argument follows similarly to the "f being bounded" case.

EDIT:
Ah this is possible:
Lemma: Every uncountable subset A of R has a bounded uncountable subset.
Proof. Suppose for the sake of contradiction, it did not. Then We can construct a sequence of sets [-n,n] ∩ A, all at most countable, whose union, (which again must be at most countable) gives A, a contradiction. Q.E.D.

Thus in the proposition above, even if f was unbounded and non-constant, then by our earlier work we could show its range must be uncountable. But now we can find a bounded subset of the range that is uncountable, and apply the same logic for when f was bounded, to achieve the conclusion that continuous functions from Co-countable to Usual topology must be constant.
What a convoluted proof.

[prime elbow]

If a ≠ b, then there exists disjoint open set U and V such that U contain x and V contains y

Now take inverse image of U and V

But in R with co-countable topology, every open set intersect with each other

[hot pivot]

Oh...

Very slick. I'd never think of that.

Perhaps I should more heavily consider leveraging the topological properties of the spaces when solving problems from now on. Otherwise the solutions might get needlessly complicated.

[rancid umbra]

another argument not appealing to that last fact which may be more natural:
if a != b in the image of f, then we can separate a and b by open subsets A and B of R_usual

f^{-1}(A) and f^{-1}(B) are disjoint subsets of R, each with countable complement.
so in particular, f^{-1}(A) must be a subset of R \ f^{-1}(B), a contradiction

[rancid umbra]

i think that doing what you did and thinking about the problem was fine. you gained a lot of intuition for the problem and would have eventually gotten it. the idea to bisect and use the nested interval property was creative (although, some bits i think are incorrect)

[prime elbow]

Yes

[prime elbow]

i think for part c, take f: R -> N, with both discrete topology.

it is local homeomorphism beacuse for every x in R we can take open set {x} and V is {f(x)}, then f_| {x} is homeomorphism.

it is not homeomorphic because R is not countable
right?

[tough hamlet]

what is f

[hot pivot]

{\bf Exercise.} Describe what properties a function $f : X \to Y$ must have in order to be continuous if $Y$
has the discrete topology. \
{\bf Attempt.} Intuitively, it's easier for functions to be continuous if the domain is discrete, or if the co-domain is indiscrete, but it becomes harder and harder to find continuous functions where the co-domain gets finer and finer. In this case, where the co-domain is as fine as it can be (the discrete space), this is a sort of ``boss-mode'' for continuity, since it should be extremely hard for a function to achieve the revered (revered?) status of continuity in this setting. At first I thought maybe constant functions would be it. Another thing we can do is, we can vary the topology on $X$ (I'm not sure if the exercise is only mandating we talk about $f$ and not $X$). If $X$ is also discrete, then $f$ can be anything, and still be continuous. But how about a situation where $f$ is neither constant, nor is $X$ discrete? \
The only example I can think of is something like this:\
Say $X$ was the Sorgenfrey line, and $f$ be the floor function (or something analogous that maps each interval $[n,n+1)$ to some different constant each time).
This should be continuous since for any open set $G \subseteq Y$, (i.e., for \emph{any} set in $Y$), whichever $y_1,y_2\ldots$ reside in will produce their appropriate Sorgenfrey pre-images. Sorgenfrey base sets are both open and closed so this should work.\

So it looks like, $f$ can range from the constant function to anything as long as we choose the topology on $X$ to be fine enough. In particular, there exists a non-constant function on a non-discrete space to a discrete one that is continuous. So no such characterisation should be anticipated.\
Am I way off base here?

[gentle osprey]

Chaos22

[tender halo]

that is a lot of text for nothing, just use the definition of continuity together with the fact of Y is discrete

f is continuous iff all fibers are clopen

which is pretty restrictive on X

namely, X has to be the sum topology of f's fibers

thats about it

[hot pivot]

Ah that explains the comment for this exercise:
"This vague-seeming question will come up again quite late in
the course."
Considering they haven't talked about fibres or "sum topologies", I assume my background isn't sufficient to approach this question yet.

[tender halo]

a fiber is just a preimage of a single point

i imagine they expect something akin to "all fibers are clopen"

[prime elbow]

Real-analysis- Carother

[alpine hound]

Can we determine what sets are dense in the cofinite topology?

[gritty widget]

Any infinite set intersects all cofinite sets(why?)

[alpine hound]

Thank you lol.
In my head this part made sense but the second direction didn't. Now I get it, thanks

[gritty widget]

Np

[waxen oxide]

In image, I am wondering why $\bar{V}$ is disjoint from $C$.

[gentle osprey]

Spice

[paper wedge]

V is disjoint from W

W containing C

@waxen oxide

[waxen oxide]

Yeh but, $\bar{V}$ contains $V$

[gentle osprey]

Spice

[paper wedge]

W contains C

if an element is in C then it must be in W but W is disjoint from V

@waxen oxide

[waxen oxide]

I get that by, why can't element on the boundary of V be in W

[paper wedge]

V and W are disjoint by definition

so i thikn what ur asking is what happens if this element is a limit poitn

a limit point of V

[waxen oxide]

Yes

[paper wedge]

what do u think

regarding that both C and cl(V) are compact sets in Y

[waxen oxide]

Assuming x is a limit point of V, since W is open, and x in W, it contains a neighborhood of x, within W. This means that this neighborhood is disjoint from V, a contradiction. This doesn't use compact properties

[paper wedge]

x is assumed to be in V not in W

if y is a limit point of V

then every nbd of y intersects V at a point other than y

but now y is in C and hence y is in the open set W

and hence we should be able to fit an entire nbd of y in W

but that wouldn't intersect V would it

[waxen oxide]

if x is in V, it is trivially in the closure of V

[paper wedge]

does what i said above make sense

to you

[waxen oxide]

You just repeated what I said

[paper wedge]

oh yeah lmao

[paper wedge]

ur starting with x in U tho so x can't be in C

any other element can't be in the intersection by what i said above which is really the same argument u said

[waxen oxide]

I forgot about the property where you can form open subsets around elements of open subsets

[paper wedge]

yeah

np tho

[waxen oxide]

This is not true in general actually. However, if we take W itself as the disjoint neighborhood of y from V, we are done.

[paper wedge]

it is true

if you have an open set and you have a point in that set, there exists a nbd around that point that is contained in your open set

[waxen oxide]

I meant $\subset$, you mean $\subseteq$, I was wrong

[gentle osprey]

Spice

[paper wedge]

yeah np

[dry seal]

When are limit point compact spaces sequentially compact?

Is being T1 enough or does it need to be Hausdorff?

Is there some other condition that is more minimal?

[rancid umbra]

metrizable comes to mind

[novel acorn]

You need very strong assumptions

Basically this only holds for metric spaces

Which are T6

A more general statement that holds is that a topological space is compact if and only if every net has a convergent subnet

This is because the idea of a sequence can be pretty pathological in general topological spaces

But nets are better behaved

[tender halo]

thats way too much

asking the space to be sequential and T2 is enough

although sequential is arguably a pretty strong condition

[dry seal]

So 1st countable T2 is enough

[tender halo]

it is

[dry seal]

Ok that's neat

I'm noticing a lot of countably compact theorems can be used to bridge these gaps, does anyone know of some good literature for this? I have never interacted with this notion before

[tender halo]

you can read the relevant chapters of Engelking

uhh 3.10

[dry seal]

Thanks 🙏

[sinful cloak]

T1 and first countable is.

[dry seal]

I see, since you just need to be T1 to be countably compact, and then first countability gets you to sequential compactness from there

Neat

[waxen oxide]

Say some metric space $(X, d)$ is given the metric topology. Given some $x \in X$ I am wondering how to show $B_d(x, \varepsilon)$ is connected for each $\varepsilon \in \mathbb{R}$

[gentle osprey]

Spice

[tender halo]

why would it be connected

in, say, Q, none of the balls are connected

[waxen oxide]

You are right I specifically mean the space $(\mathbb{C}, d)$ where $d$ is the usual pythagorean distance function

[gentle osprey]

Spice

[tender halo]

the easiest way is probably to show that its path connected

[iron bolt]

yup. balls in normed spaces are convex, convex sets are star-shaped, star-shaped sets are path-connected and path-connected sets are connected

that's the chain of implications I like to remember it as

[tender halo]

alternatively, balls in R^n are homeomorphic to R^n and thats connected

[hexed steppe]

you can just use a straight line path to prove convex sets are path connected

no need to invoke the term star shaped sets

[unreal stratus]

Yeah lol like you can do this by definition

[cloud kindle]

Finally got around to looking at this, exactly what I needed, thank you! Also seems like a cool book I might need to give it a skim at some point

[silver ridge]

I'm really struggling to work with RP^n. Second countability is inherited from R^{n+1} \ {0}, so we just need to show it is hausdorff and that it is locally euclidean. I'm currently struggling to show it is hausdorff. There is one main theorem given in the text for this. Let X be a topological space, R an equivalence relation on X. Then if the projection X -> X/R is open, then X/R is hausdorff iff R is closed in X x X. I've also tried to show RP^n is hausdorff more directly and I've failed both times. I would appreciate any hint on how to progress 🙂

[rancid umbra]

for hausdorff, have you tried drawing a picture?

you can reduce it to the fact that any two lines through the origin in R^{n+1} can be contained in double-cones whose intersection is the origin

going through all the details is a bit cumbersome

but it’s doable

[silver ridge]

Yeah I see the vision

Details now I guess

I guess if I take the midpoint line in some sense that ought to work

[rancid umbra]

yea. the picture in R^2 extends to R^{n+1}

i would try it for that first

[silver ridge]

Thank you

[waxen oxide]

I am wondering why the Ursyhon lemma implies the existence of such functions. I thought regularity was a weaker condition than normality

[gritty widget]

Wait I might me misunderstanding

nvm

[waxen oxide]

Actually I think I found a proof that a regular space $X$ with a countable basis is normal.

Let $A,B$ be disjoint closed sets contained in $X$. Then by regularity, for each $x \in A$ we have a neighborhood $U$ of $x$ such that $\bar{U} \cap B = \emptyset$. Since all of these neighborhoods contain a basis element which contains $x$, we have a countable covering of $A$, call it $U_1, U_2, \ldots$. Similarly we have a countable covering of $B$, call it $V_1, V_2,\ldots$.

Now define $Y_n = U_n \cap \bigcap_{k=1}^n X\setminus \bar{V_k}$. Clearly ${Y_k}$ forms an open covering of $A$. And similarly define the open covering of $B$, ${Z_n}$.

We now claim $Y_m \cap Z_n = \emptyset, m,n \in \mathbb{Z}^+$. We have if w.l.o.g $m \geq n$ then

$Y_m \cap Z_n \subseteq Y_m \cap V_n \subseteq Y_m \cap \bar{V_n} \subseteq U_m \cap X\setminus \bar{V_n} \cap \bar{V_n} = \emptyset$.

[gentle osprey]

Spice

[cerulean oriole]

This looks good. I'm just not sure why {Y_n} covers A.

[waxen oxide]

Take $x \in A$. Then there must exist some $U_i$ such that $x \in U_i$. Furthermore, $x \in X \setminus \bar{V_i}$ for each $i \in \mathbb{Z}^+$ since $\bar{V_i} \cap A = \emptyset$ by definition.

[gentle osprey]

Spice

[cerulean oriole]

Ah, I thought only V_i ∩ A = ∅ for some reason.

Thanks!

[waxen oxide]

Does this work:

Take $x \in \mathbb{C}$

Take $\varepsilon \in \mathbb{R}$. Define $S_{\varepsilon} := {(\varepsilon cos(\theta), \varepsilon sin(\theta)) : 0 \leq \theta \leq 2\pi}$. Then we have

$B(x, \varepsilon) = \bigcup_{\epsilon \in \mathbb{R} : \epsilon < \varepsilon} S_{\epsilon}$.

But for each $\varepsilon \in \mathbb{R}$ we have $S_{\varepsilon} \cong \mathbb{R}$. Thus, $S_{\varepsilon}$ is connected, and the union of connected sets is connected. Thus $B(x, \varepsilon)$ is connected?

[gentle osprey]

Spice

[hexed steppe]

circle is not homemorphic to R

[waxen oxide]

yeh I realised that lol

[hexed steppe]

and the union of connected sets need not be connected

i think the simplest proof is what others already suggested

i know

actually you meant in R^2

obviously i understand that lol

again the union of nonempty connected sets in a metric space need not be connected

consider two disjoint connected sets

and take their union

[waxen oxide]

Ok, but they all have a point in common

Actually they dont

[gentle osprey]

Spice

[ruby delta]

to save some time, $A \sqcup B$ is commonly used to denote the disjoint union of $A$ and $B$. So you can just write that instead of "$B = C \cup D$ where (...) and $C \cap D = \emptyset$"

[gentle osprey]

HChan

[unreal stratus]

(There is an issue that disjoint union means disjoint union topology too, though this doesn't matter when stuff is open)

[rancid umbra]

at this point, you may as well just cite that a path-connected space is connected without convoluting your proof using contradiction and other definitions.

on that same note, how do you know that f is a path in B? (it is, i am just saying that it is something you need to justify)

[gentle osprey]

Chaos22

[opaque scroll]

This is called being sequentially closed, and does not imply closed in general. But does so for certain nice spaces.

https://en.m.wikipedia.org/wiki/Sequential_space

As an example, if you let w be the first uncountable ordinal and let X = w+1 with the order topology, then w as a subset of X is sequentially closed, but not closed.

[hot pivot]

Is there an easier/elementary example? Or do we have to resort to these weird ordinal-style constructions to find such a pathological topology?

[opaque scroll]

The cocountable topology works

Not sure you find that more or less pathological

[hot pivot]

Co-countable is smth I understand. I don't know anything about ordinals. Thanks!

Ah yes, this is easy to see. Since the convergent sequences in the co-countable topology are the eventually constant ones, and can only converge to that constant. (right?)
So smth like the interval (0,1) from R-co-countable, isn't closed (because it's uncountable), but every convergent sequence inside it must converge within (1,5).
Very nice.

Hint for this please.
I'm convinced that this is impossible. Feels like if X can be embedded in Y, and Y in X, then surely they must be homeomorphic...

[opaque scroll]

I'm not sure how great of a hint this is for you, but for example the interval (0, 1) is homeomorphic to R.

But the embedding of (0, 1) into R is much smaller than all of R. So if you had some subset of R that contained (0, 1) it would have an embedding of R.

[rancid umbra]

this may be useful

[grizzled lagoon]

this may be a dumb question, but is there a simple way to prove #2?

[opaque scroll]

Hard to come up with a simple way to prove it, but you can come up with some pretty simple counter examples.

[grizzled lagoon]

why is that? also what do you mean by being able to come up with simple counterexamples?

[opaque scroll]

The exercise is asking you if the statement is true.

The statement is not true

[grizzled lagoon]

oh 😭 😭

i will try to think of a counterexample

[grizzled lagoon]

are you able to nudge me in the right direction, i cant seem to come up with anything

[opaque scroll]

Well, it's a little hard to give hints that isn't just saying the solutions.

But I guess try to make S as simple as possible.

The simplest nonempty subset you can think of

[grizzled lagoon]

i tried experimenting with singleton sets like S = {vec{0}} if thats on the right track

intuitively though i just dont see how we can get a counterexample since the closure in euclidian space should just mean changing the > into a >= right... unless im missing something

[opaque scroll]

But you're not taking the closure in Euclidean space. You're taking the closure in S

[grizzled lagoon]

ah okay so heres what i came up with, does this work?
[S = {\vec{0}, \hat{i}}]
Choose $x = \vec{0}, r = 1 > 0$

Then ${y \in S: d_2(x,y) \geq r} = {\hat{i}}$, $cl({y \in S: d_2(x,y) > r}) = \phi$ since $B_{r/2}(y) \cap S \backslash {y} = \phi$ so no element in $S$ is an accumulation point

[gentle osprey]

Evil Bean

[tender halo]

depends on what you mean by elementary, but there are a lot of natural examples of non sequential spaces

say, spaces of functions with the topology of pointwise convergence

there are natural spaces where there are no convergent sequences besides eventually constant ones

so every set is sequentially closed

but the space itself of not discrete of course

[iron bolt]

oh, do you have an example? I don't think I've seen that before

[tender halo]

sure, the Stone-Cech compactification of a discrete space is one such example

i think i knew of another but i forgot

also note that existence of a non trivial convergent sequence in a Hausdorff space is equivalent to the existence of an embedding of the countable Fort space (that is, the one point compactification of the naturals) into our space

which is kinda funny imo

[hot pivot]

Ah so a later question had us working with (0,1) [0,1], and I just realised that these two would work perfectly.
Man that question tripped me up.

[hot pivot]

elementary at my level: smth I'm likely to already have been exposed to early on in a basic point set topology course. But yes, I understand how that may not be a precise description. Fortunately, the co-countable topology example given above was exactly what I was looking for.

[iron bolt]

it's super neat. once you realise that convergent sequences correspond to maps from that space, it doesn't much effort anymore iirc to prove that spaces are sequential iff they're a quotient of a metric space / first-countable space

that is to say, the sequential spaces form the coreflective hull of both metrisable spaces and first-countable spaces in Top

[hot pivot]

a) and b) were straightforward, but I can't see how this helps us with c).
I guess I could cheat a little by using some analysis knowledge, and say that [0,1] is not homeomorphic to the others because it is compact, but we haven't covered compactness yet in the course. Further, I don't really see how to show that [0,1) is not homeo to (0,1).
What I think the question wants us to do, is somehow prove that the intervals have a different number of cut points, or prove that one has a non-trivial clopen subset whilst another does not. But right now I can't find any cut points, nor can I find a non-trivial clopen subset of any of these intervals. Is this indeed what we need to show, or am I way off track here?

[prime elbow]

Observe the cuts in (0,1) and [0,1)

[prime elbow]

If I cut a single point x from (0,1) then it looks like (0,x) u (x,1) now think, does it have non -trival clopen set or not?

[hot pivot]

Proof verification for (b):

[whole sapphire]

that's correct, but you still have to show f is open.
Let A be open in X and define E and V in the same way.
Now argue that the restricted map f|E : E--->V is a homeo and conclude f(A) is open in Y.

(Side note, completely ignorable, the subtle application of axiom of choice in your proof can be avoided by the pasting lemma again. For each x in X, define E_x as the union of all open nghds U of x for which there exists an open V in Y making f|U : U-->Y a homeo, and define V_x as the image f(E_x); the pasting lemma ensures f| E_x : E_x ---> V_x is a homeo.)

[hot pivot]

Ah I pasted a third image, doing the "f is open" part.

[hot pivot]

Extra justification for this line: "image of E_x ∩ A is open in V_x" : This is true because, f restricted to E_x is a homeomorphism to V_x. And homeomorphisms are open, and E_x ∩ A is open in E_x, since A is open in X. Thus the image is open in V_x.

[dry seal]

I'm going down a rabbit hole of different notions of compactness, there ended up being way more than I expected. Does anyone know of such a notion which is not equivalent to regular compactness in metric spaces? This seems a bit absurd but it would produce a fun counter example I imagine

[prime elbow]

take $S_N = {\sum_{i=0}^{i=n} |x_i|^2}^{\frac{1}{2}}$, then for all $i$, $|x_i| \leq S_n$ for all $n\geq i$. Hence $|x|_{\infty} \leq |x|_2$

[gentle osprey]

Notknow🙇

[prime elbow]

Is it correct?

[young stone]

yes

[young stone]

What do you mean? Are you asking for a notion of compactness in spaces less general than a metric space that is not equivalent to compactness for all metric spaces?

[tender halo]

i mean lindelof-ness is not equivalent

but i am unsure if you are willing to call it a "compactness notion"

[dry seal]

You know what that does kinda fit the bill, thanks

[tender halo]

only goes one way though, compact metric spaces are lindelof

[dry seal]

I mean a notion of compactness which does not immediately imply compactness in metric spaces (since in metric spaces basically everything is equivalent)

[dry seal]

This is exactly what I was looking for then, thanks again

[prime elbow]

Thank you

[hot pivot]

re: proof w/o AoC: by the pasting lemma, wouldn't that make f|E_x continuous, not necessarily homeo?
btw i did include a proof to show "f is open."
Interesting AoC catch though. Should I avoid using AoC with my proofs? All I know about it is that it's controversial (though not inconsistent). And since point-set-topology feels very intimately tied to working with sets, I perhaps should avoid using AoC as much as possible?

[gritty widget]

Point-set topology without AoC is a trainwreck.

Subsets of second countable spaces need not be second-countable without AC

AC is famously equivalent to Tychonoff’s theorem

[tender halo]

yeah without AC it gets weird fast

[cerulean oriole]

Point-free topology moment

[hot pivot]

for (c)
As a crude example, let a constant function f take a discrete space X to ParticularPointTopologyR (at 7), such that f(x)=7. For each x in X, f is a homeo from {x} to {7}.
Is there a more interesting example of local homeomorphisms? Preferably one that motivates their definition in the first place? Since it looks rather contrived.

[tender halo]

consider the quotient map R -> S1, x |-> (cos(2pix), sin(2pix))

i.e. wrapping the circle

[gritty widget]

Manifolds

[waxen oxide]

Given $f: X \rightarrow Y$, if $f$ is continous, and $X$ is compact, then does $f$ being injective imply that it is an embedding of $X$

[gentle osprey]

Spice

[waxen oxide]

Ohh, because every compact subspace of a Hausdorff Space is closed

If $F$ is an embedding, then is it an embedding into $\mathbb{R}^N$ for some $N \in \mathbb{N}$, if so why?

[gentle osprey]

Spice

[rancid umbra]

is F the map X —> R^{n + nm}

if not, could you be more specific about the domain and codomain of F?

[waxen oxide]

The domain is a compact m-manifold, the codomain is given as above, like a point in $x$ in the codomain would look like $(x_1, \ldots, x_{2n})$ such that $x_1,\ldots, x_n \in \mathbb{R}, x_{n+1}, \ldots, x_{2n} \in \mathbb{R}^m, m \in \mathbb{N}$. I am assuming there is some kind of additional homeomorphism

[gentle osprey]

Spice

[rancid umbra]

the codomain is homeomorphic to R^{n + nm}

[waxen oxide]

why though

[rancid umbra]

it’s the same as asking why R^n x R^m is homeomorphic to R^{n+m}

[waxen oxide]

I didn't know that

[rancid umbra]

you should prove it!

[waxen oxide]

lol thats optimistic

Oh, actually I see it

[rancid umbra]

sorry, made a typo

[waxen oxide]

For example, $R^2 \times R^3$, $((x_1, x_2), (y_1, y_2, y_3)) \mapsto (x_1,x_2,y_1,y_2,y_3)$?

[gentle osprey]

Spice

[rancid umbra]

yea into R^5

[oblique sentinel]

Is every set in a finite topological space with the discrete topology both open and closed?

[tranquil cosmos]

yes, and this even goes for discrete spaces wholesale

[oblique sentinel]

So showing that if X is finite and hausdorff it has to have the discrete topology is kind of trivial since we have to prove every singleton is open.

[exotic hemlock]

any ideas on what to say for the 2nd part? i see that this metric is bounded by \pi whereas the eucliden metric is unbounded but im not sure what other interesting observation im suppose to make

[opaque scroll]

I guess given the above text a natural question would be if the identity function on R is uniformly continuous for example

[prime elbow]

It looks complicated, any easy way?

[cerulean oriole]

I think you need the Cauchy-Schwarz inequality to do this.

[prime elbow]

yes

thanks Raghuram, i was forgot to use Cauchy Schwarz theorem

[plush crescent]

What's the dif between connected, path connected and arc connected?

[warm kettle]

Did you read their definitions?

[plush crescent]

I thought I understood the dif but then I saw this example on wolfram, how's that there's no inverse? I'm assuming taking {a, b} as the trivial top means that you have T={empty,{a},{b},{a,b}}

[tranquil cosmos]

Nah, it means the topology is {ø, {a,b}}

Indiscrete topology rather than discrete topology

[plush crescent]

I realized this after, yeah it's so trivial I don't even consider it

I think I get why theres no inverse, bc either collapes to a or b however it's defined

It has nothing to do with geometric intuition even though the idea is geometric

But again only when understood it's geometric, I'm just contradicting myself lol

[tranquil cosmos]

Nah, you can define paths like the above where one endpoint lands at a and the other at b

However maps back from the indiscrete space are problematic

[plush crescent]

Yes paths for sure, but I see why they're problematic

[cerulean oriole]

If I is a directed set, let I^+ be I with a new point ∞ adjoined, such that I is a discrete open subspace and a set containing ∞ is open iff it contains {j : j ≥ i} for some i ∈ I. It's true that a function f from I^+ to any topological space X is continuous iff f(∞) is the limit of the net f_I, right?

[cerulean oriole]

This space is Hausdorff iff I has no maximum (equivalently, maximal) element (if I has a maximum, ∞ is just a specialisation of it). This space is compact if for every i ∈ I, {j : j ≰ i} is finite. In particular, if my previous message was correct, any k-closed subset of a topological space is closed under limits of nets indexed by I without a maximum element such that {j : j ≰ i} is finite for all i ∈ I (a condition somewhat stronger than quasi-well-ordering; I can't think of any examples noticeably different from ℕ).

[whole sapphire]

It's not right to say "the" limit right?
But it seems true?

=> Let f:I^+ -->X be contn and U be a nbhd of f(∞).
The preimage of U is an open nbhd of ∞ and thus contains all j ≥ i for some i.
Hence f is eventually in U.

<= Let f converge to f(∞) and fix an open set U in X.
Let j be in the preimage of U. If j= ∞, then U is an open nbhd of f(∞), hence there is some i such that {k : k ≥ i} is an open nbhd of j contained in the preimage of U.
If j is not infinity, then {j} is an open bbhd of j contained in the preimage of U.

[cerulean oriole]

Thanks!

[cerulean oriole]

And BTW I did come across one more such I: the collection of finite subsets of an infinite set.

[heady skiff]

Can somebody help me see why this comment helps us show that every nonempty open subset of S^{n - 1} contains an open subset D whose complement is an (n - 1)-cell?

[cerulean oriole]

IG every non-empty open subset of S^{n-1} contains a small ((n-1)-dimensional) disc centred at one point, and the complement of this in S^{n-1} is homeomorphic (say, by stereographic projection from the centre of the removed disc) to a closed ((n-1)-dimensional) disc?

[gentle osprey]

Spice

[iron bolt]

no for simple reasons

think about the case K = Ø and U = X. then K is compact, contained in U and U is open like you asked

but if it was true that any continuous function f : X → Y was uniformly continuous on U, this would mean that every continuous function is uniformly continuous

which I'm sure you've seen counterexamples of

[cerulean oriole]

Is it true that for any topological spaces X, Y, Z, composition [Y, Z] ⨯ [X, Y] → [X, Z] is separately continuous wrt the compact-open topologies? If F(K, U) := {f : f(K) ⊆ U}, then for any f: X → Y, the preimage of F(K, W) under precomposition by f is F(f(K), W) and for any g: Y → Z, the preimage of F(K, W) under postcomposition by g is F(K, g^{-1}(W)), right?

[heady skiff]

Yeah, this is something along the lines I was thinking of

[tranquil cosmos]

Why is this construction supposed to be intuitive?

I suppose the idea is something like "embed X everywhere possible and check the biggest closures of its image you can find"

But why take a product then?

Likewise, the actual definition is also pretty esoteric at first glance

[tender halo]

I think a better (classic) way of looking at SC is that the compactifications of a space form a complete upper semillattice wrt the order \rho X < \tau X iff there is a function f: \tau X -> \rho X such that f | X = id_X (note that that function is necessarily surjective and carries the complement of X in \tau to the complement of X in \rho)

being a complete upper semilattice, it necessarily has a maximal element, which is the SC compactification of the space

the construction using products is that same thing except you throw everything and the kitchen sink at it

[opaque scroll]

I mean, if you want every map from X to compact Hausdorff K to factor through the compactification, this seems like the most natural way to do it.

Like you every map factors through the product just by the projection maps, you don't need everything so just take the image, and then you need to take the closure to make sure it's compact.

[tender halo]

the reason unit interval shows up is because of the separation properties of tychonoff spaces and the fact that the Tychonoff cube is universal for compact spaces

in fact a compactification \tau is the SC compactification iff every function X -> I is extendable to a function \tau X -> I

re "why take products" consider the diagonal theorem, the intuition is that the more maps you take into a diagonal, the more separated the right hand side becomes

and SC is about adding all the limit points the space can have (for that perspective, consider the ultrafilter construction)

[tranquil cosmos]

This is not immediately obvious to me...

[tranquil cosmos]

Oh the projections are a really really good point

[tender halo]

for a family of compactifications c_i X consider the diagonal X -> \Pi c_i X

the closure of the image of X is the least upper bound of c_i's

[tranquil cosmos]

Ah

[tranquil cosmos]

The product construction is pretty slick then

[opaque cloud]

what does "coarse-ness" mean when it comes to topology? I'm assuming it's basically the "Size" of the elements of the topology (just intuitively, I know we don't have a way to measure distances in a topological space)

so would the discrete topology be the finest topology ever and the trivial topology the coarsest topology?

[spare light]

Inclusion of topoloy(open subsets)

[tranquil cosmos]

Indeed

If you have a set X, the topologies on X are partially ordered

[opaque cloud]

I have to learn more about partial orders

[tranquil cosmos]

And "x<y" means x is coarser or y is finer

[opaque cloud]

inch-resting

[tranquil cosmos]

So coarse and fine are only relative terms

[opaque cloud]

https://math.stackexchange.com/questions/3541203/partial-order-on-the-set-of-all-topologies-minimal-and-maximal-elements-on-this

I see that doesn't seem to hard to understand

the partial order that is

[tender halo]

its the usual subset poset, just on the set of topologies

there are interesting things to be said about the poset sometimes, for example the compact topologies are minimal Hausdorff topologies (but of course a minimal Hausdorff topology need not be compact)

[tranquil cosmos]

Anyway, back on stone cech

What are some good examples on it that aren't trivial or incomprehensible? Also, iirc, it shows up in func ana as the mirror of an operation of C* algs, do you remember which?

[tender halo]

the points of the stone cech compactification of X correspond to the maximal ideals in the ring of real continuous functions on X

in fact that is one of the ways you can construct \beta X

\beta N is a very important set theoretic object too

\beta N is probably the most accessible example?

it has some fun properties

for example, every infinite closed subset of \beta N has a subspace homeomorphic to \beta N

also no non trivial sequences converge in \beta N

so its not sequential in an extreme way

[tranquil cosmos]

Hmm, I was thinking more about stuff like, say, open subsets of Euclidean space

[tender halo]

wdym exactly

[tranquil cosmos]

Open disk, real line

Do those have easy SCCs?

[tender halo]

nope

[tranquil cosmos]

What does then, that isn't a compact space already

[tender halo]

they have N as a closed subspace, so \beta R necessarily contains \beta N

and \beta N is not a pleasant object

[tender halo]

omega_1 has omega_1 + 1 as SCC

only has one compactification which is the Alexandroff compactification which is the SCC

[tranquil cosmos]

This is so damn wild

SCC is insane glitch concept

[tender halo]

its really fun yeah, probably my single favorite object in maths

[tranquil cosmos]

adds multiple copies of the very well-behaved interval, a paragon of compact spaces, to N: nope, doesn't tame it one bit

adds just an uncountable bunch of points: s-sorry for every being difficult

[tender halo]

this is due to the fact that every continuous real valued function on omega_1 is eventually constant

[tranquil cosmos]

I assume the universal map $\beta\omega \to \omega+1$ just maps all the new points to the point at infinity

[gentle osprey]

PKThoron

[tranquil cosmos]

Also man, I thought bN introduced a new limit point for every imaginable subsequence of N, but this implies the opposite

[tender halo]

yes, because maps between compactifications of X that fix X map residues to residues

let me find the relevant theorem

[tender halo]

it introduces a new limit point for every ultrafilter on N

and a sequence can be extended to an ultrafilter in a lot of ways

so the reason why you cant get to a limit with sequences is that they are not enough to specify a point in \beta N

[tranquil cosmos]

Neat

So every sequence hovers around a whole bunch of points as its limit points?

While being discrete

Which sounds like the opposite of a well-separated space, yet the entire thing is Hausdorff

[tender halo]

well, consider the sequence which is just N, it hovers around the whole of \beta N

[tranquil cosmos]

Crazy

[tender halo]

because you can extend that sequence to any ultrafilter

[tranquil cosmos]

Also the fact that N is open in it

N shouldn't be open in anything!

[tender halo]

that is just the fact that N itself is a member of any ultrafilter

[tranquil cosmos]

I assume bN is a very, very disconnected space

When I was younger, I tried grasping nontrivial ultrafilters on N

[tender halo]

it is in fact extremally disconnected

[tender halo]

i.e. the closure of every open set is clopen

[tranquil cosmos]

(For once, I'll define N as not including 0)
Came up with $\bN \supset 2\bN \supset 4\bN \supset 8\bN \supset \dots$, which generates a filter that isn't contained in any principal ultrafilter

Uhh how do the superset thing

[tender halo]

\supset

[gentle osprey]

PKThoron

[tranquil cosmos]

Thanks

I assume this filter can be extended to $\beth_2$ ultrafilters lol

[gentle osprey]

PKThoron

[tranquil cosmos]

Since N is very self-similar, I assume bN also is

[tender halo]

a compactification \alpha of X is equivalent to the SCC iff every pair of completely separated subsets of X has disjoint closures in \alpha X. if X is normal, the same is true about disjoint closed subsets

[tranquil cosmos]

Oh so that's why bR is so crazy, (0,1) and (1,2) are disjoint open, but their closures intersect

So bR has to "invent" a whole load of points to say "actually their closures differ"

[tender halo]

so in N, consider a set A and N \ A, they are disjoint and closed - so their closures in \beta N are disjoint too, but their union is the whole \beta N

[tranquil cosmos]

Mhm

[tranquil cosmos]

This reminds me a lot of the étalé space of a sheaf

[tender halo]

? no why

their closures intersect and they will intersect in \beta R

i was talking about completely separated and disjoint closed subsets not disjoint open

[tranquil cosmos]

Ah my b

[tender halo]

also side note about limits: take a function f: R -> I, when does a limit of f under x -> +-inf exist?

consider the filter-base (-inf, a) \cup (a, +inf). the abovementioned limit exists iff every ultrafilter that extends that filter-base converges to the same point under f iff f is extendable to a function S1 (as \omega R, the alexandroff compactification of R) -> I

limits x -> +inf correspond to the filter base of (a; +inf), limits to -inf correspond to the filter base (-inf, a)

[glossy spade]

How do i prove part b. I do know that f^-1( r , infinity) belongs to τ is a lsc for every r belonging to euclidean topology

Thanks in advance!🙏

[heady skiff]

What am I missing? Since h is a homeomorphism and U is open, shouldn't h(U) = V be automatically open?

[tranquil cosmos]

Let X be $([0, \infty) \times \bR) \cup (\bR \times {0})$, endowed with the subspace topology of $X \subset \bR^2$. Then $(-4,-3)$ is open and homeomorphic to $(3,4)$, but the latter isn't open

[gentle osprey]

PKThoron

[prime elbow]

What is here (-4, -3)?

[tranquil cosmos]

The interval

[prime elbow]

But it is in R^2?

[tranquil cosmos]

I omitted the ×{0}

[prime elbow]

I ses

[heady skiff]

I see, thanks

Where I was wrong in my reasoning?

[tranquil cosmos]

Homeomorphisms respect open sets within U and V, relative to U and V

[heady skiff]

But isn't U open in U?

[tranquil cosmos]

Yeah

But if there's an ambient space X, like in your situation

Then U and V may lie in it differently

Particularly, one might be open in X and the other not

[heady skiff]

Ah okay, so they're talking about relative to the subspace topologyarising from being a subset of S^n

[tranquil cosmos]

Yep

[heady skiff]

gotcha thanks

[plush folio]

You already got an answer, but just in case: h being a homeomorphism means that it sends open sets in U to open sets in V. Now h(U) = V being open in V is trivially true (regardless of h), but it doesn't tell us whether V is open in S^n

[hot pivot]

I don't find it intuitive that the Cartesian product of two topological spaces forms only a basis of the product topology, and is not the entire topology (despite having proved the basis part rigorously).
Can someone give me an example (let's say, in R^2), of a set that is clearly open in R^2, but it cannot be expressed as the cartesian product of two open sets of R?

[prime elbow]

What will you get if you take a product of two open intervals ?

A rectangle, right?

[hot pivot]

Just as you said that, it struck me that an open disc would probably work.
I'm just not yet sure it's impossible to write as the cartesian product.

[prime elbow]

Now take the intersection of two rectangles, are they rectangle?

I suggest you look at Munkres they give a reason why we are taking product of open sets as a basis

[prime elbow]

Yes open disc also work

Take open disc x^2 + y^2 = 1

Now think can you write as U×V, where U is an open and V is open

[hot pivot]

This is where my mind's at. Intuitively it looks impossible, but I'm trying to see if I can precisely articulate why this is so.

[prime elbow]

Yeah try to proof

[plush folio]

You don't even have to think about topology, you can't write the open disc as a cartesian product AxB regardless of whether A and B are open

It's just set theory

[prime elbow]

Yes

[hot pivot]

Ah is it because, (0,1) and (1,0) are in the set, but (1,1) is not?

[prime elbow]

Actually the point is (1,0) in A×B, (0,1) in A×B but (1,1) is not in {(x,y) | x^2+y^2≤1}, here you can modify your argument as open disc

[prime elbow]

Yes but since open disc doesn't contain (0,1) so you can modify it

[hot pivot]

Ah right, good catch! Yeah the modification seems straightforward. (0.9,0) etc... Awesome!

[prime elbow]

Yes

[exotic hemlock]

im having touble seeing how the inverse function would be continuous here

if i were to just manually calculate it out, i get $f^{-1}(x)=\frac{1 \pm \sqrt{1+4x^2}}{2x}$

[gentle osprey]

somethingwrong

[tranquil cosmos]

Have you checked continuity near 0?

I.e. you choose a sign and check whether left and right limits are the same

[exotic hemlock]

thats the problem im having, im not sure how to check continuity at x=0 for this, do i just show that for any $\epsilon$ neighborhood of $f(0)$, there exist $\delta>0$ so if $x\in(-\delta,\delta)$ implies $f(x)$ lies inside

[gentle osprey]

somethingwrong

[exotic hemlock]

i guess for points not $0$, we get continuity since our function is rational which are known to be continuous as long as denominator is non-zero right?

[gentle osprey]

somethingwrong

[tranquil cosmos]

You check the limits $x \to 0^-$ and $x \to 0^+$

[gentle osprey]

PKThoron

[tranquil cosmos]

Orrrrr

Just check whether $f^-^1$ is continuous via IFT

[gentle osprey]

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

[whole sapphire]

Hey, I wanted to prove this particular claim for sequences indexed over an ordinal.
Is this claim true? If so is the spirit of this sketch, when written carefully, a valid arugment?

[whole sapphire]

edit, i don't think im right.
here is an attemtped proof.
I think all mysteps are correct except the last sentence in the proof.

I am thinking this theorem is false, but can't come up with a counter example.

[gritty widget]

Here “cluster point” means “every neighborhood of x’ contains infinitely many elements of the a-sequence”, right?

Not sure if the definition is any different for sequences indexed by an arbitrary ordinal.

[whole sapphire]

not exactly that.
here are teh defns i am currently using.

An ordinal sequence x : A--->X has a cluster point x' iff for every open nbhd U of x', for every i<A, there exists j>=i such that x(j) is in U.

An ordinal sequence x : A--->X converges to x' iff for every open nbhd U of x', there exists an i<A such that x(j) is in U for all j>=i.

[gritty widget]

Ic

[whole sapphire]

I am thinking of modifying these defns to make the proof work.
(For example: Convergence would mean it just has to be omega-eventually in any nhd, meaninig there is an "ininital segements" worth of the sequence in the nbhd wihtout any guarantee about what happens in the limit ordinal "after"l.)

But I'm desperate to make it work with these defns.

[gritty widget]

Consider the cocountable topology on $\omega_\omega$, and any take any injective $\omega_\omega$ sequence that doesn’t have 0 in it. Take x=0. Every neighborhood of 0 contains all but countably many members of the sequence. Given any $j\lneq\omega_\omega$, the sequence “cut off at j” is clearly of lesser cardinality than $\omega_\omega$, so it must contain some members after j(since it contains $\aleph_\omega$ many of them). So 0 is a cluster point.

However, no subsequence converges to 0(the only sequences that do are the eventually constant sequences).

Let $U$ be some open set of 0. Then, consider the set $V=U\setminus{\omega, \omega_1, \omega_2, \omega_3, …}$. This is an open set, since it’s the cocountable topology.

Clearly, there is no ordinal m such that $x_b\in{V}\forall{}b\geq{}m$.

[gentle osprey]

NAT Enthusiast

[gritty widget]

@whole sapphire

Shoot I said j<omega1

just rewrite that in your mind to omegaomega

I’ve edited this too much already

[whole sapphire]

Wow this is brilliant! Thank you so much :)

[gritty widget]

Np

Also just to clarify a bit, I had to go through this same thought process with the cofinite then cocountable topology on R first

Took me a while lmao

I usually get frustrated after awhile of trying to solve a problem and failing, I’d assume most people do, but always try not to

[whole sapphire]

i feel the same way. I feel silly trying to prove this for half a day.
I'm stealing your counter example! Shelfing it under the "motivation of nets and directed sets"

[gritty widget]

This is the first time I’ve actually used topology(instead of just algebra) for topology in quite a while

Feels like riding a bike for the first time in a decade

Might write some notes and whatnot on high-cardinality(or maybe low cardinality) topology stuff

So thank you for the question

Theorem:a countable space M with points $x_0, x_1, x_2, …$ is path-connected iff there is a continuous surjection $F:\mathbb{R}\to{}M$

Suppose there is such a surjection.

Then, let $x_n, x_m$. Since it’s a surjection, there’s some real number $p$ such that $F(p)=x_n$, and $q$ such that $F(q)=x_m$. Assume wlog $p\leq{}q$. Then, $F|[p, q]$ is a path from $x_n$ to $x_m$.

Suppose M is path connected. Then, let $p_{n+1}$ be some path from $x_n\to{x_{n+1}}$. Define $F_0(r)=x_0\forall{r}\lneq0$, to be a function from $(-\infty, 0)$ to $M$. and define $F_n(r)$ on $(-\infty, n)$ by induction, for integer n.

Suppose you have $F_n$. Then, let $F_{n+1}$ be $F_n$ on $(-\infty, n)$, and, for r some number between n and n+1, let $F(r)$=$p_{n+1}(r-n)$. This is a continuous function, because it’s simply path concatenation(when considered as a function from [0, n) to M, which is extended to this function by letting it be constant on the non-positive part).

Then, let $F_\infty$ be defined piecewise by $F_\infty(r)=F_{floor(r)}(r)$.

By the pasting lemma(which also applies to arbitrary collections of open subsets covering the space), this is a continuous function.

[gentle osprey]

NAT Enthusiast

[gritty widget]

Surjectivity follows from the fact x_n enumerates M

Does a similar proof for “every $\aleph_1$ sized space is path-connected iff there’s a continuous surjection from the long line to it” work? It seems like it should, but I’m tired rn so I’m not gonna check myself. I’ll just update with results tomorrow.

[gentle osprey]

NAT Enthusiast

[tranquil cosmos]

For the construction of Stone-Čech, you do pay attention to size issues by just focusing on [0,1]. Why don't we do something similar for condensed sets? ( @cedar pebble )

[cedar pebble]

people absolutely pay attention to size issues around condensed sets

[tranquil cosmos]

Yeah but inaccessibles are a bit hamfisted, aren't they

Can't you do a natural cutoff?

[cedar pebble]

not at all

but also this isn't really the issue

[tranquil cosmos]

Like for stone cech it's something like |PPX|

[cedar pebble]

there is the standard issue of talking about something like the category of sheaves on a large category like that of profinite sets without imposing some size cutoff

in practice what one typically does is restricts to a subcategory of "\kappa-small profinite sets" for a suitable cardinal \kappa

kappa does not actually have to be so large to hit most examples that you encounter in day to day mathematics. If you need to expand the size, you can always just replace \kappa with some larger \kappa' and pass to the next larger category

usually \kappa is taken to be some uncountable strong limit cardinal and there are plenty of these regardless of how you feel about inaccessible cardinals

[tranquil cosmos]

Ah so it can be singular

[cedar pebble]

it's also worth mentioning that people have recently started to favor working with "light profinite sets" which give you quite a small category with nice properties that still hits all the useful examples

[tranquil cosmos]

Like $\beth_{\omega}$ sticks out as a good choice here

[gentle osprey]

PKThoron

[cedar pebble]

light profinite sets are those for which the indexing set in the inverse limit of finite sets is at most countable

[tranquil cosmos]

Oh dear... people worked with uncountable ones?

[cedar pebble]

well yeah

in practice you sort of don't need to

if you like to think of the test spaces in terms of totally disconnected compact Hausdorff spaces, this amounts to adding the word "metrizable"

of course there are plenty of things that aren't metrizable, but lots of examples we really care about fit this bill

there are tradeoffs to either approach of course

if you like the philosophy that most of the theory of topological spaces sits nicely in the condensed setting, then imposing metrizability seems a bit restrictive

on the other hand it makes things much easier to work with and it's not so restrictive that you lose access to most of the examples you care about

I guess for those uncomfortable with most cardinals "countable limit of finite sets" is pretty good

another nice characterization is if you like Stone duality, then the light profinite sets are those which correspond to countable Boolean algebras

[tranquil cosmos]

Limit set comprehension! Only allow countable operations! Make power sets and the real line objects of higher type! Abolish the pairing axiom!

[cedar pebble]

I mean you still encounter uncountable things but yeah in practice the actual data defining these things is very small

maybe the only downside is that you are forced to think of "standard" things in somewhat bizarre ways sometimes but that's just an issue of people getting used to thinking this way

[tranquil cosmos]

Say

In algebra, you sometimes encounter "size issues" as well, and you solve them by reducing to compact support, N^(N) for example

Not sure I got that across correctly but you get the point

Do people look at sheaves on the cat of profinite spaces with "compact support"? (Or countable support, \lambda-support...)

[cedar pebble]

not really

certainly not for the reason you mention

you should keep in mind that while these convenient size restrictions are made at the level of the test category of (light) profinite sets, one does not make such size restrictions once passing to categories of sheaves

this is where the usual machinery of universes/inaccessibles starts to kick in

I sort of dislike calling these size issues since the only issue is trying to avoid the standard maneuvers for these things, there are no size issues if you just use standard ZFC and epsilon more

[tranquil cosmos]

I somehow failed to consider that

But also somehow, sheaves on small categories don't bother me

Even though they're class-size as well

The category of sheaves i mean

[cedar pebble]

yeah that's fine usually the main size issues you encounter are if you try to talk about sheaves on large categories (in which case you make size restrictions and expand them at will) or when you iterate this construction (in which case you need universes/inaccessibles)

classes don't really fix the latter issue

[tranquil cosmos]

For example the category of all categories bothers me

Cause it seems like some Russell shit

[cedar pebble]

there again you would want to impose some size restrictions

[cedar pebble]

amusing remark: this isn't actually a paradox

https://mathoverflow.net/questions/476170/foundations-and-contradictions-of-scholzes-work-the-category-of-presentable-in

this sort of reflection principle is incredibly useful and it lets you avoid a lot of otherwise painful maneuvering around definitions of (\infty,n)-categories if you ever need to work with those

[unreal stratus]

Lol this definition came up in smth im reading today

[unreal stratus]

Tbf I think this is orthogonal to what was said

[unreal stratus]

I mean here note that what really is meant here is "the large category of small categories" where large and small correspond to different sizes of sets in some sense (often formalised using Grothendieck universes)

But there is also nG's important point there that shows even this isn't really necessary ig

[unreal stratus]

Specifically like a paper used this approach to (∞,n)-categories which was quite nice

[cedar pebble]

Not completely

[unreal stratus]

Yeah I have changed my mind

[unreal stratus]

Was kinda funny lol like

Studying deformations of En-monoidal categories can be done using (∞,n)-categories lol

To me the latter is way more complicated lmao

Or at least less familiar

[tranquil cosmos]

Ah, thank goodness

Lol

I usually envision Cat as the category of all locally small categories even

[cedar pebble]

Yeah that is reasonable

[tranquil cosmos]

Rip topoi though or condensed sets

[cedar pebble]

You can always enlarge it’s not a big deal

Most things just start life in the small setting and get bumped up in size by standard constructions

I don’t think I’ve ever encountered a situation where I’ve needed more than like three universes/inaccessibles

[tranquil cosmos]

Oo

[cedar pebble]

Such a mild thing to worry about

[tranquil cosmos]

Inaccessibles are cool

[cedar pebble]

Yeah I used to be intimidated by this issue until I realized that aleph_0 is also inaccessible if you forget axiom of infinity

Some large cardinal axioms are unsettling but “there exists infinitely many inaccessibles” is as mild as these get

[tender halo]

V_omega is the only real model of set theory

[cedar pebble]

V_\omega_CK