point-set-topology

[night heron]

My autism makes me particular ig lol

[opaque zodiac]

there definitely are mathematicians who are super picky about this kind of thing

[robust drum]

I think you get less picky the more math you do generally

But yeah I definitely agree it depends on context

[opaque zodiac]

well I'm thinking like historically

and in the context of model theory

Tarski always wrote things in excruciating precision

for instance

[night heron]

Doing algebraic topology rn and still just as picky

Even more picky actually

[robust drum]

I used to be very picky and I got very annoyed by eg Hatcher’s writing style but I’ve been doing low dim topology research this summer and it chilled me out a lot

Reading a lot of papers where we describe isotopes by just saying shit like “push through the ball to eliminate intersections”

There are certainly those who will always be picky but you can only go so long in math doing things like treating different set theoretic constructions of the same object as different because the underlying sets are different

[paper wedge]

i am never good at this

like the deformation retraction arguments or

"gluing along the bla bla"

it's pretty hard g

for me too

[robust drum]

Yeah but even something like the proof of Brouwer’s fixed point theorem if you actually try to write a formula for the map induced by assuming there’s no fixed point and projecting radially outward

It’s disgusting it’s like 4 lines and some really shitty polynomial

[paper wedge]

like the map from the point inside untill it hits the circle u mean

yeah

[robust drum]

Yeah

[paper wedge]

yeah idk if that

if thats normal

[robust drum]

So like it’s good to look it up and see that some nerd out there has actually done it but like

[paper wedge]

if i dont like the

deformation retraction

shit

arguments

[robust drum]

If you actually try to be insanely picky about things and explicit

You’ll cure yourself of wanting to do it

[quartz horizon]

I guess im worried that it won’t be correct

[paper wedge]

i really dc i just dont know like

[robust drum]

Like I once looked up the homeomorphism from S^1 x S^1 to the surface of revolution torus in R^3 and it’s also really gross lol

[paper wedge]

if thats how ppl do things

like i remember

reading gullimans pollack

and he had this picture of

transversality

it's like

u need to use ur visual intuition reallyu

[robust drum]

Yeah

[paper wedge]

rather than the map

[rancid umbra]

how else are you doing things in alg top?

or any geometry?

[robust drum]

I mean I think I’m biased because I’m doing a lot of low dom stuff but doing everything with formulas is just too much pain

[paper wedge]

yeah i thought it was limited honestly like

[rancid umbra]

or topology?

[paper wedge]

like i like the more algebraic or combinatorial side of things

imo

but i like when the algebra has sense so

thas why i wanna learn

[quartz horizon]

Wait it is?

[robust drum]

If I’m remembering right

Cause you need to actually parameterize the surface of revolution torus first

[rancid umbra]

what coords u using haha

[robust drum]

And then it’s not bad

I didn’t actually try this myself tho

I just remember seeing some stack exchange post that did it

[rancid umbra]

(s,t) |—> (s,t) bam, done

[robust drum]

And it looked kinda ass

[robust drum]

Sure but I guess I would say the tough part is proving those coords indeed parameterize the shape you get by rotating a circle about an axis

[rancid umbra]

yea i suppose.

with regards to the stuff you guys were talking about with being thorough, i think it’s a good skill for when you are suspicious or want to see something for yourself. it would take me ages to get through anything if i had to be as thorough as i was in intro to RA or LA courses

[quartz horizon]

Yeah I think like

I would prefer being able to see it for myself

[rancid umbra]

yea. most of the times if i’m reading a text book or a paper, i’ll read over it and then come back to it later

[quartz horizon]

I think this is why I struggled when I first got into algebraic topology

There were lots of “intuitive” operations being done on topological spaces

And I couldn’t connect that to the rigor I went through when first studying topology

[pearl trail]

Am I approaching this correctly? If so, I just need to find an epsilon and I’m done? I am thinking epsilon = max{e1, e2}. Can someone help me correct any mental model issues or provide a hint to nudge me in the right direction?

[quartz horizon]

Hm so

First, do you know how to prove $d : X \times X \to \mathbb{R}$ is continuous?

[gentle osprey]

Pseudonium

[quick crane]

The mental model for problem 1 is you need to show that when $d(x, x_0)$ and $d(y, y_0)$ are small, then $|d(x, y) - d(x_0, y_0)|$ is small.

[gentle osprey]

L

[quick crane]

The execution is by using the triangle inequality a few times

[quartz horizon]

Well like

Yeah I guess so

[quick crane]

In fact, $|d(x, y) - d(x_0, y_0)| \leq d(x, x_0) + d(y, y_0)$

[gentle osprey]

L

[quartz horizon]

It’s a general theorem that if T is “topologization” as applied to metric spaces, then a function f : M -> N is metric-continuous if and only if f : TM -> TN is topology-continuous

[quick crane]

So d is Lipschitz from X x X to R

[quartz horizon]

(I.e. T is full and faithful)

[quartz horizon]

Right though strictly we need to show d is continuous from X_d x X_d -> R, where the concept of Lipschitz doesn’t make sense

I was asking whether they could show the metric space version, or whether the issue was more translating into topological spaces

[quick crane]

That's a matter of showing that continuity of maps between metric spaces with the induced topologies is equivalent to the $\varepsilon$-$\delta$ definition.

[quartz horizon]

Yes that’s what I said here

[gentle osprey]

L

[quick crane]

it's pretty much true by definition, since balls form a basis of the topology

[quartz horizon]

And strictly that the product metric induces the product topology also needs to be shown

I.e. that T(X x X) is the same as TX x TX

[quick crane]

true you need to consider what topology is on X x X or how it can be metrized

[quartz horizon]

That is something which used to annoy me a little about metric spaces

In terms of how many metrics you can choose on the product

I mean it is cool

But it was not fun showing that like

(M x_2 N) x_infty R was homeomorphic to M x_1 (N x_5 R)

Where M x_p N is the pth product metric on M x N

[quick crane]

most of these follow pretty easily from equivalence of norms on $\mathbb{R}^n$

[gentle osprey]

L

[quick crane]

the various product metrics can be shown to be equivalent

[pearl trail]

back...checking your hints.

[pearl trail]

I don't think so. Let's see. Consider d(x) in [0, infinity). We WTS the preimage d^-1(x) is open. Consider x = (x1, x2) in X \times X. The set U in X \times X is open if x in B(x, epsilon) = {y in X \times X | d(x,y) < epsilon} subset X \times X. Consider z in X \times X open. Then

d(x,y) <= d(x,z) + d(y,z) < 2epsilon.

How's this?

[quartz horizon]

What is Y exactly

[pearl trail]

The codomain of d

[quartz horizon]

So… the real numbers?

[pearl trail]

well, [0, infinity)

Since d is a distance function

[quartz horizon]

I think you should just use $[0, \infty)$ then

[gentle osprey]

Pseudonium

[quartz horizon]

Rather than Y

[pearl trail]

Agreed. Any other issues?

[scenic ridge]

This was a bit long ago, but I do understand how the construction works, just not why this is a sufficient condition. Could you explain a little more why forming a bijection between continous maps out of the nondecreasing subset of [0, infinity)^n to continous maps out of [0, infinity)^n which are S_n equivarient proves that [0, infinity)^n/S_n is homeomorphic to the nondecreasing subset of [0, infinity)^n

[pearl trail]

I need to do this for x = (x1, x2). I think I only did it for one of them technically

[pearl trail]

isn't this another way of saying that there is an open ball that contains both x and y?

[pearl trail]

$|d(x, y) - d(x_0, y_0)| \leq d(x, x_0) + d(y, y_0) < \epsilon + \epsilon$

[quartz horizon]

So have you heard of the universal property of the quotient?

[gentle osprey]

GhostTheSavage

[scenic ridge]

yes

but i odnt see how this directly applies

are you just substituting [0, infinity)^n/S_n with the nondecreasing subset of [0, infinity)^n in the commutative diagram

[quartz horizon]

I’ve shown that the nondecreasing subset satisfies the universal property of the quotient

[scenic ridge]

so this?

[quartz horizon]

And thus must be uniquely homeomorphic to it

[scenic ridge]

ah ok

so if it satisfies same universal property it is homeomorphic

ic

[quartz horizon]

Essentially we’re using the yoneda lemma

[quartz horizon]

In a unique way, yes

[quick crane]

My mental model is to first understand what metric is on $X \times X$, then prove continuity of $d : X \times X \to \mathbb{R}$ using the $\varepsilon$-$\delta$ definition of continuity.

[gentle osprey]

L

[quick crane]

The first inequality here is the important quanititative part, then the second is applying whatever metric you chose for $X \times X$, e.g. $d((x, y), (x', y')) = d(x, x') + d(y, y')$.

[gentle osprey]

L

[pearl trail]

when we speak about closeness of d(x, x0) and d(y, y0), I follow. But I get lost when we combine them

[night heron]

Are you familiar with the topological definition of continuity?

You can define openness using the metric here

[quick crane]

1. Define an appropriate metric on $X \times X$. 2. Write down what it means for $d : X \times X \to \mathbb{R}$ to be continuous in terms of the $\varepsilon$-$\delta$ definition. Then verify it holds.

[gentle osprey]

L

[quick crane]

The reason I avoid open sets is because it will come down to $\varepsilon$-$\delta$ in the end.

[gentle osprey]

L

[night heron]

The exercise seems to be hinting at the topological definition so your method is better

[pearl trail]

Did you arrive at it like this:

Consider the distance $d(x,y)$. By the triangle inequality, we have $d(x,y) \leq d(x,x_0) + d(x_0,y)$.

Now apply the triangle inequality to $d(x_0, y) \leq d(x_0,y_0) + d(y_0,y)$. Combining both, we have $d(x,y) \leq d(x,x_0) + d(x_0,y_0) + d(y_0,y)$. We also have $d(x0,y0) \leq d(x0,x) + d(x,y)$.

Finally,

$$
\begin{aligned}
d(x,y) & \leq d(x,x_0) + d(x_0,y_0) + d(y_0,y) \
d(x,y) - d(x_0,y_0) & \leq d(x,x_0) + d(y_0,y) \
|d(x,y) - d(x_0,y_0)| & \leq d(x,x_0) + d(y_0,y)
\end{aligned}
$$

[gentle osprey]

GhostTheSavage

[night heron]

Man, typing latex on mobile is such a pain

[scenic ridge]

wait ive gotten confused again, what exactly do you mean for like a set Y to satisfy the universal property of the quotient?

Isn't the statement true for every space?

[quartz horizon]

So what I mean is the following

Suppose we have a topological space X, and an equivalence relation R on it

We say that another topological space Y “satisfies the universal property of the quotient” iff there’s a natural bijection between continuous maps Y -> Z and continuous maps X -> Z which respect R

(In fact this definition works even if R is not an equivalence relation)

By “respect R” I mean that whenever xRy, we necessarily have f(x) = f(y), so f converts the relation on the input to equality in the output

It turns out that, if such a space Y exists, it is unique up to unique homeomorphism

The quotient topology on X/R is an explicit construction of a space Y satisfying this

[scenic ridge]

is there a difference between natural bijection and bijection

[quartz horizon]

Yes

[scenic ridge]

what

[quartz horizon]

So, have you heard of naturality before?

[scenic ridge]

no

[quartz horizon]

I see, then let me try to explain it

[scenic ridge]

is this category theory

[quartz horizon]

Yes

So, for every topological space Z, we need a bijection between continuous maps Y -> Z, and continuous maps X -> Z respecting R, right?

[scenic ridge]

yes

[quartz horizon]

On the face of it this is quite a weak condition

After all bijections are just about cardinality

So it seems like I’m just saying the cardinality of the set of continuous functions Y -> Z is the same as the cardinality of the set of continuous functions X -> Z respecting R

And it seems impossible for this to specify a space uniquely up to unique homeomorphism, right?

[scenic ridge]

yes

[quartz horizon]

Naturality is what fixes this

The idea is, we have some assignment of a topological space Z to a bijection Hom(Y, Z) -> Hom_R(X, Z), where Hom(Y, Z) is continuous functions from Y -> Z, and Hom_R(X, Z) is continuous functions from X -> Z respecting R.

What we can do is investigate how this assignment behaves as we vary Z. So for a different space W, we get a different bijection - it’s now between Hom(Y, W) and Hom_R(X, W)

The idea of “naturality” is that this assignment is what you might call “relational” - if there’s a way to relate inputs, so a continuous map Z -> W, then there’s some corresponding way to relate the bijection for Z with the bijection for W

To make this precise, we can draw a commutative diagram

[quartz horizon]

Here’s the diagram

Sorry I’m on my phone atm and it is, uh, quite hard to use quiver

Here’s the idea

We can “transpose” a continuous map f : Y -> Z to a continuous map f^T : X -> Z

And similarly for W - we can “transpose” a continuous map h : Y -> W to a continuous map h^T : X -> W

In both cases, the transposed maps respect the relation R. We can do this because we’re assuming the existence of a bijection between Hom(Y, Z) and Hom_R(X, Z)

And a bijection between Hom(Y, W) and Hom_R(X, W)

But now suppose we have some way of relating Z and W, by a continuous map $g : Z \to W$

[gentle osprey]

Pseudonium

[quartz horizon]

Then there’s two ways to get from f : Y -> Z to a map X -> W

We can first transpose to get f^T : X -> Z, and then compose to get g o f^T : X -> W

Or we can first compose to get g o f : Y -> W, and then transpose to get (g o f)^T : X -> W

[scenic ridge]

yea

[quartz horizon]

Naturality asks these processes to give the same result - that is, g o f^T = (g o f)^T

It’s our way of relating the bijection for Z, and the bijection for W

[scenic ridge]

but what exactly is the transpose?

[quartz horizon]

So, we had a bijection between Hom(Y, Z) and Hom_R(X, Z) right?

That means there’s an invertible way to take an element of Hom(Y, Z), and make an element of Hom_R(X, Z)

[scenic ridge]

oh thats transpoe

[quartz horizon]

I’ve called this the “transpose”

[scenic ridge]

i see

and is g any function

or does it have to be continous

[quartz horizon]

Any continuous function

Because Z and W here are topological spaces

So the correct way to relate them is with a continuous function

And what naturality says is “if there’s a way to relate the inputs, there’s a corresponding way to relate the outputs”

[scenic ridge]

ok I see it just needs to saitisfy g f^T = (g f )^T right

[quartz horizon]

Here the “input” is a topological space Z, and the output is the “transpose” map for Z

[quartz horizon]

Yep exactly

This is what is meant by a “natural” bijection

And most ad-hoc bijections you choose won’t satisfy this

But if you have a space Y with a “natural transpose” for every space Z, then you have something satisfying the universal property

This is what makes it significantly stronger than just requiring equal cardinalities

We didn’t check this condition when I gave my universal property proof

But it’s not hard to show it works

[left eagle]

Im stuck on part 2. Can someone help explain how d being continuous helps proving that T_d is the coursest topology such that blah blah?

[rancid umbra]

hint: for x in X, set F : X --> R as F(y) = d(x,y) with respect to T

you may also want to recall what a subbasis is. or a basis

[left eagle]

Oh I see we get F is continuous because its just a restriction of d to {x} times X. and then from there you can just get all e-balls centered at x open in T

[rancid umbra]

yea, exactly that

[left eagle]

Heres a proof on stackoverflow that A subspace of a metrizable space is metrizable. On the last block of text, the two bases arent necessarily equal because hes only considering e-balls centered at some y in Y?? B_Y contains e-balls centered at some x not in Y right?

[opaque zodiac]

yeah but that isn't a problem

[left eagle]

why not?

[opaque zodiac]

uh well they won't be balls but you can show that they're contained by balls centered on points in Y

using the triangle inequality

[left eagle]

ye i suspected that but my dumbass couldnt figure out how to use the triangle inequaltiy to prove that

[opaque zodiac]

If $d(x,y) < \varepsilon$, find a $\delta > 0$ such that $d(x,y) + \delta < \varepsilon$. Then we'll have that $B_\delta(y) \subseteq B_\varepsilon(x)$.

[gentle osprey]

Exomnium (MSC2020 03C66)

[left eagle]

ok but dont you need to find a delta such that $B_\delta(y) \subseteq B_\epsilon(x) \cap Y$

[gentle osprey]

elonmosqito96

[opaque zodiac]

Well $B_\delta(y)$ computed in $X$ won't be a subset of $Y$ necessarily, but $B_\delta(y)$ computed in $Y$ is the same thing as $B_\delta^X(y) \cap Y$

[gentle osprey]

Exomnium (MSC2020 03C66)

[left eagle]

wait what im confused

[opaque zodiac]

So let me write it like this: $B_\delta^Y(y) = {z \in Y: d(y,z) < \delta}$ and $B_\delta^X(y) = {z \in X : d(y,z) < \delta}$. The point is that $B_\delta^Y(y) = B_\delta^X(y) \cap Y$.

[gentle osprey]

Exomnium (MSC2020 03C66)

[left eagle]

oh so $B_\delta(y)^Y = B_\delta^X(y) \cap Y$ and since $ B_\delta^X(y) \subseteq B_\epsilon^X(y)$, $B_\delta(y)^Y \subseteq B_\epsilon^X(y) \cap Y$

[gentle osprey]

elonmosqito96

[opaque zodiac]

yes

[left eagle]

ok ty

[plush folio]

I don't understand this comment. If two topologies T1 and T2 on a vector space V are homeomorphic, then T1 and T2 contain exactly the same sets, right? So what other kind of sameness are they talking about?

[opaque zodiac]

I think maybe they're distinguishing between it being witnessed by some self-homeo and it being witnessed by the identity

[plush folio]

Ok, I see 👍

but in practice it wouldn't make much difference, right? Since sets are unordered, they're not "more same" if the homeomorphism is the identity?

[opaque zodiac]

uh well

I could hypothetically see it mattering

[junior adder]

Why can we just casually do the highlighted bit

[opaque zodiac]

The complement of C is a non-empty open subset of S^2

every non-empty open subset of S^2 contains an open subset that is homeo to an open ball in R^2

because S^2 is a manifold

[junior adder]

oh i must've skipped the part that proved S^2 is a manifold

ty

[opaque zodiac]

uh well you can show that directly pretty easily

well

it depends on how sloppy you're being

but a cap of the sphere is hoemomorphic to a disk

although I guess I am just saying 'it's a manifold lol'

[junior adder]

aight well I feel comfy just taking it for granted for now, it sounds like a super fundamental fact anyhow

o shit

https://math.stackexchange.com/questions/1821031/how-can-i-prove-that-sn-is-a-n-dim-manifold it's just stereographic projection

[opaque zodiac]

yeah that's the thing about super explicit objects, you get super explicit proofs

of things like this

[junior adder]

btw do u think the algebraic topology portion of Topology by Munkres is worth reading or should I just move to a book dedicated to AT? I think I'm gonna plow ahead and read the stuff on Jordan separation regardless but idk otherwise

[opaque zodiac]

I have no idea

[junior adder]

bet ty

[limpid fern]

it's fine ish

[iron coyote]

ohh interesting

intuitively, is that the 1/5th top of the sphere?

[opaque zodiac]

well take any intersection of a plane with a sphere, the connected parts that aren't in the plane will both be disks

[iron coyote]

i see i see

on that stack echange question linked above, what is the meaning of these particular maps, and why is S^n quotient'ed with 1,0,0, and V quotiented with (-1, 0, 0..)

[left eagle]

Im confused (2) showed that the topology generated by the two metrics are the same. And part 3 wants me to show that they arent the same???

[left eagle]

Oh ok ic. for some reason i thouyght that d and bar d were equivalent iff they proudce the same topologies

from lecture notes

[left eagle]

also for X = R would that mean d is the standard |x-y| metric?

because Im not sure how to procede if d is just any metric over R

[limpid fern]

i believe you are requested to produce a metric d such that d and d bar are not equivalent in the case of R specifically

[left eagle]

oh ok then wouldnt the standard metric over R work?

for d

[limpid fern]

It might

Give it a try

[sullen nimbus]

What is a non trivial homeomorphism from [0,1] to itself that is not f(x)=1-x

[bold sphinx]

the identity

[sullen nimbus]

Oh wait x^2

Non trivial

Nvm sorry for bothering

[sullen nimbus]

So I believe I need to take a closed set C in X and g(C) is closed in Z, now I have f(h(C)) is closed

And f is onto and cts

So how to get that h(C) is closed

Cause if i pullback using f^-1 wouldn't that be larger than h(C)

[sullen nimbus]

Wait

I just realised that it should be h(f(C))

h(C_y) =h(f(f^-1(C_y)))=g(f^-1(C_y)) and that is closed

Lmao i didn't even realise that I was reading it wrong 💀😭

[quartz horizon]

Yeah this works cause f(f^-1(C)) = C whenever f is surjective

[sullen nimbus]

Also isn't this equivalent to the axiom of choice ?

I mean the statement that onto functions have a one sided inverse

[opaque zodiac]

yes

[sullen nimbus]

So is this statement also equivalent to the axiom of choice?

[quartz horizon]

Wait

It is?

[sullen nimbus]

Idk

[quartz horizon]

I’m just talking about preimage and direct image here

I didn’t realise you needed choice for that

[sullen nimbus]

Yeah

For all sets ?

[quartz horizon]

So it’s not this statement I think

Unless I’m mistaken

[sullen nimbus]

No i mean for singletons

ff^-1(x) is x right

[quartz horizon]

Yes

But here if f : X -> Y

[sullen nimbus]

So one sided inverse of onto function?

[quartz horizon]

Then f^(-1) is not a function from Y to X

It’s a function from the power set of Y to the power set of X

The preimage

[sullen nimbus]

No i mean closed sets cand be singletons too

Sometimes

[quartz horizon]

Yes

But I’m not defining a function $Y \to X$ this way

[gentle osprey]

Pseudonium

[sullen nimbus]

Yeah but for that thing to work

It needs AOC?

[quartz horizon]

Yes to define a function Y -> X from this you probably need AOC

But I don’t think the statement about preimages and direct images requires AOC? But I could be wrong

[sullen nimbus]

Now I don't think does too, but I am sleepy

Well I was doing an excercise from top book

I don't think I should care

[ebon galleon]

Surjective <==> has a section (rightinverse) requires choice, yes. You shouldn't need choice to prove the statement you want: all you are using is that for any c in C, f^{-1}(c) is nonempty, which is typically how one would define surjective lol.

It's the difference of choosing an x in f^{-1}(c) for an individual c (to show c in f(f^{-1}(C)) ) versus choosing an x in f^{-1}(c) for every c to get a section

[sullen nimbus]

Alright thanks guys

It's a bit more clear now

[pearl trail]

Can someone help with 2.2? Pretty lost

[opaque zodiac]

Well one way to do this would be to show that for any point x and any d-ball B centered on x, there's a \bar{d}-ball B' centered on x such that B' \subseteq B (and then the same with d and \bar{d} switched

[pearl trail]

How's this? If we do T_d subset T_dbar. Then consider a U in T_d and x in U. Then there exists e > 0 s.t. B_d(x,e) subset U. Then for any y in B_d(x,e), we have d(x,y) < e. Now consider dbar. Let delta = e/(1+e), this follows from d(x,y) < e => dbar(x,y) = d(x,y)/(1+d(x,y)) < delta. Then we have y in B_d(x,y) and dbar(x,y) < delta, thus y in B_dbar(x,delta).

[proper fiber]

So some observations I made are R/A has to be dense

And there are some instances where A can be dense too

So like defining A to have a constant value and R/A to have another constant value won't work

[opaque zodiac]

That should work for one direction, but you also need to show inclusion of dbar-balls in d-balls

[proper fiber]

Ok so I have a gut feeling that if that set a is countable then maybe we could somehow define this as some kind of sum function

[opaque zodiac]

Do you know what $X \setminus \overline{U}$ means?

[gentle osprey]

Exomnium (MSC2020 03C66)

[inland thistle]

not really, I might be lacking some prerequisite knowledge, and I dont know how the complement of the closure of some open set is infinite. In order to show this, we probably need to pick some finite open set U and with the closure of U, which is finite too?

Im sure the fact that X is an infinite Hausdorff space comes into play at one point, but I can't see anything now

I first want to understand what this question is about

[opaque zodiac]

there won't necessarily be any non-trivial finite open sets

I mean uh

what do you mean by 'about'?

so like consider R as a topological space

(0,1) works for R in this instance because the closure of (0,1) is [0,1] and the complement of that is infinite

[inland thistle]

actually pls disregard this part

I just wanna know how one would solve this question

because I have no idea at this point

[opaque zodiac]

do you just want a solution or do you want a hint?

[inland thistle]

might want a solution for this one

[opaque zodiac]

oh well first of all I'm sure they meant 'non-empty open subset' because otherwise {} works

[inland thistle]

Yes, I believe so, I ignored that case too

[opaque zodiac]

Okay there's a lemma we need first:

\textbf{Lemma.} Suppose $U$ is a finite open subset of a Hausdorff space. Then for any $x \in U$, ${x}$ is clopen.

\emph{Proof}. Let $x_0,\dots,x_{k-1}$ be an enumeration of $U$. For each $i < j < k$, let $V_{i,j} \ni x_i$ and $V_{j,i} \ni x_j$ be disjoint open neighborhoods. Let
[
W_i = U \cap \bigcap_{j < k,, j \neq i} V_{i,j}.
]
Note that by construction $W_i = {x_i}$. Furthermoreover, note that since this is a finite intersection of open sets, it is open. Therefore each ${x_i}$ is an open set.

Now we just need to show that each ${x_i}$ is a closed set, but this follows from the fact that in any $\mathrm{T}_1$ space, singletons are closed. $\square$

Now to actually do the problem. Suppose $X$ is an infinite Hausdorff space. Fix $x,y \in X$ and find disjoint open neighborhoods $U \ni x$ and $V \ni y$. We have that $y \in X \setminus \overline{U}$. If $X \setminus \overline{U}$ is infinite, then we're done. Otherwise, we have that ${y}$ is clopen by the lemma, whereby $X \setminus \overline{{y}} = X \setminus {y}$ is infinite and so ${y}$ is the required open set.

[gentle osprey]

Exomnium (MSC2020 03C66)

[unreal stratus]

Also like U is finite and T1 (since X is) hence discrete

[opaque zodiac]

uh yeah

[tender halo]

thats like, hard

[opaque zodiac]

I think 'finite T1 spaces are discrete' is supposed to be one of those facts that you get exposed to in a class about this stuff

[unreal stratus]

lol

U=empty set

takes pedant hat off

[crude verge]

I was wondering if someone had a resource that could direct me to understanding this lemma. This stack exchange thread stated that if U is a connected open subset of R^n, and S is a closed subset of U, then U - S is path connected if dim(S) < n - 1, where dim(.) is the Hausdorff dimension. I was wondering if someone had a textbook that confirms this because this lemma seems too good to be true.

[plush folio]

I know that every subspace of a Hausdorff space is Hausdorff. Is this true for all the separation axioms? Ie. the separation axiom is inherited by any subspace?

[tender halo]

No

T0-T3.5 are inherited as well as perfect normality

so the only failure is normality

its actually somewhat non trivial to construct a normal space that isnt hereditarily normal

one example is take two one-point compactifications of discrete spaces, one of cardinality N and another of cardinality 2^N, then their product is normal but not hereditarily normal

[plush folio]

Interesting, thanks

[austere dirge]

lmao i have this exact question on my hw this week

gotta exit the channel before i get an academic offense

[limpid fern]

it would have to be nonempty open subset U

Since it is a part a it's probably intended for use in something else

I'm going to guess and say it might be used to show that X has a discrete subspace

[iron coyote]

its the quotient of X (maybe a metric space?) by some action group U?

wait

oops

i didint read the above message

reading fail

[iron coyote]

its the quotient of X with the closure of U i think

Heres my drabby attempt:
Let ( X ) be a finite-dimensional Hausdorff space.
By induction on the dimension of ( X ), for ( \dim(X) = n ), choose a closed subset ( A \subset X ) with ( \dim(A) = n-1 ) and an open subset ( V \subset A ) such that ( A / \overline{V} ) is infinite.
Then there exists an open subset ( W \subset X ) such that ( W \cap A = V ) and ( \overline{W} \cap A = \overline{V} ).
Hence, ( X / \overline{W} ) is infinite, as it contains ( A / \overline{V} ) as a subset.

[gentle osprey]

Ayunipear

[gritty widget]

Furthermoreover

[wind bronze]

Apparently, the category of open sets (and their inclusions) does not provide enough information to recover the original topological space (unless the topological space is sober).

However, I don't see how this is true. A topological space is entirely determined by the set of its open sets, and the category of open sets contains this set of open sets (as objects), plus the inclusions between them.

Any help would be appreciated.

[wind bronze]

I meant set of open sets, not of subsets*.

[quartz horizon]

So consider indiscrete spaces

They all have the same poset of open sets

up to equivalence

[unreal stratus]

Ig this depends what you mean by "the category"

If you have "the" category then sure you can just take the underlying set of the biggest object but if you are only interested in the category up to equivalence then you can't recover the space for the reason pseudonium mentions

[wind bronze]

Uhh, I don't understand.

[wind bronze]

Ooh

[quartz horizon]

yes this is a good way of phrasing it

[unreal stratus]

If X is indiscrete and nonempty then the poset (equivalently: category of opens) is isomorphic to just like {0 < 1}

Which isn't enough to recover X since that could be an arbitrary set

[wind bronze]

Okay, wait, I'm confused again now.

[wind bronze]

So, if we are interested in the category of open sets up to equality, we can determine the topological space up to equality. ~~If we are interested in the category up to isomorphism, then we can determine the topological space up to isomorphism.

But if we are interested in the category up to equivalence, then we cannot recover the space, because different indiscrete topological spaces might have the same category of open sets (up to equivalence), no ?~~

(Just to make sure I understood)

[quartz horizon]

Even up to isomorphism breaks things

The poset of opens of an indiscrete space is always isomorphic to {0 < 1}

[wind bronze]

Ohh, okay.

Thank you both, I understand now!

For some reason, by "indiscrete" I thought of a topology which is not discrete, not of the trivial topology.

Thank you!

[wind bronze]

A ringed space is defined as a topological space equipped with a sheaf of rings. But what exactly does "sheaf of rings" mean ?

Does it mean that:

• it is a sheaf F over the category of open sets of the topological space, where, for each open set O, F(O) is a one-element set (where the element is a ring) ?
• or that the sheaf F actually is not a Set-valued presheaf, but rather a Ring-valued presheaf which behaves in a particular way (which I don't know) ?

The first variant seems off. It doesn't require any coherence conditions between the ring of an open set, and the ring of an open subset of it (you would expect the ring of the open subset to be a sub-ring of the "original" ring, no ?). And the second variant is definitely incomplete.

[red yoke]

Not sure what "F(O) is a one element set" means

[wind bronze]

F is a sheaf, and thus a presheaf. It maps every object to a set. What I'm saying is that that set should have only one element - a ring.

[red yoke]

A presheaf of rings on a space is a contravariant functor from the category of open sets to Ring

[unreal stratus]

[Though usually with a commutativity assumption on the rings]

The sheaf condition remains the same. That is, a presheaf of rings is a sheaf iff its "underlying" presheaf (of sets) is a sheaf

[wind bronze]

Underlying as in, forgetting the ring structure and only caring about the underlying set ?

[unreal stratus]

Yup

Slightly fancier, but you can also view a (pre)sheaf of rings as a "ring object" in the category of (pre)sheaves, which can be a helpful perspective

If that means nothing to you you can ignore it lol

[red yoke]

A ring object in the the category of set-valued presheaves?

[wind bronze]

I see, it makes sense now. I didn't understand how the sheaf condition could be phrased for presheaves which are not Set-valued. Thank you both!

[wind bronze]

Unfortunately, it doesn't. 😦

[wind bronze]

Wait, now I'm curious what exactly you mean by "ring object" here. A model of the Lawvere theory of a ring in the category of presheaves ?

[red yoke]

Did you look up nlab

[wind bronze]

No.

I mean

I've read that nLab article not long ago, so it still is a hot topic for me, haha.

[quartz horizon]

I guess like

[red yoke]

I think the equational definition of ring object coincides with sheaf / presheaf of rings

[quartz horizon]

You can phrase all the ring axioms with commutative diagrams

Alternatively, a ring object is an object X for which Hom(-, X) factors through the category of rings

together with the forgetful functor from Ring to Set

this works cause of yoneda

[red yoke]

By factors do you mean like
Set(-, R) factors as S → Set(S, R) ∈ Ring

[quartz horizon]

Yeah

[wind bronze]

This is quite hard to understand. But I'll think about it.

[quartz horizon]

Why is it hard to understand?

Essentially I’m saying a ring object is one where “maps into it form a ring”

[wind bronze]

I only see the one-direction implication. Namely, if it is a ring object, then yes, maps into it form a ring. But does it go the other way around too ? Like, do the maps into an object form a ring if the object is a ring ?

Ohhh, wait, it makes sense actually.

[quartz horizon]

The other direction is from yoneda

So long as your category has products

[wind bronze]

I don't see how Yoneda's lemma is aware of the ring structure of the hom-sets.

[quartz horizon]

I meannnn

You have a ring multiplication

Hom(-, X) x Hom(-, X) -> Hom(-, X)

Then if you have products that’s equivalently

Hom(-, X x X) -> Hom(-, X)

Bam apply yoneda

There’s your morphism X x X -> X

Etc etc

[red yoke]

Maps into R naturally form a ring

[quartz horizon]

Ye

[wind bronze]

I was thinking more that this implies the maps from the terminal object to the ring object have a ring structure. And since maps from the terminal object are esentially "elements of the ring", that's like saying that the elements have a ring structure - that our object is indeed a ring.

[quartz horizon]

Sure if the terminal object represents the forgetful functor

You only really need the forgetful functor to be representable for your argument

[wind bronze]

Oh

Wait, so

A ring object R has maps R x R -> R. By Yoneda's lemma, those are actually maps Hom(-, R x R) -> Hom(-, R). Since the Yoneda embedding preserves limits, these maps actually correspond to maps Hom(- R) x Hom(- R) -> Hom(- R). And since it is a functor, all relations and ring axioms (commutative diagrams) which were satisfied in the original category will also be satisfied in the presheaf category. Thus Hom(-, R) is a ring object in the presheaf category, no ?

[quartz horizon]

Ye

You also get things like cogroup objects this way

“Maps out of them naturally from a group”

Like the spheres in the homotopy category

[wind bronze]

Hmm, haven't heard of cogroup objects before. Interesting.

I understand now. Thanks everybody for the help.

[inland thistle]

just one small question, f being topologically continuous does not necesarily mean that f is bijective right

thanks

also if we are looking at f^{-1}(V) where V is some subset of co-domain, and the corresponding element in the domain does not exist for each y, then f^{-1}(V) is just empty right?

[wind bronze]

Yeah, if the domain has no elements that get mapped to V, then the preimage of V is the empty set.

[inland thistle]

Also another question, unless otherwise stated, for the questions in topology, do we assume the space is equiped with discrete topology?

Also what is the usual topology on N (the set of natural numbers)? Do we usually assume the discrete topology of N for it?

[inland thistle]

thanks

[wind bronze]

No - the discrete topology is just a special case (and perhaps one of the most boring and uninteresting ones) of a topology.

If the topology of a topological space is not stated, then it is usually because the underlying set has a "natural" notion of topology in it - eg. if you hear "topology of a <insert metric space>", then you can assume it is the metric topology.

And the set of natural numbers has no conventional / standard topology.

[inland thistle]

and the goal is to show this map from n to some principal ultrafilter is a homeomorphism onto its image

I even forgot how to show something is an imbedding, but I assume we need to show it is a continuous open map onto its image, am I right about this?

so for this, I was wondering what are open sets of N, because I think I need to take some open set in beta-N, but then I need to show its pre-image is also open but in N

I've somehow managed to not know what the open set in N is like

for N right?

I see thanks for your input

[tender halo]

this question is about Stone Cech compactification of N, here N indeed has discrete topology

[plucky veldt]

typically the topology on any subset of R or R^n is the Euclidean topology unless otherwise noted

the subspace topology in N induced from R with the Euclidean topology is discrete

because for each natural number there is an open set in R that contains only that number

in other words, every natural number has non-zero distance (1, actually) from all other natural numbers

[inland thistle]

Thank you so much

[austere dirge]

stone topology

friendly advice: careful not to get AO'd

for posting our psets verbatim online

anyways second one follows actually by HW VII removed exercise

use the ultrafilter definition of compactness

compact iff every ultrafilter converges

ohh

i misread

my bad

yea its the discrete topology

[inland thistle]

oh good point, thanks for that, I will remove the picture from the server now

[iron coyote]

i dont think anyone would steal a single practice problem from an ultrahard super complex specific field on a discord serve

rbut precaution are goof

god

good

[austere dirge]

it is not an ultrahard super complex specific field it is literally a first course in topology and the pset is for grades

[inland thistle]

btw is this for like every sequence has a convergent subsequence kind of thing?

I personally think its better to use open cover finite subcover definition for this

[austere dirge]

.

if u check out pset VII Q2 (c) it used to be iff

[inland thistle]

oh also what is the removed exercise?

[austere dirge]

he changed to if

[inland thistle]

oh I see, didnt know that

[austere dirge]

if u prove iff thats the easiest way

imo

[inland thistle]

yeah that makes sense

[austere dirge]

btw i can send u a discord where the rest of the class is

or like

a lot of them

if u need

[inland thistle]

oh really, that would be helpful

[austere dirge]

okay i dm'd

[gritty widget]

is this channel new?

wasnt it just top-algtop before

[tender halo]

got split

[inland thistle]

This channel was created to help students who are taking mat327 at UofT this summer

I think our prof's name is something like Daniel Wilches

but he seems to love set theory too

it seems like the course is purely focused on point-set side of topology

and not covering any component of algebraic topology sadly

[sonic crane]

@valid raft oh u went to uoft?

Thats cool

I went to ubc

For undergrad

[inland thistle]

I wonder why this server is also very dominated by (depressed) UofT students lol

thats def an option yea, I was thinking about that too

I might as well do diff topology if I could

but I dont think too deeply about what's uncertain for now

what do you mean by cancelled?

💀

true I think we can get back to the topology at some point

but it does not change the fact that this channel is for UofT MAT327 summer students

[austere dirge]

as a very wise discord user—who just so happens to be a set theorist—said, "theres enough point set topology to fill a full course, we dont need to fill it with algebra propaganda"

im enjoying the direction dcw is taking the course, but also i plan to read hatcher

anyways regarding the topic of the channel, is showing that a construction of stone cech satisfies the universal property is sufficient to prove it's compact hausdorff?

or do you have to prove its compact hausdorff then show it satisfies the universal property?

the objects in our category are like morphisms from X -> K for a fixed topological space X, and compact hausdorff K, so it seems that to show a morphism X -> Y is initial in this category we'd first have to prove that it's actually in the category, and thus Y is compact hausdorff, right? i am a beginner to category theory so im not sure

[quartz horizon]

I believe you would have to show Y is compact Hausdorff, yes

[fringe thorn]

no, it was born from the ashes of my thread

(but hello, fellow UofT students )

[lusty trench]

The “algebra propaganda” is useful, though. Dare I say, more than most of the silly point-set esoterica like Warsaw circles and Hawaiian earrings.

For 95% of people, the usefulness of point-set topology is based on the premise that geometric spaces are topological spaces with extra structure.

[opaque zodiac]

I don't really think of Warsaw circles and earring spaces as set-theoretic esoterica

They're pretty concrete spaces (compact metric spaces) and their behavior has nothing to do with set-theoretic phenomena

The earring space even showed up in my research naturally

[opaque zodiac]

As evidence of this, note that as compact metric spaces, they're sober, and so all of their properties are just as much present if you think of them as locales rather than topological spaces

In any case I don't think it's really constructive to call these examples 'silly'

[gaunt linden]

I thought the Warsaw circle's main claim to fame was as algebraic esoterica: a path-connected space that is not contractible, but whose homotopy groups are all trivial.

[unreal stratus]

This is how I was introduced to it

[plain raven]

are analysts included in the 5%

[plain raven]

Nice slipping in the "constructive" comment immediately after talking about pointless topology of locales

I see what you did there

Also Exo is completely right and dismissing Warsaw circles and Hawaiian earrings as "silly point-set esoterica" is a serious philosophical misstep. They're concrete and straightforward geometric objects. I wouldn't bother piling on if it was just Eduardo saying things like this, I think many algebraic topologists who work primarily with CW complexes develop tumors that cause them to think CW complexes are the only geometric objects.
Ruling out the Warsaw Circle and Hawaiian earrings as "pathology" amounts to ruling out shape theory itself as a legitimate subfield of topology

[gaunt linden]

I know something called "shape theory" exists because Wikipedia redirects the Warsaw circle to it, but it's not particularly informative about what it is. Is "shape" a technical concept? And if so, how is it defined?

[umbral panther]

A shape is roughly a formal inverse limit of homotopy types. Technically, it would be better to take homotopy after take pro-objects

The Hawaiian earrings is an intersection of finite CW complexes. Its shape is the formal inverse limit. This probably comes up in Alexander duality, where you have a duality between singular and cech cohomology

The other place shape theory comes up is that the “end” of a manifold has a shape, not a weak homotopy type, nor a strong homotopy of non cw spaces

For example, if you remove a cantor set from the plane, you can recover its shape as the end of the manifold. If you remove the Hawaiian earrings from the plane, you get a disconnected manifold with boring end. But if you remove it from R^3, the shape of the end is the shape of the earrings

[warm hedge]

is it true that every open nonempty subset of a polish space contains a comeager subset of the polish space ?

[opaque zodiac]

no

(0,1) as a subset of R

can't be comeager because it's not dense

[warm hedge]

true...

[opaque zodiac]

every dense open set is comeager

[warm hedge]

is there a chance only by this image to be able to understand why the red lined thing is true ? 😛

[opaque zodiac]

What is the statement being proven?

[opaque zodiac]

what is the superscript delta?

[opaque zodiac]

forall star is for all g in a comeager subset, right?

[warm hedge]

yeah

and the delta nonmeager

[opaque zodiac]

what book is this?

[warm hedge]

invariant descriptive set theory by Gao

[opaque zodiac]

what's a regular topology again?

is that one of the separation axioms?

ah

[warm hedge]

yeah this

[opaque zodiac]

ah and this is all happening on a group

or rather a G-space

G has a fixed Polish(?) topology?

[warm hedge]

yeah exactly and the weird U is a base of this topology

[opaque zodiac]

X is the set being acted on by G?

[warm hedge]

yes G acts on X

also the weird A is a collection of analytic sets on X

[opaque zodiac]

it's \mathcal{U} and \mathcal{A}

they're in calligraphic font

[warm hedge]

yeah true

my problrm is that for some reason V must be comeager on G or something but i dont understand why

[opaque zodiac]

I think all that's happening is that

$\forall^\ast g \in V (g\cdot y \in A^{\Delta U_1})$ just means that the set ${g \in V : g \cdot y \in A^{\Delta U_1}}$ is comeager in $V$. The proof establishes that it is in fact all of $V$, which is comeager in $V$.

[gentle osprey]

Exomnium (MSC2020 03C66)

[warm hedge]

ohh so it means comeager in V how could have i understood the difference between comeager in V and comeager in G ?

[opaque zodiac]

uh

where is the forall star notation defined?

[warm hedge]

oh let me find it

[opaque zodiac]

also like you were saying if it needed to be comeager everywhere, the notation wouldn't make much sense

[opaque zodiac]

yeah I think that last sentence of the first paragraph

you're supposed to go 'aha, forall star x \in V must mean we're applying the notation to V as a subspace of X'

but it's certainly not spelled out well

[warm hedge]

ahhh yeah

i missed that

thank you so much

[warm hedge]

so to be clear, this also means nonmeager / comeager in the sets that are after Δ and * ?

[opaque zodiac]

Yes

[warm hedge]

thanks

[worn root]

im taking topo in the fall, would anyone be able to recommend any good video lectures? building intuition, and esp visual intuition, would be nice too. thank you!

[gaunt linden]

I'm afraid none of that makes much sense to me ...

[umbral panther]

Start with the end of a manifold. It isn’t a space. If the manifold is KxR, the end should, in some sense be two copies of K. What invariants of K can you recover? Can you recover its homotopy type? In fact, yes. But what the end is more complicated, like in the plane minus the cantor set or the whitehead manifold?
https://en.m.wikipedia.org/wiki/End_(topology)

[gritty widget]

Hello

Can someone help me understand how these two sets are homeomorphic?

[gentle osprey]

Les (literally Hrithik Roshan)

Les (literally Hrithik Roshan)

[opaque zodiac]

They're not?

No subset of R is homeomorphic to [0,1]^N

[red yoke]

Are you trying to ask whether the Cantor set is homeomorphic to {0,1}^N?

[barren magnet]

Looking at Topological bases; they seem like a ‘free construction’ so is there a way to view them as an adjunction? If the poset of topologies on a set X is the poset of subframes of P(X) then would it work out to be a left adjoint to the forgetful functor to the poset of sub-meet-semilatices of P(X) or something?

[quartz horizon]

Yes it’s definitely an adjunction

A basis is a subset of a topology T iff the topology generated by the basis is a subset of T

That’s the adjunction

It’s a closure operator

[barren magnet]

Yeah that makes sense - and basis need to already be closed under intersection right?

So is meetsemilattices the right category for this adjunction?

[red yoke]

Maybe they have to be sublattices of some power set?

[barren magnet]

Yes sorry that’s what I meant by sub-meet-semilatices

[quartz horizon]

Wait what are those

[barren magnet]

Posets with finite meets

[quartz horizon]

What’s a meet?

[barren magnet]

Greatest lower bound or intersection

[quartz horizon]

Ah so like a product?

[barren magnet]

It's the decategorification of a product in other words

YEah exacly

[quartz horizon]

Ok cool

So finite limits then

Then yeah I guess so

[barren magnet]

This lemma in Munkres seems to desribe the action on morphisms

But what confused me was that I thought that there woiuld have to be something about preserving meets

[quartz horizon]

I’d imagine these just follow from the universal property

Namely if T’’ is another topology for X

Then B is a subset of T’’ iff T is a subset of T’’

That’s the adjunction

[barren magnet]

Yeah that makes sense

I was jus trying to work out what category the adjunction was actually between

[quartz horizon]

Sure

I guess it’s kind of unnecessary

The universal property is what matters

[barren magnet]

Yeah - I have a (maybe bad) habit of pretending my brain is a proof assistant when trying to learn maths

So I wanted to know the categories involved before parsing it as a universal construction

[red yoke]

Subbases / bases don't need to have meets

[barren magnet]

Oh

I thought the differnce between a base and subbase was that the former had meets?

[quartz horizon]

Yeah I guess I just think of it in terms of representability

[quartz horizon]

Also yes I thought this too

[red yoke]

Every finite intersection of basis elements is an arbitrary union of basis elements

That's the requirement for being a basis

[gaunt linden]

Consider for example this base for the usual topology on R^2: "all open balls whose centers have rational coordinates and whose radius is a (possibly negative) power of 10".

[red yoke]

The topology generated by a subbasis is indeed free
Edit: subbasis

In the sense of being left adjoint to the forgetful functor from the category of topologies on X to the poset category P(P(X))

[quartz horizon]

Yep that makes sense

[quartz horizon]

Which is essentially this

[barren magnet]

Ahh okay that helps - I think I had just got tripped up on the definition of a basis

I thought it had more structure than it really did

[red yoke]

Probably the same with bases, for the category of bases on X instead of P(P(X))

[gaunt linden]

Since a basis is a subbasis too, you can still use P(P(X)) :-)

[barren magnet]

Yeah my confusion was that I thought the category of bases was the same as the category of sub-meet-semilattices. But that's not right.

[barren magnet]

This helped with that - thanks

[tender halo]

Is there an easy way to see that without referring to inductive dimension or something of the sort

[red yoke]

Connectedness

[tender halo]

fine imagine im asking about R^100

instead of R

https://tenor.com/view/possum-eating-food-smacking-lips-lips-smacking-fruit-gif-27120401

[gaunt linden]

A circle can embed into [0,1]^N, but not into R (thanks to the IVT), and thus not into any subset of R either.

[red yoke]

S^100

[quartz horizon]

This is really nice

[gaunt linden]

Thanks.

[gritty widget]

Yes.

[gritty widget]

C is the Cantor set.

[tender halo]

easiest proof, personally:

[gaunt linden]

Actually your C is the complement of the Cantor set within [0,1].

(No, not even that).

[tender halo]

yea its got the border points

[tender halo]

take a family of continuous functions f_i that maps i_th coordinate of C to 2i/3^i

take their sum (which is continuous because the limit is uniform)

the sum is bijective and continous, C is compact so its a homeomorphism

[austere dirge]

it seems that the written set is just [0, 1/3] U [2/3, 1]

but obviously he meant the cantor set

[gaunt linden]

No, since n=1, k=1 adds [1/3, 2/3] to the union.

[austere dirge]

huh. strange

well now im curious to figure out what this is

[gaunt linden]

(The upper limit of k is also too small to get even all the gaps in the Cantor set -- or alternatively the "3k" spacing needs to be significantly more intricate).

[gritty widget]

ok imma be real

idk what they were doing but the resource i used said that is how u write the cantor set in union notation

im just trying to do the proof for cantor set <---f---> [0, 1]^N where f is a homeorphism

[unreal stratus]

You mean {0,1}^N I guess

[alpine nest]

That would be much easier to show (because it's true)

[alpine nest]

How does this embedding work, by the way?

[tender halo]

Embed into [0;1]^2 as a square

[gaunt linden]

Um, e.g. t -> (½+½cos(t), ½+½sin(t),0,0,0,....)
Am I missing something?

[alpine nest]

No, I just blanked.

It indeed is that simple.

Somehow I forgot that [0,1]^2 is sort of a subset of [0,1]^N

Not really a subset

[tender halo]

i was about to akshually

[alpine nest]

But it's very naturally "in there"

[opaque zodiac]

Every compact metric space embeds into [0,1]^N

[alpine nest]

Sounds vaguely familiar

[sonic crane]

Whats the reason for why topologies that have one-point sets being closed are “not interesting”?

[opaque zodiac]

They're not?

[unreal stratus]

Well lol this is gonna be hugely context-dependent. e.g. in more geometric areas you may work exclusively with such spaces

[opaque zodiac]

It's one-point sets being open that makes the topology trivial

[sonic crane]

It just says so in munkres lol

Sorry

I meant one point sets being not closed

[opaque zodiac]

Ah

[sonic crane]

Ok yeah if every one point set is open then the topology is trivial and thats not interesting

[tender halo]

its just an unnatural situation from a geometric point of view

they are interesting, but for topologists these kinds of spaces are generally not naturally arising

[unreal stratus]

Well not being closed is not the same as open ofc

[tender halo]

but there are contexts where they arise naturally, Scott topologies or Zariski topologies are T0 but generally not T1

[tender halo]

every Tychonoff second-countable space embeds into I^N

which includes compact metric spaces

consider the countable basis of cozero open sets, say they are cozero for a set of functions f_i

they are kinda like coordinates inside your space

so if you send a point x to (f_0(x), f_1(x), ...), then you'll get your embedding

[gaunt linden]

Schemes are not interesting, got it.

[sullen nimbus]

I am still on the nonempty part

I have no clue on how to produce an element in this intersection

[opaque zodiac]

If your definition of compact is the open cover definition, you're going to need to do a proof by contradiction to show that the intersection is non-empty

[sullen nimbus]

Yes it's the open cover definition

So if the intersection is empty then i don't actually see a problem

But also like i can't come up with am example or sth close even

[tender halo]

wait is it even true if compact means quasicompact

yeah guess not, take like infinite trees with right order topology, they are quasicompact but have infinite descending chains of compact subtrees

[tender halo]

compact subsets of Hausdorff spaces are closed, take complements, enjoy

[sullen nimbus]

But it's not given in the problem that they're hausdorf

I thought of taking complements to form an open cover

But like they just write that it's a topological space

[tender halo]

its just false without it

so you can assume they are Hausdorff

[sullen nimbus]

Is there any simple counterexample

Like dosent have to be very simple

But something standard ig

[tender halo]

i just said one

i mean you can make it even simpler

take N where open sets are rays [n; ...)

it is obviously (quasi-)compact

[i; ...) is a sequence of quasi-compact sets with empty intersection

honestly i just think your text means compact to be Hausdorff

[sullen nimbus]

I looked up quasi compact

[tender halo]

as it should

it is the morally correct option

[sullen nimbus]

I haven't heard of it before but I guess it's stronger than. Compact?

[tender halo]

no, compact is quasi-compact+Hausdorff

[sullen nimbus]

Uhhh

[tender halo]

or some texts say that compact and quasi compact are the same thing

[sullen nimbus]

Very confused rn

[tender halo]

quasi compact is just "every cover has a finite subcover"

[sullen nimbus]

Oh

[tender halo]

compact has different definitions depending on what you are reading

[sullen nimbus]

So it is stronger than compact?

[tender halo]

it either means the same thing as quasi-compact

or it means quasi-compact and Hausdorff

[tender halo]

no, compact is stonger

[sullen nimbus]

Ok thanks for this counterexample

[sullen nimbus]

Compact is every open cover has a finite subcover?

And quasi is the more general one?

By stronger I mean the more general one

[tender halo]

uhh by cover i meant an open cover of course

[sullen nimbus]

💀

Oh okay

Then it's the same thing

[tender halo]

i mean its an obviously nonsensical def otherwise, cuz a cover made out of individual points wont have a finite subcover

unless your space is stupid

[tender halo]

well in your text i imagine compact includes Hausdorff

[sullen nimbus]

It's a quals question

[tender halo]

cuz otherwise the problem doesnt work

[sullen nimbus]

They never mentioned it

Ok anyways thanks a lot 🙂

[crude verge]

https://math.stackexchange.com/questions/2730111/can-a-set-of-hausdorff-codimension-2-disconnect-a-connected-open-set This thread claimed that if U is an open subset of R^n, and S is a closed subset of U with Hausdorff dimension at most n - 2, then U - S is path-connected. Following the proof of the top answer, it makes sense; however I've been having trouble finding such a claim in any topology textbook. I usually do not like to use such general lemmas in my research without having an authority source supporting it, even if I have a proof, particularly when it involves topics I lack experience in (i.e. hausdorff dimension). Would anyone know a source that supports this claim or something similar?

[opaque zodiac]

If $F$ is a regular closed set and $U$ is a regular open set, is it true that $\mathrm{int}(F\cup U)$ is regular open? I'm assuming not, but I can't think of an example right now.

[gentle osprey]

Exomnium (MSC2020 03C66)

[opaque zodiac]

oh wait

no

I think it is

[brittle rapids]

quasi means "sort of"

[unreal stratus]

I mean that doesn't really help here where quasicompact has a very specific meaning

[gaunt linden]

Namely "compact".

[brittle rapids]

at the very least this tells you that it's weaker

[ebon galleon]

Well for some people compact includes Hausdorff, and for others it doesn't. So for the first group it's weaker, for the second group it's just compact lol

[austere dirge]

compact including hausdorff is baffling

[twin siren]

is this wrong

[unreal stratus]

It is badly written

It should say that f is continuous iff ( for all open V in Y, f^-1(V) is also open)

For some reason they put V first

Also like people just say continuous

Saying globally continuous is a bit odd in that being continuous "is local", as in you can check it on an open neighbourhood of each point

[tender halo]

compact sets that arent closed are more baffling

[opaque zodiac]

This is just a really old minor religious war in mathematics. Topology that was influence by (iirc) French algebraic geometry uses the compact-implies-Hausdorff convention. It makes sense in the context of AG because a lot of Zarski spaces are (quasi-)compact but not Hausdorff

It's kind of like how some people seem to think literally no mathematicians ever seriously use the word 'clopen' but then in some fields it's completely standard

[grave solstice]

bruh I need a sanity check

Consider the topological space prod_(i=1)^infty {1,2,3,4,5,6} with the product topology, where {1,2,3,4,5,6} has the discrete topology. Is prod {1,2,3} closed in this space?

[opaque zodiac]

Yes

[grave solstice]

yeah okay

[opaque zodiac]

but why six points tho

[limpid fern]

maybe they want to use tychonoff? finite space = compact

[umbral panther]

Prod {0.1} is the cantor set. It is T_1, so points are closed. By the universal property of products the map from the six element set to the 2 element set induces a continuous map of topological spaces. The preimage of the closed point is a closed subset

[grave solstice]

for {1,2} my question is not as juicy, because you would just get a point. I obviously wasn't going to pick a prime, and 4 felt a bit boring.

[iron coyote]

is this the math server

[red yoke]

Indeed it is, and you are in the point set topology channel

[brazen flame]

Wouldn't X'xY' and XxY be uncomparable in some cases?

like consider this:

[quartz horizon]

Yes the notation is a little confusing here

Really it should be $(X, \mathcal{T})$ and $(X, \mathcal{T}’)$

[gentle osprey]

Pseudonium

[quartz horizon]

It’s the same underlying set, but different topologies

[brazen flame]

Wait so, what does

X and X' denote a single set in the topologies T and T'
mean?

[viral terrace]

let X and X' be some open sets from T and T' respectively?

[brazen flame]

but in that case, the product topology of X' and Y' cannot always be compared to the product topology on X and Y

[viral terrace]

but T' and U' are subsets of T and U

so, every unprimed element from the base of the respective product topology belongs to the primed base

[brazen flame]

can you point out what's wrong in this counterexample

T',T,U',U taken in order

[fading vale]

Where is this a counterexample to the claim?

[brazen flame]

shouldn't have used counterexample but point being, i think X'xY' and XxY cannot be compared in some cases

[fading vale]

Right, but why do the topologies you wrote yield incomparable product spaces?

[viral terrace]

you need to exhibit an open set from X\times Y that isn't in X'\times Y'

[brazen flame]

cause they've got different elements
{2,3}x{b,c} should be finer than {1,2}x{a,b} according to the problem
but for that to happen, every element of {1,2}x{a,b} must be in {2,3}x{b,c}, but that isn't the cause ({1},{a}) exists

[fading vale]

I think you are maybe confused about what "finer" means here

The exercise is asking you to prove that if every open set in X is open in X', and every open set in Y is open in Y', then every open set in X x Y is open in X' x Y'

where X and X' are the same set with two different topologies, and likewise for Y and Y'

{2, 3} x {b, c} is a subset of the product, not a topology, so it cant be finer than anything

[brazen flame]

im confused, I thought X and X' are some single sets belonging to the topologies T and T'

so X is an element in T

[fading vale]

No, X and X' are two different topological spaces

they have the same underlying set, but the topology in X' is finer than on X

The wording in the problem is somewhat ambiguous

but that is what they mean

[viral terrace]

I think they are open sets viewed with subspace topology, because it's said that taken from the topologies

[brazen flame]

gotcha, thanks

[brazen flame]

OH, X and X' denote a single set means that they denote the same set with topologies T and T' on it right?

[fading vale]

Yes

[brazen flame]

thanks!

[prime elbow]

Every space has maximal connected subspace, right?

Because the singleton set is connected

[unreal stratus]

With the exception of the empty space :)

[prime elbow]

Yes

[unreal stratus]

(If connected => nonempty in ur definition)

But yeah

This is the existence of connected components

[prime elbow]

Yes

[unreal stratus]

You can consider the union of all connected subspaces containing a given point and considering singletons means this is a union indexed by a non-empty set

[prime elbow]

So if there exists maximal connected subspace doesn't mean the space is connected

[tender halo]

wdym? there always exists one

for connected spaces its the whole space

for totally disconnected spaces each point is a maximal connected subspace

[prime elbow]

Yes

[unreal stratus]

Maximal, but not "maximum"

[prime elbow]

Yes I understand maximal under inclusion order, right?

[unreal stratus]

Yeah

[manic seal]

Man, Munkers is great

the proofs are so easy to read it's surreal

[limpid fern]

munkers moment

[tough summit]

I have no idea if this counts toward topology at all, but I am a first year uni student trying to explore deeper into more advanced maths, and I was wondering if anyone could read my attempt to prove that the unit circle carved into R^2 is homeomorphic to the same circle with radius 2. It would be great if I could be told whether or not my proof is correctly constructed. Thanks

[austere dirge]

u didnt prove its bijective properly

bijections are two sided inverses

so f o f^-1 and f^-1 o f are the identity

[tough summit]

Oh I see thanks, that's easy fix, does the continuity parts look good though?

[austere dirge]

the rest seems fine

small nitpick though

what u wrote is not the definition of continuity

unless ur doing metric space topology only

it still works, but calling it "by definition" is dubious

[tough summit]

Oh okay that makes sense

So is there is a different definition for continuity in topology?

[rain ether]

preimage of open sets is open

[sonic crane]

Why for finite products the two topologies are the same?

[unreal stratus]

Well they only differ when there are infinitely many options for alpha

The "except" clause never kicks in

[real granite]

(1) is true. The sequence of distinct points that lies in A n B will also lie in A (because that's how intersections work), so x is a limit point of A and a limit point of B, therefore x in A' and x in B', i.e. x in A' n B'.

(2) feels true but I'm not sure how to prove it. x in A' and x in B', so there is a sequence of points x_n (distinct from x) that lie in A s.t. x_n converges to x, and there is a similar sequence y_n in B that converges to x.

(3) is true. Let x in (A u B)'. Then there is a distinct sequence x_n in A u B that converges to x. At least one of A or B must contain infinitely many terms in this sequence, so x is a limit point of at least one of A or B, i.e. x in A' u B'. Conversely, let x in A' u B'. Then x in A' or B'. So there is a distinct sequence x_n in A (or y_n in B) s.t. x_n converges to x. But x_n in A in A u B, and y_n in B in A u B, so in any case there is a distinct sequence contained in A u B that converges to x, thus x is a limit point of A u B.

Please would someone check this.

[viral terrace]

(2) is quite false. Take two sequences with different members converging to the same point. The intersection is empty.

[real granite]

I did consider this but wouldn't it be vacuously true?

E.g. A=(-∞,1) and B=(1,∞). x_n=1-1/n converges to 1 from below, whereas y_n=1+1/n converges to 1 from above.

Clearly A and B are disjoint, so their intersection is empty, but if you can speak about the tallest person in a room that is empty, why can't you speak about a sequence in ø converging to 1?

[red yoke]

∅ does not have sequences

[viral terrace]

(2) asks if you have the sets of limit points coincide. You have the empty set on one hand and non-empty on another.

[red yoke]

Although every sequence in ∅ converges to 1

∀xp(x) does not imply ∃xp(x)

[real granite]

Hmmm

OKie

[viral terrace]

it's almost as if asking if closure distributes over intersections

[real granite]

Ye. What about 1 and 3? Was I right about those?

[viral terrace]

Yes

[real granite]

Last thing. If A and B are connected, and A and B have non-empty intersection, is A n B connected?

I think the answer is no, if you think of a horshoe shape and an upside-down horshoe shape. Each horshoe is connected, but the intersection will be two disjoint regions. This is my poor attempt at drawing what I mean

[viral terrace]

that's right, you have a counterexample

[real granite]

sweeet

[broken nacelle]

MUNKRES

MUNKRESSSSSS

RAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA

[tender halo]

munkres is too intuitive and well written, i feel like it doesnt quite represent the clunkiness of classical point-set topology and therefore its a bad textbook

[broken nacelle]

is this a serious take?

[tender halo]

10% serious

i feel like munkres doesnt talk about a lot of ugly results

its one of the best undergrad textbooks that exist, sure

[tender halo]

and classical point-set topology is like 50% ugly results

[limpid fern]

counterexamples in topology moment

[real granite]

Possible silly question:
The definition of a closed set is that it contains all of its limit points, C' in C. Intuitively, it seems that every point in the interior of C would be a limit point, so could you also say that a set is closed iff C=C'?

[opaque scroll]

Usually limit point means points accumulated to by other points of C, so C' doesn't contain the isolated points of C.

[real granite]

Wdym by isolated?

Also yes, I was misremembering the definition, which is about balls containing points outside of the set, and was thinking instead of the sequential version. If you use the former, then obviously for a point in the interior, sufficiently small balls will contain only points in the set itself, so interior points cannot be limit points.

Uh wait

That's wrong I think

Yeah I was thinking this meant every ball about x in A' must contain points outside of A, but this is not at all what the definition is saying

So I return to what I said originally, which is why interior points can't be limit points

[tender halo]

interior points usually are limit points

isolated means there is a nbhd that doesnt intersect the rest of the set

[real granite]

You mean something like (0,1) u {2}, then 2 is an isolated point?

[alpine nest]

The singleton {1} is closed in R with the Euclidean topology, but isn't equal to its set of limit points (which is empty)

[real granite]

This is a good point

[alpine nest]

Or in your example, 2 is in the closure of your set, but isn't its limit point

[tender halo]

yea

[real granite]

Well... isn't the empty set a subset of every set, so trivially you have C'=ø in C

[tender halo]

empty set is also not a nbhd of anything

[alpine nest]

Yes, a set is closed if its set of limit points is its subset

[real granite]

Oh wait yeah I'm being dumb

Yes I see your point now

[alpine nest]

Inclusion yes, equality no

A set that's equal to its set of limit points does in fact have a name, they're called perfect sets.

[real granite]

And, guessing here, perfect sets are sets without isolated points?

[alpine nest]

Closed sets without isolated points

(0,1) doesn't have isolated points but isn't perfect

Because it has two limit points that don't belong to it

[real granite]

Ye

[real granite]

How does the highlighted sentence work?

My understanding is that $n_k$ has to be strictly increasing, so $\forall N_{\epsilon} \in \bN, \exists K_{\epsilon}\in\bN,\ \text{s.t.}\ k>K_{\epsilon} \implies n_k>N_{\epsilon}$.

But it seems as though he is saying $m=n_k > N_{\epsilon} \implies k>K_{\epsilon}$, and I'm not entirely sure how he is getting this.

[gentle osprey]

Douglas

[real granite]

Actually, is this just a consequence of n_k being strictly increasing, so the implication goes both ways?

[viral terrace]

Hmm, $n_k > N_\epsilon$ does imply $k > K_\epsilon$.

[gentle osprey]

Capy

[real granite]

Hmmm. To use the triangle equality, don't we need n_k > N_e as well as n>N_e?
In the highlighted bit, he conditions what he says on n>N_e and n_k>N_e, but then in the next sentence he forgets about the latter condition.

[real granite]

And is this only because n_k is strictly increasing?

[viral terrace]

strictly increasing (that part of subsequence defn) and because limiting diameter of a set isn't weaker than limiting distance from a point by the same half epsilon

[real granite]

@viral terrace What about this?

[viral terrace]

he says all inequalities for n, k, n_k are working in that sentence.

initially, N(epsilon) is for n, but you can use it for nk and k > K follows

[real granite]

Oh I see... so because the statement speaks about "for all n>N", then that applies specifically to the n_k as well?

[viral terrace]

yes, he uses it like that saying "n>N and nk>N, so that k >K", but it's not entirely correct, because it gives another K' with the property that if k > K', then d(xnk, x) < eps/2.

[real granite]

I need to go somewhere, but I will try to have a proper think about this when I get back.
Thanks for the help.

[viral terrace]

good luck

[proper fiber]

I was able to show that Y is closed and discrete

But I am not having any progress regarding finding f

[limpid fern]

it helps to draw a picture. essentially we're saying if the sorgenfrey plane is metrizable you should be able to put disjoint balls around each point on the y = -x line

in fact you can already put disjoint basic nbhds around each point in Y

[limpid fern]

oops, forgot to ping

[proper fiber]

I see

Thank you

I was able to do it in the meantime

I think the conclusions I drew were similar

[proper fiber]

After this I think we get a contradiction using countability

[civic juniper]

Can someone explain why hausdorff hypothesis is required. Isn't it enough that the natural projection map X->X/G is open? For since continuous open maps preserve compact neighbourhoods, we will have the proposition?

[tender halo]

does your definition of local compactness include Hausdorffness perchance?

[civic juniper]

Well, the definition I was using is that a space is locally compact if each point has a compact neighbourhood i.e. a compact set around it containing one of its neighbourhood. But may be, this book (Bourbaki) uses different. Let me check.

Aha, you were right

[tender halo]

there are a lot of different definitions of local compactness that are not equivalent if you dont assume hausdorffness

so often people assume it in the definition in order to not say hausdorff over and over

[civic juniper]

Got it. Something to keep in mind. But did not think local compactness has various definitions. Thanks

[tender halo]

wikipedia lists 5 non-equivalent ones

[pearl trail]

Anyone around to nudge me on this direction of the proof? I am wondering if there are finite x_n inside U since n in N and the natural numbers are countably infinite

[vocal wharf]

recall that in the definition of convergence, there is a for all quantifier on the U in x \in U \subseteq X

[real granite]

I am back.

Your point here is that we're "scaling" $\epsilon$ so to get $x_{n_k}$ to within $\epsilon/2$ (note the half) of $x$ then $k$ needs to be such and such amount bigger than it needed to be for $x_{n_k}$ be within $\epsilon$ of $x$. Is that correct?

[gentle osprey]

Douglas

[real granite]

Actually, he starts out with a factor of a half anyway, so there isn't really any issue with scaling idt

[viral terrace]

my point is that if $N$ is such that $d(x_n, x_m) < \varepsilon/2$ for all $m, n > N$, then the first $K$, such that $n_K > N$ is good enough to have $d(x_{n_k}, x)<\varepsilon/2$ for all $k > K$, if $x$ is a limit point.

[gentle osprey]

Capy

[viral terrace]

though, I probably incorrect about strictness of the last inequality: $d(x_{n_k}, x) \le \varepsilon/2$ is definitely true

[gentle osprey]

Capy

[real granite]

Well yeah but I'm saying I don't know where this K' business is coming from. You can use the same K can't you?

[viral terrace]

K's can come from two places. One is K being smallest that n_K > N for some N and the author got K from the convergence definition

[real granite]

I am not following you. The proof in the notes makes sense to me without making (minor) corrections

For reference

[viral terrace]

he definitely means that n_k > N implies k > K

and that K came the convergence definition, that is, he limited d(x_{n_k}) by epsilon/2 and got that K

oh well, anyway, just take $N = \max{N_\varepsilon, n_K}$ and it would be flawless to me

[real granite]

Yes, this follows from n_k being strictly increasing, but that also means n_k > N <==> k > K.

I don't see the distinction you're making between K and some other K' (see here #point-set-topology message). I think everywhere he writes K, he (correctly) refers to the same K.

[gentle osprey]

Capy

[real granite]

Anyway...

What would happen if you altered the definition to be $c\in[0,1]$ and $T$ s.t. $d(Tx, Ty) < cd(x,y)$?

[gentle osprey]

Douglas

[viral terrace]

hmm, I'd try to construct a pathological example with $d(Tx, Ty) < d(x, y)$ on $[0,1)$.

[gentle osprey]

Capy

[real granite]

Apparently it has something to do with geometric seriesa

[tender halo]

just have the contraction be unbounded in c

well "unbounded"

[viral terrace]

for [0,1), it's simply a convex function

anyway, the idea of the contraction is that it shrinks diameter of a set by at least some scale different from 1

so, $c\in[0,1]$ wouldn't lead to good results. One has to have $c\in[0,1)$

[gentle osprey]

Capy