point-set-topology

[alpine nest]

In fact it can be taken as a definition of connectedness, that the only clopen sets are the two trivial ones (emptyset and whole space)

[warm kettle]

So the answer seems to be that it’s an arbitrary connected component, right?

[wise mist]

i kinda feel that those are called regular open for now

[alpine nest]

Any clopen set, it doesn't have to be a connected component.

On a clopen set the interior and closure operators don't really do anything.

I'm not sure offhand if clopen sets are the only ones which satisfy this #point-set-topology message, but they do satisfy it.

[wise mist]

but for infinite intersection this is going to be singleton so we are gonna loose the openness

is this fine ?

[alpine nest]

What is a regular open set?

[alpine nest]

Ah OK, I haven't heard that term before, but fair enough.

[warm kettle]

The set of integers in the Euclidean topology would also work, I imagine

[wise mist]

cl(int(A)) just becomes A

[wise mist]

me neither, this problem set talks about it

[alpine nest]

It doesn't, it has empty interior.

Unless you consider that set as a space in its own right, in which case it's a discrete space

[warm kettle]

Yes, and it’s closed, so we get the empty set on either side of the equation. But I might be wrong

[alpine nest]

Ah yes, true. The interior of its closure is also empty

[alpine nest]

Why do you say that the infinite intersection is going to be a singleton? Does it follow from something that was established earlier?

[warm kettle]

Nowhere dense sets should work then. That’s a very interesting question 👏

[wise mist]

oh that's contextual to the problem

sorry, i tried to produce that

not in general, wording has been all over the place

[alpine nest]

Just as a warning, you aren't going to get a singleton as intersection of countably many open dense sets in R

As the problem states, such an intersection has to be dense, which a singleton isn't

You can lose the openness, but you won't lose the denseness

[wise mist]

alr i will give it another try

[alpine nest]

Also note that it's asking you for an example, i.e. for a specific sequence of open dense sets, whose intersection isn't open. You can't prove in general that the intersection isn't open, because it might be (for example if all the sets are the same)

The problem just asks you to show that it doesn't have to be open

[wise mist]

(a_i, b_j)
then take a seq of ai and increseringly converge it to a

[alpine nest]

Those intervals aren't dense in R

[wise mist]

oh right, how dumb of me

(a,b) in [a,b]

with R usual

[alpine nest]

You're supposed to find open dense subsets of R with the usual topology

They need to be dense in R

That's what "Open dense subset of <space>" means

[wise mist]

then keep deleting one points and they stil are dense

[alpine nest]

Its closure has to be the whole space

[wise mist]

yeah delete all the Qs eventually

desnse but not open anymore

[alpine nest]

Yep, that will work

[wise mist]

or we can say whole R then deleting the Qs

alr, did u have anything else in mind to offer ?

pls do so

need more wankey based arsenal

[alpine nest]

I mean, that's fine, enumerate the rationals as r_1,r_2,..., and let A_n = R without {r_1,..r_n}

They're open dense, and their intersection is the irrationals, which aren't open

In general open dense sets are fairly weird and unintuitive (unless they're complements of countable sets like in this case)

For example the complement of the Cantor set is open dense

[wise mist]

i am thiking if it has some relation with no where dense while taking the compliment or not

[alpine nest]

Another example of open dense sets also starts with rationals: let $\bQ = r_1,r_2,\ldots$, and let $A = \bigcup_{n=1}^\infty (r_n-\frac{1}{2^n},r_n+\frac{1}{2^n})$. That set is clearly open, it's also dense in $\bR$ (because it contains all the rationals), but it's not all of $\bR$, because the total length of those intervals is $1$.

[gentle osprey]

Outsider

[alpine nest]

So its complement is a (fairly huge) closed nowhere dense set.

[wise mist]

right!, i might not be digesting all of this now, kinda exhausted, but i will go through all of this again

[alpine nest]

Correction, the total length of those intervals is 2, but either way, by replacing the 1 in "1/2^n" by something smaller, you can ge the total length to be as small as you want.

[alpine nest]

And yes, open dense sets are closely related to nowhere dense sets, since there's a sort of duality between them via taking complements.

Most theorems about open dense sets have equivalent/dual forms that talk about nowhere dense sets

[wise mist]

i think i have seen similar wordings on my screen today

[alpine nest]

I expect that's what your course is gearing towards, the very excellent Baire's theorem.

[wise mist]

im not in any school

on my own

[alpine nest]

Well, whatever material you're studying

[wise mist]

@prime elbow gave me this

so im solving(trying to) this carefully

[wise mist]

did u mean baire category theorem, i think i had seen it in the past when i was doing some metric flavoured topology, but i dont remember anything

[alpine nest]

It's problem 20 in the same sectio nof that PDF (although only in the version for R, then there's the fully general version in section 12)

[wise mist]

boyz, tomorrow will be a rough day

[elfin fable]

I'm reading something that says that the projection function of the product topology is continuous, open and surjective

But doesn't continuity and openness imply homeomorphisms? Why would they word it that way?

What do you mean?

But doesn't continuity and opennes → homeomorphism?

Oh it has to be a bijective application

Okay you're right

Thanks!

Yeah true, quite limited

Yeah that also makes the surjection clear

[prime elbow]

I think there is misprint in 3.30, I think they meant if E is closed then d(x,E) = 0 implies x in E.

[plush folio]

Agree, x in E implies d(x, E) = 0 is true for any subset E

[hasty goblet]

My Prof gave us some topology questions, and I have some doubts about one of the questions. Can I ask it here (no one is responding on my help channel), or is that not allowed?

[plush folio]

Yes, just ask

[hasty goblet]

so I'm not sure I understand the questions
for (a) what does it mean to identify X with I^I?
for (b) isn't that just the definition of pointwise convergence?
and for (c) is it false because the Ascoli theorem requires equicontinuity?

If it means X is equivalent to I^I then the statement is true by tychonoff's theorem right?

[hasty goblet]

for part (a)

[opaque scroll]

Identify X with I^I just means we write X instead of I^I, because as sets they're the same thing.

(b) That is the definition of point-wise convergence yes, so then the exercises reduces to showing that convergence in the product topology is equivalent to point-wise convergence.

[hasty goblet]

okay

thank you

[sharp panther]

what is an example of connected and compact manifold with Euler characteristic of 3?

[gritty widget]

Trying to solve 3 (self-study). I'm stuck on the reverse direction - it's clear that $d(A, B) \leq \epsilon$ and that strict inequality holds if $A$ and $B$ are compact by EVT, but how does one go about it in general? (A solution online just says "because $A$ is closed," so I'm assuming I'm missing something fairly simple...)

[gentle osprey]

Shovel

[opaque scroll]

So if P is the projective plane and nP is the n-fold connected sum of P with itself, then nP has euler characteristic 2-n.

So X = 3P has characteristic -1 and Y = 5P has characteristic -3. Hence XxY has characteristic 3. And XxY is a closed 4-manifold

[sharp panther]

Great, thank you

[sharp panther]

Let X be a topological space and A, B second countable subspaces such that X = A \cup B. Is X second countable?

I am trying to answe the same question for locally connectedness as well

[iron bolt]

I don't know about second-countability, but ||for first-countability and local connectedness the answer is no||

||for counterexamples B can be something as simple as a single point||

[prime elbow]

I can say every singleton set is open therefore it is discrete topological space, and then any mapping from X to Y is continuous

[wise mist]

im not getting this wording, where is Y

[prime elbow]

In general

[tender halo]

No

take the quotient R / N

its not even first countable despite both R \ N and the point N collapsed to both being second countable

local connectedness is easy just take a finite discrete spacd

[rancid umbra]

yes

[small obsidian]

A and B each have a countable basis. Given a countable basis for each, can you construct a basis for A U B? Is there a way to construct a countable basis always?

[warm kettle]

Let's say I have two homeomorphic spaces X and Y. If I define two equivalence relations on them—A and B respectively—and my homeomorphism f preserves equivalence classes of those relations, i.e. x A y <=> f(x) B f(y), then is it true (and if so, clear?) that f induces a homeomorphism between X/A and Y/B?

[frigid sapphire]

what exaclty does $\omega\cup\aleph_1$ mean? i'm reading a book on model theory and it defined the power of a language by $\omega\cup|\mathscr{L}|$ but i havent really studied set theory and dont know what union means here with a cardinal number like that

[gentle osprey]

Emma

[opaque scroll]

Well, cardinals are just specific kinds of ordinals, and the ordinals all contain eachother.

So it could just be a funny way if saying the max of omega and |L|

[frigid sapphire]

that seems to be the case

[ebon flower]

bruh was max(omega, |L|) not good enough

[opaque scroll]

Gotta bring down that symbol count somehow

[prime elbow]

If x_n is a cauchy sequence such that it is not convergent in metric space X, then { x_n | n in N } is a closed set, right?

I am not sure but if it is not closed then there exists x in X\{x_n | n in N } such that there is a sequence in {x_n | n in N } which converges to x, say y_n converges to x.

Then maybe we can find the subsequence of y_n which is a subsequence of x_n then it contradict x_n is not convergent.

Is it correct?

Can I say if every open ball of X is complete then X is complete?

Let x_n be a cauchy sequence in X then take eps = 1 there exists N such that d(x_n, x_m ) < 1 for all n,m ≥ N, thus x_n in B(x_N, 1) for all n≥N.

Since every open ball is complete therefore x_n is convergent in B(x_N, 1) implies x_n is convergent in X.

Correct?

[prime elbow]

Got it

[iron bolt]

a sequence can have convergent subsequences without being convergent

example: ||x_n := (-1)^n in R||

in that example {x_n | n in N} is still closed, but it doesn't have to be

[prime elbow]

but it is not cauchy

[iron bolt]

oh, right. my bad

I missed that part

[boreal heath]

hi i need help! is this statement true? im stuck.

let x_n —> x (unique).
let S={x_n | n is in |N}U{x}.
is S closed?

here X is just a topological space

[finite token]

On the set {1,2,3}, the topology {{}, {1,2}, {3}, {1,2,3}} the sequence 2,3,3,3,... converges uniquely to 3 but {2,3} is not closed

[boreal heath]

oh wow thank you! now that i think about it, are we always able to extract a closed set from the image of a sequence union its limit? in the example you gave, if we take the subsequence 3,3,3,... {3} is a closed set

[opaque scroll]

If you add the condition Hausdorff, then this is true

[boreal heath]

i have a strong feeling that first countability implies it as well, just struggling to write down a proof

this is because you can characterize closed sets once you have first countability

[opaque scroll]

The example {1,2,3} is first countable

[opaque scroll]

This isn't necessarily true.

Take [0, 1] with the topology whose open sets are (x, 1]. Then xn = 1/n converges uniquely to 0, but no subset is closed.

[boreal heath]

im doing a lot of translating, but when i say first countability, i mean a countable basis of neighborhoods, are you familiar?

[opaque scroll]

Yes

[opaque scroll]

This example is also first countable

(even second countable)

[boreal heath]

okay, thank you so much, you're being very generous with your time, i still have one more question though.

i found this over at math stack exchange, and i am aware that there's a mistake in the first few lines but i don't think it matters, as we only care about if a sequence converges then it does so inside the set. my question is, where's the mistake?

[iron bolt]

"whenever a sequence in A has a limit, that limit also lies in A" is not equivalent to A being closed

it's a condition called sequentially closed instead, because, well, it's a bit like closed but about sequences

[boreal heath]

when we assume first countability it is

[iron bolt]

ah, yup

[boreal heath]

yeah

[iron bolt]

I don't really understand the proof in the screenshot - do you have a link to the full context?

[silver umbra]

given an open subset U of R^n, is it possible to cover U with countably many compact sets such that the interiors of the compact sets form an open cover of U?

[boreal heath]

i do, but the original question assumes a metric space, but that specific response (as far as i'm aware) doesn't it only uses the result from first countability, here:

https://math.stackexchange.com/questions/125299/prove-that-a-set-consisting-of-a-sequence-and-its-limit-point-is-closed

[iron bolt]

I see. I didn't know what exactly y_n was, that's why I got confused

the subtlety you missed is that this proof also uses uniqueness of limits

[boreal heath]

which i have assumed right?

[iron bolt]

I think it would need uniqueness of limits for all sequences in {x_n} ∪ {x}

because even something like a constant sequence y_k := x_0 could converge to something other than x_0 itself

[silver umbra]

ok wait in that case

i'm getting a very confusing contradiction

in a problem im working on

can i ask u abt it in the alg-top channel

[boreal heath]

but would that really matter that much? we only want the limits (however many) to stay inside S?

[iron bolt]

I mean, take this example

the constant sequence (2) converges to 1

that's why {2,3} isn't closed - its closure also contains 1

[boreal heath]

i totally get you, i know first countability isn't enough, i just want to know why that specific proof is wrong, sorry for being annoying lol

[iron bolt]

oh, no worries

I think the proof kinda falls into the "not even wrong" category for me

because it just doesn't give the details that would be required to show A is sequentially closed

I mean, to show that you need to take a sequence y_n in A with a limit point y, and show that y belongs to A

actually, nevermind, the answer is accurate - but just skips a lot of details

because they just say that y_n must be either eventually constant or converge to x

[boreal heath]

okay i promise ill stop after this one, but if the proof is accurate wouldn't that be a problem? like, you know we just found a counter example but this proof works? am i missing something

[iron bolt]

the proof just states that this holds in metric spaces, but the hard work is proving it

and in general topological spaces you can't prove it because it's false

I mean, try to derive it yourself, I think it's not very easy

[boreal heath]

okay i gotchu, thank you so much holy 😭

[iron bolt]

the way I would prove it in metric spaces is: ||if y_n has a subsequence that converges to x, then y_n must converge to x to, which would mean that y=x. if y_n has no such subsequence then you can show by contradiction that y_n becomes eventually contained in a finite subset of {x_n}, and every point outside that finite set has an neighbourhood that is disjoint from it||

but as far as I can see that relies on (1) uniqueness of limits of all sequences that converge to x, and (2) the fact that for every finite set, any point outside that set has a neighbourhood that is disjoint from it

(2) holds in all T1 spaces, but for (1) I think you actually need something like Hausdorffness

and of course, in order to reduce the proof to showing that A is sequentially closed in the first place you need that X is sequential (i.e. something like first-countable) too

[boreal heath]

yeah, that's what i'm trying to show, i'll assume T2 from the beginning and show it's closed, like jagr2808 indicated first. thank you once again for the help, and for everyone else too :)

[prime elbow]

i want to prove the equivalence of Cantor Intersection property and completeness.
I proved that completeness implies Cantor Intersection property.

now, i have to prove Cantor Intersection property, let x_n be a cauchy sequence now take set E_n = cl( { x_k | k>=n } ). It follows Cantor intersection property therefore intersection over N E_n is singelton set.

how to proceed now?

[prime elbow]

got it

I know this is not true for discrete metric space, so there is a reason to consider R with usual metric space, and in R bounded sets are totally bounded.

Any hint?

[opaque scroll]

Feels like there is some kind of typo here, because the condition that diam(Fn) is a bounded sequences doesn't make that much sense.

Like the sequence is decreasing, so it's always bounded by diam(F1). So it would just mean requiring that diam(F1) is finite.

Which you can do easily without changing d much

[prime elbow]

I see

[sacred heart]

I think the true condition is that $diam(F_n) \to 0$ as $n$ goes to infinity

[gentle osprey]

TimourX

[sacred heart]

it would make more sense

[unreal stratus]

Yes

[prime elbow]

Yes that's a true condition but can I get a counterexample such that diam(F_n) is boundary and intersection over N F_n is empty?

[prime elbow]

I proved one direction.

I have to prove that if E has an infinite subspace such that every subset is open and closed then E is not totally bounded.

Now I used contrapositive here, let E is totally bounded.

Since E is totally bounded so we have every sequence has a Cauchy subsequence.

Let E has infinite subspace which has every subset open and closed, make a sequence of distinct elements from that infinite set, and find contradiction.

Any hint?

[prime elbow]

I am thinking of showing this metric space to homeomorphic to subspace of R^n with Euclidean metric

This is homeomorphic to (0, ∞) with Euclidean metric, right?

[quick delta]

Homeomorphic isn’t enough
You need isometric, but that’s true as well

[sacred heart]

No it will always be non empty, and if diam(F_n) goes to 0, you can prove that the intersection is a singleton

[opaque scroll]

I don't think this is true, like consider E = {1/n : n in N} with its usual metric.

E is totally bounded, infinite, and every subset is clopen

[prime elbow]

Yes got it, thank you

[prime elbow]

Yes

[prime elbow]

Right

[marble apex]

does anyone here have a good understandign of JJK and gojo's domain expansion?

(yes, this has relation to topology)

If so, I'd like to question its validity and whether it would defy topology and by which theorems

[rain ether]

The topology channel description used to be

(see ⁠get-advanced-access to use this channel) point-set topology (topological spaces), algebraic topology (homotopy/homology/etc.), geometric topology, anime

[sand hill]

Hi I was working on a problem in Munkres to do with local compactness

I managed to solve the problem through an argument I think works

but I dont understand how the resut we prove doesn't contradict this corollary

am I missing something?

Is this just because of how strangely the rationals behave as a subspace of R?

[tender halo]

How does it contradict it?

Q is neither open nor closed in R

[sand hill]

Yeah fair enough I figured it out the moment I asked I was under the impression it was open as it wasnt equal to its own closure but I didnt look too deep into it

thanks

[paper wedge]

use the definition directly

and the fact that a compact subset of a hausdorff space is closed

the definition tells u that for any point u can find a compact subspace that contains a nbd of that point

[hollow geyser]

24 hours with no topology

[paper wedge]

here is a topology problem: compute the homology of a knot complement

[tender halo]

here is a topology problem for chat

take an infinite product of compact spaces, prove that continuous functions that only depend on finite amount of coordinates are dense in C(R) with sup topology

[opaque scroll]

What does it mean for a function to only depend on a finite amount of coordinates?

Like it factors through the projection onto some finite set if coordinates? Okay that makes sense

[red yoke]

It is possible to convince employers it does

There is also high correlation between studying topology and being statistics-literate

[ebon flower]

topological data analysis is a fairly active field of research

[quick crane]

somewhat. It's better to study (mathematical) finance and learn topology when you need it

[ebon flower]

tbh topology is one of the best-suited branches of maths to just being absorbed on the go when you need it

[red yoke]

Point-set?

[ebon flower]

mainly pointset yeah (thats the channel were in)
i dont know enough algtop or geometry to comment on the rest

[red yoke]

Do people do topological data analysis outside of metric spaces

[quick delta]

I don’t think I ever formally learned point set topology
I just absorbed it

[opaque scroll]

I know there are some techniques were you have a space equiped with several continuous functions to R, then you look at various intersections and unions of preimages.

This gives you like nice filtrations of your space, but I guess it's not so relevant if your space is just a discrete dataset. But it can be used at other states where you have a more continuous space, or for theoretical purposes.

[tender halo]

like uhh enthusiasm i guess

[civic verge]

QIt is neither closed nor open, since there will always be an element of x that belongs to I (irrational) so it is not closed, so if Q is open then the complete is closed which is false.

[cyan pewter]

just learned that the derived set is actually called the "derivative of a set" in german
that hurts

[sonic crane]

Topological data analysis?

[sonic crane]

You need to understand the basics of topology and some abstract algebra. From what i know the main tool is called “persistent homology” which is a way of detecting “holes” in your data by representing ur data as a point cloud

[sonic crane]

Youre welcome!

[pallid comet]

hi! i have a question that might be stupid

let A be the plane R^2 with the closed unit interval removed. let B be the plane with the half-open unit interval removed. are A, B homeomorphic? probably not, right? can we show this with point set topology?

[tender halo]

one of them is locally compact and the other is not

[keen gust]

since for each x in A there is an open set U containing x such that U is contained in A as well, can I somehow show that the union of all these U equals A and say since U is open, their union is open and so A is open in X

or is there any simpler way

[tender halo]

thats more or less the proof

[still vortex]

If the intersection of A and B is empty, prove that the intersection of int(cl(A)) and int(cl(B)) is empty.

[outer crescent]

seems like a little definition manipulation, do you recall what the interior of the closure of a set is?

[still vortex]

I think it's the same as int(A) but I can't prove it

[outer crescent]

Well lets start with recalling the closure of a set, and maybe some properties of a closed set?

[still vortex]

Closure of a set is all the points whose every neighborhood intersects with the set.

[outer crescent]

so maybe that gives you something to think about with respect to the interior of that set, and the intersection of Int(Cl(A)) and Int(Cl(B))

[trim spade]

A= rationals, B= irrationals. The interior of their closures is R, so do not have empty intersection

[still vortex]

I just realised the problem says they are open.

[opaque scroll]

If A is open then A is a subset of int(cl(A))

[keen gust]

Is there only one possible partition of a set?

Oh wait nvm

So let's say I have a set which is the closed interval A = [0, 2] , and I want to partition it

D = { [0, 1] , [1. 2] }

This would be a partition right? Since [0, 1] union [1, 2] = [0, 2]

[opaque scroll]

They're not supposed to overlap, but something like
{[0, 1), [1, 2]} or {[0, 1], (1, 2]} would be partitions

[keen gust]

Ah yes they need to be disjoint

Thanks

[keen gust]

How to show "same topology"?

[hoary breach]

tell the people here if you've tried anything

[keen gust]

i dont understand what it means by generating same topology on M

does it mean that there will be some continuous map?

[opaque scroll]

A topology on M is a collection of subsets of M called the open sets.

For a metric space, the topology generated by a metric has open sets unions of open balls

[keen gust]

i understand that much

but what is "same topology"

[opaque scroll]

Same topology means exactly that. That the topologies are the same.

Identical, equal, the same.

[keen gust]

Now, I'm quite confused how do I show they are the same topology

[opaque scroll]

The usual way to show that two sets are equal is to show that the first contains the second and that the second contains the first.

So you would have to show that every open set in one topology is also open in the other and vice versa

[keen gust]

Alright thanks

I'll try to do that

[civic verge]

int(A)=A

[keen gust]

Proof:\
$(\implies)$ Assume $d$ and $d'$ generate the same topology on M. Then any $B_r^{(d')}(x)$ is an open set in the topology generated by $d$. This means for every $r > 0$, there is some $r_1 >0$ such that:\
$B_{r_1}^{(d')}(x) \subseteq B_r^{(d)}(x)$\
Similarly, any $B_r^{(d)}(x)$ is an open set generated by d'. This means for every $r > 0$, there is some $r_2 > 0$ such that:\
$B_{r_2}^{(d)}(x) \subseteq B_r^{(d')}(x)$\
\
Therefore, these two conditions are satisfied if $d$ and $d'$ generate the same topology.
\
$(\impliedby)$ Suppose for every $x \in M$ and for every $r > 0$ there is some $r_1 > 0$ and $r_2 > 0$ such that\
$B_{r_1}^{(d')}(x) \subseteq B_r^{(d)}(x)$ and $B_{r_2}^{(d)}(x) \subseteq B_r^{(d')}(x)$.\
We want to show that $B_r^{(d)} = B_r^{(d')}$.\
So we have to show 1. $B_r^{(d)} \subseteq B_r^{(d')}$ and 2. $B_r^{(d)} \subseteq B_r^{(d')}$\
\

1. For every $r > 0$ we have some $r_1 > 0$ so that $B_{r_1}^{(d')}(x) \subseteq B_{r}^{(d')}(x)$. It follows that $B_{r_1}^{(d')}(x) \subseteq B_{r}^{(d)}(x) \subseteq B_{r}^{(d')}(x)$. Thus $B_{r}^{(d)}(x) \subseteq B_{r}^{(d')}$. \
   \
2. For every $r > 0$ we have some $r_2 > 0$ so that $B_{r_2}^{(d)}(x) \subseteq B_{r}^{(d)}(x)$. It follows that $B_{r_2}^{(d')}(x) \subseteq B_{r}^{(d')}(x) \subseteq B_{r}^{(d)}(x)$. Thus $B_{r}^{(d')}(x) \subseteq B_{r}^{(d)}$. \

Combining both cases, we have $B_r^{(d)} \subseteq B_r^{(d)}$ and 2. $B_r^{(d')} \subseteq B_r^{(d')}$. Therefore, $B_r^{(d)} = B_r^{(d')}$. Since this is true for every $x \in M$, $d$ and $d'$ generate the same topology on $M$. Q.E.D

These proofs are so tedious

I probably have lots of typos

[gentle osprey]

Slender

[keen gust]

Can anyone critique this

[hoary breach]

in the converse you say

we want to show that B(...)=B(...)
why?

think of the unit ball in R^2 under the l1 and l2 norm, one is round, the other not

(they both generate the Euclidean topology; this is something you prove after this problem)

[hoary breach]

https://www.geogebra.org/m/pyxfvyyk

[keen gust]

Ah right

Rather I would want to show that some subbset U containing these balls would be equal

[hoary breach]

how would you normally prove equality of two sets

bro got banned... 💀

[keen gust]

Uh

Show that one contains the other

[hoary breach]

topologies are sets too so do that here

[keen gust]

Yea but how

[hoary breach]

whats an element in the topology generated by d?

[keen gust]

An open set

[hoary breach]

show that thats also an element in the topology generated by d'

[keen gust]

Mb that's what I meant

Meant open set U

Not subset

[civic verge]

Hi guys, this demo is on the road.

[rancid umbra]

you should be inducting on the size of Y if you want a formal proof.
also note: Y cannot be empty

[civic verge]

The size is induced by the counter set, I think it would be more than enough, since being a bijection, then the cardinality In is equal to the cardinality of

no?

[rancid umbra]

what is the ‘counter set’?

[civic verge]

I_{n} sry my english is bd

[rancid umbra]

you can replace Y with some I_n, which is basically what varphi does.
but you still have not described how to define phi

induction essentially abstracts this for you

[civic verge]

That's where I have my doubt, because I have not been able to make this diagram work, I consulted a little and did not find much, so I asked chat gpt for an idea and he told me that I could use the pigeon principle, which clearly I do not know about it, so without having knowledge of it,

[rancid umbra]

if Y has n elements, can you see why we can simply replace Y with I_n = {1, 2, …, n}?

[civic verge]

Yes

[rancid umbra]

great! then induct on n.

Base Case: if n = 1 and X is an infinite set, can you find a surjective function f : X —> I_1?

Inductive Hypothesis: Assume that for some natural number n, if X is an infinite set, there is a surjection f : X —> I_n.

Inductive Step: Show that if X is an infinite set, there is a surjection X —> I_{n+1} using the Inductive Hypothesis

[civic verge]

And how could this help me in the demonstration?

It is the phi? part, i.e. the final step for the case of In in X?

[rancid umbra]

this is different all together. do you absolutely need to prove the statement this way?

however, as i said before, any formal proof of this statement is going to require induction

[civic verge]

Not really, I just had that idea that way, hehe.

[civic verge]

I'm nt familiar with the inductive hypothesis, I would like to know how, it is a little late where I am, I will go to sleep, if there is no problem I will write to you privately so you can explain me how it is really what you are trying to tell me, thank you very much for help.

[sand hill]

what stops us from modifying this argument to show that subspaces of normal spaces are normal (which isnt generally true unless the subspace is closed)

Considering the converse of Urysohns lemma holds

am I missing something

[alpine nest]

Try to write down the modified argument and see where it fails

[sand hill]

I dont understand where it fails. One place I thought it'd fail is that $\bar{A} \cap Y$ is not necessarily closed unless $Y$ is too, but this doesn't seem to matter for the result as we construct a continuous function from $X \to [0,1]$ that separate some disjoint closed subsets $\bar{A}$ and $\bar{B}$ and then restrict that function to $Y$, which im pretty sure preserves it continuity, and as $\bar{A} \cap Y \supset A$ and $\bar{B} \cap Y \supset B$, our function still successfully separates the sets

[sand hill]

perhaps im looking in the wrong place

or maybe I just missed the failure

[alpine nest]

Isn't the existence of that function perfect normality? Normality on its own should be "every two disjoint closed sets can be separated by open sets"

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

[sand hill]

yeah but the converse of Urysohns lemma holds, and that states that a set is normal if you can separate disjoint closed subsets with a continuous function

perfect normality as I understand it is when you can not only separate the sets with a continuous function but do it such that f(x) = 0 iff x \in A and f(x) = 1 iff x \in B

[alpine nest]

I think the issue is in showing that the closures of A and B in X are disjoint (even if A and B are disjoint closed sets in Y)

[sand hill]

Wow yeah that should be it

it wouldnt matter if Y was closed but if its open we've to treat the larger sets

[alpine nest]

https://math.stackexchange.com/a/220486

[sand hill]

thanks dk how I didnt see that

[iron bolt]

for the future: I think one really good way to resolve issues like this where you can't find the mistake in a proof of a known-to-be-false statement is to take a counterexample, and go through the proof on it

at some point you will get obviously false statements, abd that helps you narrow down which part of the proof is problematic

[hollow geyser]

Is there a term for a "one sided homeomorphism"? Say $X$ and $Y$ are spaces, and there is some bijective function $f :X\to Y$ where $f$ is continuous, but $f^{-1}$ may not be continuous. Is there a word for this? (It may not even require $f$ be a bijection, but perhaps just surjective or injective?)

[gentle osprey]

SWR

[alpine nest]

"Embedding" springs to mind (those are injective continuous maps)

[hollow geyser]

Oh thanks. I'll look into that definition

[alpine nest]

Oh wait, I think an embedding should also be a homeomorphism onto its image

So it's a bit stricter than just being continuous and injective

[hollow geyser]

That name sounds like what I am trying to think of

[hollow geyser]

oops or maybe not

I'lll just wiki search homeomorphism and see if they mention any weaker ideas

[opaque scroll]

Not strictly a topology term. But a bimorphism in a category is a morphism that is both mono and epi, which in the case of topological spaces would be a continuous bijection.

You also have the terms section and retraction, which are morphisms that have one sided inverses, but not necessarily twosided inverses. And in the case of topological spaces you have some stronger ones like deformation retract or strong deformation retract which are retracts which give nice homotopy equivalences.

Similar thing might also be weak homotopy equivalence, though this is a quite a bit weaker again.

[azure harbor]

hi guys, im completely new to this, a question i had is that could there be a point in a topological space X such that the only open set containing it is X ?

[opaque scroll]

There can indeed. Maybe you're able to cook up an example yourself where X is some small set.

[azure harbor]

thx !

[ebon flower]

taking the indiscrete topology on any set, this is the case for every point

[fringe thorn]

<@&268886789983436800>

[barren haven]

I've been referred here from #category-theory lmao, I'd appreciate some help (more info after the original message)

[empty grove]

Show that the map into A × B is continuous

[barren haven]

done!

[empty grove]

Then the map into the fibred product is automatically continuous because that's a property of subspace topologies

[barren haven]

from AxB?

[empty grove]

Corestriction (shrinking the codomain while still containing the image) preserves continuity

[empty grove]

No, the map that you want

f

[barren haven]

hmmm

[empty grove]

So the claim is

[barren haven]

how do I show that

[empty grove]

If f: X → Y is continuous and B contains im(f), then if you view f as a map X → B then it is continuous

If you give B the subspace topology

[barren haven]

I see, thank you!!

so we don't even need alpha and beta for the question?

What's the point of saying it's a fibre product...

[empty grove]

They show up in showing that the map to the product is continuous

It's combining two steps. Taking the product and taking the subspace

[barren haven]

Cartesian or fibre

[empty grove]

Cartesian lol

Wait sorry

[barren haven]

I don't think I did use them

[empty grove]

Alpha and beta show up in defining the subspace as a set

Fibred products are ways to construct spaces of "coherent pairs"

The maps alpha and beta specify the "coherence"

[barren haven]

true, but they could've phrased the question for any subspace, right?

[empty grove]

Maybe looking at a fibred coproduct (pushout) would be more instructive

[empty grove]

Yes

[barren haven]

alright, thank you so much

[azure harbor]

hey guys, im completely new to this and reading through janich's book. so why is the traingle inequality needed to "insert" a strictly contained delta ball ?

[azure harbor]

i guess that we take the difference between xy and xz and regard that as a guaranteed choice of delta? then the delta ball would certainly lie within ball centered around x but with d(x,z) radius ( due to traingle inequality) ?

oh i think i got this via searching .

[warm kettle]

Why is R^2 not homeomorphic to R^2 \ {(0,0)}?

[paper wedge]

i dont think its easy to see without basic algebraic top

[prime elbow]

One is connected and the second one is not connected, right?

[warm kettle]

Both are connected

[prime elbow]

Sure?

[warm kettle]

Ugh, that's bad news 😵‍💫

[warm kettle]

Yep

[prime elbow]

Yes

[warm kettle]

For example, if I want to connect (-1, 2) and (1, -2), I can go from (-1, 2) to (1, 2) and from there to (1, -2)

[prime elbow]

Yes

Btw Thingoln are you an undergraduate student?

[warm kettle]

Yep

I'm in my second year

[prime elbow]

So in your college they teach topology in second year

Can I ask which university?

[warm kettle]

University of Warsaw

[prime elbow]

Oh group, ring and field in same semester

[warm kettle]

Yeah, but I don't think they go into much detail until abstract algebra 2. My undergraduate course seems to generally focus on analytic subjects

But you can choose algebraic electives if you're a fan of pain that area

[prime elbow]

Oh

It looks so good

[warm kettle]

I enjoy it a lot! My profs are all very nice people too!

[prime elbow]

Is that course in english or in other language?

[warm kettle]

Polish

[prime elbow]

I see

[plush folio]

The idea is that every loop in R^2 can be continuously shrunk down to a point, but in R^2 \ {(0, 0)} you can have a loop around the origin that cannot be shrunk down. We say that R^2 has a trivial fundamental group, while the fundamental group of R^2 \ {(0, 0)} is Z (the integers), and any homeomorphic spaces have the same fundamental group. To actually prove this requires some algebraic topology, but the intution is not too difficult to understand

[dawn tundra]

R^2 \ {(0,0)} is not simply connected

[prime elbow]

What is simply connected?

[plush folio]

Simply connected means it has trivial fundamental group, ie. every loop can be shrunk down to a point

[prime elbow]

Okay

Thank you

[paper wedge]

not only that king i think u also need the space to be path connected.

[plush folio]

Yeah, good point I think when a space isn't path connected the fundamental group depends on which base point you use, so usually we just assume the space is path connected when talking about a fundamental group

[warm kettle]

We haven't discussed fundamental groups yet, but after skimming through the article on simple connectedness, it makes sense. I'll go back to this when I've learnt more about some basic alg topo and try to prove it formally. Thank you guys for the help 🙏

[novel plank]

Given a topology on X x Y, is there a natural way of defining a topology on X and on Y?

[opaque scroll]

Sure just take the quotient topology with respect to the projections

[novel plank]

That's just the final topology right?

[opaque scroll]

Yup

[alpine hound]

Let $(X,\tau)$ be a topological space. Let $f:X\to X$ be defined as $f(Y)=\overline{Y}$. Is $f$ continuous?

[gentle osprey]

𝒢𝒾𝓃𝑔𝑒𝓇 𝑀𝒶𝑔𝓂𝒶

[alpine hound]

A function is continuous iff $f(\overline{A}) \subset \overline{f(A)}$. But both of them are equal to $\overline{A}$ so I want to say yes, I am just unsure of myself

[gentle osprey]

𝒢𝒾𝓃𝑔𝑒𝓇 𝑀𝒶𝑔𝓂𝒶

[alpine nest]

f isn't well-defined, arguments of a function from X into X should be elements of X, not subsets of X

[alpine hound]

I make special mistakes

[iron bolt]

I will argue that if you did put some reasonable topology on P(X), f should be discontinuous just based on vibes

I don't know of any reasonable power set topology though

[iron bolt]

to be more specific: say you put a reasonable topology on P(X) for some space X, such that the closure operator cl : P(X) → P(X) is continuous

since f is reasonable, the complement operator X \ - : P(X) → P(X) should probably be continuous as well. but the interior operator int : P(X) → P(X) can be written as X \ cl(X \ -), so it should then be continuous too

when means that in particular int(cl(-)) should be continuous

but now if X is Hausdorff and separable, you can find a sequence (a_n) such that each initial segment of it has closure with empty interior, but the closure of the full sequence is the full space

which means that that the initial segments of your sequence, viewed as a sequence in P(X), can't converge to the full set {a_n} of your sequence because otherwise the constant sequence at the empty set would have to converge to the full space X

so... does not seem super reasonable to me either way

maybe I'm wrong though and there are some other reasonable properties such a topology can have, I don't know

[tender halo]

https://math.stackexchange.com/questions/4311465/is-there-a-neat-way-to-prove-that-r2-is-not-homeomorphic-to-r2-0-0

[wise mist]

form circles

R^n\ {0} is path connected through (r)S^n-1

[warm kettle]

Great proof, thanks!

[novel plank]

Can different topologies on X and Y give the same topology on X x Y?

[ebon flower]

no
suppose X, X' and Y, Y' are two different topologies on the same sets, and the identity map XxY->X'xY' is a homeomorphism, and ι:X->XxY is an inclusion map which fixes some y and maps ι(x)=(x,y), and π:XxY->X' is the projection map, then since ι and π are continuous, so is their composition f:X->X'. Do the same in the other direction to find a continuous g, and then note that g is the inverse of f, to find that f is a homeomorphism.

[novel plank]

Thank you 🙏

[azure harbor]

hey guys , here is the universal property of product topology given in "categorical point set topology". i am thinking around how well this is coped with categoric theoretic product. i think that uniqueness/exsitence of f and the composition structure are all guaranteed by considering underlying sets and their product. thus we only need to prove the set wise product could be " lifted" to product between topological spaces ?( continuous etc )

[unreal stratus]

Pedantic answer: yes, if one is empty

[ebon flower]

oh sure then my proof fails because there exists no inclusion map

[unreal stratus]

But yeah otherwise I'd do exactly what you did heh

[unreal stratus]

You need a lifting to spaces, but it needs to still initial among all topological spaces with that property, which is why you put the smallest topology with continuous projections

[elfin epoch]

hello guys, Im trying to write proofs that are easy to follow and understand on each step. I've started with the heine-borel theorem, can someone take a look and give your opinion on it?

[iron bolt]

I only briefly skimmed it, but as far as I can see you're proving that every continuous map from a compact metric space to a metric space is also uniformly continuous, right?

that's not what I know as the Heine-Borel theorem

Heine-Borel is the statement that subsets of R^n are compact iff they are closed and bounded, or a generalisation of that

as for the proof itself, I think it's always hard to judge whether a proof is written up well because it very much depends on the target audience

people who are somewhat new to the material and/or haven't done a lot of proofs yet would probably benefit from a long proof with lots of explanations like you have done here

while people who are already familiar with many similar proofs would probably just like to see it proven as concisely as possible

I think you did a reasonably good job at writing for that first audience though

[novel plank]

Is that the same topology as the one generated by the image of open sets under the projection map?

[iron bolt]

it's the topology consisting of all sets whose preimages are open

in general those don't necessarily agree

[novel plank]

Yeah, that's what I suspected. Then is it possibleto find a basis for that quotient topology with respect to the basis of the original space?

[iron bolt]

oh, that's difficult

I don't know for sure, but I think probably not. at least not in any 1-1 way

because iirc there exist second-countable spaces with quotients that are not second-countable

so to go from a basis of a space to a basis of the quotient, you might have to increase the cardinality

example: R is second-countable, but R/Z (as in, R where Z is collapsed to a point, not the quotient group where every coset x+Z is collapsed) is not even first-countable in the point corresponding to Z

[novel plank]

idk if this helps, but for some context, I'm considering sets X and Y, and a topology t on X x Y. My goal is to find a topology tX for X and a topology tY for Y such that the product topology tX x tY is the same as t on X x Y. jagr suggested to take tX and tY to be the final topology with respect to the projection maps

in this setup, can we describe a basis for tX and tY with respect to a basis for t?

[novel plank]

or even, in this setup, will this be true?

[iron bolt]

not every topology on X ⨯ Y is a product topology though

so what you're trying to do can't be done

if it is a product topology, then you can do something like taking the quotient topologies under the projections to recover the topologies on X and Y

but there's no guarantee that that's the case

[novel plank]

do you know of any result that gives equivalent conditions for a topology on X x Y to be a product topology?

[iron bolt]

not really

I also don't know what you're trying to do, so I don't know what kind of condition would be helpful to you

one thing I can say is that when you have a topology t on X ⨯ Y, define tX and tY as the quotient topologies under the projection maps and then t' as the product topology of tX and tY, t' is always at most as fine as t, and equal iff t is a product topology

so there's probably something like a Galois insertion between all topologies and product topologies on X ⨯ Y in place there, idk if that's helpful to you

or Galois coinsertion? I always mix up my directions

[hollow geyser]

Is there a name for points on graphs of a function that have no neighborhoods where the function is locally injective?

i.e. y=x² at (0, 0)

[languid edge]

Let X and Y be topological spaces. Is there a commonly used notation for the set of continuous functions X --> Y?

[iron bolt]

C(X,Y)

[rain ether]

non-degenerate critical points, maybe

[hollow geyser]

If you can type, the script version of C is more often used: $\mathcal{C} (X,Y)$ you can also use it define the set of continuous functions on a real interval like $\mathcal{C}([a,b])$

[gentle osprey]

SWR

[hollow geyser]

Hmm. Never heard of that 😦

[rain ether]

umm, like points where the derivative is equal to zero, but excluding functions like y = x^3 at (0, 0)

[hollow geyser]

Derivatives may not exist for these functions though

[rain ether]

yeah, such points might also be critical points

[hollow geyser]

Okay but what about spaces that have absolutely zero differential properties?

[languid edge]

thanks!

[rain ether]

I am not sure about the purely topological case, but I have seen some paper where the term "critical point" is (synonymously) used to refer to a point where the function is not locally injective

[rancid umbra]

singularity or singular point comes to mind. i think the wiki page offers several interpretations across different areas of math

[median sand]

Are the (open) cells of a CW-complex open as subsets? Wouldn't that imply that the complex is their coproduct?

[red yoke]

No

A cell is open iff it does not intersect the boundary of any higher dimensional cells

[sonic crane]

why do anime pfp make u wanna vomit

[alpine nest]

It is an attempt to to signify one's superiority by disparaging a perceived out-group

[sonic crane]

Fr

[languid edge]

What is the topologist's sine curve?
1️⃣ Graph of sin(1/x) on 𝐑 \ {0} + the point (0, 0)
2️⃣ Graph of sin(1/x) on 𝐑 \ {0} + the segment (0, -1) to (0, 1)
3️⃣ Graph of sin(1/x) on 𝐑⁺ + the point (0, 0)
4️⃣ Graph of sin(1/x) on 𝐑⁺ + the segment (0, -1) to (0, 1)
5️⃣ Graph of sin(1/x) on (0, 1] + the point (0, 0)
6️⃣ Graph of sin(1/x) on (0, 1] + the segment (0, -1) to (0, 1)

I ask since everybody seems to define it differently

[paper wedge]

its the closure of one of these

[languid edge]

if you want it to be a closure then it's one of 2️⃣ 4️⃣ 6️⃣

[rancid umbra]

real life use case of topology

[hollow geyser]

If $X,Y\subseteq \bR$ and $f$ is continuous and bijective, is $f^{-1}$ continuous?

[gentle osprey]

SWR

[red yoke]

No

[hollow geyser]

What is a counterexample?

[red yoke]

N → {0} ∪ 1/N

Or N → any countably infinite nondiscrete subset of R

[hollow geyser]

By 1/N, I assume you mean {1/1, 1/2, 1/3, 1/4...}?

[red yoke]

Perhaps 1/(N+1)

[hollow geyser]

Let me specify more clearly my definition of continuity: If $V$ is open in $Y$, then $f^{-1}(V)$ must be open in $X$. But I am also requiring $V$ be open in $\bR$.

[gentle osprey]

SWR

[hollow geyser]

Or better, let's say X and Y must be open subsets of R

[red yoke]

Then yes

[hollow geyser]

huzzah

[red yoke]

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

[hollow geyser]

Is this a property of completeness? Or is this possible in Q too?

[red yoke]

No this is a property of R^n

[hollow geyser]

I did a proof sketch before looking into the idea, but I used IVT

[hollow geyser]

So, compactness maybe?

I see I see

yeah just read your article

Well I proved it for R with regular topo

100 years late, but I tried my best

[hollow geyser]

This function is Q-continuous, right?

[red yoke]

Yes

[hollow geyser]

Is there any $f :\bQ\to\bQ$ that is bijective and continuous but $f^{-1}$ is not continuous. My first instinct is yes, but I can't really think of a specific function (not that I thought the above was a counterexample, I'm just trying to think of what Q-continuity can actually mean)

[gentle osprey]

SWR

[rancid umbra]

continuous in the sense you mean here?

or as a function Q to Q

[hollow geyser]

No. Continuous in the space of Q's open sets

[hollow geyser]

yes

[red yoke]

Yes

First note any two open intervals in Q are homeomorphic

[hollow geyser]

for this example, do you mean specifically to each other, or each to Q, are do I need to care that all three are homeomorphic?

[tender halo]

they all are homeomorphic to Q

[hollow geyser]

yes, but I am asking which case I actually care about for this problem

[red yoke]

The former

Next, note that any clopen interval in Q is homeomorphic to an open interval

[hollow geyser]

a clopen interval would be irrational endpoints, yes?

[red yoke]

Take a clopen interval (right), split into infinitely many open intervals (red blue) and an endpoint (black), and rearrange (left)

[red yoke]

No

The closed end has to be rational

[hollow geyser]

How is that closed then?

[red yoke]

Something like (-π, 5]?

[hollow geyser]

What about $(1, \sqrt2)$?

[gentle osprey]

SWR

[red yoke]

That's open

[hollow geyser]

is it closed?

no, right?

[red yoke]

No

By clopen I mean (] or [)

[hollow geyser]

is this open? Is 5 an interior point?

[red yoke]

Where the [ end is in Q

[hollow geyser]

Ahh. I thought clopen=closed+open

[red yoke]

Oof I should use half open

[hollow geyser]

Okay, so you say that $[1, \sqrt2)$ will be homeomorphic to $(1, 2)$?

[gentle osprey]

SWR

[red yoke]

Yea

[hollow geyser]

I'm not seeing it

[red yoke]

^

[hollow geyser]

I know it's my R brain, but I'm just missing it

[hollow geyser]

Oh. You are saying such a function as this will be your homeomorphism?

[red yoke]

Yea

The smaller and smaller intervals on the right will map to intervals closer and closer to the center on the left

[hollow geyser]

okay I'll marinate on that later, but I'll just take it as true for now

[red yoke]

Now split Q into open intervals (n+π, n+1+π)

And map each to (n, n+1]

This is continuous and bijective

[hollow geyser]

okie

[red yoke]

But not homeo

[hollow geyser]

how do we show that f^-1 is not continuous?

Because of n+1]?

[red yoke]

Yea

[hollow geyser]

I see

[red yoke]

If we try to invert it then n+1 should map to n+π+(1/2) via this diagram

But then continuity from right side fails

[hollow geyser]

Okay. I'll need to think about your open to half-open homeomorphism but I see what you are doing

In theory, we can simplify this to $(-\infty, 1)$ and $[1, \infty)$, could we not?

[gentle osprey]

SWR

[red yoke]

The open to half open map?

[hollow geyser]

yes

[red yoke]

Yea ig

[manic seal]

There exists a bijection between [0,1) and (0,1) but that bijection will not be continuous

If we are using standard topology that is

That leads me to guess that there does not exist a homeomorphism between the two sets

Nvm this is not standard topology

[hollow geyser]

yeah this is what is slowing me down

[rugged tree]

Hello there! Suppose X a topological space and an equivalence relation, lets call it E, why is the quotient projection q: X ---> X/E an open map? The topology defined on X/E is s.t. U in X/E is open if q^-1(U) is open in X

I've been stuck in this problem for few days

[iron bolt]

the reason you're stuck is that what you're trying to prove is false

it is nice when quotient maps are open, but they only are in specific scenarios

like the projections from products, or quotients under finite continuous group actions for example

if you on the other hand look closely at the quotient ||of R that collapses [0,1] to a point||, you'll see that the quotient projection isn't open

[dire dove]

||I fear that's just R, you probably need to collapse [0,1) or something like that||

[iron bolt]

||the quotient is indeed homeomorphic to R, but the quotient map isn't open||

that makes it easy to visualise

[dire dove]

My bad, you're right

[rugged tree]

Thanks dude

Can I confirm the solution of a problem? It basically states that if there is a continuous function f: X ---> Y (obviously, X and Y are topological spaces) and there is a set A s.t. f(A) = Y, and the restriction of f to A is a quotient map, then f is a quotient map
What I did is basically, if f restricted to A is quotient map, then there is an equivalence relationship s.t. there is a homeomorphism between A/E and Y, s.t. the diagram commutes and f = f' composed with q, where q is the projection on X/E.
Now, since f(A) = Y and f(X) = Y, create an equivalence relationship as follows: If x in X and a in A are s.t. f(x) = f(a), then x is related to a. Now, since the projection on E is equal to A, it follows that f is a quotient map.

[sharp juniper]

f(x) = 2x + 1 . what is f(3)

[hoary breach]

#prealg-and-algebra

[rugged tree]

But this is not the trivial topology?

[iron bolt]

it is not

the quotient space you get is homeomorphic to R - the quotient projection could also be understood as f(x) = (x if x ≤ 0, x-1 if x ≥ 1, and 0 otherwise), with 0 corresponding to the equivalence class [0,1]

you can check that that map is a quotient map, i.e. f^-1(U) is open if U is

but it isn't open: ||the open interval (0,1) gets mapped to {0}||

[rugged tree]

But you are sending the entire real line to a single point, this cannot be homeomorphic to the real line

[iron bolt]

no, I'm sending [0,1] to a single point

[rugged tree]

Ok, but how this single point can be homeomorphic to the entire real line?

[iron bolt]

it is not, the quotient space is

maybe I should be more clear

for any space X and subset A, you can define an equivalence relation on X such that x ~ x' iff x = x' or (x ∈ A and x' ∈ A)

the quotient space X/~ is then commonly denoted X/A, and said to be the quotient space of X where A is collapsed to a point

i.e. in the case of X = R and A = [0,1], this space would contain one point for every x ∈ R\[0,1], and one point corresponding to all of [0,1]

that's the quotient space I was talking about

[rugged tree]

i think I'm understanding now, I just need time to think, but thanks dude

[rugged tree]

What do you think about this sol.? I just skipped one step at the end, because I need to compose the equivalence that i made with the quotient map @iron bolt

[rugged tree]

AH, o, now I fully understand, this is homeomorphic to the real line, because any open set in there, is an open set in the real line, but the [0,1] is transformed to a point. And the image of any open set inside [0,1] is closed. Ok

[iron bolt]

tbh I think you're slightly overcomplicating it 😅

I would have argued directly in terms of open sets - since f is continuous you already know that for all U ⊆ Y the preimage f^-1(U) is open if U is, so you would only need to show that if f^-1(U) is open U is too

[rugged tree]

Okay, it is really direct, because f restricted to A is .... Ok, dude you are brilliant, congratulations!

[iron bolt]

I've just done this for a while already, lol

it's totally normal to slightly overcomplicate things sometimes when first learning a field

over time you get more intuition for when certain kinds of arguments are needed and when not

[rugged tree]

I'm taking this master class of topology and I'm having a hard time with the quotient topology and Urysohn's Lemma

[rugged tree]

Okay, it's like advanced calculus, I look back at my old solutions and now I see really simple ways to solve the same problem that I had a hard time trying to proof on hahaha

[shut moat]

is there a standard example of a regular space for which C(X) consists only of the constant functions?

this was surprisingly hard to google lol

[iron bolt]

by regular, does the text mean regular and Hausdorff or just regular?

I don't know of an example either way, but I'm curious

[shut moat]

the book takes the convention that regular spaces are hausdorff

[iron bolt]

huh. wild that such a space can exist then

completely evades my intuition

any such example definitely has to be connected and non-compact/non-Lindelöf, i.e. in particular uncountable

but that's about all I can say

[languid edge]

Wait what, isn't the identity function always continuous

[uneven bronze]

Theorem: Let $E$ be a separable locally compact metric space. Then there exists an increasing sequence $(L_n)$ of compact subsets of $E$ such that, for every $n\in\mathbb N$, $L_n\subset L_{n+1}^\circ$ and $$E=\bigcup_{n\geq 1}L_n=\bigcup_{n\geq 1}L_n^\circ.$$ \

Proof: Let $(x_p)$ in $E$ be a dense sequence in $E$. Let $I$ be the set of all pairs $(p,k)$ of positive integers such that the closed ball $\overline{B}(x_p,2^{-k})$ is compact. ... \

Question: Silly question maybe, but what ensures that $I$ is not empty?

[gentle osprey]

psie

[uneven bronze]

I'm a beginner at this and I'm having doubts about whether a closed ball is compact in metric spaces other than Euclidean space.

[iron bolt]

it is. I think C(X) there means C(X,R), not C(X,X)

i.e. continuous functions to the real numbers

[iron bolt]

in general, closed balls are not compact

see for example closed balls in I think every infinite-dimensional Banach space

but you can check that a metric space is locally compact iff small enough balls are compact

or more precisely, iff for every x there exists an ε > 0 such that for all r < ε, \overline B(r,x) is compact

[uneven bronze]

ok 👍

[alpine nest]

My googling yielded this, maybe it will be helpful? https://math.stackexchange.com/a/1047452

[warm kettle]

I was thinking today how to prove that a closed two-dimensional disc D^2 is not homeomorphic to S^2. My reasoning was this:

• I assume there exists a homeo f : D^2 -> S^2.
• I take a diameter I of D^2. It divides D^2 into two connected sets. It means that f(I) also divides S^2 into two connected sets. I itself is connected, so f(I) is as well.
• This is the step I don't know how to justify yet, but I want to say that f(I) would have to be homeomorphic to S^1 to be able to divide S^2 into two connected components because f(I) is path connected since I is.
• That is a contradiction because if that's true, I could remove any point from f(I) without making it unconnected, but it's not possible with I.

Any tips how I can get that homeomorphism to S^1? Or something else that would help there

[iron bolt]

I think you would have to do something algebraic-topology-flavoured either way, so you might as well do it more directly

[earnest basalt]

i think using Borsuk Ulam is the easiest way

[iron bolt]

||you can remove a point from D^2 and get a space with nontrivial fundamental group, while with S^2 you can never do that||

[iron bolt]

oh, that's also clever

should require a roughly a similar amount of background knowledge I think

[paper wedge]

the spaces are not homotopy equivalent

[iron bolt]

yup. showing that directly requires homology or π_2 though, that's what I was trying to avoid

hence the argument about removing a point

[paper wedge]

I see

[warm kettle]

It seems pure point-set topology is not very powerful :(

Thank you guys for the help 🙏

[paper wedge]

pure point-set topology is more set-theoritic and logic

very diff flavor than "usual" topology

not many ppl study point-set topology for its own sake but rather u just need the tehcnical stuff to be able to do things in "usual" topology

like proving a manifold has partitions of unity etc

[ebon flower]

well analysis is just point set topology with a metric

[plush folio]

Hmm, would it be possible to prove this by arguing that D^2 is homeomorphic to [0, 1]^2 and S^2 is homeomorphic to [0, 1]^2/~ where ~ identifies the boundary to a point? Then if D^2 and S^2 are homeomorphic then the quotient map [0, 1]^2 -> [0, 1]^2/~ would have to be open, which it clearly isn't

not sure if it works, I guess they can be homeomorphic even though the quotient map isn't open?

[iron bolt]

yup. a space can be homeomorphic to a non-open quotient of itself

[plush folio]

I see

[iron bolt]

and I mean, point-set topology is powerful in the sense that it can distinguish between spaces like D^n, S^n and R^n - they're all not homeomorphic

it's just that proving that they're not homeomorphic is hard, and that's where algebraic topology comes in

constructing better topological invariants than point-set topology alone can provide

[warm kettle]

Fair enough. I'll try to explore alg topo a bit soon because I keep running into problems like this 💀

[iron bolt]

the fundamental group is really nice I think - I think it'll be worth learning about it, even if you don't want to get into more complicated stuff like homology

very simple idea, easy to visualise, and beautifully related to covering space theory

[warm kettle]

Sounds really awesome!

[urban zinc]

for more info, look up "Heine Borel spaces"

[urban zinc]

here, you're using the fact that E is locally compact; take a look at the definition of locally compact that your book uses again

[ebon flower]

one example of a space where they are is the space of holomorphic functions on an open set

[finite token]

one example of a space where they aren't is Q

[red yoke]

A metric space is compact iff complete and totally bounded

[frigid rose]

Is there a set that admits a non-discrete topology in which every non-empty closed set has a non-empty interior?
Obviously it can't be T1 because tjat would mean that all the singletons are open but could a topology like that have some "interesring" property?

[opaque scroll]

Consider a closed set that is not open. Then it's boundary is a non-empty closed set with empty interior.

So such a topology must have every open set be clopen. It doesn't have to be discrete though, trivial topology also works, or any disjoint union of spaces with trivial topology.

[frigid rose]

Ahh i see

Thanks!

[shut moat]

omg tysm

this is wildly stronger than the initial claim too

[shut moat]

can you make sense of the order relation being used here?

the notation is very strange

[white ginkgo]

So the first part is OK, but I haven’t figured out how to show f is not homotopic to g relative to {-1, 1}.

Intuitively, it makes sense. I think one has to use that R² \ (0, 0) is not contractible, and that the concatenation g * f (where we view f and g as paths) is a generator of the fundamental group. But I don’t quite see how to proceed. Hints?

[quick delta]

If f is homotopic to g, then f g^-1 is homotopic to gg^-1

(Both rel -1, 1)

[pallid comet]

sorry for interrupting the discussion - may i ask a simple question?

if A is a closed set in a compact T2 space, is (the union of all connected components that have a nonempty intersection with A) closed?

[gritty widget]

I dont see how to use compact here tbh

[gritty widget]

Have you thought of anything yet

Ok I think I can say yes

And I think I see why compact matters but idk what T2 is for here

[pallid comet]

so from some earlier theorems i just learned, i was thinking about identifying every point in each component with each other (essentially viewing each component as a point)

and this new space would also be compact T2

and we need to somehow prove that the identification map is closed?

[gritty widget]

I was thinking something along the lines of finite subcovers and connected components are closed

[gritty widget]

Youre thinking of showing U->U/~ is closed?
Where U is that union of connected components.

[pallid comet]

yeah i am too

showing the map is closed is basically the same thing so ..

WAIT NAH IM DUMB. showing the map is closed is way easier, i just havent thought about that approach.

continuous maps from compact spaces to t2 spaces are closed

[gritty widget]

Closed map lemma

[pallid comet]

sorry for the question >.<

like actually why does this keep happening. i get stuck on something for a few days and figure it out like 2 hours after asking the question here

thank you anyway!

[gritty widget]

Can you prove closed map lemma tho

[pallid comet]

yup

[gritty widget]

Let X,Y be metric space such that f : X -> Y is continuous on every compact subset of X, then f is continuous on X.

Let x_n converges to x then {x,x_1,x_2,..} is compact set so f is continuous on it implies f(x_n) converges to f(x).
is there any problem?

[red yoke]

Counterexample: any discontinuous function on Q

[gritty widget]

how? i see the problem but i don't get where i am wrong?

[warm kettle]

Why is {x,x1,…} compact?

[gritty widget]

why this is true when i take X is an interval of R and Y is R

[gritty widget]

take that open set U which contain x then it will contains all points but finitely many

[red yoke]

Sorry my brain is not loading

This seems correct

[warm kettle]

Right

Nice

[gritty widget]

in problem they take X = I inteval and Y = R and they use same argument now i am generalise that

[keen gust]

Anyone who's willing to coorganize a Munkres topology study group with me? ( Preferably someone who's proficient in topology)

If so please dm me

[sand hill]

might be a stupid question but every cell complex is a quotient space of the disjoint union of each cell right?

where the characteristic maps of each cell in the cell decompositions restricted to the boundary of each cell serve as the attaching maps

and then you could say furthermore that a cell complex that satisfies closure finiteness but whose topology (which is still the quotient topology) is weak in the closed cells in its decomposition is a CW complex

im just trying to make sure i have intuition straight

its all kind of confusing

[rancid umbra]

i think that sounds right. i’m a bit rusty but that feels right. all of this is somewhere in hatcher and lee’s ITM

[gritty widget]

can i join that one?

[sand hill]

yeah lee just explained the characteristic map very loosely and hatcher didnt seem to spend much time on complexes, at least in chapter 0 which is where id expect to find them

thanks for confirming

[civic verge]

Hi guys a question, this definition is telling me that S is a topological space of Y, so where is the fact that Y is a topology of X?

[granite crane]

S is the relative topology to set Y
likewise \tau is the topology of set X

(i guess this tho)

[civic verge]

I just answered this way

$Y\subset X$, $(X,\mathcal{T})$ then $(Y,\tau)$

[gentle osprey]

Homology

[thorny agate]

$(Y, \tau)$ doesn't make sense because $\tau$ (or $\mathcal{T}$ as written in the picture you sent) is a topology on $\mathbf{X}$, not $Y$. So what you wrote doesn't ``type check'' so to speak.

[gentle osprey]

Spamakin🎷

[thorny agate]

in fact if $Y$ is a strict subset of $X$, then there is an open set in $\mathcal{T}$ which \emph{cannot} be an open set of $Y$ regardless of whatever topology you put on $Y$, namely $X \in \mathcal{T}$

[gentle osprey]

Spamakin🎷

[civic verge]

Thanks for the clarification, I just wrote it that way to get out of the way, I understand its meaning now, that's what's important.

❤️

[sharp aurora]

if I have a projection map $\pi:\mathbb{R}^2\to\mathbb{R}$ where $\pi(x,y)=x$ is $\pi$ forward continuous?

[gentle osprey]

sushi

[rancid umbra]

does forward continuous mean that it takes open sets to open sets?

[sharp aurora]

yes

[iron bolt]

it is, yes. projections of products to their factors always are

but the property is usually called "open"

[white ginkgo]

Ah, I think I see how to take it from here. Thanks!

[mellow bridge]

i guess this question is probably appropriate

how do we define the topology over a matrix group?

if i understand correctly, matrix groups can be seen as manifulds

i just dont get where the topology comes from/what theyre manifolds over

[plush folio]

M_n(R) can be identified with R^(n^2), by just considering each entry of the matrix as a coordinate, so we can give M_n(R) the topology of R^(n^2) through this identification. There are also various matrix norms you can use which AFAIK will all give the same topology too

Not sure if you can use matrix norms to induce a manifold structure actually 🤔

[iron bolt]

a matrix norm is just any norm on R^n⨯n, right? and that is a finite-dimensional vector space, so all norms are equivalent

so they indeed all induce the same topology

[plush folio]

yeah, the topology would be the same, but I'm not sure if you can induce a smooth structure through a norm. I think you would need the correspondence with R^(nxn) either way

[iron bolt]

showing that all matrix groups are topological groups is easy then - just show that matrix multiplication is continuous and that taking inverses of invertible matrices is continuous. then GL(n,R) is topological, and so all subgroups of it must be too

for Lie groups you actually need to show for each group that it is a manifold

that is nontrivial

there are even matrix groups that are not Lie groups, like the subgroup of SO(2) consisting of all rotations by rational angles

so you really do need a proof for each specific group

[plush folio]

Isn't there a theorem that says that a closed subgroup of GL(n, R) is a Lie group? That would save a lot of work

[iron bolt]

https://en.wikipedia.org/wiki/Closed-subgroup_theorem

true. that's cool

[civic verge]

ye continuous

[cedar jungle]

Show that a closed subset of a regular space is equal to the intersection of all open sets that contain that closed set.

[paper wedge]

maybe try to use the definition of every nbd has contains a smaller nbd and its closure

[cedar jungle]

but how does that help?

[paper wedge]

well ig that was stupid

my bad

it follows much more easier from the first definition

[tender halo]

isnt any subset of T1 space the intersection of all open sets that contain it

[paper wedge]

@cedar jungle do it by contradiction

[rancid umbra]

@cedar jungle
one inclusion is clear, i.e., if C is a closed subset, then C is contained in the intersection of every open set containing C.

for the reverse inclusion, if x is a point in every open set containing C and x is not in C, how can you use the definition of regular to reach a contradiction

[paper wedge]

suppose you have an element in the intersection of all open sets that contain the closed set

but is not in the closed set

how can u seperate things so u get a contradcition

[iron bolt]

can't help but point out that this is just a problem statement, not a question

[gritty widget]

Lol I was gonna say the same but obviously it was a question

[mellow bridge]

ah i see

thanks!

[median sand]

A video I was watching about CW-complexes stated that the attaching/characteristic maps are not "part" of its structure, only the filtration is and the condition that each step in it be a cell attachment, e.g. how [-1,1] is a 1-cell attachment to {-1,1} either via the identity map or via the negation map. But how can we then speak of the cells of a complex, if these are the images of the characteristic maps? Should I check manually that given a fixed filtration any collection of attachment/characteristic maps produces the same cells?

[median sand]

Actually nvm, the n-cells are simply the components of X^n\X^n-1, aren't they?

[paper wedge]

yes

[cedar jungle]

Find an example of a topological space that is T4 but not T2

[opaque scroll]

How are you defining T4? Usually the definition includes T2 (or T1), but if you remove that you could just take the trivial topology.

[cedar jungle]

We defined normal spaces as T4 + T1

every two disjoint closed sets of X have disjoint open neighborhoods

[opaque scroll]

In a T1 space points are closed

So then T2 is a special case of T4

[cedar jungle]

So if space is not T1 then points are not closed

[opaque scroll]

That's right

[cedar jungle]

so any space with a trivial topology works?

Because you have no disjoint pairs of closed sets

[opaque scroll]

Yes, you could also do something like X = {0, 1} with open sets Ø, X, {0}

[cedar jungle]

thanks

[cedar jungle]

Show that an open subspace on an Euclidean space has countably many connected components.

[urban zinc]

what have you tried

[cedar jungle]

nothing really, I don't know even where to begin

[urban zinc]

What do you know about Euclidean spaces that has to do with countability

[cedar jungle]

Every Euclidean space is R˘n with the Euclidean metric. It is not countable. But it contains a subset Q^n which is dense and countable

[urban zinc]

Yup

In particular, it has a countable basis for its topology

Do you know what that means?

[cedar jungle]

yes

[urban zinc]

Okay, can you give an example of a countable basis for R^n?

[cedar jungle]

(a,b)^n, a<b, a and b rational?

are you there?

[urban zinc]

I'm not entirely sure what you mean by that

Could you write it out in set-builder notation?

[cedar jungle]

I don't know what that is

[urban zinc]

Nvm

[cedar jungle]

I meant (a,b) as an open interval

[urban zinc]

Are you saying like take an interval (a,b) and take a Cartesian power?

[cedar jungle]

yes

[urban zinc]

(a,b) x (a,b) x ... x (a,b)

[cedar jungle]

yes

[urban zinc]

That's not going to work

[cedar jungle]

why not?

[urban zinc]

You need to allow it to be a different interval for each dimension

[cedar jungle]

oh ok

what about open balls?

[urban zinc]

But otherwise yea that's correct

[urban zinc]

Yeah that would be another example of a countable basis if you just take open balls centered at rational points w/ rational radius

[desert vortex]

just let your set = {x in X | x is bla bla etc etc} this is set builder notation

[cedar jungle]

ok, thanks

[urban zinc]

Okay so now how can you use the countable basis to conclude there are at most countably many components

[cedar jungle]

every component can be written as a union of countably many sets from the basis

[urban zinc]

Why? And how does that tell you that there are at most countably many components?

[cedar jungle]

because the basis is countable

[urban zinc]

Which question are you responding to?

[cedar jungle]

this one

[urban zinc]

There's two questions in that message

Which one?

[cedar jungle]

the first. "Why?"

[urban zinc]

Okay, how do you know the components are a union of basis sets?

[cedar jungle]

component is a union of all connected subsets of a space that contain a chosen point, by definition

[urban zinc]

Yes, what does that have to do with the basis sets?

[cedar jungle]

uhhh

[urban zinc]

Haha it is true that all of the components are unions of basis sets in this case, but note that all you actually need to prove is each component contains a basis set

Try proving that first

[cedar jungle]

okay

I can just take an open ball that has a center inside the component and that has a small enough radius

[urban zinc]

Yeah why does that exist?

As in why is there a small enough radius

[cedar jungle]

Q^n is a dense subset of R^n so if there are real numbers then there must be also rational numbers in between

[urban zinc]

What does that have to do with the component?

[cedar jungle]

every component must contain a point with rational coordinates

[urban zinc]

Why?

[cedar jungle]

because Q^n is a dense subset of R^n

[urban zinc]

What does that have to do with the component?

I think we're getting sidetracked here, what info in the problem have we not used yet?

[cedar jungle]

I don't know

[urban zinc]

Repost the problem again

[cedar jungle]

Show that an open subspace on an Euclidean space has countably many connected components.

so we have an open subspace

[urban zinc]

Yes!

We haven't used openness yet

[cedar jungle]

okay

if it's open it has to contain a set from the basis

[urban zinc]

Can you be more precise? For every point in an open set, ...

[cedar jungle]

there is a set from the basis that contains that point

[urban zinc]

and that set from the basis is a subset of what?

[cedar jungle]

it's a subset of the component

[urban zinc]

No

Or at least we haven't proven that yet

What's the definition of basis for a topology?

[cedar jungle]

a group of open sets is a basis of topology if every set from topology can be written as a union of those sets

[urban zinc]

Right so then for any open set U and every x in U, there's a basis set which contains x and is a subset of what?

[cedar jungle]

it's a subset of U

[urban zinc]

Yes perfect

So let's call the open subspace in the problem U

We have x, contained in some component of U

What can we conclude?

[cedar jungle]

there is a basis set which contains x and is a subset of U

[urban zinc]

Yup! and what we'd like to do now is show it's a subset of the component as well

But how do we do that

[cedar jungle]

components are open

[urban zinc]

Why?

[cedar jungle]

I'm not sure

[urban zinc]

Let's step back a sec, what was the definition of component?

[cedar jungle]

component of a point x from X is a union of all connected subsets of X that contain x

[urban zinc]

so how can we show this basis set is a subset of the component of U that contains x

[cedar jungle]

a basis set is connected?

[urban zinc]

Yup! in this case

why?

[cedar jungle]

Euclidean space is locally connected

[urban zinc]

yup

so now we've proven each component contains a basis set

how do we show there are at most countable components?

[cedar jungle]

Euclidean space has a countable basis for the topology

[urban zinc]

we know the basis is countable yeah

so why are there at most countable components

[cedar jungle]

and every component contains a basis set

[urban zinc]

so in other words there's an injective function from what set to what set?

[cedar jungle]

so there are fewer or equal components that basis sets

there is an injective function from basis sets to components

[urban zinc]

other way

[cedar jungle]

oh

there is an injective function from components to basis sets

[urban zinc]

yes! from the set of components to the basis

why is this function injective?

[cedar jungle]

a basis set can only be contained in one component

two diferent components are disjoint and therefore cannot contain the same basis set

thanks

[urban zinc]

you're welcome!

[chrome plank]

Little question

The book I'm using stated the following proposition:
Let f:S->T be a continuous injective mapping, then, if T is Hausdorff, so is S

So I was wondering, is the reverse true? That is, if S is Hausdorff, can we assure T is as well?

[finite token]

No

Consider the identity map from X with the discrete topology to X with the indiscrete topology

[chrome plank]

Hm, alright

[quick delta]

Intuitively, an injective mapping can only “control” its image
So you have no chance of a global property like Hausdorff-ness transferring
(Technically things are slightly more complex than this in general, but it’s a half decent intuition)

[civic verge]

Which topic is studied for the first time, basis of a topology or subspace?

[iron bolt]

neither depends on the other, so it doesn't really matter

bases are useful to understand things like the product topology a bit more concretely, while the subspace topology is just the topology induced on a subset of a space by the inclusion

[finite token]

Well, the property of not Hausdorff-ness would hold

Not sure if that counts as a "global property" though

[opaque scroll]

Being Hausdorff is a property of the form "for every pair of points in the space blabla"

So non-Hausdorffness is a property on the form "there exists a pair of points blabla".

So it makes sense that if you only consider some subset, then Hausdorffness still holds, but if you don't examine every pair you can't detect if it fails.

[finite token]

maybe but it's not like every statement beginning "for all x (or x,y) in T, ..." has this property

[opaque scroll]

How do you mean?

[finite token]

well I can construct a silly counterexample

I can't think of any non-silly ones that aren't instead like for every open subset of T or something

[opaque scroll]

I could take a silly counterexample

[finite token]

for all x in T, T-{x} has exactly 10 elements

true of any topology with 11 elements but not true of its subspaces

[opaque scroll]

Well, it is true unless you change what the symbol T means

Like if S is a subspace then it is true that
for all x in S, T-{x} has 10 elements

[finite token]

But that's the point despite starting with a forall condition the insides for Hausdorff still depend on the topology

So we are allowed to change T to be the new topology we're examining

If for every pair of disjoint closed sets C,D in T, there are open sets separating C,D, then T is said to be normal

But a subspace of a normal topology need not be normal

[opaque scroll]

That's true, you can create new closed sets when going to a subspace

But the main point still holds for all of these. Just knowing that it holds on some subset doesn't give you hope of proving it for all x

[finite token]

There aren't many properties I can think of that start with for all x in T

so I can't think of a better counterexample then the 11 point thing

[opaque scroll]

You can always add "for all x" to any statement of course

[finite token]

does that make it a global property?

[opaque scroll]

I think "global property" is just used to describe anything that doesn't descend to (open?) subspaces

[finite token]

descend?

[red yoke]

A property of topological spaces is often called local if it holds on the space iff it holds on some open cover

[finite token]

So Hausdorff-ness is a local property?

[red yoke]

No

Line with two origins has a cover (the two lines) of open Hausdorff subspaces

[finite token]

Can the two origins not be separated by open sets?

[red yoke]

They cannot

[finite token]

Haven't you just given such a separation in constructing the cover?

ah

both open sets contain both origins

ok

hmm

[red yoke]

A map f: X → Y is separated if the diagonal ∆f: X → X ×Y X is a closed subspace (∆f is always injective)

[finite token]

Oh wait nvm

[red yoke]

A space X is Hausdorff if X → · is separated

By the above we see that separatedness is not local on the source (on X)

But from defn it is easily seen to be local on the target (on Y)

[finite token]

I'm not sure I grasp what makes separateness a property of topological spaces rather than a property of the map between them

[red yoke]

A property of maps is local on the source if it holds for a map iff it holds for restriction to an open cover of the source

And local on the target if it holds for a map iff it holds for a map iff it holds for restriction to the pullback cover of an open cover of the target

[finite token]

Hmm

[ebon flower]

well in that case they wouldnt be hausdorff right?
i think theyre just not disjoint

[finite token]

Ye I figured it out eventually

For some reason in my head I was imagining the open cover would automatically form a separation but I forgot about disjointness

All right interesting, I can at least understand the definition of local

[red yoke]

Arguably the whole point of a topology is to define what it means for something to be local

[finite token]

[grave fossil]

topology has many points

closure, cohesion, convergence, etc

[gritty widget]

Pointless topology doesn’t!

[deft swallow]

What’s the point of “covers” especially “finite covers” in topology? Couldn’t you always just pick a cover that is equivalent to the set you are trying to cover meaning every set had a finite cover?

My book claims “A subset of R^n where every open cover of A has a finite subcover is called compact”

This isn’t obvious to me, is there an example you know of?

The open interval is all real numbers between 0 and 1 here, right?

Ahh okay I get it now, thank you so much

[deft swallow]

For this problem couldn’t you take two sub covers, one when n is positive and one when n is negative with the same formula?

Which is a finite sub cover

[iron bolt]

as cover is a collection of open sets - which open sets exactly are you suggesting?

[urban zinc]

that's not a subset of the original cover

also the handwritten note is not correct there

the intervals are centeted at n, not the origin

[deft swallow]

{D(n,1) where n >=1 } and {D(n,1) where n <=1 }

[deft swallow]

Why is it not*?

Also this is the definition of D(x,y) in case that’s not a standard thing

[urban zinc]

those are two subsets, neither of which is finite

maybe you're thinking of the union of them? that would be a subcover, but not a finite one

[deft swallow]

Ohh I thought it was only two sets of infinite magnitude rather than infinite sets if that makes sense

I’m getting confused about what’s allowed to be infinite: the number of sets or the magnitude of the sets

But I think I am confused because my answer is a set of sets rather than set of elements from R

[urban zinc]

So, an open cover is a set of open sets

A subcover is a subset of that which is also a cover

Which means it is also a set of open sets

Compactness says that you are always able to find a finite subcover, so a finite number of open sets from the original open cover which still cover the space.

So in this example it's asking if there's a finite number of values of n such that when you take the union of D(n,1) over those values of n, you include all of R

[deft swallow]

That makes sense, so when you pick a subcover you have to pick sets from the original cover and you can’t group a bunch of those sets into a single set

[urban zinc]

yea exactly