point-set-topology

[limber wyvern]

Hi, im starting with topology and I wanted to know if I have the concept of "base of neighborhood" really attached, as my uni book is a bit clunky:

So, I understand it as: a minimum set from which every other neighborhood is created, like the base of a vector.

[steep wedge]

it doesnt have to be minimum, and 'a' instead of 'the'

[limber wyvern]

Why isn't the minimum?

Is not like the minimal set from which every other is created?

[quasi forum]

the point is that the same topology can have multiple bases, and it very well may be the case that one is smaller than the other

for example the typical choice for the base of the topology on R^n is just all balls B(x, r) where x is in R^n and r > 0

but you can also choose x in Q^n and only choose r to be a positive rational and this base still generates the usual topology on R^n, even though it is countable and therefore much smaller

[limber wyvern]

I see

just like the bases in vectorial spaces

okey

apart from that, is my understanding correct?

[warped helm]

i would advise against thinking about it in analogy to vector space bases

the prototypical example for a “good” neighborhood base for a point x in a metric space would be all the balls B(x,1/n)

the point is that they tend to be more tractable to work with than considering every neighborhood

and you can relate it back to all the neighborhoods since you know how the basis elements sit inside them

[kind marlin]

i think one useful way to think about the commonality between bases is that they are generating sets

a basis for a vector space generates the space by taking all possible linear combinations of vectors in the basis

a base for a topological space generates the space by taking all possible unions of sets in the basis

one difference between the two is that a vector space basis also needs to be "minimal", in that removing any vector from the set should result in failing to generate the vector space. we don't really have this requirement in topology

in general it's useful to reduce larger systems to generating sets, as long as you can prove that properties of the generating set also apply to the larger system

[limber wyvern]

Thanks for the answer 🙂

[rancid umbra]

i agree with the advice that others have given previously, that you shouldn't try to relate bases in topology and bases in linear algebra for the most part.
however, there is some sense in which they are the same. this may be a bit complicated and is not explicitly used very much, but it expresses the relationship.

fix a topological space X. a basis for a top space X is a collection of subsets which cover X and are such that for any two basis elements B1 and B2, and any x in the intersection of B1 and B2, there is a B3 in the collection such that x is in B3 and B3 is a subset of B1 and B2.

there is a (posetal) category Basis(X) of collections satisfying the two properties above.

similarly, there is a (posetal) category Topologies(X).

there is a free functor Basis(X) -> Topologies(X) and a forgetful functor Topologies(X) -> Basis(X) which forms a free-forgetful adjunction.

a basis for some topology on X freely generates that topology, just like a basis for a vector space freely generates that vector space.

[dark crypt]

this is set theory, not topology, but i am struggling with problem 45. in the solution they claim that |Zg(αᵢ, n)| ≤ n, but all i see is that |Zg(αᵢ₊₁, g(αᵢ, αᵢ₊₁))| ≤ f(αᵢ, αᵢ₊₁) ≤ n. and we can have |Zg(αᵢ₊₁, g(αᵢ, αᵢ₊₁))| < |Zg(aᵢ₊₁, n)| right?

[dark crypt]

(i think i see how to fix the proof — if the sequence f(αᵢ, αᵢ₊₁) is bounded, find the maximum value n that occurs infinitely many times in the sequence g(αᵢ, αᵢ₊₁). then work on some final segment where g(αᵢ, αᵢ₊₁) ≤ n. then find some k such that g(αₖ, αₖ₊₁) = n and large enough that |Zg(αₖ, g(αₖ, αₖ₊₁))| = |Zg(αₖ, n)| > n. this will be the contradiction.)

[hollow pawn]

how do I show that a space X is Hausdroff if and only if {(x,x): x \in X} is a closed set in X×X?

[tender halo]

by definition more or less

write down the definition of what it means for the complement of the diagonal to be open

[crisp lintel]

if x is not equal to y, then (x,y) is in the complement of the diagonal

[limber wyvern]

Hi!, I'm back to ask if my understanding is correct.

In this case I need to check if my understanding of adherent point is correct:

For me, a adherent point is a point which is arbitrarily close to a subset A of X, in which the intersection between A and the neighborhood of X is not the empty set

Am I right?

[rancid umbra]

your intuition sounds like it’s on the right track but your explanation is wrong. a point x in X is an adherent point of A if every open neighborhood of x meets A.

so from the point of view of the topology on X, or when viewed under the lens of the open subsets of X, the point x adheres to A, or appears to be glued/stuck to A.

[steep wedge]

according to this intuition of topology an adherent point of a set A would a point such that you cant distinguish it from A.

[tranquil cosmos]

what does adherent mean

$x \in \partial A$?

[gentle osprey]

PKThoron

[steep wedge]

That's a stronger statement then adherent

That means x is in the boundary of A
Or intuitionwise, x is indistinguishable from A and x is indistinguishable from complement of A.

[tranquil cosmos]

then $x \in \bar{A}$?

[gentle osprey]

PKThoron

[limber wyvern]

I see

thanks

[limber wyvern]

Im back xd

Im still facing difficulties understanding the concept of adherent

Even when trying to draw it in R^2

So, I have a point x in X, and is an adherent point of A if every open neighborhood of x meets A, but this means, the nerighborhood is arbitrary, so it can be as big as X, or as small as x, so, if a subset A exists then everypoint of X is an adherent point of A?

I know this is not true

but it's where my reasoning leads me to

[warped helm]

well no

for example x = 2 is not an adherent point of A = (0,1) as a subset of R

as the neighborhood (1.5, 2.5) does not intersect A

[opaque scroll]

There's a difference between "every neighborhood meets A" and "some arbitrary neighborhood meets A"

[limber wyvern]

mmm I see

I think i got it this time lol

[heady skiff]

Can I have a hint for why Q is not locally compact? Let's take the point 0 as an example; if Q was locally compact, there would be a compact subspace C of Q with 0 \in C. Now C can be covered by a finite union of open sets of Q, thus C is finite. I'm having trouble progressing further from this though

[quasi forum]

why is C finite exactly?

[heady skiff]

Oh yeah that's wrong

[quasi forum]

heres an example that gives you an idea of what the problem is

say you have a sequence of rational numbers converging to sqrt(2)

so 1, 1.4, 1.41, 1.414, ...

you can show this set is not compact by choosing open intervals centered at each of those points which are sufficiently small so as to all be disjoint

any open set in Q which contains 0 would have to contain a sequence like that, converging to some irrational number

[fierce mesa]

Hi. Can anyone give me a hint as to how to do part (c)?

The definition of an equivalent metric that I am introduced to is given in the second image.

I am not particularly sure how to set up my proof... How would I show that ball with a metric defined by $d_{(p)}$ is contained in another ball?

[gentle osprey]

Average Math Student

[ruby delta]

what do you have right now?

[fierce mesa]

Well, so far, if I keep $x = (x_1, x_2)$ and $r$ fixed, then I have that $B_1$ is the set of $y = (y_1, y_2)$ such that:
[ d_{1}^{p}\left(x_{1},y_{1}\right)+d_{2}^{p}\left(x_{2},y_{2}\right)\le r^{p} ]
And similarly for $B_2$, we want to show that there exists a $\rho$ such that:
[ d_{1}^{q}\left(x_{1},y_{1}\right)+d_{2}^{q}\left(x_{2},y_{2}\right)\le\rho^{q} ] Is contained in $B_1$.

Aside from that though, I don't know what I should be doing after this.

[gentle osprey]

Average Math Student

[ruby delta]

you know, you have control over p and q as well

[quasi forum]

what does the question mean by uniformly equivalent

[ruby delta]

you can WLOG set p=1

[quasi forum]

that all the metrics are equivalent regardless of the choice of p?

[ruby delta]

it is then sufficient to show that q is always equivalent to p=1

[fierce mesa]

Oh uniformly equivalent, I missed that part. Yeah it should be that the $\rho$ is independent of $x$ used.

[gentle osprey]

Average Math Student

[quasi forum]

ok fair enough, makes sense

[fierce mesa]

Ah okay yea that makes sense... If every q is equivalent to 1 then it is just equivalent to each other. Okay, let me work with that. Thank you!

[ruby delta]

(uniformly) equivalent

[limber wyvern]

Hi, I have a question regarding solving the exercises of topology, so, I understand the theory but I'm not able to do any exercise of my book. I'm starting out so common exercises are : Verify that this is topological space, test that two bases generate the same topological space

But I'm not able to do them even though i know the theory, any advice?

[quick delta]

Why aren’t you able to do them?
Where do you get stuck?

[limber wyvern]

Im not able to even start the problem xD

I mean, I know what to do, but I don't know how to express it formally

[quick delta]

Look at the definitions and any relevant theorems

[limber wyvern]

for example, testing that something is a topological space, it has to pass the 3 axioms, and well, the triviality axiom is easy, but to test for the union and intersection axioms is pretty hard for me

[quick delta]

What are you struggling with with checking them?

[limber wyvern]

I don't know how to prove it

[warped helm]

take an arbitrary collection of open sets in your space and see if their union also falls under the definition of open in your space

[limber wyvern]

mmm

let me make an exercise and test that xD

[tall crown]

Which book are you following?

[warped helm]

do it for metric spaces

[limber wyvern]

its my uni book

its custom made

It comes from this books

Is my reasoning right?

sorry if something in english is wrong, I had to translate it from my language

[warped helm]

"arbitrary intersection" is not the correct axiom

you're supposed to verify closure under finite intersections

[limber wyvern]

interpret it as that

the literal translation from spanish is that

[warped helm]

"arbitrario" is still arbitrary

[limber wyvern]

oh yeah

u are right

Is this also right?

Im sorry but my book doesn't have a solutionary😅

[warped helm]

more or less

you technically didnt list every possible finite intersection but i dont recommend

[nova fjord]

I have a silly question about CW complexes.

I'm trying to show that a map from a CW complex into any topological space is continuous iff its restriction to every skeleton is continuous.

Restrictions of continuous maps are continuous, so suppose I have a map $f \colon X \rightarrow Y$ where X is a CW complex and Y is a top space. If V is a closed subset of Y, I need to show $f^{-1}(V)$ is closed.

By the weak topology, it suffices to show that $f^{-1}(V) \cap \bar{e}^q_i$ is closed for an arbitrary cell $e^q_i$.

I can take a q-skeleton $X^q$ containing $e^q_i$, so by assumption $(f|_{X_q})^{-1}(V) = f^{-1}(V) \cap X^n$ is closed. But I'm not sure how I can use this to argue about how its intersection with each cell in the skeleton is closed?

[gentle osprey]

Eternal Way

[opaque scroll]

A cell is a subspace of the skeleton

[nova fjord]

Does that necessarily imply the intersection with each cell in the skeleton is also closed?

[opaque scroll]

I mean, by definition of subspace topology, the closed sets are exactly what you get by intersecting with closed sets

[nova fjord]

Ah right, the weak topology definition wants the intersection with each cell to be closed in that cell

This wasn't clear with the way my book worded the statement of the weak topology

[nova fjord]

But wait, how come I can't apply this reasoning directly to X^q as a subspace of X to conclude f^-1(V) is closed in X?

[opaque scroll]

I think you can, or how are you defining the topology on X?

[nova fjord]

the weak topology right? A set F of X is closed iff for any cell e^q_i, the intersection F \cap \bar{e}^q_i is closed (this is what's in my book at least)

[opaque scroll]

So then f^-1(V) closed means the intersection with each cell is closed.

Since X^q also has the weak topology the intersection with X^q is also closed

[nova fjord]

Do the subspace topology and weak topology on X^q coincide out of curiosity?

[opaque scroll]

Yes

[nova fjord]

Ah very nice

But I suppose the spirit of the problem is to use the weak topology

[ripe creek]

https://www.roblox.com/games/122578990384115/point-set-topology-RP

[balmy briar]

roblox is banned in my country what am i missing with the point set topology rp

[nova fjord]

Is there a simple example where the cartesian product of two CW complexes does not satisfy the weak topology axiom?

[opaque zodiac]

Iirc counterexamples to this need to be pretty big

Like continuum-many n-cells sharing an edge

[balmy nexus]

there's a weak topology on CW??

don't they know we already use weak topologies in other contexts???

[hidden abyss]

I thought countably infinite cells in both factors already gives you an issue?

Nevermind I remembered the classical example wrong

[opaque scroll]

Whose "they" here? Are you saying it's bad that the weak topology appears more than once in math?

[tall crown]

I am going to start studying topology from Bourbaki

[brazen condor]

Im curious if there is any like, useful point set theoretic topology studies

There's a few areas that spring to mind

Continuum theory
Wild topology

Cech cohomology I think is of a point theoretic flavor

Curious if anyone has more input

[balmy nexus]

yes 🥸

[cosmic mirage]

how so?

[brazen condor]

i haven't really studied it but i think it's more used for 'badly behaved spaces' (ie stuff that require you to do point set shenanigans), like I've seen it used for stuff like the sierpinski carpet in wild topology

At the very least it isn't as 'combinatorial' as simplical compexes or CW Complexes

[cosmic mirage]

huh i see

yeah i dont know much about this either, my impression is that for poorly behaved spaces people usually work with sheaf cohomology

[tall crown]

Any hints for the given problem?

[novel acorn]

What's the problem here?

[tall crown]

I tried to find the sets A and B that satisfy the given condition, but I couldn’t.

[crisp lintel]

I would try to find open dense disjoint sets

or maybe not disjoint that's impossible

but if you can find sets with A, B dense in R then the first one is B, second is A, fourth is R

[tall crown]

It is not possible

[crisp lintel]

Ah I guess the intersection of open dense sets is dense so you wouldn't be able to distinguish 3 and 4

I'd probably start just playing with unions of intervals then

basically if you can find sets where any two are distinct

You can combine them by shifting them suitable far from each other and taking unions

[tall crown]

I tried using the union of intervals, but it didn’t help

[burnt tendon]

It’s possible to find an example where the difference between closure(A \cap B) and closure(A) \cap closure(B) is just a single point

And you can do it with a couple of intervals

[rancid umbra]

no, i don't think so.
|| if cl(A \cap B) U {x} = cl(A) \cap cl(B) for some x not in cl(A \cap B) then {x} is open in cl(A) \cap cl(B), and so is an isolated point of cl(A) \cap cl(B). ||
|| for an isolated point to be in the closure of a set, it must be in the set itself (since it can't be a limit point) so x must be in A and B, which is a contradiction. ||

[novel acorn]

Take (-1,0) and (0,1)

Then the closure of A cap B is empty so the closure is empty

While the intersection of their closures is {0}

[rancid umbra]

ah, let me see where i am going wrong

[burnt tendon]

Yeah uh, cl(A) \cap cl(B) does not appear to be the closure of a set in my example

[rancid umbra]

cl(A) \cap cl(B) is closed, so it is equal to it's own closure

[burnt tendon]

Mm, touché

I still stand that my example is correct

[rancid umbra]

what example?

[novel acorn]

Just because x is an isolated point in the intersection doesn't mean it's an isolated point in each set on it's own

[burnt tendon]

Check your DM

[novel acorn]

Why tf dm them lol

[rancid umbra]

thanks for pointing that out

[burnt tendon]

I am not willing to post the example while Noether is still working on the problem

[rancid umbra]

you can spoiler it

[rancid umbra]

what about if you demand that the intersection is non-empty

[burnt tendon]

Check the same DM as before…?

[rancid umbra]

i don't quite understand the double brackets

if they even mean anything

are you drawing the intersection?

okay, i think i understand now, lemme look

[burnt tendon]

Here is the example for reference

[novel acorn]

you can still take something like A = (-3,-2) U (-1,0) and B = (-3,-2) U (0,1)
Then cl(A) cap cl(B) = [-3,-2] U {0} while cl(A cap B) = [-3,-2]

[rancid umbra]

yea, was just about to type something like this

cool

[novel acorn]

these four aren't distinct tho?

oh wait

I see what you did

okay this is a clever solution

[burnt tendon]

It first came to me to ||have open sets that partitioned the line, so that the closure of the intersection would be empty but the intersection of the closures would be the shared boundary||

Then to get four distinct intersections I bodged the situation by ||introducing a small overlap||

At some point in the past I worked out an example that solves Kuratowski’s closure-complement problem — given a subset A of a topological space X, show that there are at most 14 sets obtained by taking repeated closures and complements of A, and show that this bound is optimal (it’s in Munkres somewhere)

So if you want a tougher problem along the same theme as the previous, there’s one to try

[rancid umbra]

this feels like more of an algebra problem

also, this problem never really appealed to me; i have never seen it used other than as an exercise

i guess this?

[tall crown]

That exercise is from Bourbaki

[tall crown]

Let X be a topological space. Consider the collection of all sets that are complements of closures of subsets of X, where the closure is taken with respect to the given topology on X. Is this collection the same as the given topology on X?

[rancid umbra]

what do you think so far?

[tall crown]

I think this is true

[rancid umbra]

yea, it is.

[storm wadi]

What do we know about all closures?

is my clue

[rancid umbra]

this is a vague clue

lol

[storm wadi]

I guess, I don't know how much more of a clue you can give though without giving the answer entirely

its like two lines, assuming I'm not being dumb

[rancid umbra]

maybe add "in relation to open sets"

[storm wadi]

True, that adds some direction once they recall the important property of the closure. Thanks, always open to how I can offer more constructive clues

[tall crown]

I am considering the collection of all closures of subsets of $X$, and I am trying to show that it forms a topology by closed sets. I have already proved that it is closed under finite unions. Now I am trying to show that it is closed under arbitrary intersections. For that, it suffices to prove that
[
\overline{A \cap B} \subset \overline{A} \cap \overline{B}.
]

[gentle osprey]

Emmy Noether

[rancid umbra]

there is a more straightforward way to go about this

[tall crown]

Okay, what is the way to do that?

[rancid umbra]

see the hint above

rephrasing the hint, recall the definition of a closed set

[tall crown]

This is a different question; I have already proved the previous one.

[rancid umbra]

oh

i see

hmm haha same hint applies

[tall crown]

Sorry, I don’t see how this helps me

[rancid umbra]

a set is called closed if it is the complement of an open set

[tall crown]

Yes

[rancid umbra]

if we take a finite collection C1, ..., Cn of closed sets any collection C_i of closed sets

expand out \bigcup_{i = 1}^n C_i \bigcap_i C_i

if each C_i = X - U_i for some open set U_i

[tall crown]

But no topology is given

[kind marlin]

|| can't you just argue that all closed sets equal their closures, and all closures are closed sets, so the set of closures of subsets of X is just the set of closed sets in X, meaning the set of complements is the set of open sets in X ||

[rancid umbra]

@tall crown then how are you considering all closures of subsets of X?

[kind marlin]

oh is this a diff question

nvm

[rancid umbra]

okay, i misread

you are doing intersections

[storm wadi]

yeah different question but that is the way for the first one i think yes

[tall crown]

Here, I made a mistake: this is not a closure. It is just a function that maps a set $M$ to $\overline{M}$ with some property

[gentle osprey]

Emmy Noether

[rancid umbra]

okay, so problem 9

[tall crown]

Yes

[rancid umbra]

what is B(X)?

[storm wadi]

this inclusion follows from the fact that the intersection of two closed sets is closed right?

and obviously AnB is contained in \overline AnB

[tall crown]

I am reading Bourbaki for the first time, and the notations are confusing. As far as I understand, $B(X)$ is supposed to denote all neighborhoods of $X$, but that does not make sense if a topology is not given. What if we treat it simply as the power set of $X$?

[gentle osprey]

Emmy Noether

[tiny obsidian]

It is indeed meant to be the power set; whether this is a mistake, or bourbaki has some (unexplained) reason to use this notation here anyway I cannot say

[fierce mesa]

Hi! I'm a bit stuck on this question (again), so I would like a few more hints. \

I'm doing Q3c. I know that for some arbitrary $p$, I have $d_{(p)} \le d_{(1)}$, which means that for every ball defined under $d_{(1)}$, there will exist a ball defined under $d_{(p)}$ contained in the previous ball. \

I'm not sure how to prove the other way, that is that for every ball defined under $d_{(p)}$, there will exist some ball defined under $d_{(1)}$ contained in this ball.

[gentle osprey]

Average Math Student

[crisp lintel]

for uniform equivalence id recommend comparing the p-metric to the max metric (i.e. just the max of both metrics), then compare the max matrix to the 1-metric.

and if you combine those you should get the constants you need

[quasi forum]

thats likely easiest. alternatively you can show directly that d_1 <= C_p d_p for a suitable constant C_p > 0 depending on p

in particular you can take C_p = 2^{1-1/p}

[hollow pawn]

would this be the topology generated by the preimage of every basis element under f_\alpha for every X_\alpha?

[quasi forum]

you can just take it to be the preimage of all open sets i think because every open set is a union of basis elements by definition

but yeah thats correct

this construction is strongly related to weak topologies btw

[warped helm]

its not given that X_alpha has a topology generated by a basis

[quasi forum]

i mean if you have any topology its trivially generated by a basis no

just take the basis to be all open sets or something

its useless but technically its not wrong

[hollow pawn]

okay thanks, I'll do the construction with all open sets

[fierce mesa]

Hi! Wdym by max metric? I mean, you only mentioned the p-metric here... did you mean using the max metric on the p-metric and the 1-metric?

[fierce mesa]

Hi! How'd you derive this C_p? Could you give me a hint?

[crisp lintel]

I mean max(d_1,d_2)

you could go directly but at least to me this is a bit more straight forward

[quasi forum]

ill admit i nuked the bound by invoking jensens inequality to get that oddly specific constant

[fierce mesa]

Ohh okay

[quasi forum]

the max metric imo is more natural to use, but if you really want that exact bound you may be able to derive it through some sort of derivative argument

[fierce mesa]

I'll have a play around with that then, thx 👍

[fierce mesa]

I see

[fierce mesa]

I was thinking that too but the expression is a little bit dodgy. Maybe I will attempt that as a last resort.

[quasi forum]

fait enough, play around and see what you get

[balmy nexus]

(big hint): try to show that all norms are equivalent on R^2 first

[tall crown]

Thank you

[quick crane]

This is basically a question about $p$ norms on $\mathbb{R}^2$. You can explicitly find the optimal constants $c, C$ such that $|x|_p \leq C|x|_q$ and $|x|_q \leq C|x|_p$ for all $x \in \mathbb{R}^2$.

[gentle osprey]

L

[heady skiff]

Can somebody please explain how the fact that for all $\epsilon > 0$, the set $[-\epsilon, \epsilon] \cap \mathbb{Q}$ is non-compact implies that $\mathbb{Q}$ is not locally compact?

[gentle osprey]

okeyokay

[heady skiff]

Never mind, I didn't read the definition of local compactness close enough

[rancid umbra]

there are a few of them. which one are you using?

[kind egret]

If $\mathbb{Q}$ is locally compact, and if $x$ is rational, then there exists a compact neighborhood of $x$, i.e. a compact set $K$ and a positive real number $\epsilon$ such that $[-\epsilon+x,,x+\epsilon] \subset K$. This means that $[-\epsilon,,\epsilon]$ is compact, for it is a translation of a closed subset in a compact set.

Then you have a contradiction

[gentle osprey]

La Chouette Aveugle

[kind egret]

better like that

I've by the way seen somewhere that $\mathbb{Q}$ is the only countable metric space which is 0-dimensional and nowhere locally compact (up to homeomorphism)

[gentle osprey]

La Chouette Aveugle

[midnight umbra]

Q is also the unique countable metric space without isolated points

[kind egret]

ah yes that's it

this has surprising consequences, such that $\mathbb{Q}^n$ is homeomorphic to $\mathbb{Q}$, or that $\mathbb{Q} \cap [0,,1]$ is also homeomorphic to $\mathbb{Q}$

[gentle osprey]

La Chouette Aveugle

[warped helm]

to disprove the statement you have to prove that a specific point doesn’t have a compact neighborhood

[thick vigil]

Am i missing anything for this solution?

[tall crown]

I don’t understand what it means to show that a set is inductive. In the previous exercise, it says that in a Kolmogorov space, if it has no isolated points, then every non-empty set is infinite.

b part

[hollow pawn]

how would I show that $(x_1-\epsilon , x_1+\epsilon)\times\dots(x_n-\epsilon,x_n+\epsilon)\times\dots$ is not the same as the epsilon ball $B{_\bar\rho} (x,\epsilon)$ in the uniform topology where $\bar\rho$ is the uniform metric (in $R^\omega)$

[gentle osprey]

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

[warped helm]

the point with y_n = x_n + epsilon(1-1/n) is in the product but not the ball

|y_n - x_n| = epsilon(1-1/n) of which the sup is epsilon

[limber wyvern]

Hi, so, I'm starting to study topology, and I find it a bit of a mathematical nonsense, I mean, why is it important to define a topology?, why it is important to define a topological space? I do understand the theory behind this concepts, but I don't understand why they are important

[quartz horizon]

one nice motivation is that a topology is exactly what remains invariant when you consider homeomorphisms between metric spaces

there are lots of equivalent metrics you can put on the same space, e.g. think about product spaces

but so long as they're equivalent, you get the same topology

so even if you only care about metric spaces, it's helpful to think in topological language

(indeed most of the topological spaces you'll meet in an introductory course are metrizable, but you consider them as topological spaces)

[limber wyvern]

I see, thank you

[quasi forum]

historically point set topology developed out of trying to generalize a lot of concepts from real analysis

stuff like extreme value theorem, intermediate value theorem, continuity etc

the big question at the time was how to do it in a way that is useful and applies to most branches of math, even in spaces that lack the nice structure of R^n

[cyan pewter]

there is a nice paper in historia mathematica which tells the story of how the historical things were formalized over time

https://www.sciencedirect.com/science/article/pii/S0315086008000050

[quasi forum]

i think in topology especially it is important to not forget the origins of the subject

because it feels very unmotivated until you see how useful it is

[cyan pewter]

i wonder how differently topology would be viewed if we stuck with defining a topology via its closure operator

[brazen condor]

What happens in the history of math is that something that originally had a rougher but more clearly motivated definition becomes boiled down more and more because that is the more efficient and general way to work. Topological spaces are one such example. I don't think you should be expected to have an actual concrete intuition to such a broad definition, this is just the happen stance of history that this is the 'right' approach.

I'll say that broadly speaking no one really 'cares' if something is alone a topological space, but more what other properties you can add to it. So don't force yourself to 'care' about the most general notion of a topological space and taking the examples of like a set of 3 elements and making one non hausdorff and no no one cares about that. One cares about whether one is first countable, connected, compact, hausdorff, properties that better let us use it.

Topological spaces are very loosely a way of describing the structure of 'regions' of a space. Regions is a very broad concept (what counts as a region)? I think in a sense it is clear that a point in R^2 is too 'specific'. The use of a region is they give an easier defined structure to speak about points in relation to other points. Think about how in epsilon-delta the question of convergence is answered by points lying in epsilon and delta balls: no other information is needed . The choice of a topological space is the choice of 'regions' you want to be able to use to describe an area in your space. With this epsilon delta framework generalized (now using the alloted regions as the information about where a limit lies), this lets you create new notions of convergence and continuity.

Of course, the topological space notion is so broad that basically anything could be one and the set of regions could either be the entire power set or solely the set itself (and empty set). So its less like a topological space alone 'tells' you a lot.

You get more information from other properties

[cosmic mirage]

lol I was trying to come up with examples of non-metrizable topological spaces for smtg recently and was struggling so much

like the examples I had were all too complicated for the specific context

[brazen condor]

Urysohn metrization (and Nagata-Smirnov metrization) are good conditions to say that basically nice enough topological spaces are metric spaces

[cosmic mirage]

huh I've never heard of this. That's messed up

luckily metrics are convenient only in areas of math I don't care about 💀

[unreal stratus]

So are topological spaces

[cosmic mirage]

LMAO

[brazen condor]

For instance:

Let us say I have a topological space that lets me separate two points with two non intersecting regions. I have now in a sense given my space more structure, I've let 'smaller' regions enter the picture so as to be able to separate any two points. This is the Hausdorff property.

Let us say I want a way to say for me to say whether a point x is 'closer' to y or z. One idea for a structure is effectively you have regions that 'increase' and 'decrease' in size. Say a countable many of them U_n in a descending order: U_1 ⊂ ... ⊂ U_n ⊂ U_n+1 ⊂... each containing x and in a manner the intersection of all the open sets if x. If I say have y in U_n but y not in U_n+1 but z is in U_n+1, then in a sense I'd want to say x is closer to y than to z. This definition alone has some issues (do open sets containing x need to have an intersection equal to x, also a countable amount?). If you noticed, this type of reasoning is similar to metric space convergence using ball B(x,1/n), and to do this would require the first countable property and hausdorff property.

The point I want to make is that point-set topology is about studying what properties we should put on these regions and what they entail, and the definition of a topological space is just the bare minimum requirement for that.

[quasi forum]

i think a relatively simple toy example is the space of binary sequences with the box topology

nagata smirnov is like the theorem for metrization iirc but the hypotheses for urysohn's theorem are easier to understand

[unreal stratus]

Indiscrete topology on two points lol

[quasi forum]

yeah i guess that works too

[cosmic mirage]

well I also wanted useful examples 💀

does the cofree topology even do anything

[brazen condor]

One question that may come up is if we care about loosely associated this notion of regions, why do we treat in say the euclidean topology only the 'open sets'. Why not just take the closed sets or something? You can but you get all subsets of R^2.

I'm still trying to find a good way to explain this very vague intuition but one way of thinking of why there is arbitrary unions in topological spaces and not arbitrary intersections is that we expect to have arbitrarily small regions, so an intersection of these just gives a single point: not useful. Another way of thinking about it is that an open set is like an inequality: it is an easy to verify piece of information. If x is in the open set (-2,2), I can verify it by simply taking a decimal expansion. If I see the first digit if 2 or -2 I already know it is not in that . If it is 1,0, or -1 I immediately know it is in that open set. I can't do the same with closed sets: if x is in [-2,2], if I see 2.000000000 it might still not be in there or not depending on the next digit. So effectively, to consider elements of closed sets, I need 'exact', effectively infinite information. So it defeats the whole point of using regions to 'simplify' the amount of information I need. Another point: if I say know the distance from x to a point y has d(x,y)<r_1, r_2, ..., r_n, then I can form r=min(r_1,...,r_n) and conclude d(x,y)<r. I can quite easily conclude x is in this region. This idea doesn't work in infinite case: I can know d(x,y)<1+1/n for each n but I can't know d(x,y)<1.

I think on some level this is actually the fundamental reason why topological spaces are used and this connection with this kind of computable kind of intuitionistic logic, but I can't put my finger on how this exactly works. But this is my rough explanation of why we use finite intersection.

this answer here explains a bit about this POV
https://math.stackexchange.com/a/3854718/400654

And in the context of 'torus = coffee cup topology', the reason this idea of region is important is it allows us to effectively talk about the global overall structure of these regions, their connectivity on a fundamental level. Such connection properties don't change if I simply take a torus of playdoh and stretch it. A metric space would effectively be including extra unnecessary information of this connectivity structure, so if we want to study just these 'global structure properties', topological spaces are the right setting.

ngl i think graph theory is very good for getting this intuition: what matters there is the connectivity between the vertices, not the actual distance or how they are arranged, and hence why graph theory is in a sense the first form of topology.

Also it's worth noting there are other definitions of topological spaces, like this one which to me seems to much more immediately generalized metric space theory

[kind egret]

One way I explain myself the open sets is the following one : we want to answer the question "what does it mean to be in the neighborhood of a point x ?". But we realise that there is no answer for that, at least no absolute answer. So we definie the neighborhood basis, such that one system describes a way for points to be close to each other, and another describes another way for points to be close to each other, such that two different system can bear different informations.

For me, the real deal in a topology, is the neighborhood basis. The open sets are just "super-neighborhoods" because they are exactly the subsets of the space which are the neighborhood of any point they have. Thoses specials neighborhood, in that case, allow us to describe all of the others neighborhoods, so that's why they're interesting

Then the closed sets : nothing's more important than their border. Indeed a closed set is just it's interior + it's border, and the border is contained inside of it (that's another définition of closed subsets). For me the only important thing to understand closed subsets is to understand their border.

And for the axioms, to take an arbitrary union mean that we can add anything, the loss of precision is unlimited. On the other hand, to do an intersection is to look for more precision, and we can reach points where by taking again and again, we're not look at something that describes us a neighborhood, because it is now too small.

For more formal reasons, topologies would be way more easy to understand without the finite intersection axioms, so it would'nt be interesting

[limber wyvern]

wow

thank all for the answers

they really helped

[hollow pawn]

how would I show sums, products, and (well defined) quotients of convergent sequences converge to the sum, product and quotients of their limits (respectively) using the open set definition for convergence in R?

i could show equivalence with epsilon-N but idk if that's how my textbook wants be to do that

[warped helm]

that is kinda the point

balls in R form a neighborhood basis for your point

so this is still topology

[hollow pawn]

okay, then I'll go ahead with that

ty

[quartz horizon]

conceptually i think of this in terms of proving that sum, product and quotient are continuous operations

[hollow pawn]

would using the sequential definition of continuity on those operators along with the fact that the sequence (x_n, y_n) converges to the limit of each component work?

[quartz horizon]

yep!

i like this because i feel like it separates out the distinct parts of solving your original problem

one part is showing that (x_n, y_n) converges to the limit

and the other part is showing that the 2-argument function you're considering is continuous

and these are kind of independent of each other, right?

[kind egret]

If you want to prove those facts with sets only, then you'll have to use the definition of the open sets of R, which uses epsilons and inequalities. So in the end, it always comes to the same, even with the epsilon-N stuff. As it has been said, when you use the epsilon-N, you implicitly proove that the sum or the product from RxR to R are continuous

[hollow pawn]

it's because the epsilon-N proves the sequential criterion in a way, right?

thanks for the responses

[kind egret]

we use the epsilon-N to verify the sequential criterion, yes, and in general the original definition of continuity implies the sequential criterion, but the fact that the sequential criterion implies the original defintion of continuity is only true in metric spaces, and is a consequence of the axiom of dependant choice

[hollow pawn]

does the dotted arrow in general denote anything different from the other arrows

[queen prism]

you use it when you want to say that the given maps p, q together induce a (unique) map f making the diagram commute (i.e., f o p agrees with g)

[quartz horizon]

Usually having the solid arrows causes you to get a dotted arrow

[kind egret]

Cat eaters like those arrows

I was talking about category theorists, not about kittens 🫣

[quartz horizon]

category theory mentioned

[hollow pawn]

how would I do part b? i thought of trying something with the quotient map but i couldn't figure anything out

G here is a topological group

[opaque scroll]

A hint could be that multiplication by g is a homeomorphisms, hence takes closed sets to closed sets

[hollow pawn]

sorry could you elaborate

[opaque scroll]

Okay, first, what are the closed sets in G/H (like what's the definition)?

[hollow pawn]

if p is the quotient map, then those sets whose preimage under p is closed

[opaque scroll]

And what is the primage of a one point set?

[hollow pawn]

for {xH} x and all points y such that xy^-1 is in H

[quartz horizon]

Right

[quartz horizon]

This is actually just $xH$ I believe

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

Wait

I think you mean x^-1 y in H

Otherwise you get a right coset?

[hollow pawn]

oh right

[quartz horizon]

But yeah

Do you know whether or not xH is closed?

[hollow pawn]

nope

[quartz horizon]

.

[hollow pawn]

oh the mapping q:H-->gH?

[quartz horizon]

No

[hollow pawn]

what then? 🥲

[quartz horizon]

Multiplication by x is a homeomorphism

This may help

[hollow pawn]

q(a)=ax for any a in H and some a in G?

[quartz horizon]

left-multiplication by x

[hollow pawn]

right

thanks for the help

[fast yew]

#1433590201741082624 message need help with 5b: show that product topology and a certain metric topology coincide

[foggy quartz]

Question2: Let $(X,T_x)$ and $(Y,T_y)$ be topological spaces. Let ${C_\alpha}{\alpha \in A}$ be a collection of closed subsets of X such that their union is equal to $X$. Let $f: X \rightarrow Y$ be a function. Assume that $f|{C_\alpha}$is continous for every $\alpha \in A$

[gentle osprey]

you_are_me

[foggy quartz]

I dont get what that last sentence means like what is the notation f|_Calpha a

is it like $f: C_\alpha \rightarrow Y$ is continous? is that what they mean?

[gentle osprey]

you_are_me

[foggy quartz]

C_alpha equiped with the subset topology ofc

[unreal stratus]

What have you tried?

(This is a common question and I think it's probs best not to spoil it lol)

[foggy quartz]

nono i dont have a question regarding how to solve it i just have a question regarding the notation

[unreal stratus]

Oh sorry I didn't read properly aha

[foggy quartz]

i havent tried anything yet because im not sure what the notation in the last sentence means

[unreal stratus]

Restriction

So yes what you said – compose with the inclusion

[midnight umbra]

yes

[unreal stratus]

And give it the subspace topology etc

[midnight umbra]

you sometimes see a symbol that looks like an L for restriction too

[foggy quartz]

oki thank you then i can solve it i believe

[viscid blade]

I didnt know obama studied math

[quasi forum]

hes very talented

[zealous berry]

https://www.youtube.com/watch?v=k4RRi_ntQc8

[tepid briar]

The topology itself is always a basis. Are there any common subbases that contain the topology its generating?

[brazen condor]

i don't quite know what you mean by subbases containing the topology, its the other way around (every subbasis element is itself open)

[unreal stratus]

i'm confused - the topology geneerated by a subbasis always contained the subbasis (by definition of "generates")

[tepid briar]

Yeah, its a weird question, since I don't think it should occur in most reasonable situations. I'm wondering if there is a common subbase S for a topology T, where S = T.

Obviously any topology can be a subbase of its own topology, but the question is a well-known or common subbase S, such that the topology generated T = S.

[brazen condor]

it would definitionally need need to be a topology already if S = T

so you'd just be choosing any topology S

using it as subbases to generate a topology T which is just S

[tepid briar]

Right, I feel like if a subbase was like that, it would've been mentioned.

[heady crest]

i mean not always

most the times subbases are just implied to be smaller in order to generate a topology T

though there are some special cases in which the topology itself is so small that you cant really pick a small generating set (well u could its just trivial to do so) , so in this case you could imply the subbase is equal to the topology

[heady crest]

i would read into indiscrete topologies if you want to see these cases but this could be applied to any coarse topology, like i said earlier it just becomes trivial to do so :/

[opaque scroll]

I guess an example that comes to mind is that the Zariski topology on SpecR is usually described by the basis of distinguished opens D(f), but if R is a PID, then these are actually all the opens.

[viscid blade]

Any ideas on proving pth connected space is connected using contradiction?

[quasi forum]

suppose for contradiction that the space is not connected, join a path between two points in your separation and use connectedness of the unit interval to get a contradiction

[prime elbow]

what is pth connected space?

[radiant stone]

i think it's path connected space (but typo)

[viscid blade]

How

[quasi forum]

ill start the proof

suppose $X$ is path connected but not connected, so $X = A \cup B$ for disjoint non-empty open sets $A, B$. choose $a \in A$, $b \in B$ and join them by some path $\gamma: [0,1] \to X$ with $\gamma(0) = a$ and $\gamma(1) = b$. since $[0, 1]$ is connected, $\gamma([0,1])$ must also be connected.

[viscid blade]

Yea i figured all that

[gentle osprey]

KraySovetov

[quasi forum]

you have your separation with A and B

how does that apply to $\gamma([0,1])$

[gentle osprey]

KraySovetov

[viscid blade]

Im guessing i need to use continuity here

[quasi forum]

you already used continuity

that was the statement that [0, 1] connected implies the image of gamma is connected

[viscid blade]

Sure

[quasi forum]

how can you make a separation of the image of gamma using A and B

if you can do this you get the contradiction

because we already know it has to be connected by continuity

[viscid blade]

I split the functiom into the points that are in A, and the ones that map to B

[quasi forum]

yes

$\gamma([0,1]) = (\gamma([0,1]) \cap A) \cup (\gamma([0,1]) \cap B)$

[gentle osprey]

KraySovetov

[quasi forum]

why is this a contradiction

[viscid blade]

Ohhh wait

No

I was going to say union of connected is connected

But doesnt make sense

[quasi forum]

yeah thats clearly false

any two singleton sets

you only know the union of connected spaces is connected if their intersection is non-empty

anyway once you have that and note \gamma([0, 1]) is connected youre basically done

[viscid blade]

I dont get it

[quasi forum]

again

why is this a contradiction

[viscid blade]

Idk

[quasi forum]

ok

you are aware that if f is continuous and E is connected, then f(E) is connected yes

if $A, B$ are open in $X$ then $\gamma([0,1]) \cap A$ and $\gamma([0,1]) \cap B$ are open subsets in the subspace topology of $\gamma([0,1])$

[gentle osprey]

KraySovetov

[quasi forum]

this is by definition

so $\gamma([0, 1])$ is being written as a union of disjoint open subsets in its topology

[gentle osprey]

KraySovetov

[quasi forum]

which means its not connected

[viscid blade]

Ohhhh

Right

I completely didnt see that

Thanks kray

[quasi forum]

np

[storm wadi]

I think no, but I can't think one would construct an open cover of an infinite dimensional space in general? What even are the open sets in C0? Apologies if they are silly questions, this is the first question I have come across in my class that talks about spaces of functions rather than finite dimensional ones.

I wonder if there is a result that reduces this to something easier than constructing an open cover

[midnight umbra]

i think sequential compactness would be the right way to go

[storm wadi]

Alright, thank you. I'll have a think

[midnight umbra]

||if you know that in an infinite dimensional space compact sets have empty interior its also a little trivial||

[storm wadi]

My knowledge of function spaces in general is just really poor, I'm too used to working with just Rn where things are 'easy to visualise'. How do you intuitively consider the interior of an infinite dimensional space?

[midnight umbra]

I genuinely don't think there is a nice way to visualise C[0,1]. I mean, we can only really visualise R^n for n ≤ 3 anyway

so when you step up from 3 basis vectors to, well, uncountably many, I don't think there's much to be done except to kind of pretend it looks like R^2 lol

[storm wadi]

Yeah that's true I guess. Would you be able to help me understand why the interior of a compact set in a function space is empty? As per your spoiler. As I say we never really explicitly discussed such spaces in class, at least from what I remember.

I haven't taken anything on functional analysis yet if that is relevant

[midnight umbra]

at least the proof I know uses a lemma from functional analysis known as Riesz's Lemma

[storm wadi]

Okay no worries then haha

[midnight umbra]

https://en.wikipedia.org/wiki/Riesz's_lemma

tl;Dr if you take an open ball you can find countably many disjoint open balls inside it

so the open cover by the set of countably many open balls has no finite subcover

[storm wadi]

Ah right, that makes sense.
Going back to the sequential compactness argument, I'm not really sure how to construct a sequence in B with no convergent subsequence.

[crisp lintel]

you want some sort of sequence that never really settles down in one spot

there are probably a number of choices

or one that oscillates faster and faster

my advice would be to try and construct a sequence that converges pointwise to zero

but each has norm 1

then there can't be a convergent subsequence

[storm wadi]

max(0, 1-xn)?

Why do we require uniform convergence for sequential compactness over pointwise convergence? Is it just a definition?

[quasi forum]

pointwise convergence is bad at preserving properties that you like to have in functions, such as continuity

i wouldnt say pointwise convergence is bad, but in your case youre looking at continuous functions

if you only insist on pointwise convergence then the space wouldnt even be closed, because there are sequences of functions that converge pointwise to a discontinuous function

[storm wadi]

Right yeah that makes sense

[quartz horizon]

It becomes better in measure theory right?

[quasi forum]

i mean pointwise convergence still has undesirable properties but the theorems you get in measure theory allow you to tame those kinds of functions yes

stuff like dominated/monotone convergence theorems and egorov

[storm wadi]

although this isn't pointwise convergent to exactly zero, it is sufficient right because the pointwise limit is unique and f_n(x) is continuous?

[quasi forum]

yeah and the pointwise limit is discontinuous

[storm wadi]

Yeah exactly.
I made such a hash of this question lol, I feel like I need to go do 20 questions on pointwise and uniform convergence.

[midnight umbra]

pretty easy example is f_n = x^n. any limit f of a subsequence still needs f(1) = 1 and f(x) = 0 for x<1, which isn't cts

@storm wadi

[storm wadi]

Ah right yeah, of course

[south kraken]

i am confusing myself to hell an dback trying to use the universal property of the subspace topology to show that if X is a top. space, Y is a subspace of X, and A is a subset of Y, that the subspace top. on A from X is the same as the subspace top. on A from Y..

or would it perhaps be better to show this just using the topologies themselves?

[quartz horizon]

This example is simple enough that you can just use the topologies themselves

It would more just be practice working with universal properties if you wanted to do it the other way

I mean once you’re comfortable enough with universal properties the proof with them becomes just as fast

[south kraken]

yeah, i (think) i did it correctly with just the topologies themself

but i would like to understand how to utilise the universal properties in a way that i'm not completely lost when i see them

[quartz horizon]

Ok so what is the universal property of the subspace topology, to your understanding?

[south kraken]

to my understanding, the universal property of the subspace topology is that for a topological space X, and its subspace A, if you have another topologicla space Y and a map f from Y to A, that for f to be continuous, the composition of the inclusion map from A to X, and the map f, has to be continuous

[quartz horizon]

Yes, there are alternative ways to state it

Here’s my preferred version

Continuous maps $Z \to A$ naturally correspond to continuous maps $Z \to X$ whose image is contained in $A$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

In that you can interconvert between these

So given $f : Z \to A$ continuous, you can compose with the inclusion map to obtain $\iota \circ f : Z \to X$, which is continuous and has its image contained in $A$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

The universal property allows you to go in reverse

If you have a continuous map $g : Z \to X$ whose image is contained in $A$, then you obtain a continuous map $g^T : Z \to A$, shrinking the codomain

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

This satisfies $\iota \circ g^T = g$ so that the processes are inverses of each other

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

Does that make sense?

[south kraken]

mm somewhat

[quartz horizon]

What parts feel unclear?

[south kraken]

well, i can state how i understand what you just said, and you can correct me if you see anything i missinterpreted

So, what i take from this is that you have these three topological spaces with maps going between them.
So if you have a continuous map f from Z to A, and given the inclusion map that from the (alternate) definition of the subspace topology, is continuous, you can get a continuous map from Z to X by composing the two. Though, here, why is it important that the image is contained in A? or is that just the result of the definition of the inclusion map?

And from the other part where you have a continuous map g from Z to X, the transpose(?) of g where you end up with g^T being a sort of restriction on g?

[quartz horizon]

For the latter it’s called a “corestriction”

Since you’re shrinking the codomain, not the domain

[south kraken]

that does make sense yeah

[quartz horizon]

Also, it’s important that the image is contained in A because this doesn’t apply to arbitrary maps

If you have an arbitrary continuous map Z -> X, there’s not an obvious way to produce a continuous map Z -> A

Only when the image of your map is already contained in A does it make sense to shrink the size of the codomain

[south kraken]

Yeah that does make sense

[quartz horizon]

In general you can shrink the codomain to the image of your map using the subspace topology

But yeah the condition is imposed to ensure there’s a two-way equivalence

Indeed this is what the subspace topology “does”

You have a collection of continuous maps into X satisfying some condition

In this case, their set-theoretic image being contained in A

And what you do is represent such maps by continuous maps into A satisfying no extra conditions

The original conditions you were imposing just get baked into the definition of the topology on A

Do you see what I mean?

[south kraken]

and the original conditions being the continuity and image containment? or am i misunderstanding

[quartz horizon]

Yes exactly

There’s no image containment condition for the maps Z -> A, only continuity

[south kraken]

right.. that does make sense i suppose

[quartz horizon]

Would you like to now see how to prove your original statement with universal properties?

[south kraken]

could i tell you how i tried to solve it first?

[quartz horizon]

Sure

[south kraken]

so my argument being that we set the space Z equal to A with the subspace top. inherited from Y, and let A have the subspace top. inherited from X - we already have the inclusion map from A to X, so we want to find a map from A to itself s.t. the inclusion composed with this map is continuous.
The natural choice falls to the identity map, so we set id: (A, T_(AY)) to (A, T_(AX). We then see the composition of the inclusion with the identity to just be the inclusion from (A, T_(AY)) to (X, T) - but here i think i might not get a continuous map?

and i'm not sure how to argue that it should be one

[quartz horizon]

Having a snack but will get back to this afterwards

[south kraken]

^^

enjoy the snack

[quartz horizon]

Ok I’m back

Lemme introduce some notation to help

Let’s say $A_1$ is the set $A$ with the subspace topology from $X$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

And $A_2$ is the set $A$ with the subspace topology from $Y$ from $X$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

As you suspected, we want to show the identity map is continuous both ways

[south kraken]

right

[quartz horizon]

Let’s start with $A_1 \to A_2$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

The identity map here is continuous if and only if the composite map $A_1 \to Y$ is continuous, by the universal property

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

Now we do know that the inclusion map $A_2 \to Y$ is continuous - however, as you saw, this doesn’t immediately imply $A_1 \to Y$ is continuous

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

I think this is where you got stuck before

[south kraken]

yeah, it is

[quartz horizon]

However, we have more we can do

The composite map $A_1 \to Y$ is continuous iff the corresponding map $A_1 \to X$ is, by using the universal property for $Y$ as a subspace of $X$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

And this time we do know the map is continuous!

[south kraken]

wait, so A_1 to X is the inclusion again right?

[quartz horizon]

Essentially, we take the inclusion map A_1 -> X, corestrict to Y, and then further corestrict to A

[quartz horizon]

Indeed

Because the composite of the inclusion A -> Y with Y -> X is just the inclusion A -> X

[south kraken]

that sounds about right

[quartz horizon]

This shows the identity map A_1 -> A_2 is continuous

Now we just need to check that for A_2 -> A_1

Using the universal property, this is iff the inclusion map $A_2 \to X$ is continuous

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

However, we can write this map as a composite $A_2 \to Y \to X$, and those individual maps are continuous by definition of the subspace topologies

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

Thus, being a composite of continuous maps, this map is continuous

And so we can corestrict to A to show A_2 -> A_1 is continuous

[south kraken]

huh

lemme just draw those diagrams rq

[quartz horizon]

Mhm, sure

[south kraken]

i realise this looks horrendous, but would it be smt like this?

or am i misunderstanding something

[quartz horizon]

Why do you have j^-1?

[south kraken]

I didn't see what other map i should've used instead - as both k and j are inclusions into X, i thought it would be natural to use the preimage of the inclusion back to Y, but that is maybe not always continuous?

[quartz horizon]

Well the preimage isn’t even a function from X to Y right

[south kraken]

yeah, i mixed up cuz i had in my head that j^{-1}(X) = A ∩ X which doesn't actually help

xpp

[quartz horizon]

It is true that you can corestrict k to Y

[south kraken]

Yeah, i just don't quite see what map i should use there instead

[quartz horizon]

Well you use the universal property for Y as a subspace of X

[quartz horizon]

.

https://en.wikipedia.org/wiki/Corestriction

One thing you get used to in cat theory is that there are a lot of operations you can do on functions that aren’t just composition

[south kraken]

ah, i haven't really heard of (or used) corestrictions in my classes

but that does make sense

[quartz horizon]

Corestriction is one such operation

And I think it’s conceptually helpful to use it

[south kraken]

oh for sure

[quartz horizon]

Incidentally this is where the “closure” axiom for groups comes from, if you know what that is

[south kraken]

in this case the composition would be k \circ j^Y?

[south kraken]

i am taking abstract algebra this semester as well yeah

[quartz horizon]

I would denote it as $k |^Y$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

k co-restricted to Y

[south kraken]

but the message is the same then

[quartz horizon]

A good example is the group of invertible matrices

You have matrix multiplication as a map $\text{Mat}(n, K) \times \text{Mat}(n, K) \to \text{Mat}(n, K)$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

You then restrict this to invertible matrices, so $\text{GL}(n, K) \times \text{GL}(n, K) \to \text{Mat}(n, K)$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

Closure is precisely what you need to co-restrict this to a binary operation on $\text{GL}(n, K)$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

Namely, you check the image of this map is contained in GL(n, K)

I.e. the product of two invertible matrices is also invertible

[south kraken]

huh, interesting

[quartz horizon]

It’s the kind of operation that’s often invisible

You use it implicitly without thinking

But in more “geometric” contexts like topology, it becomes something that requires effort to prove

What’s maybe surprising is that this works even at the level of manifolds

If you have a smooth map $M \to \mathbb{R}^3$ whose image is contained in the unit sphere, you can co-restrict it to a smooth map $M \to S^2$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

This is not at all obvious since the way you check smoothness is genuinely different for R^3 and S^2

But nonetheless this statement is true

[south kraken]

that is.. very interesting

[quartz horizon]

One thing I like about cat theory is that it makes these implicit operations explicit

[south kraken]

because you have to generalise it so much?

or?

like, the reason it becomes explicit i mean

[quartz horizon]

In part yes

I think it’s just that in general cat theory tends to be quite explicit

For example, there’s a difference between knowing that two spaces are isomorphic, and actually having a specific isomorphism between them

Which is more to do with how they’re isomorphic

Cat theory tends to care more about the latter

[south kraken]

huh, i always had the impression that it was the former

[quartz horizon]

Interesting, that’s surprising

[south kraken]

tbf i haven't had that much exposure to category theory

[quartz horizon]

Because in my experience cat theory treats things like isomorphism as data rather than a property

Not a yes/no Q

[south kraken]

curious - i do think i remember watching a video by gpp where he states something about the yoneda lemma

but i'm not so sure

[quartz horizon]

Oh yoneda is a beautiful result

It deserves to be called the fundamental theorem of cat theory

[south kraken]

i am planning on taking homological algebra next semester together with our rings and modules course - we unfortunately don't have many topology courses because of funding cuts

[quartz horizon]

Oh cool, idk much homalg myself but I’m trying to get there

[south kraken]

we have two additional ones i could take, alg. top. 1 and 2, that focuses on homology/cohomology and the other that focuses homotopy theory and chain complexes

if i understood the thing correctly

the problem being they go every other year now

and next year would be starting off with alg. top. 2

which is the homotopy one

[quartz horizon]

I’m currently trying to learn algtop myself

Maybe the folks in #alg-top-geo-top could help you out

[balmy nexus]

theyre a bit disjoint

depending on how theyre taught you might struggle through

[south kraken]

That's what i thought too - and considering I'll be taking homalg and rings and modules, it would be a bit rough i think

[warped shore]

Interesting

Maybe wrong channel but I'm curious to hear about

[quartz horizon]

Sure sure, this is what I like to call the categorical perspective on equality

[warped shore]

Something something HoTT? Or no

[quartz horizon]

It’s tangentially related to HoTT but more elementary than that

The example I always like to use is counting

2 + 7 is the same as 7 + 2, they both equal 9

But to a small child they’re not the same

Not cause they’re bad at counting, but because they’re genuinely different processes

For 2 + 7 you do 2, 3, 4, 5, 6, 7, 8, 9

For 7 + 2 you do 7, 8, 9

It’s the same result, but two different ways of getting to it

For example, it’s a lot easier for them to do 99 + 2 than 2 + 99

[warped shore]

So commutativity is the property? What's the data

[quartz horizon]

Oh don’t think in terms of property and data for now, that comes later

My point is that there’s a sense in which 2 + 7 is equal to 7 + 2, but there’s also a sense in which they’re different

[warped shore]

Data as in the "path" to equality?

The proof?

[quartz horizon]

.

[warped shore]

Oops

[quartz horizon]

Please listen to what I’m saying

[quartz horizon]

And it’s the sense in which they’re different that actually gives the sense in which they’re the same meaning

It’s why saying “a + b = b + a” is something you have to prove in Peano arithmetic, and a nontrivial property of addition

Otherwise you’d just be saying something like “x=x”

Which, while true, isn’t especially helpful

The categorical perspective on equality, then, recognises that equality isn’t necessarily a yes/no Q

Instead, there often multiple senses of equality you can use - things might be equal in one sense, and distinct in another

So it’s quite helpful to specify how two things are equal, in what sense you’re considering them equal, rather than just saying they are or aren’t equal

And so equality gets upgraded to “data” rather than just a property

[tranquil cosmos]

like two fundamental groups on the same space but at different points!

[quartz horizon]

Mhm

[quartz horizon]

To answer this, HoTT often considers many different “levels” of equality

And indeed the data of an equality is given by a path

[quartz horizon]

This could be, for example, a specified isomorphism between $A \coprod B$ and $B \coprod A$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[warped shore]

Yeah

[quartz horizon]

Yes, essentially - the data specifies how two things are equal

[tranquil cosmos]

you can build spheres with this

[quartz horizon]

As another example, it’s a generally agreed upon human right that all humans are created equal

But of course that’s not literally true, some people have different needs than others

So in one sense people are equal, and in another people are distinct, and that isn’t a contradiction

The senses just become more or less relevant in different contexts

@warped shore does that make sense?

[warped shore]

Yeah I got the concept in concrete contexts but didn't consider it categorically

[tranquil cosmos]

\begin{tikzcd}
\bullet \arrow[r, bend left] \arrow[r, bend right] &\bullet
\end{tikzcd}

[gentle osprey]

PKThoron

[quartz horizon]

fr

[warped shore]

Property vs data is like cobordant vs cobordism, diffeomorphic vs diffeomorphism, etc respectively

[tranquil cosmos]

countable vs bijection to N

the second is lowk so much better because it spares you a lot of choice

[quartz horizon]

This is like a piece of missing syntax in ordinary mathematics

$\exists$ by default is the property version

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

We don’t really have a syntax for “exists with a specific witness”

In type theory you have the sigma type for this

[quartz horizon]

This is part of why the “axiom of choice” is a “theorem” in type theory

It’s because by default the sigma types are “exists with a witness”, for which choice is easy to prove

On the other hand the axiom of choice refers to “exists without a specified witness”

[brazen condor]

This argument feels more pertaining to type theory than category theory, however.

I suppose if i interpret the connection well enough is that the categorical structure lets us take multiple paths to the same point (commutative diagram) and that structure of each of those independent paths can be thought of as a different choice of computation to be considered technically different, and type theory gives the higher structure to encode such a notion.

[warped shore]

yup

[warped shore]

wdym theorem

iirc it's still a blackbox

[quartz horizon]

It’s called the type-theoretic principle of choice

[warped shore]

Choice on what

types?

in zfc the aoc is smth like, given any family of sets, you can pick elements from each set

idk what the type-theoretic analogue is besides encoding exactly that

[pallid swan]

well we can use a second existential clause for that

∃x, ∃y, y witnesses φ(x)

[brazen condor]

i think the argument is more so a reframing of the systems in question negate the need for axiom of choice.

for instance, this is point-set relevant: locales (a lattice with the same sort of intersection and union properties as a topology except there is no underlying set), the tychnoff theorem holds as a theorem without axiom of choice. in regular point-set topology, tychnoff theorem requires axiom of choice. the 'axiom of choice' comes up when it comes to showing that every locale can be given as a topology on some set.

the 'axiom of choice' comes in when things have to be 'sets'

[pallid swan]

i.e. φ(x, y)

alternatively you can "skolemize"

for every existential formula you can replace it by a witness-providing operation

for arity 0 these are called henkin constants i think

[warped shore]

i see

but then as soon as sets come into the picture, as in point-set topology...you need aoc?

[pallid swan]

actually yeah i think skolemization is automatic with axiom of choice but is not without it

so i think the non-type theory way to view this would be "the axiom of choice lets you skolemize your existential statements"

[warped shore]

skolemizes the statements where you use aoc ig

feels a little circular idk

[brazen condor]

mathematical logic is a spiral, not a circle

ie for instance

the 'axiom of choice' in ZFC would as stated just be some syntax but you use axiom of choice on the outside to do things with that syntax

i think axiom of choice is used in the completeness theorem.

yeah it is

[pallid swan]

im probably not saying things in a good way pedagogically tbh, sorry if this is true

[brazen condor]

id roughly view it that way yeah

[warped shore]

😬

[pallid swan]

i will write up an explicit example of what i mean

consider the following two statements:

• ∀n, ∃f, f is a function that embeds the set {0, 1, 2, ..., n} into X
• ∃f, f is an injective function that sends any non-negative integer n to X

AoC is required to prove these are equivalent conditions on X

i think skolemization might have been a bad concept to bring up cuz i think you can still skolemize the former bullet point, you just can't do it uniformly

sorry about that

[warped shore]

lwk i cant tell the difference between the statements

:(

[pallid swan]

yeah they're equivalent with choice

[warped shore]

how come, whats the justification using aoc

[pallid swan]

to construct the latter function, we need to make a series of arbitrary choices, one for each n ∈ ω = {0, 1, 2, 3, ...}

the former bullet point guarantees we won't run out of elements to choose at each "step"

an explicit example of how this can fail without choice is the "socks set", consider a countable family of indistinguishable socks. if you could embed ω into it, then you could construct a choice function on the set of pairs of socks in the image, which must be infinite

(sorry for derailing the category theory discussion btw... i didn't mean to do that, hopefully this is a fun side tangent at least)

[urban zinc]

(a weaker version, fwiw)

[warped shore]

oh oh

[brazen condor]

My argument is going to sound more elementary but the sense in which I think categories is opposed to structures and not so concerned with the information in other ways.

On a conceptual, I feel like math structures come first, then the suitable morphisms and extra structure. Category theory wants you to 'fix' the structure and the morphisms apriori, which to me does not match some of the fluidity of certain areas of math. For instance, if you are in a nice land of Hausdorff spaces, adding just a bit of structure can completely mess up our categorical structure: product of Hausdorff spaces are Hausdorff, product of normal spaces aren't (same with subspaces).

When things are stated categorically, say colimits and limits, it is of a 'there exists this object with these morphisms and these properties' but it does not tell us either the existence of said object or what it will look like. (existence can be given by existence of (co)products and (co)equalizers) so that's neat) and the coproduct of specific categories tend to have different constructions. The coproduct of sets is a disjoint union, of abelian groups it is finite formal sums, of gropus it is the free product, of topological spaces is disjoint union, of based spaces is their wedge sum. and say you can have vector spaces, coproduct and product are the same: add normed structure, product completely goes away. all of these morphism specific properties are highly contingent on the finer s\tructure in question. (category of smooth manifolds and topological spaces in particular are badly behaved)

[pallid swan]

i think overall i like that category theory encourages being principled about what data you care about, so it can go both ways, i suppose

you can go super heavy type theory pill and record every single way in which 3 + 7 is isomorphic to 7 + 3

or you can just regard it all as the same

[brazen condor]

And in some cases the universal property you want for something doesn't even make sense (because it is 'mixing' morphisms between two different kinds of objects via a set theoretic map). For instance this universal property characterizing free groups on n generators (im sure you can rework this in the proper language but my point is not all universal properties are thought this way).

[pallid swan]

to add onto pseudo's point though, natural transformations often feel like "you need to provide an explicit witness that makes sense as part of your data" to me

[pallid swan]

yeah, i suppose it motivates the idea behind a forgetful functor, cuz it often just goes unwritten otherwise

i.e. you can't have a map from a set to a group but you can from a set to forget(a group)

on the other hand yea you're right some properties are just flat-out versal, i.e. not universal, there are many distinct witnesses, i.e. algebraic closure of fields

at which point it feels like it's not a category theory construct anymore

[brazen condor]

actually i kind of notice this with the concept of natural isomorphism that is often imagined as between classes of objects (like there is a natural isomorphism from V + W to W+V (+ is direct sum it won't copy pasta), but the natural transformation language would require you to imagine a functor from one category to another so one would take the identity category and then make this correspondence)

Though in language like say algebraic topology where one is inherently working on some form of Top and mapping into Ab through homology, it makes perfect sense to call two groups in Ab from homology naturally isomorphic in the normal natural isomorphism sense between functors (interpreting the homology on a space as a functor on that space).

idk i just feel in most contexts categories are to the aid of some mathematical structure and not the fundamental starting point

I do think though in certain areas it is quite important to find a 'good category' in which you can do things

because when your methods are benefited categorically category theory is a great aid and I think this is a huge crux of certain areas of algebraic topology (I don't know much about this area)

[novel acorn]

well this is how it started but these days there are many, many constructions that just aren't possible without categorical reasoning

[brazen condor]

I agree

eg Grothendieck topology
Though I'm sure there are more examples in algebraic topology if i knew more

[pallid swan]

true

[unreal stratus]

Yes I would say this is sometimes an underrated thing and becomes particularly important when you have homotopy theoretic things or higher categories

[south kraken]

i am confusing myself again, trying to show that a certain quotient space is equal to the digital line topology on the integers

So essentially i have $$(\mathbb{R}, \mathcal{T}_{std})$$, and then a map $$\pi : \mathbb{R} \to \mathbb{Z}$$ such that $$ \pi(x) = {x\text{ if } x \text{ is an integer, or n if } x \text{ is in } (n-1, n+1) \text{ and n is an odd integer }}$$

I've found the quotient topology, but i am very unsure of how to peocede to show they are equivalent

is it possible maybe to just show double set containment?

upe

[gentle osprey]

pax_ignis

[rancid umbra]

what is the digital line topology?

[south kraken]

it's defined by the basis $$B(n) = { {n} \text{ if n is an even integer, and } {n-1, n, n+1} \text{ if n is odd } } $$

[pallid swan]

double set containment on the family of opens?

[gentle osprey]

pax_ignis

[south kraken]

and n in the integers

[pallid swan]

this feel like the convention is flipped from the quotient map you gave earlier

[south kraken]

it might be

[pallid swan]

doesn't matter ultimately but

[south kraken]

yeah it was

singletons are odds

i forgor

[pallid swan]

yeah i think a good strat would be like, to prove an open of one is an open of the other and vice versa

if that's what your saying

or er maybe show they have the same basis or something

[south kraken]

the task specifically says to show that the quotient topology on Z induced by $\pi$ is equal to the digital line topology on Z

[gentle osprey]

pax_ignis

[south kraken]

yeah

[pallid swan]

it's definitely true as written

[south kraken]

cuz we have the basis of both now i assume

[south kraken]

yep, so i just need to figure out how to approach this - it would be weird not to have anything with the universal property, but i guess that comes in a later chapter

so double set containment for the basis was it?

[south kraken]

just to say, they have introduced the universal property

it's just not used in any of the tasks yet

[pallid swan]

it's also homeomorphic to the topology that AT people on graphs, on the bi-infinite line graph

idk if that's helpful for the exercise but it's how i visualize it

[pallid swan]

yea should work

idk im spitballing

[south kraken]

it's not as helpful when those things haven't been introduced to us yet

[pallid swan]

ah, are you aware of what im alluding to

[south kraken]

not at all

[pallid swan]

ah alr wasn't sure based on your phrasing

[south kraken]

yeah fair

[pallid swan]

oh yea the uniprop is how i would prove it

at least think about it

nvm

[south kraken]

i am just uncertain about how i could use it

[pallid swan]

i take it back

[south kraken]

at least one way

oh?

[pallid swan]

i just usually think about quotient topologies by like

the basis is given by sets whose preimage is open

i guess that is the uniprop in a sense

just more concrete

[south kraken]

well, i guess

[pallid swan]

wait nvm

[south kraken]

i have the basis though

and i know the basis for the digital line top

[pallid swan]

ok fixed

this is gonna sound kinda silly but i also think of it as open sets which dont change when you take their image and then preimage

like i just imagine open sets that are closed under expanding them to fill the entire fibers

idk if that's helpful but that's my excuse for why i fumbled my words a little, i conflated the perspectives a little

[south kraken]

and the fibres would be the preimages?
it's not a term I'm used to so..

but okai - so I'll have to do double set containment

mm

[pallid swan]

yea

it's like the fancy word

[south kraken]

i know you could probably use the uniprop somehow, and just slapping the identity on there and show it's continuous

[south kraken]

ah

fancyness

[pallid swan]

well fiber has a connotation of preimage of singletons

i sorta think of it as a huge partition of the entire space into these subsets

[south kraken]

right 👀

[pallid swan]

quotient maps

that is

[south kraken]

all the glue going to waste xpp

[pallid swan]

ohhh i see what you mean

yeah could work

[pallid swan]

i eated it

[south kraken]

I'm just uncertain how i can show the identity is continuous - because you got $\pi \circ Id$ right

[gentle osprey]

pax_ignis

[pallid swan]

ultimately both proof methods will be essentially equivalent

[south kraken]

but as of now you don't know if that is even continuous

[pallid swan]

once you get to the nitty gritty

[south kraken]

fsir point

[pallid swan]

like to prove identity is continuous, is to do an inclusion argument on the basis

identity function on the underlying sets i assume

[south kraken]

so i have to do the thing I'm trying to prove to show the inclusion is continuous

[pallid swan]

yea

[south kraken]

mm

that do make sense

[south kraken]

i mean, the basis elements for each n are equal, i just need to find an argument that isn't super wishy washy to show that the identity takes each n to itself

in the different topologies

cuz essentially i can claim for both topologies that for n even, the basis elements is of the form blabla, and for n odd, blabla

right

[pallid swan]

yeah

[south kraken]

bjt that isn't yet strong enough to say it takes n to n

or

[pallid swan]

tbh I'd draft something and then try to rewrite it to be cleaner

[south kraken]

i mean i guess

my problem is just that atm i don't really see a way to show equivalence

[rancid umbra]

is this supposed to be read \pi(x) = x if x is an integer, and \pi(x) = n if x is in (n - 1, n + 1) for an odd even integer n?

[south kraken]

yes, but swap odd to even

i did a fuckup

[rancid umbra]

yea, the easiest, most straightforward way is to show that the two topologies agree.

using quotient maps and universal properties is detracting

[hollow pawn]

for (a), i was able to show A had a single point, but idk how to show that point is the unique fixed point

[kind marlin]

A is a subset of each A_n, so f(A) is a subset of f(A_n). f(A_n) is a subset of A_n since we are taking images of smaller and smaller subsets, so f(A) is a subset of A_n, for every n. that means f(A) is a subset of the intersection of all A_n, which is just A. since A has one point and f(A) has one point, f(A) = A

[hollow pawn]

i see, thank you!

[kind marlin]

actually a more clean way to write this is \

$f(A) = f(\cap_n A_n) \subseteq \cap_n f(A_n) \subseteq \cap_n A_n = A$

[gentle osprey]

snowflake

[warped helm]

if theres another fixed point q, then d(p,q) = d(f(p), f(q)) \leq a d(p,q)

conclude:

[hollow pawn]

got it, thank you

[dense laurel]

i wonder how various schools' point-set "topology" courses compare: what texts are used, what topics are covered, etc.

in particular, do many students cover metric spaces as part of their topology courses? is this the first introduction to metric spaces?

at my previous school, studied metric spaces as part of "analysis" courses first and didn't say much about metrization during topology, which used Armstrong. currently, at a different school, we use Munkres and I think it's many students' first exposure to metric spaces.

[cosmic mirage]

my undergrad did metric spaces in analysis and a tiny bit in topology. We used munkres as well

[storm wadi]

yeah we did metric spaces in analysis, before topology.

[frozen sparrow]

Here, you can do both the Metric Spaces course and the Point-Set Topologyat the same time as both of their requirements is a course on Real/Multidimensional Analysis. I myself didn't do that as I thought the Topology course makes more sense after the Metric Space one but you do mention what a topology is, that metric induces a topology, and some other basic topological notions in the Metric Space course.

The book we used for Metric Spaces is from a former professor in the uni, but it's probably used sparingly. The Topology book is Munkres though.

[south kraken]

here we had metric spaces baked into the topology course

so we started with metric spaces before moving on to the topological spaces

[urban zinc]

Yea the topology course at my uni does metric spaces first too, seems more logical that way

[south kraken]

oh for sure

[tiny obsidian]

I had a separate metric spaces course followed by a topology one

[frozen sparrow]

Our Metric Space course is compulsory bot the Topology course is an elective.

[rancid umbra]

i never had a proper point-set course, it was baked into the two analysis courses i took, and then used throughout the rest of the math curriculum. later i ended up taking differential topology, but this assumed knowledge of point-set

[south kraken]

diff top was removed from our course plan because money issues - instead they merged lie algebra, diff top and some other course to make differential geometry

iirc

[kind marlin]

in my topology class we used munkres and covered metric spaces

[brazen condor]

This problem comes from May's algebraic topology book. I want someone to check for any oversight or see if they know a faster way to do this (though maybe spoil it since it do want to figure this out if something is wrong)

Let a space $X$ be called weakly Hausdorff if for every compact Hausdorff spaces $K$ and map $g:K \rightarrow X$, $g(K)$ is closed in $X$.

The goal is to show that $g(K)$ is Hausdorff.

Proof:

let $g(x)$ and $g(y)$ be two points in $g(K)$. Evidently $y$ is compact hausdorff so by the condition, $g(y)$ is closed in $X$. Then there exists an open set $U$ of $g(x)$ that does not contain $g(y)$. The issue is $g(y)$ may lie in the boundary of $U$.

First, let $h: K \rightarrow g(K)$ be given by $h(x)=g(x)$. First note that $h$ is an open map. To see this, let $A$ be open in $K$. Then $K-A$ is closed in $K$, hence compact Hausdorff, so $g(K-A)=g(K)-g(A)$ is closed in $g(K)$, hence $g(A)$ is open in $g(K)$.

Since $h$ is surjective and open, it is closed.

Since $K$ is regular, there exists an open set $V$ so that $x \in V$ and $\overline{V} \subset g^{-1}(U)$. Then, $h(\overline{V}) \subset U$. Since $h$ is closed, $h(\overline{V})=\overline{h(V)}$, so $\overline{h(V)} \subset U$. Note that from above, $h(V)$ is open and thus an open set separating $g(x)$ from $g(y)$

From above $g(K)-\overline{h(V)}$ is open and contains $y$ (since y is not in U), so there exists an open set $W$ in $g(K)$ so that $h(V) \cap W$ is disjoint, and $W$ contains $g(y)$.

This completes the proof.

One sec

Im typing this on phone thats a bad idea

[random verge]

Can you clarify what you’re ahowing

Showing*

I just see a definition I think

[brazen condor]

Edited

[random verge]

Okay here are my hints, it can be simpler. Let me check to see if I know how to write a spoiler: ||test||

[midnight umbra]

im going to be upfront and say that im feeling too tired to be fully helpful, but id like to note that you cant distribute $g$ over set differences unless you know $g$ is injective

[gentle osprey]

Sarah!

Convergant

[midnight umbra]

for example, $g(x)=|x|$, $A={-1,1}$, $B={1}$. then, $g(A\setminus B)=g({-1})={1}$, but $g(A)\setminus g(B)={1}\setminus {1}=\varnothing$

[gentle osprey]

Convergant

[random verge]

Two facts, that can be proven, but are well known can be used here to keep things cleaner. 1. ||the image of a compact space under a continuous map is compact|| and 2. || a compact subspace of a Hausdorff space is closed||. Finally ||a closed subspace of a Hausdorff space is Hausdorff||. Granted it’s been a while since I have thought about this. So someone please correct me if it’s irrelevant

[crisp lintel]

That doesn't really help because the image isn't Hausdorff

or the codomain rather

[midnight umbra]

we arent even given that g is cts

[crisp lintel]

I assume g is cts otherwise it's obviously false

[brazen condor]

g is continuous

[midnight umbra]

i see

[random verge]

Yeah I think that’s the main issue here, X isn’t assumed to be Hausdorff it’s weak. My mistake

[brazen condor]

Yeah that was dumb of me

Well hang on because

Gah I see the problem

I think I have a fixed but ill wait until I can write it down

Some nonsense with considering all pre-made points seems essential as I originally thought

Oh I have ab idea

[gentle osprey]

Sarah!

[queen prism]

til all functions commute with unions

[brazen condor]

You could give me 10 years of math training and im never gonna remember the rules

I do think I have a fix tho

[rancid umbra]

like f(A U B) = f(A) U f(B) and f^{-1}(A U B) = f^{-1}(A) U f^{-1}(B)?

[queen prism]

I just know f^-1 commutes with everything

[rancid umbra]

yea, preimages commute with everything, the images don't generally commute with intersections

that is the only bad one

and i only remember that one

lol all the others just work as you expect

[queen prism]

images also don't commute with set differences though this is (probably) a corollary of not commuting with intersections

[rancid umbra]

yea

[quasi forum]

you can write every set difference as an intersection so yeah

[rancid umbra]

A - B = A n B^c

[brazen condor]

I looked up the answer, it does use normality of K theres just one trick on top of that.

The chances of solving a problem as you fixate more and more on it diminishes woth time...

[rancid umbra]

i did not know that compact Hausdroff implies normal

but points are closed if X is weakly Hausdorff

so Hausdorff follows from normality

neat

[brazen condor]

Its a good proof to try and work out

[brazen condor]

Not quite

[brazen condor]

In general two really useful facts to remember are that

Closed subsets of a compact space are compact
Compact subsets of a Hausdorff spaces are closed

One key result the is that if f:X->Y is a bijective map from a compact space X into a Hausdorff space Y, then for is in fact a homeomorphism

[quasi forum]

the proof of this fact isnt super hard but its a nice exercise in applying compactness

[rancid umbra]

yes, this is a fun proof after working it out

[rancid umbra]

sorry, i was getting things confused here. yes, there is one more trick that you have to employ using the facts you mentioned.

[rancid umbra]

i don't know why i mentioned X...

[brazen condor]

One result I showed (that ended up not used in this part of the proof... I think) is that such a map is a quotient map onto its image.

[rancid umbra]

you do have to use that points are closed in X for X weakly Hausdorff tho.
so like, if you take g(x) and g(y) in g(K), then points are closed in g(K) since g(K) is closed. now the fibers A and B of g(x) and g(y) are closed. by normality, you get disjoint opens U and V around A and B. K - U is compact Hausdorff since it is closed, and so its image is also closed in g(K), likewise for K - V. g(x) is in g(K) - g(K - U) and g(y) is in g(K) - g(K - V), and these are disjoint opens.

[rancid umbra]

yea, i had proved this before you mentioned the normality hint, but i couldn't find a good spot to use it

[plush salmon]

is it better to write the separation axiom satisfied by a space as "T_n" or refer to it by the name (Kolmogoroff, Frechet, Hausdorf, etc.)

[brazen condor]

it probably depends from book to book but I think by name is preferred like normal, regular, and Hausdorff (also you will never see T_2 it will always be called Hausdorff).

[plush salmon]

ty!

[tiny obsidian]

Some people do call it T2, "never" is far too strong, althpugh I agree there is a very strong preference for 'Hausdorff'. Also some authors do not treat the name and Tn for n > 2 as synonyms (e.g. I learned T3 as being T2 + regular, which is not the same as just regular), and think T0 and T1 are more common than whatever the names for them are

[iron bolt]

Mathlib calls it T2Space unfortunately 🥲 but I agree that "Hausdorff space" is a lot more widely used and probably better

T1 spaces I would rather call T1 than Frechet though, because Frechet space already refers to first-countable complete Hausdorff locally convex vector spaces

[silent garnet]

why is R special topologically?? Almost every topological space (manifolds, CW complexes, Spheres, Spectra) and operation/object (Suspension, join, homotopy, all the algebraic invariants) is defined using R in some way.
but all the definitions of R i know (complete ordered field, sequences of rational numbers) dont provide intuition about why R would be so useful in topology.

[warped helm]

it's THE continuum

[silent garnet]

i mean R with euclidean topology

as a set almost every space is a continuum ahahah

[opaque scroll]

It's 1d locally compact Hausdorff and contractible.

Those are all nice topological things

But I think your question is sort of the wrong way. People started out studying R and things built from R, then defined topology

Not the other way around

[warped helm]

can you elaborate on this?

[silent garnet]

but if you do go the other way around, you will end up with a lot of nice things being related to R

[warped helm]

i'm asking because i don't really see a natural way to consider Q a continuum for example

wdym? for example a manifold is defined using R^n

[opaque scroll]

I mean R is not a continuum in the topology sense.

The continuum is also the name of the cardinality of R, but not super relevant

[silent garnet]

yeah Q isnt a continuum.
but cardinality of things like manifolds is continuum as far as i know

[warped helm]

what does continuum in the topology sense refer to? i learned from munkres that R is considered a "linear continuum"

[iron bolt]

R is kind of the field that all of physics happens over, at least at first glance. C becomes very relevant later too, but everything you yourself can measure? that's rational/real

[opaque scroll]

I'm not sure what you're saying.

If you build things out of R then they will be related to R, sure...

[opaque scroll]

Compact connected metric space.

"Linear continuum" would be the order theory thing.

[warped helm]

ah ok

why is compactness demanded?

[iron bolt]

I mean, you can't measure things to infinite precision so you could argue that everything you measure is rational, but the real numbers are connected, the rationals are not, and that makes them a lot more useful for e.g. modelling the kind of continuity you encounter in physics