point-set-topology

[alpine nest]

Although I'd also express my puzzlement at why you chose that particular route

[wise mist]

3.20 still gives me pain

forgot about it by doing algebra

[wide kayak]

it was a lot shorter than the usual diagonal proof 😅

[alpine nest]

I mean, fair

It's basically a one-liner

[craggy sage]

This reminds me in my Galois Theory course I basically proved S_n acts transitively on the set of roots of an irreducible polynomial if there are n many of them, then used that theorem to solve a different question

[wide kayak]

yup. I argued about it and the prof wouldn't budge. Luckily I passed anyway (barely)

[tranquil cosmos]

Simply prove the category theorem as well

[craggy sage]

My instructor said "you solved a problem I was gonna give in in thw next assignment"

[wise mist]

for me its IVT and something with connectedness

ahhh math is circular

[alpine nest]

Never mind, this is not true, you can build a counterexample starting with the fact that if C denotes the standard Cantor set, then C+C = [0,2]

I must have been thinking of something else

[wise mist]

good to know about the wrong answer for it is not material less

[alpine nest]

(also that fact about Cantor set is fairly neat and an interesting exercise)

But now I can't stop wondering what was that fact about translates of a first-category set that I was misremembering.

Ahh, it was this: #point-set-topology message

You can translate a first-category set so that the translation contains no rational numbers.

[craggy sage]

Power move

[opaque scroll]

Hmm, is that argument circular? I don't remember how you prove BCT

[wide kayak]

I didn't think so, but if that's the case I definitely accept getting zero for it

[opaque scroll]

I don't think it's circular, but you are kind just hiding some steps in a proof that wasn't given

Still deserves at least partial credit imo though

[wise mist]

this problem set asks me to prove prove special case of BCT or R1
and that gives me nightmare

[rancid umbra]

if Y is reducible, it can be written as two non-empty, proper, closed subsets (in Y) C U C’ = Y n (D U D’) for D,D’ closed in X. further, since Y = Y n (D U D’) , then Y = D U D’, so Y is closed in X

[opaque scroll]

You sure you want it to be a disjoint union here?

Anyway, if Y is contained in a closed set, then so is the closure of Y.

[opaque scroll]

You're a bit quick to assume Y is equal to DuD'

[rancid umbra]

thanks

[acoustic fog]

Do any of you know of a book on general topology that is about 2000-2500 pages? I have been looking for it but to no avail. It is 4 volumes I think and it is available online. It has a green-yellow cover.

[mighty hull]

The Bourbaki General Topology books?

[acoustic fog]

No it is much more modern

[mighty hull]

Beats me, then!

[sinful cloak]

2000 to gasp 2500 pages?

Freely available?

That must be an awesome reference.

[acoustic fog]

I have found one that is more advanced and it is like 3700 pages but it contains more than general topology

[acoustic fog]

That one was certainly a good resource but I can't find it

[sinful cloak]

We need more pages.

[sinful cloak]

Link?

[acoustic fog]

https://friedl.app.uni-regensburg.de/papers/1at-total-public-october-14-2022.pdf

[civic verge]

Topology Stefan Friedl

[acoustic fog]

That is the book I have linked to but I don't think it is the same as the one I am searching for. But it might be, I don't know

If I remember correctly the book I am talking about was for general topology only

[inland thistle]

is there a set theory channel here?

I have a question regarding set theory, but I could not find it anywhere

and I want to ask a question here since this channel is probably most relevant

I simply would like to know any paper or youtube video that explains about the history of axiom of choice, or how we can stop worrying about using the axiom of choice

[opaque scroll]

There is #foundations

[inland thistle]

ok thank you for letting me know

[opaque scroll]

As for how to stop worrying, that's easy, just stop. Other than the fact that it was added in later, AC isn't that different from the other axioms of ZFC. At the end of the day you just pick an axiomatic system that lets you do whatever math you want to do.

[inland thistle]

ok thank you so much

[iron bolt]

also, I think one key fact to be aware of regarding choice is that while it has been proven that choice does not follow from ZF - i.e., it really is an extra axiom, not just a theorem - it has also been proven that if ZF is consistent, ZFC is too

i.e. it's guaranteed that you can't prove anything contradictory or false with choice - you can just prove some things that in ZF were neither true nor false but just independent of ZF

that's what sets it aside from large cardinal axioms and such, where assuming their existence via an extra axiom does indeed carry a small risk of making your theory inconsistent even when ZFC is not

[woven jackal]

$A$ is closed if it contains all its limit points, that is $A = A \cup A'$. If we take an arbitrary convergent sequence in $A$, such as ${ x_n } \rightarrow \alpha$, and we can prove that $\alpha \in A$, then have we proven that $A$ is closed?

[gentle osprey]

bot

[woven jackal]

In other words, true or false: if $A$ contains all the limit points of arbitrary sequences within $A$, then $A$ is closed.

[gentle osprey]

bot

[unreal stratus]

This is correct under the assumption that you work with first countable spaces

But not in general

[woven jackal]

Yeah I'm working with metric spaces so I'm good

[unreal stratus]

Epic style

[woven jackal]

Awesome.

[unreal stratus]

Yeah this is a good exercise to do

[woven jackal]

Aka, I can extend this in metric spaces to "bounded sequences" since every bounded set will contain a convergent subsequence

[unreal stratus]

Well that isn't true in general for metric spaces

[woven jackal]

Is it true for any bounded sequence in R^n?

[unreal stratus]

Yes, Bolzano-Weierstraß

[woven jackal]

THanks

[alpine hound]

I need to prove that every Noetherian topological space is also compact. I saw there is a solution on MSE but it is a bit from what I had in mind, so I think there might be a mistake in my solution.

Let $U:={U_i}{i\in I}$ be an open cover for Noetherian top.space $X$. Using axiom of choice, we can choose some $U{i_1} \in U$, and we can choose another $U_{i_1}\ne U_{i_2}\in U$ and so forth. Now, we can look at the chain ${S_j}{j\in I}$, where:
$$
S_j=\cup\limits{k=1}^j U_{i_k}
$$
This is an increasing sequence of open sets, thus it has a "maximum", so there is some $n\in\mathbb{N}$ such that:
$S_n=U$ so we got our finite cover.

Is this wrong?
In second thought maybe I am correct, because Zorn's lemma(which they used in MSE) is equivalent to the axiom of choice.
Any help would be great, thanks!

[gentle osprey]

𝒢𝒾𝓃𝑔𝑒𝓇 𝑀𝒶𝑔𝓂𝒶

[alpine hound]

Actually, the proof is almost identical to the one in MSE, except that I don't use Zorn's lemma. Is this proof wrong?

[desert vortex]

I dont think this is correct because how do you know theres a countable subset of U that covers X ?

youre assuming that the U_(i_k) cover X for k = 1,2,3..

I guess you can say, assume that you can choose $U_{i_k}$ so that $U_{i_k}$ will contain a new point not contained in any $U_{i_j}$ for all $j < k$ that is every next set in the sequence you choose contains an entirely new point not contained in any previous set. If this cant be done for all infinitely many k then we get a finite subcover and we are done so assume we can choose such $i_k$. Then use your argument to show this is impossible because eventually $S_{k+1} = S_{k}$.

[gentle osprey]

wasd

[unreal stratus]

They didn't assume that, but proved it

[unreal stratus]

Yeah this

[peak dove]

The set of all closed intervals is a basis for the discrete topology right? (I realize that it’s quite obvious how it’s a superset of the set of singleton sets and all sets are clopen, but I was just thinking about how we have the usual topology [which can be expressed as open intervals] as a subset of the lower and upper limit topologies [which can be expressed as half open intervals] as a subset of the discrete topology [which can seemingly be expressed as closed intervals].)

[unreal stratus]

Indeed, as it contains all singletons, which form a basis

[desert vortex]

cool beans 😎

[lilac musk]

Not always. Consider the topology on [0,1] with all closed intervals as a basis. Clearly {0} and {1} are not open sets.

More generally, the set of closed intervals is not a basis for the discrete topology on a dense linear order with an endpoint.

[gritty widget]

Degenerate closed intervals.

[1, 1]={1}

[0, 0]={0}

[lilac musk]

Ah
Yeah then you obviously have a superset of the set of singletons

[wise mist]

Show that every second countable topological space is both separable and first countable

the resource never mentioned anything like seperable

what is the canonical interpretation for that

[gritty widget]

Hint for separability:pick some countable basis of the space:every open set intersects this collection in at least one point of one basis element.

[gritty widget]

A separable space is a space with a countable dense subset

[gritty widget]

For first countability, the countable basis is literally also a neighborhood basis(every neighborhood of a point x contains an open subset U which is a union of basis elements)

[wise mist]

ok i will not look at any hint tho i appreciate the input

i will after solving my myself

[wise mist]

thanks im gonna use this, oh my memory is trash nevermind

i was indeed introduced to it

[gritty widget]

Easiest example to remember a separable space is Q and R imo, fwiw. The rationals are countable but every real number can be approached to arbitrary precision without leaving Q

[wise mist]

128gigs of memory with 32 gigs of ROM

[craggy sage]

Yeah, my bad. The definition doesn't require disjointness, and I somehow made myself believe it was the case. I had the answer long ago but yeah I thought it was incorrect 😭

[peak dove]

Btw, even if we weren’t allowing the trivial/singleton intervals, it would still at the very least be a subbasis because any singleton set would be given by the intersection of two with an overlapping endpoint.

[lilac musk]

{0} is not obtainable as an open set in the situation I described

[peak dove]

Ah that’s true.

[median sand]

If
A → X
∩ ∩
Y → Z
is a pushout and A is closed and a (strong) deformation retract of an open neighborhood in Y, how to show the same for X in Z? I've seen some special cases involving compactness and Hausdorff conditions, but I assume this must hold in full generality.

[scenic garden]

How do we know that a torus is not simply-connected?

[gritty widget]

I’ll answer in #alg-top-geo-top

[fringe thorn]

besides the formal answer provided, it’s pretty intuitive that any loop around the ‘equator’ of the torus cannot be shrunk to a point

the same is true for any revolving circle loop

[alpine hound]

The definition of a Noetherian topological space is that every chain stabilizes or that every countable chain stabilizes?

[red yoke]

Every increasing sequence of open sets stabilizes

[scenic garden]

Hmm... it's not intuitive to me

[alpine hound]

So Countable chain, yes?

[red yoke]

I'm not sure what chain means

[lilac musk]

The two are equivalent

[alpine hound]

Why is that?
More specifically, how can one prove that if every countable chain stabilizes then every chain stabilizes

[lilac musk]

Suppose there exists an uncountable chain that does not stabilize. Then one can find a countable subchain that does not stabilize

[alpine hound]

Fair enough.

[alpine hound]

So I don't understand @desert vortex comment about why my proof is incorrect.

[opaque scroll]

How are you defining a chain stabilizing such that this question even makes sense?

[alpine hound]

I guess a well order on some infinite set of unions

[opaque scroll]

Okay so you define chains to be well ordered, and then stabilizing is just that there is some point after which all elements are equal?

[desert vortex]

I just harped on some irrelevant detail that is ultimately not very important because its obvious

[alpine hound]

Yes

[alpine hound]

So my proof is fine?

[desert vortex]

basically, apart from some details that you neglected

[alpine hound]

So can you expain it to me again, because I didn't understand what you wrote earlier

I never assumed that U_i_k for k=1,2,3 is covers X

[desert vortex]

first of all your proof was wrong because you cant choose U_(i_1) not equal to U_(i_2)

because your open cover can consist of the same set over and over again

so you say ok assume we CAN choose an infinite sequence of sets that contain new points or else if we cant do this then we have an open cover by contradiction

its basically contradiction proof

do you understand?

[alpine hound]

Yes

But by new points, you don't mean that U_i_1 sect U_i_2=empty
You mean that U_i_1 sect U_i_2=V where V is a PROPER subset of U_i_1 and U_i_2

For all 1,2 LOL

[desert vortex]

no I mean simply that we choose $U_{i_{k+1}}$ so that $(\bigcup_{1 \leq i \leq k} U_{i_k}) \cap U_(i_{k+1})$ is not empty

bruh

[gentle osprey]

wasd

[desert vortex]

because else we have a finite open cover if we cant choose an infinite sequence of $i_k$ like this

[gentle osprey]

wasd

[alpine hound]

Ohhh

Ok

Now I understand

ok

[desert vortex]

😄

basically covering all your bases, two cases: we can find an infinite such sequence, or we cant, if we cant we have an open subcover, if we can then we use noetherian property on the $\bigcup_{1 \leq i \leq k} U_{i_k}$ to show it cant be. done

[gentle osprey]

wasd

[alpine hound]

I don't understand the "if we can" case

Then we do what I suggested?

[desert vortex]

because if we can then we have an infinitely strictly increasing sequence of open sets

[alpine hound]

So my approach works only for "if we can" case, that is what you are saying?

[desert vortex]

basically yeah

[alpine hound]

And just to be clear, I know(now) that chain stabilizes and countable chain stabilizes means the same, but my approach DIRECTLY uses which of them?

If my question makes sense

[desert vortex]

dont know

[alpine hound]

Lol

[desert vortex]

good luck! 😄

[alpine hound]

Thanks haha

[opaque scroll]

Countable

[timid pier]

are x+x^3 and 2x topologically conjugated?

[gritty widget]

Never heard that term before, not sure.

[iron bolt]

probably means they're conjugated with respect to some homeomorphism R ≃ R, right?

it sounds plausible, but I don't know exactly either

[timid pier]

yeah

[iron bolt]

my guess would be that all functions of the form x + f(x) for for f strictly increasing are topologically conjugate, but it's been too long since I last thought about dynamics to remember details

[naive trench]

Quick question, border of a compact set is compact? Considering C^n

[unreal stratus]

yes, as the boundary of a closed set is a closed subset

[iron bolt]

I just realised - maybe you should ask this in #dynamical-systems? I suspect you might get a better answer there

since topological conjugacy is kind of a notion of isomorphism of dynamical systems

[warm hedge]

probably a stupid question but could we say that every cauchy sequence converges somewhere ? What i mean is that for every complete space of course we can say that, but if we take a non complete space, could we say that the cauchy sequences converges but at points outside of the space ?

or to rephrase. Can we make any metric space complete (without change its metric) by "adding" some extra points to the space ?

[gritty widget]

this is called a completion of a metric space

[unreal stratus]

Lol everyone answering simultaneously

This is a very good quesiton though!

[warm hedge]

i will check it

[warm hedge]

when we are doing something like that we are changing the metric, right ? or it isnt necessary ?

also could we do it the other way around too ? Like take a metric space and "remove" some points to make it complere ?

[gritty widget]

The original space is isometric to a dense subspace of the the complete space made by the completion, as far as I’m aware, so it doesn’t change the metric.

[woeful coral]

If I have a sub-base that generates a topology tau, for every open set can i find an element in the subbase which contains this open set?

i dont see the argument if it is true unfortunately

specifically this. this was given as a remark, but I don't know how to argue this

[gritty widget]

not typically, no

[woeful coral]

is there an easy counter-example?

[gritty widget]

The set of open intervals in R is a basis and hence a sub-base

[woeful coral]

Perhaps this is true for linear topology?

[gritty widget]

Wait no

Not the set of rational open intervals

That works

Then

Wait no

I’m tripping lmao

Shoot

The set of rational open intervals of length less than 1, yeah

Then (-pi, pi) is not contained in any basis element

[woeful coral]

damn

so in a general topological space one can't use base neighbourhoods to talk about convergence?

one has to use all neighbourhoods?

[gritty widget]

I’m not entirely sure what you mean. If some net converges to a point, then it must eventually belong in every basis element containing that point, too

[woeful coral]

I have the following definition:

$x_\alpha \longrightarrow x$ if every neighbourhood contains a tail of $x_\alpha.$

[gentle osprey]

Brayden

[gritty widget]

Yeah

[woeful coral]

okay

so how do I show that this definition is equivalent to:

given a subbase $B$ $x_\alpha \longrightarrow x$ if every subbase of x contains a tail of $x_\alpha.$

[gentle osprey]

Brayden

[gritty widget]

Do you mean every element of some subbase of x?

[woeful coral]

yes

exactly

not some

we have a subbase already

[gritty widget]

Every open set around x is by definition an arbitrary union of finite intersections of subbase elements. If two subbase elements contain a tail, their intersection contains the smaller tail, at least, for “normal” nets, e.g. say a sequence or generalized sequence.

Not sure about in general

And unions obviously contain them

All you have to do to prove this is prove finite intersections of sets each containing a tail contains a tail

[woeful coral]

okay I agree about intersection, just take the common majorant over the net

[woeful coral]

what does proving all this do though?

[gritty widget]

Every open set around x is a union of finite intersections of subbase elements

Hence a union of sets containing tails

They must then too contain tails

[woeful coral]

so if you have the first definition, since every element of subbase is open you are done.

ohhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh

a finite intersection of subbase elements has tail taken by choosing the common majorant over all right?

and then arbitrary union just take any tail

[gritty widget]

I’m not sure what a majorant is

[woeful coral]

a net has the property that if a and b are in the net then there is c >= a and c >= b

[gritty widget]

Also, to be clear, what’s the definition of net here? Preordered or partially ordered? Directed?

[woeful coral]

directed set

is it not always treated as a directed set?

[gritty widget]

Directed set sometimes means partially ordered, not preordered, etc.

[gritty widget]

Oh then yeah it applies to all

[gritty widget]

It always means directed though afaik, so it doesn’t really matter here

I just haven’t done stuff like this in awhile lol

[woeful coral]

ah no worries

thanks for the help

[gritty widget]

Np

[naive trench]

Thanks

[pallid comet]

hi topology! can anyone help me with proving this (8.60)

i was thinking along the lines of

proving that points p and q can be connected by a path in R^2 minus finitely many disjoint open discs

and that Cantor set can be covered by disjoint open discs (and so we can reduce to a finite cover)

[pallid comet]

so the proposition i'm currently stuck on is:

for every cantor set B in the plane and 2 points p, q not in B, there exists a covering of B by pairwise disjoint open discs, none of which intersect p or q.

can anyone help me with this? or is there a different, simpler approach?

(i also included the next theorem as well in the screenshot to show that i can't use that yet - and will also have to prove that later)

[cosmic pier]

This is very interesting! Here is my plan of attack. First i am going to prove that if I am given an embedding of a compact interval to the plane, then the complement of its image is still path connected and it's image has empty interior. Assuming we have that result, since the Cantor set is the intersection of a numerable collection of compact intervals, the complement of the image of the given embedding is a union of complements of the images of those intervals. The intersection of those complements is non empty (because of Baire's theorem for complete metric spaces) and therefore the complement of the image of the Cantor set has to me path connected, as it is the union of path connected sets with non empty intersection. Furthermore since the complement of the image of the Cantor set is open and path connected, it is path connected by polygonal paths.

My proof of my first assertion, namely that the complement of the image of a compact interval via an embedding stems from the belief that i can approximate (uniformly) the embedding by a sequence of injective piece-wise linear functions. The complements of the images of those piece-wise linear maps is path connected and therefore the complement of the image of the original embedding also has to be path connected.

I hope it makes sense. Sorry for not writing down explicitly the construction but i don't have enough time, though as i said i found your problem extremely interesting.

I forgot to give a justification as to why the image of a compact interval via an embedding needs to have an empty interion. I guess it is for the same reason as to why the real line and the plane are not homeomorphic.

[tranquil cosmos]

My ad hoc proof of that first claim is

• I is homeomorphic to its image J, because it's an embedding
• thus J is contractible in R2, perform a homotopy contracting it
• then you get R2 minus a point, which is path-connected

It does rely on compactness of the interval tho, cause you could injectively map [0,1) to the circle and then the complement isn't path connected

[cosmic pier]

Yeah, i guess you can see that more clearly when you try to write down the homotopy you suggested.

[gritty widget]

Dimension is a topological invariant, and it’s well known that any subset of R^n has dimension n iff the interior is non-empty(and an embedding is a homeomorphism to the image, hence preserves dimension)

[cosmic pier]

Yeah I pointed this out in my justification. Though the invariance of dimension is too strong of a result since we work in the plane.

[gritty widget]

technically, you don’t need IOD

[pallid comet]

hmm- wouldnt the step where you used the fact that the cantor set being the intersection of compact intervals to show that the complement of the image is open require extending the embedding of the cantor set to an embedding of the interval though

(which is also what im trying to do but so far havent gotten anywhere yet)

[cosmic pier]

Why would that be?

[gritty widget]

Cantor set is compact

Image is compact(hence closed)

[pallid comet]

sorry - please correct me if i'm wrong, but here's what I was thinking when I read your plan:

lat h be the chosen embedding of the middle-third Cantor set C into the plane.
I was thinking that your argument would go something like:
Since C is the intersection C_0 intersect C_1 intersect C_2 intersect ..., the image h(C) would be h(C_0) intersect h(C_1) intersect ... (and hence the rest of the argument)

but that would require h to be defined on [0,1], not just C

[cosmic pier]

Ohhhhhhh yeah okay you are right

[pallid comet]

so I was thinking

maybe I could extend h to [0,1] by filling in the gaps between points of the cantor set with paths?
formally, if we wrote [0,1] \ C as a countable collection of disjoint intervals (a_i, b_i), we could define h on (a_i, b_i) to be some arc connecting h(a_i) and h(b_i)

and since it's countable we could use induction to make sure the arcs don't intersect

[cosmic pier]

yeah that sounds good

[pallid comet]

but then i realized that those arcs also can't intersect h(C) and now i'm pretty much back to square one (smh, after like 2 hours of writing the proof)

[cosmic pier]

Which arcs?

Ahhh i see

[pallid comet]

h on intervals (a_i, b_i)

[cosmic pier]

so that you preserve injectivity?

[pallid comet]

yeah

[cosmic pier]

I was very stoked with my solution, but i guess topology crashes your dreams every time

[pallid comet]

indeed

[cosmic pier]

what was your initial idea?

something to do with open discs?

[pallid comet]

oh yeah it was initially

let p, q be two points outside h(C). i was trying to cover h(C) by pairwise disjoint *open discs that don't contain p or q

[cosmic pier]

oh right

[pallid comet]

so we could reduce that to a finite cover and be done

[cosmic pier]

yeah yeah i see

Is that the same as claiming that each point of the image is an isolated point?

[pallid comet]

one open disc can (actually, must, but that's beside the point) contain multiple points

[cosmic pier]

yeah

otherwise you would have problems with compactness

[pallid comet]

but i dont think this idea is very good actually since it feels too dependent on geometry

[cosmic pier]

I was thinking something like Tietze's (spelling) extension theorem, but then you would have to make sure that the extension is injective

oh i am an idiot

it aplies to real functions. You could probably apply it coordinate wise but i dont think that leads anywhere

[pallid comet]

is that short for "i solved the problem and its like really simple" or "topology crushed my dreams again"

.oh

i mean tietze works for R^2 too

[cosmic pier]

yeah

but i think it is too general of a result

i don't see a way to alter the construction given in the proof so that it is injective as well

simply because you start with the cantor set

I might be wrong though

Though, if you use Tietze you might be able to salvage the arguments i gave

i think

[pallid comet]

hmmm i just discovered that the embedding can be extended to an embedding of [0,1] is actually a named theorem! https://en.m.wikipedia.org/wiki/Denjoy–Riesz_theorem

[cosmic pier]

I was thinking that maybe you could salvage some of the results about the images of intervals that are used to construct the cantor sets that i mentioned above, but i dont think the cantor set is "dense" enough for that

Oh it appears that we are done then?

The image of the cantor set is totally disconnected right?

[pallid comet]

if we apply this theorem then yeah! but im still looking for a more elementary proof though

the proofs of danjoy riesz appear to be way beyond the scope of the book i'm reading

and judging by the fact that the authors made this an exercise, there's probably some clever solution

[cosmic pier]

One other idea i had, but i guess was too geometrical was that if we were to assume that for every point of the image there was an open disc such that the complement of the embedded image of the cantor set, in that open disc, was not path connected we could probably deduce that injectivity fails

by considering the "end points" of the image, but again that is too hand wavey and intuitive

If you were to permit an idea from complex analysis, perhaps i have a solution. Let us assume without loss of generality that (0,0) is not a point of the image. Then, by Tietze we can find a continuous extension of our embedding on [0,1], call it H. I work with the assumption that i can somehow construct H such that (0,0) is not in its image. Then i can find two continuous functions R and φ such that H(t) = (R(t)cos(φ(t)),R(t)sin(φ(t))). H cannot be closed, since both 0 and 1 are in the Cantor Set.

[rancid umbra]

i mean, you can just do this for the standard embedding into the x-axis using the next theorem (8.61). then approximate the path via small enough p.w. linear paths after a homeomorphism

[cosmic pier]

yeah, but i think it was mentioned that we cannot use the next theorem for the solution

i think

[rancid umbra]

i suspect they are independent of each other

so it shouldn’t really matter

but if you aren’t allowed to do that, oh well

more complicated arg

after reading through, i don’t see chloe saying anywhere that they cannot use theorem 8.61

[cosmic pier]

here?

[rancid umbra]

ah. reading is so hard

[pallid comet]

nope :C

[pallid comet]

HMM sorry i will need some time to unpack this

[cosmic pier]

yeah sure, i am just throwing ideas

[north dust]

proof is enough?I just have some doubts abt choosing h
Theorem.Let E' be set of all limit points of E,then E' is closed.
Proof.
Let x be limit point of E'.By def of limit point we can find q for any ε neighbourhood such that q is in E'.
Therefore
d(x,q)=ε-h/2>0 where ε>h.
Because q is in E' that means its limit point of E therefore there exists point p ≠ q in any neighbourhood of q such that p is in E.
Let h/2 be its neighbourhood,which implies d(p,q)<h/2.
To show that x is limit point of E,first we show p≠x
d(x,p)≥d(x,q)-d(p,q)>ε-h/2-h/2=ε-h>0
now apply triangle inequality again
d(x,p)≤d(x,q)+d(p,q)<ε-h/2+h/2=ε

[prime elbow]

i think it is correct but there exist point p not equal to q in any neighbourhood of q such that p is in E, it is wrong you have to say for any neighbourhood of q there exist p in that neighbourhood not equal q such that p is in E

[north dust]

Yeah forgor

[frail edge]

hello! i'm working on origami and came across this paper from Robert Lang. Do someone know if a proof exists for the statement highlighted ?

[granite slate]

Why is $\tau = {(a, b) | a, b \in \mathbb{R}}$ second countable but not $\tau = {[a, b) | a, b \in \mathbb{R}}$ for instance if $q, p \in \mathbb{Q}$ isnt the basis $\tau = {[q-p, q+p)}$ a countable basis?

[gentle osprey]

Dompa

[opaque scroll]

How would you write [sqrt2, 5) as a union of such sets for example?

[granite slate]

oh okay so the last basis isnt a basis for the same topology?

[opaque scroll]

Yeah, a basis is only a basis for one topology

[gritty widget]

Can’t you just pick some decreasing sequence of rationals, say, ${q_n}$ that converges to $2^{1/2}$, and then take the union $\bigcup_i{[q_i, 5)}$?

[gentle osprey]

NAT Enthusiast

[opaque scroll]

You can do that, but the union won't be [sqrt2, 5)

[gritty widget]

Oh lmao

Idk how I missed that

thanks

[alpine hound]

Let $f:X\to Y$ be a dominant map, and let $U$ be dense in $X$, is it true that $\text{Cl}(f(U))=Y$?

[gentle osprey]

𝒢𝒾𝓃𝑔𝑒𝓇 𝑀𝒶𝑔𝓂𝒶

[alpine hound]

I have an idea.
$\text{Cl}(f(U))=\text{Cl}(\text{Cl}(f(U)))\supset\text{Cl}(f(\text{Cl}(U)))=\text{Cl}(f(X))=Y$

[gentle osprey]

𝒢𝒾𝓃𝑔𝑒𝓇 𝑀𝒶𝑔𝓂𝒶

[alpine hound]

I am just not sure if \text{Cl}(f(U)) \supset f(\text{Cl}(U)) is correct

[opaque scroll]

You can try to prove it.

A hint could be that
f(A) < B
iff
A < f^-1(B)

[alpine hound]

Does it mean that my proof is correct? 🙂

[minor flint]

Hello everyone! I’m trying to understand second countability and the product topology. I came across the following exercise which asks me to show that the space of the real sequences(R^N) is second countable with the product topology. The way I’m thinking of proving this, without being extremely rigorous, just the thought process, would be to say that sth like the union of open balls with a radious of 1/n or the union of intervals with rational sides are countable bases in R so and open set in R^N with the product topology which has such unions for a finite amount is indeed countable(because it’s the product of a finite and a countable set) and a basis for R^N. Am I getting this correctly?

Thanks in advance

[jovial field]

Hey folks. I'm trying to show that the epsilon delta definition of continuity is equivalent to the topological one, except I'm dealing with metric spaces here.

So my question is this: in the only if direction, I assume that pre images of open sets are open blah blah. So now I have to show that given any epsilon, the associated delta exists.

Suppose I take (0, 1) and have open sets be unions and intersections of open epsilon balls. If I can have access to "any epsilon", I could choose 2. Am I just kind of stupid for thinking this? Would you just call this identical to the distance to the nearest endpoint of the interval and finish from there...

Yeah I'm probably stupid. Epsilon delta allows any epsilon because the real numbers is a big set, yeah? An open ball takes from elements of the space, not a super set. So we just assume a reasonable epsilon for an arbitrary metric space.

[finite token]

Yeah if you have a basis for the topology of every space in a product you can use them to form a basis for the product space (that basis being the set of products of basis elements where all but finitely many are the whole space)

When all the spaces have a countable basis, and the product is countable, this gives a countable basis for the product

[quick crane]

you have to find delta for every epsilon. You don't get to choose epsilon

[cosmic pier]

You have to use the fact that the collection of all open balls in a metric space forms a basis for the metric topology

[arctic mural]

On exercise 3 i don't understand why the sequence {1,2,3,...} lives in the cofinite topology. Shouldn't the sequence be something like ( {\mathbb{Z}\setminus{1}, \mathbb{Z}\setminus{2}, \ldots})?

[gentle osprey]

pingolfingo

[tranquil cosmos]

don't think so? your sequence consists of elements, not open sets, just like it does in analysis...

[arctic mural]

You are right

I was thinking the sequence should be elements in the topology

Instead of the topological space

My bad

[ruby hinge]

Hi, I'm currently stuck on a topology problem It's quite elementary I suppose, but it's not really working.

Let $X$ be a finite set with a topology $\mathcal{T}$. Prove that $\mathcal{T}'={U\subseteq X: X\setminus U \in \mathcal{T}}$ is a topology on $X$.

For axiom 2, we must prove that unions (possibly infinite) are in $\mathcal{T}'$. I did this:

Suppose for $i\in I$, $U_i\in \mathcal{T}'$. So for any $i$, we have $U_i \subseteq X$ and $X\setminus U_i \in \mathcal{T}$. The union of subsets of $X$ is still a subset of $X$. Further, we have
[
X\setminus \bigcup_{i\in I} U_i = \bigcap_{i \in I} (X\setminus U_i).
]
Because any $X\setminus U_i\in \mathcal{T}$, and $\mathcal{T}$ is a topology, it is clear that $\bigcap_{i\in I} (X\setminus U_i) \in \mathcal{T}$. Thus $\bigcup_{i\in I} U_i \in \mathcal{T}'$.
However, it does not seem to work for infinite unions, since infinite intersections of open sets are not necessarily open. So I'm stuck here.

[gentle osprey]

joel

[rancid umbra]

it is not clear that the intersection. of X/U_i is in T, since I may be infinite, while only finite intersections of open subsets are guaranteed to be open

[ruby hinge]

yes

this was my problem

[rancid umbra]

oh shoot. didn’t read the last sentence

ah right

but X is finite

[ruby hinge]

right, didnt read that part.. i feel stupid haha

[rancid umbra]

it’s good that you caught it

[ruby hinge]

So can you say that the max amount of elements is finite, namely $2^{|X|}$?

[gentle osprey]

joel

[ruby hinge]

Like that this is an upper bound

[rancid umbra]

essentially, yes.

since you have a function f : I —> T’ indexing open sets, the maximum number of unique sets that can be chosen by the index function is 2^|X|

so you can just toss all repeats and “refine” your cover

what i’m saying is, wlog assume that I is finite

[ruby hinge]

yeah alright, i'll write that down.

Thanks so much!!

[rancid umbra]

if you assume that I is a well-ordered set, then you can use the well-orderedness of I to choose a representative index for each non-empty fiber of the indexing function. the number of such fibers is bounded above by |T|, and more weakly by 2^|X|

alternatively, you can quotient by the relation i ~ j iff f(i) = f(j). now there is a unique function F : I/~ —> T’ such that F(p(i)) = f(i) where p : I —> I/~ is the natural projection. finally, one can note that F is a bijection and T’ is finite.

working without an indexing set is really the way to go since you would simply consider collections of subsets of X

[ruby hinge]

@rancid umbra I ended up writing this:
\newcommand{\topo}{\mathcal{T}}
Let $I$ be an indexing set, and $i\in I$, where $U_i\in \topo'$. We know that $X$ is finite, therefore the amount of unique elements in $\topo'$ is bounded above by $2^{|X|}$. So without loss of generality, assume $I$ is finite (if $I$ is infinite, only consider unique elements). \
For any $i\in I$, we have $U_i \subseteq X$ and $X\setminus U_i \in \topo$. The union of subsets of $X$ is still a subset of $X$. Further, we have
[
X\setminus \bigcup_{i\in I} U_i = \bigcap_{i \in I} (X\setminus U_i).
]
Because any $X\setminus U_i\in \topo$, and $\topo$ is a topology, it is clear that $\bigcap_{i\in I} (X\setminus U_i) \in \topo$. Thus $\bigcup_{i\in I} U_i \in \topo'$.

[gentle osprey]

joel

[rancid umbra]

that looks great

[ruby hinge]

So maybe just say "let the U_i's be part of a collection of subsets of X" or what?

Or is this good?

[rancid umbra]

what you have is perfectly fine.

what i was saying is this: it is equivalent to show that $\bigcup \mathcal{U} \in \mathcal{T}’$ for all $\mathcal{U}\subseteq\mathcal{T}’$

[gentle osprey]

c squared

[rancid umbra]

this avoids having to work with an index set

[nocturne matrix]

I was looking at wikipedia for a proof that the Sorgenfrey plane is not normal (https://en.wikipedia.org/wiki/Sorgenfrey_plane) and I found this, but I don't see why there are no disjoint supersets of these two closed sets; since L is discrete aren't they open as well as closed?

[rancid umbra]

there are some proofs of this online if you search for them. it’s a bit in depth

[tranquil cosmos]

$S^{-1}$ is a space so invisible even its suspension lines are invisible

[gentle osprey]

PKThoron

[tranquil cosmos]

So the only thing that gets left over of its suspension is the $S^0$

[gentle osprey]

PKThoron

[tranquil cosmos]

How's THAT for making the smash product a group

[minor flint]

Hello everyone. I’m trying to prove the following lemma about connected spaces. The problem is that I know it’s true if the family Ai has common intersection that it’s non empty. I’m not sure if I can follow the same reasoning if the sets of (Ai) are pairwise disjoint. What do you think of my proof? Does it hold?

[uneven bronze]

I feel stupid for asking, but in the Tietze extension theorem for R (the real line), can the closed subset from which we extend the function be the empty set? In the statements I've been looking at, I haven't seen it explicitly stated that this closed subset should be nonempty.

[ebon flower]

this is just the statement that there exist bounded continuous functions X->R, of which any constant function is an example

[uneven bronze]

why bounded?

[ebon flower]

the Tietze extension theorem says the extension can be chosen to be bounded if the original function is

and moreover, bounded by the bounds on the original

[uneven bronze]

ah ok 👍

[uneven bronze]

I'd be very grateful if someone could explain a step in a proof of the Tietze extension theorem. I'm watching this video for an elementary proof for the real line.

The author defines an extension g of f on R=F^c u F, where F^c is the disjoint union of open intervals (possible semi-infinite ones). The nontrivial part is verifying continuity of g on F. The author proceeds by contradiction. Assume g is not continuous at x_0 in F. Then there's a sequence (x_n) in R such that x_n-->x_0 but g(x_n) does not converge to g(x_0). They claim then that x_n is in F^c for infinitely many n, because otherwise g is not continuous on F. Why is g not continuous on F if x_n in F eventually (and why is this a contradiction)?

This occurs at around 6 minutes into the video.

[opaque scroll]

So I didn't click on this video, but just from what you said:

If xn converges to x0, then any subsequence also converges to x0.

So if xn is eventually in F, since F is closed x0 is also in F. So then you have a sequence xn' in F converging to x0 such that g(xn') doesn't converge to g(x0). Which means g is not continuous.

I guess you already assumed x0 was in F, so you don't need to use that F is closed

[uneven bronze]

thanks. The weird thing is that when they prove continuity of g on F, they assume g is discontinuous at x0 in F and conclude that xn cannot be eventually in F, since otherwise g is not continuous on F. Isn't this logic circular?

[opaque scroll]

There's a difference between g being discontinuous as a function on R and as a function on F.

For example imagine if F = {x0}.

So the assumption is g discontinuous at x0 (as a function on R), but they want g to be continuous on F (as a function on F)

So the two assumptions are actually different, even though they seem similar

That's what I gather from what you wrote anyway

Clicked on the video, and that seems correct

[uneven bronze]

ok, seems like a subtle difference there. I'm still a bit confused. f is the restriction of g on F and is continuous. Why can't we just say that g is continuous on F then?

[opaque scroll]

Well consider the example F = {0}, just the single point 0. And say g(0) = 1 and g(x) = 0 otherwise. Then clearly g is discontinuous, but the restriction of g to F is just a constant function

[uneven bronze]

Ok, this makes more sense now. Thank you. A final question maybe. If xn eventually in F, f would not be continuous on F (which is a contradiction to our assumptions). So the opposite must be true. xn in F^c for infinitely many n. But this could still mean xn in F for infinitely many n, and by passing to a subsequence xn', we'd still have a sequence in F such g(xn')=f(xn') does not converge to g(x0)=f(x0). Is this a correct observation?

[opaque scroll]

Almost, it could be that even though g(xn) doesn't converge to g(x0), the subsequence g(xn') does.

So we need this eventual sequence to guarantee that g(xn') doesn't converge to g(x0)

[uneven bronze]

ah ok. So you're saying there's no guarantee that the subsequence will not converge to g(x0), right?

[opaque scroll]

For example, say F = {1/(2n) : n in N} u {0} and say g(x) = 1 on F and 0 elsewhere.

Then the sequence xn = 1/n is such that g(xn) doesn't converge, but g(x(2n)) is just constant

[gritty widget]

Is #point-set-topology a “dead field”, so to speak? E.g. has there been any recent important work in it? If anyone has any specific examples, I’d appreciate it!

[wide kayak]

it still gets used all the time in active research areas (like linear algebra), but my impression is it's not an active research area in and of itself. I don't know of any pressing open problems in point set topology

[gritty widget]

Alright, thanks!

[tranquil cosmos]

Has there been any effort in describing negative-dimensional spaces?

Earlier I was wondering about a space S^-1 whose suspension is S^0

[hollow tinsel]

Spectra (as in stable homotopy theory, as opposed to any of the other uses of the word) give you models for a homotopy type that "suspends to S^0," among many other things, but it's probably misleading to think of what goes on in that corner of homotopy theory as "describing negative-dimensional spaces"

[rancid umbra]

are there actually spaces whose suspension is S^0

[tranquil cosmos]

I did think of spectra but I don't know that much about them

[rancid umbra]

that doesn’t seem right

[hollow tinsel]

no, there are no such spaces

[tranquil cosmos]

Do they at least formally and successfully describe S^-1?

[tranquil cosmos]

You can't achieve this in point-set or locales, that's for sure

[rancid umbra]

yea

[tranquil cosmos]

It's a bit crazy that we don't have great definitions of what a space is in the 21st century

[hollow tinsel]

In the most common/traditional model of spectra, you "describe S^{-1}" by literally just shifting indices of a grading

it's purely formal

[tranquil cosmos]

Formal is a first step!

Some part of me thinks this also necessitates defining negative dimensional rings

So leaving set theory for rings as well

Maths has stagnated on set theory for too long

Ong

[gritty widget]

I don’t see what’s wrong with “a space is a topological space”

[tranquil cosmos]

Locales are better by far

[gritty widget]

Nah, too abstract fr

[tranquil cosmos]

Well that's like saying varieties shouldn't have been replaced by schemes

AG is 3 centuries ahead of top 🙄

[gritty widget]

Idk any AG 🥱

[wise mist]

Geeked vs locked in

[tranquil cosmos]

We need to outpace AG

[hollow tinsel]

Varieties absolutely were not "replaced" by schemes

[tranquil cosmos]

Is the whole topos the space or just the heyting algebra inside it?

I kinda consider topoi like

The direct sum of a heyting algebra and some set universe with your favorite large cardinals

But idk if the sum is direct xd

[red yoke]

The subobject classifier?

[gritty widget]

anima=$\infty$-groupoid right

[gentle osprey]

NAT Enthusiast

[gritty widget]

What a weird name for them..

[red yoke]

It's a nice name

[gritty widget]

Mfw strict anima are just crossed complexes..

What’s a condensed set? Only one I see there I haven’t heard of.

[red yoke]

The category of κ-condensed sets is the site on profinite sets with cardinality < κ

[gritty widget]

the effects of categorification on mathematics is horrid !

[red yoke]

With finite surjective maps as covers

[gritty widget]

Yeah I’m not even gonna bother any more tonight I’ll just google those tomorrow lmao

[red yoke]

If you equate a profinite set with a "generalized generic convergent sequence", like how N ∪ {∞} is the "generic" convergent sequence, then a condensed set is just some data of generalized convergences

[grave solstice]

The word "space" is overused imo

[red yoke]

I think it's cool how you can define all these notions purely algebraically

Without reference to point set

And then they happen to significantly coincide with sufficiently nice topological spaces

It feels like topological spaces are too overloaded with different usages

[gritty widget]

There’s a lot of seemingly “random” equivalences like that in topology, haha

For example, finite spaces are equivalent to finite preordered sets

Finitely generated spaces are equivalent to preordered sets

[wide kayak]

this reminds me of Scholze’s work on condensed sets - not that I understand it remotely, but I think part of the idea of the program is that these are somehow a better notion than topological space, whatever that means

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

[gritty widget]

Neat!

[tranquil cosmos]

Yeah exactly, the gross amount of pathologies that emerges from the classical definition is a sign that the definition isn't quite right imo

Locales get rid of all the non-sober spaces which is already a good start

[tranquil cosmos]

What's left in my mind about the SOC is that...

• it's like an internal power set of 1
• there's an associated Heyting algebra in the topos "spelling it out"

I thiiiiink what I'm trying to say is that point-set topological spaces, locales, and on the AG side varieties, schemes... are all just "models" of some platonic notion of space (/algebraic space)

And that point-set spaces are a valiant and useful, but rather poor model

And maybe some day we'll find a model that's so good that it can naturally include S^-1 for example

[red yoke]

I've not studied the logical side of topoi very much

My only understanding is that you can use this to do logic with sheaves on the space

I somehow doubt there is a platonic notion of space

[tranquil cosmos]

Tbh i have a hunch that it's not this simple

[red yoke]

More like, topoi are a good way to define notions of space

[tranquil cosmos]

Maybe the platonic notion is the friends we made along the way

[wise mist]

Is this jung ref ?

[tranquil cosmos]

What's the persona?

The 1-skeleton?

[tranquil cosmos]

Stone is mega artificial compared to the Heyting version:

• Boolean algebras yield Stone spaces, which you barely see anywhere
• ??? algebras yield ALL reasonable spaces. And these algebras are called Heyting algebras

[gritty widget]

Just think like an algebraic topologist then..

A space is a CW complex.

[iron bolt]

I don't know how much it deserves being called S^-1 in terms of algebraic topology, but at least naively... the empty space should do, right?

I mean, it depends on your convention as to whether collapsing an empty subset of a space to a point should result in a point or not

but if it does, taking the suspension of the empty space gives you S^0

[tranquil cosmos]

Well one property I'd like from an $S^{-1}$ is that $S^{-1} \odot S^1\cong S^0$ where $\odot$ is the smash product

[gentle osprey]

PKThoron

[iron bolt]

hm, fair enough. it can't do that

or really anything else that works only for pointed spaces

[tranquil cosmos]

Forgot that the smash product needs pointed spaces lol

Smash... groupoid

[tranquil cosmos]

I remember a formal analogy between the injective hull of a module and the Stone-Čech compactification of a space

Anyone know what's up?

[gritty widget]

[wispy dirge]

how is V_n constructed here

and why is (i) true

[rancid umbra]

V1 is essentially B_{r_1}(x) for some r = r_1 > 0

from V1, you can construct V2

and so on and so forth

this is just another way to say, we will inductively construct Vn for each natural number N

to see why (i) is true, recall the definition of a limit point

[wispy dirge]

that much is clear but what about the radius of V_n

if it is just arbitrary

and not related to r ( the radius of V_1)

then how can we be sure that (i) is true for example

[rancid umbra]

it is not arbitrary. it is related to r_1

try going from V1 to V2

[wispy dirge]

in what way

[rancid umbra]

in the way that the third paragraph, "Suppose Vn has been constructed, so that..." prescribes, with n = 1

[wispy dirge]

but in this case the radius can just be arbitrary

because the intersection between V_n and P is not empty regardless of the radius of V_n

[rancid umbra]

for V1, yes, it can be any open ball centered at x

[wispy dirge]

V_n is just any neighborhood of x_n

[wispy dirge]

V_1 is an arbitrary neighborhood with center x_1 and the construction mentioned in "Suppose V_n ..." only requires that the intersection between V_n and P be not empty right ?

also from the way V_1 was constructed V_n must be a neighborhood of center x_n

thus from this, V_n can be any neighborhood centered at x_n because any such neighborhood intersects P in at least one point, namely x_n

[rancid umbra]

put V1 = B_{r1}(x1). V1 intersect P is non-empty, and contains some point x2 != x1 in P.
we need to find V2 such that
(i) cl(V2) subset V1,
(ii) x1 not in cl(V2),and
(iii) x2 in V2.

try drawing a picture

[wispy dirge]

and V_{n+1} is a neighborhood of x{n+1}?

[rancid umbra]

that is just for V1

[wispy dirge]

so the other neighborhoods apart from V_1 arent necessarily neighborhoods of points with the same index ?

[rancid umbra]

Vn will be a neighborhood of xn, if that is what you are saying

[wispy dirge]

so like V_2 isnt necessarily a neighborhood of x_2 , V_n isnt necessarily a neighborhood of x_n , etc...?

[wispy dirge]

yes thats what i am asking

[wispy dirge]

so I suppose that V_{n+1} is a neighborhood of x{n+1} no ?

[rancid umbra]

yes

V_{n+1} is not a nbhd of xn, however

[wispy dirge]

yes it shouldnt be because of ii

[rancid umbra]

right. the phrasing threw me off for a sec

[wispy dirge]

hahaha np

still i dont see how is this guaranteed

doesnt this claim make it necesary that there exists a neighborhood of x_n which contains x{n+1}

and this is not necessarily true is it ?

whats true is that every neighborhood of x_n contains infinitely many points of P

but x_{n+1} may not be one of these infinitely many points right ?

[rancid umbra]

this is the picture that you want to think of

you can check that V2 and V3 satisfy conditions (i), (ii), and (iii)

this would continue ad infinitum

[wispy dirge]

but i need to prove that this chain exists

[wispy dirge]

and with this then it isnt guaranteed is it

[rancid umbra]

each x_n is guaranteed to be a point of P

[wispy dirge]

but that doesnt mean that it is guaranteed to be a point of V_1

if x_{n+1} isnt in V_n

then (i) doesnt hold

[wispy dirge]

and nothing guarantees this

[rancid umbra]

read back over this. we are getting x2 in P cap V1 since P is perfect

[wispy dirge]

V_1 cap P can just be x_1 right

[rancid umbra]

no

[wispy dirge]

why not

ah wait

yes this doesnt work

[rancid umbra]

there is guaranteed to be another point x2 different from x1

[wispy dirge]

ok so there is x_k neq x_1 in V_1

but k is arbitrary

[rancid umbra]

xn differs from xk for all k < n

because of the induction

[wispy dirge]

yes

but still this doesnt mean that it has to be x_{n+1} that is contained in V_n right

it can be any x_k neq x_n

there are infinitely many k with x_k in V_n

but x_{n+1} is not necessarily one of them

[rancid umbra]

yes, the V_n’s are nested

is your issue the indices?

of the x_n’s?

[wispy dirge]

start with V_1 which is an arbitrary neighborhood of x_1. then this neighborhood contains infinitely many points x_i of P so that it contains at least one neighborhood of each of these points since V_1 is open.

so let one of these points be x_n

and let the correponding neighborhood of x_n in V_1 be V_n

now V_n contains infinitely many points of P and then call one of these points x{n+1}

[rancid umbra]

use \_ instead of just _

[wispy dirge]

oh ok ty this always gives me trouble hahaha

[wispy dirge]

so from here there exists a neighborhood V_{n+1} of x_{n+1} which is contained in V_n

is this the way to do it ?

[wispy dirge]

yes

it should be fine if i name the indices after the x_i's that exist right ?

so just like i did

i call them x_n , x_{n+1} etc after i guarantee that they are in the neighborhoods ?

[rancid umbra]

yes

[rancid umbra]

basically. i would be more explicit about how you find x_{n+1} using the fact that P is perfect and how you construct V_{n+1}

[wispy dirge]

what do you mean by that

so you mean that i have to say : P is perfect and x_n in P so any neighborhood of x_n contains points x_k of P neq x_n , then take one of these neighborhoods , call it V_n and let one of these x_k in V_n be x_{n+1}?

[rancid umbra]

there is a geometric argument to be made for constructing V_{n+1} from V_{n}

is what i mean

[wispy dirge]

oh

[wispy dirge]

how to do that

[rancid umbra]

pictures : )

i gtg

[wispy dirge]

i hate that : )

[rancid umbra]

good luck

[wispy dirge]

alr tysm for your help

have a nice day

[rancid umbra]

you too

[wide kayak]

does there exist a continuous one-to-one and onto function [0,+inf) -> S^1 ? I thought I've seen something like this before, where 0 maps to, say, the north pole, and then it continuously wraps around the circle one time, and does an infinitesimal approach back to the north pole. So lim x-> inf is the north pole again, but there is no x > 0 that maps to the north pole. And the image of the map is the entire circle.

This function is supposed to be continuous, but its inverse is necessarily not continuous (the circle and [0,inf) are not homeomorphic).

Am I tripping?

[gritty widget]

That is in fact continuous yeah

More specifically you can take the inclusion of [0, 1) into [0, 1] composed with the quotient map [0, 1]/{0, 1} and it’s far simpler

[wide kayak]

oh that's really nice!

thanks. I was asking o1 about this and it insisted there was no such thing, but I just tried adding that the inverse will not be continuous, and then it got it right

[gritty widget]

o1?

[wide kayak]

chatgpt o1. It's a paid version which is usually actually pretty good with math questions

[gritty widget]

This is also a good counterexample to the reversed version of “continuous bijection from compact space to Hausdorff space is a homeomorphism”

[gritty widget]

Oh, I didn’t know that was a thing. That’s pretty neat!

[wide kayak]

clearly still not great though

[gritty widget]

Yeah

[silver ridge]

I =[0,1] in R here. I don't see why X x I is a "cylinder", and I don't have any geometric intuition for why CS^n should be homeomorphic to B^n. Anyone have any intuition to offer?

[tranquil cosmos]

It is a literal cylinder for $X \in {S^1, D^2}$

[gentle osprey]

PKThoron

[tranquil cosmos]

Oh or B² as it were (I'm accustomed to D², it's the same thing)

[silver ridge]

I guess that's reasonable. Not sure why I didn't see that

Thank you#

[tranquil cosmos]

As for the cone, a quotient's job is to make several points into one

So for S1, the cylinder is morphed into, again, a literal cone by merging the "bottom" S1 into a single point

This is equivalent to taking a point P outside of the space and drawing lines from P to every point in X

[silver ridge]

I see

[tranquil cosmos]

Now take a 2D sheet of paper, draw an S1 and put the point P right in the middle

And connect it to every point on S1

[silver ridge]

Thank you 🙂

[wise mist]

Proof verification!

Should i lift the map to the product space and consider the diagonal ?

Someone at m.SE pointed out that there is no continuous difference operator!

[iron bolt]

tl;dr: you take the cone and squish it flat

[alpine hound]

Let (X,τ) be some topological space.
Is there a way to determine which sets have trivial interior?
(If it helps, I am looking specifically at the Zariski topology)

[thorny agate]

if it's about Zariski topology you'd be better off asking in #algebraic-geometry probably

[finite token]

if you mean empty interior then those are called meagre sets

in a general topological space there's not really any characterization of those

it's equivalent to having a dense complement if that's useful

[kind musk]

Part ii

[wise mist]

contradiction helps ?

did you not get the hint ?

notice the boundedness

[gentle osprey]

nastasya

[wise mist]

assume it does so, now substitute this value at the boundedness condition to see they blow up if n goes to \infinity contradicting hypothesis

[wise mist]

need help concerning [the problem](#1332828732578074754 message)

[civic verge]

name of book?

[balmy briar]

How do I approach a question like this. I thought for the smallest topology, I could consider the union, but the union only satisfies the condition for a subbasis so then I should take finite intersections and then arbitrary unions to generate a topology? Is this correct and if so what do I do from here

[unreal stratus]

Yeah exactly, ou should consider the topology generated by the union

This is then correct essentially by definition of "generated by"

[balmy briar]

In terms of notation, how can I write this topology because unions of intersections of unions is confusing me

how would I show that it is the unique smallest one too

[sinful cloak]

You can use the set theoretic hull operator around your generating family F in space X:
$$\tau = \bigcap_{\substack{A \in \mathcal{P}(\mathcal{P}(X))\A \supseteq F\\text{\text{$A$ is a topology}}}} A\ .$$

It works with any condition you can put below the intersection as long as it is preserved by intersections.

[gentle osprey]

M8327

[balmy briar]

tbh i have no idea what is a hull operator or what this means 😅

[sinful cloak]

You know intersections, right?

[balmy briar]

yes

[sinful cloak]

This is just the intersection of all topologies containg F.

It's easy to see, that the intersection of topologies is a topology.

Furthermore it is the smallest, afterall any candidate for an even smaller one is already in this intersection so there can be no smaller one.

[balmy briar]

but isn't what I want all finite intersections of sets from the topologies not the intersections of topologies?

[sinful cloak]

That's atuomatic here.

Consider a finite sequence of sets $S_1, \dots, S_n$ in $\tau$. They live in each of the topologies $A$. Henceforth, since each $A$ is a topology, you see that $\bigcap_{k=1}^{n} S_k \in A$. Since the set $\bigcap_{k=1}^{n} S_k$ is in each $A$, it is also in $\tau$.

[gentle osprey]

M8327

[balmy briar]

What is tau here?

[sinful cloak]

The smallest topology constructed this way.

We are constructing a smallest topology from a set of open sets F we want to at least contain.

[balmy briar]

im sorry but what is F again

[sinful cloak]

The generating set.

In your case the union of the input taus.

[balmy briar]

So the subbasis?

[wise mist]

Every time i think about subbasis it seems that 'this is the first time its making sense to me'

[balmy briar]

ok got it thank you

[red folio]

What is wrong with this proof?
Let $\left(X,\mathcal{T}\right)$ $A_{i} \subseteq X$ and $A'$ denote the set of limit points of A, then $\left(\bigcup_{i \in I}A_i\right)' = \bigcup_{i \in I}A_{i}'$.
Proof:
$\left(\Rightarrow\right)$ Let $x \in \left(\bigcup_{i \in I}A_i\right)'$ and suppose for contradiction that for all $A_i$ there exists $x \in U_i \in \mathcal{T}$ such that $U_{i} \cap A_{i} \subseteq {x}$, then \left(\bigcup_{i \in I}U_i\right)\cap\left(\bigcup_{i \in I}A_i\right)=\bigcup_{i \in I}\left(\left(\bigcup_{i \in I}U_i\right)\cap A_i\right)=\bigcup_{i \in I}\bigcup_{i \in I}\left(U_i \cap A_i\right)=\bigcup_{i \in I}\left(U_i \cap A_i\right) \subseteq {x}$. Since $x\in\bigcup_{i \in I}U_i\in\mathcal{T}$, there then exists $U \in \mathcal{T}$ such that $U \cap \left(\bigcup_{i \in I}A_i\right) \subseteq {x}$ and hence $x \notin \left(\bigcup_{i \in I}A_i\right)'$, which is a contradiction. Therefore $x \in \bigcup_{i \in I}A_{i}'$.
$\left(\Leftarrow\right)$ Let $x \in \bigcup_{i \in I}A_{i}'$, then $x \in A'$ for some $A' \in {A_{i}}{i \in I}$ and hence for all $x \in U \in \mathcal{T}$ there exists $x^{} \neq x$ such that $x^{} \in U \cap A$. Since $A \subseteq \bigcup{i \in I}A_i$, $x^{*} \in U \cap \bigcup_{i \in I}A_i$ and hence for all $x \in U \in \mathcal{T}$ there exists $y \neq x$ such that $y \in U \cap \bigcup_{i \in I}A_i$ and therefore $x \in \left(\bigcup_{i \in I}A_i\right)'$

[wise mist]

1. compilation error
2. ...

im kidding, sorry

[gentle osprey]

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

[wise mist]

are you not convinced by it ?

what really is bothering you !
i mean its a long jargon

[red folio]

A counterexample to this is the fact that in the standard topology of R, {x}' = ∅ but for any set A in R, A is the union of singletons of elements that are in A

[ruby delta]

I'm not convinced it's true for arbitrarily many such unions

so one side of the inclusion is clear, but a way that the other side can fail is if you can keep "escaping" into the next A_i

[wise mist]

oh now are you asking why the proof sounds yes ?

got it

[red folio]

so is $\left(\bigcup_{i \in I}A_i\right)' \subseteq \bigcup_{i \in I}A_{i}'$ true or $\bigcup_{i \in I}A_{i}' \subseteq \left(\bigcup{i \in I}Ai\right)' $?

[ruby delta]

I think I see the problem, there is bad notation

[ruby delta]

$(\bigcup_{i \in I} U_i ) \cap (\bigcup_{i \in I} A_i ) \neq \bigcup_{i \in I} U_i \cap A_i$

[wise mist]

limit of countig set ?

[gentle osprey]

HChan

[ruby delta]

it's a bad idea to use the same indexing variable

[red folio]

is this also true in general?
$U \cap \left(\bigcup_{i \in I}A_i\right) \neq \bigcup_{i \in I}\left(U \cap A_i\right)$?

[gentle osprey]

snus

[ruby delta]

you can't merge the two unions like that here, this is where the bad notation causes an error:

[gentle osprey]

HChan

HChan

[ruby delta]

that part looks correct, equality should hold there

the problem here is that the U_i's are independent of the A_i's in the union, but your notation did not reflect that

[ruby delta]

essentially the problem was that your union implicitly "forgot" that unions of the form U1 u A2 was still in the union

[crimson smelt]

I am trying to understand the proof that the Discrete Metric Generates the Discrete Topology

can anyone explain why choosing $d(x,y)\le\frac{1}{2} \implies d(x,y)=0$?

[silver spruce]

if x and y are distinct then they have unit distance from each other

[gentle osprey]

rabbits_advocate

[crimson smelt]

What about the case when $r\ge 1$?

[gentle osprey]

rabbits_advocate

[silver spruce]

then you would get the entire space X because any y (other than x) has distance 1 from x

[crimson smelt]

so in both cases I would get {x}?

[silver spruce]

if r = 1/2 you would just get {x}, yes. not the case if r >= 1, though: In that case B(x, r) = X (i.e. every point)

[crimson smelt]

Ok but then from the second case how can I show that T is the powerset of X?

[silver spruce]

the first case shows that every singleton is open in the discrete metric space, right? how can you use that fact to obtain any open subset of X?

[gritty widget]

The basis of T is the open balls, which contain all singletons

[crimson smelt]

by the union of the singletons?

[silver spruce]

yeah

[mighty hull]

Write out the definition of the discrete metric.

[crimson smelt]

but what about the second case for r >= 1 where we get the entire set X how can I show that T contains all the subsets of X?

[mighty hull]

What's the definition of the discrete metric? Genuinely, type it up here.

[crimson smelt]

Its d(x,y) = 0 if x = y else d(x,y) = 1

[mighty hull]

Ok. So, for any a in X, d(x,a) is either 0 or 1.

[gritty widget]

This isn’t a “second case”, you can choose which r to pick. If you pick r>=1, you don’t gain very much knowledge on the space.

Trying to show it’s discrete by picking r>=1 won’t work

[mighty hull]

I think he was just asking what happens when r >= 1 and why is it the whole space.

[gritty widget]

Ohhh

[mighty hull]

For which a in X do you have d(x,a) <= 1, knowing that d(x,a) is either 0 or 1 for every a?

[crimson smelt]

for all a?

[mighty hull]

Well, don't ask me! Reason it out.

[crimson smelt]

Yes for any a the distance is 0 or 1

[mighty hull]

Yep. If d(x,a) = 0 then it's less than 1. If d(x, a) = 1 then it's equal to 1. So, it's less than or equal to 1.

And if r >= 1 then d(x,a) <= 1 <= r, so it's the whole space in any case.

And if r <= 1/2 then d(x,a) can't be equal to 1, so d(x,a) = 0, which means x = a

[crimson smelt]

Ok I understand now

Thank you all

[pallid comet]

Must the closure of an open ball in a locally connected metric space be locally connected? (What if the space has to be compact?)

[cerulean oriole]

If X is a sober topological space and ~ an equivalence relation on X, can we recover ~ from the data of which open sets of X are closed under ~ (U is closed under ~ if x ∈ U, x ~ y ⇒ y ∈ U)?

[iron bolt]

nope. I think any counterexample is going to be annoying to spell out though

you can find subsets of R^2 that are counterexamples, but they're not particularly simple

[balmy briar]

can someone explain what this means i dont understand

[paper wedge]

is the collection of (-infinity,a) as a ranges (over R?) a basis?

T_5 is the topology generated by such.

@balmy briar

[balmy briar]

Yes i think its a basis?

but i dont understand what is this = {x | x < a}

[dawn tundra]

{x \in R | x < a} is just the interval (-\infty, a)

[paper wedge]

this is a set builder notation for the interval (-inf,a)

[balmy briar]

oh okay thank you

[paper wedge]

all xs SUCH THAT x is less than a. which is (-inf,a)

np

[low plaza]

I've recently started studying topology, and I noticed that the set of homeomorphisms of a topological space to itself is a group. Is this fact considered of any interest/will I study how to exploit this fact in my career?

[lilac musk]

In any locally small category, the set of automorphisms of any object is a group.

[low plaza]

Ok thanks, but is this group studied in topology? E.g. in group theory AutG (where G is a group) it's studied a lot, but I was wondering if the same happened in topology, as it looks like this group is much more complicated and almost unintelligible.

And to me doesn't seem to have any particular structure

[opaque scroll]

I don't think they come up that often, like you said they're usually pretty complicated.

But when studying fiber bundles they come up as transition functions are maps to Aut(F). Though I think usually one restricts to some simpler subgroup of Aut(F).

[gritty widget]

I’ve personally never seen it mentioned, but it could be important somewhere.

[low plaza]

Ok cool thanks

[lament vortex]

iirc deck transformations of covering spaces are automorphism groups

[iron bolt]

it does show up here and there, but like KnightWatch said, it's basically just one instance of the general pattern that automorphisms of any object in a category form a group - so not some super special insight that will allow you to do crazy new things

one place where it does show up for example is that for any topological group G and topological space X, any continuous action of G on X can be understood as a continuous group homomorphism G → Aut(X), where Aut(X) is the homeomorphism group of X compact-open topology - you might need some conditions on X for that, I don't quite remember

but the problem is that not every such continuous group homomorphism corresponds to a continuous group action

essentially because for that you want the action map G ⨯ X → X to be continuous with respect to the product topology, but only know that it is continuous in each argument, which isn't enough

so that's why it's not a super useful perspective there, unfortunately

mapping class groups are maybe a more useful example, because they're defined as quotients of the homeomorphism group and are actually quite important - but I guess they're also a bit advanced

[lilac musk]

Why do we need the 'compact-open' condition here?

[cerulean oriole]

To convert "G ⨯ X → X is continuous" (the usual hypothesis on a continuous group action) to a condition on G → Aut(X). For nice enough G and X (I believe X locally compact Hausdorff suffices), you can say G ⨯ X → X is cts iff G → Aut(X) is cts - if Aut(X) is equipped with the compact-open topology.

This is, AFAIK, one of the main reasons for the use of the compact-open topology.

[lilac musk]

Ah
Thanks

[iron bolt]

fun fact: in order for continuous maps X ⨯ Y → Z to correspond exactly to continuous maps X → (Y → Z), all you need is actually for Y to be locally compact in the sense that every point has a neighbourhood basis of compact sets

it is only commonly stated that you need Y to be locally compact and Hausdorff because there's multiple definitions of locally compact that are only equivalent for Hausdorff spaces

so asking for Y to be locally compact Hausdorff just avoids having to specify which "locally compact" you actually mean

now, in the case of the Homeomorphism group though... I'm not actually so sure locally compact is enough

because ideally you'd also want Aut(X) to form a topological group, and I'm not sure which condition on X you need to make that happen

might be something stronger like compact Hausdorff

[plush folio]

How do I show that f(t) = (t, t², t³) is an embedding of R in R^3? Showing that f is open/closed/proper directly seems cumbersome. I guess I can use invariance of domain? Kinda seems like overkill though

wait, I can just use the fact that g(x, y, z) = x is a continuous left inverse of f

[gritty widget]

Isn’t it only required that Y is core compact, not necessarily locally compact?

[iron bolt]

looking at the wikipedia article - apparently so

I only knew that locally compact is enough, not whether it was optimal

[tender halo]

another fun fact: there is at most one topology on X^Y that works in that way

so the compact open topology is in fact your only option when Y is locally compact

[cerulean oriole]

Ah, I see. I remember reading that the exact condition required for Y to be exponentiable is in fact that it be "core-compact" (every open neighbourhood V contains an open neighbourhood U such that any open cover of V contains a finite subcover of U), but I thought that was unnecessary in the original context. https://wiki.math.wisc.edu/images/Compact-openTalk.pdf (and presumably also https://martinescardo.github.io/papers/newyork.pdf that it's based on) is a very nice short write-up deriving this.

[cerulean oriole]

Incidentally, it is optimal for sober spaces (every irreducible closed set is the closure of a unique point - intermediate between T0 and Hausdorff).

[brittle otter]

I've finally unlocked the topology world

I'm already enjoying my suffering

[frail jewel]

Will my below proof work for the reverse direction:

[thorny agate]

right now the thing that is standing out to me is that you are only showing f^-1(U) is open for some open sets U, namely ones of the form U = N(f(x))

you need to show that for arbitrary open sets U that f^-1(U) is open

[frail jewel]

Thanks.
If I understand correctly, if I can show that the neighborhood of f(x) can generate an arbitrary open set U, then I am done?

[thorny agate]

Sure, depends on what you mean by generate I guess. Seems like more work than necessary in my opinion

[frail jewel]

It does seem like more work. I just couldn’t get a hang of the proof in the book at first, therefore went in this route. I’ll probably visit the proof of the book again

[gritty widget]

W

[wise mist]

for the discrete case it's easy, however need a hint

for the case when it's not discrete

i was thinking of considering an arbitrary point as a limit point and a variable open ball around it to bound all but finite elements into it...

[prime elbow]

Is it from Carother?

[wise mist]

yea

[prime elbow]

Let there are two points x≠y such that {x} and {y} are not open

Can you see if there are no such two points then we are done

Because then there are two cases,

1. Every singleton set is open, then metric space discrete.

2. There is only one singleton set which is not open, that means the remaining singleton sets are open so we can do as we are doing in 1.( No this is not correct)

Now since this is metric space so there exists open sets U and V such that x \in U and v \in V, U \cap V = \empty.

Now we can argue that U and V are infinitely set

Since {x} is not open so every open set contains x must be infinite.

Here we can consider U as an open ball of x and V as an open ball of y.

Now if U contains a finite number of elements x1,...,xN, then take r < minimum d(x, xi), here xi ≠ x, then B(x,r) = {x}.

Is it correct?

My mistake

I think we are done when we have two such points but when I have only one such point then I have to think

[wise mist]

ok i will go throught this in a bit and will let you know

[shut pier]

I used entire of this when I was in grade 10

Made aproject on Topological surfaces and applications there

Unfortunately I was question by Scientific Officers of our Research Facility

Co-ordinates of Mobius Strip?

I said R sin theta, R cos theta but could not define how R was derived

and they went away disappointed but they had to pity me and gave me a 3rd position in model building anyways

But is never really helped me in Math Olympiad thouh

[pallid comet]

(sorry for bringing my questions here again!)

does anyone have any idea to prove this? my idea was to show that if we took a separation X \ {a,b} = M | N, we would have M union {a,b} homeomorphic to an arc i.e. every point of M separates M union {a,b}

[rancid umbra]

does metric continuum mean non-empty, compact, connected, metric space?

[rancid umbra]

this idea would work. do you have any theorems saying when a metric continuum is homeomorphic to [0,1]?

[pallid comet]

yup, the previous theorem that i just proved was that any metric continuum with exactly 2 exceptional points / non-cut points (points that dont separate the thing) is homeomorphic to [0,1]

[rancid umbra]

great! then all you need to do is show that every point of M U {a,b} except for {a,b} is a cut-point

likewise for N U {a,b}

[pallid comet]

well that is what i have been stuck on unfortunately (maybe im just dumb, since this seems really easy in comparison)

[rancid umbra]

first, can you show that a and b are not cut-points of M U {a,b}?

[pallid comet]

(sorry for perhaps not being the most interlligent or thoughtful right now. i just saw the pings and am now on my phone at midnight)

[rancid umbra]

all good

[pallid comet]

i did do that

[rancid umbra]

no need to tear yourself down either 🙂

[rancid umbra]

nice

i believe that showing that every other point is a cut-point is going to rely on the fact that any two points separate X

for example, given a point c in M, the pair of points a,c separates X

[pallid comet]

i tried that and couldnt figure anything out (although that does seem like the only sensible approach)

[rancid umbra]

let $c \in M$ and let $C, \overline{C}$ constitute a separation of $X\setminus{a,c}$, i.e., $C$ and $\overline{C}$ are open in $X\setminus{a,c}$, disjoint, non-empty, and union to $X\setminus{a,c}$.$\newline$

we have $$M\setminus{c} = (X\setminus{a,c}) \cap (M\setminus{c}) = (C\cap (M\setminus{c})) \cup (\overline{C}\cap(M\setminus{c}))$$
if, for contradiction, $C\cap (M\setminus{c})$ is empty, then $M\setminus{c} = \overline{C}$, so $X\setminus{a,c} = C\cup (M\setminus{c})$. This means that $X\setminus{a} = C \cup M$ is a separation of $X\setminus{a}$, which is impossible. Thus $C\cap M$ is non-empty. Likewise, $\overline{C}\cap (M\setminus{c})$ is non-empty. $\newline$

Now, $C\cap (M\setminus{c})$ and $\overline{C}\cap (M\setminus{c})$ constitute a separation of $M\setminus{c}$, hence $c$ is a cut-point of $M$

@pallid comet

[gentle osprey]

c squared

[rancid umbra]

the only things you need to justify for this argument are that M is open in X\{a} and that the separation of M\{c} is indeed a separation

[rancid umbra]

the final thing that you need to justify is that gluing M U {a,b} and N U {a,b} back together is homeomorphic to X

since gluing together two compact intervals at their endpoints is homeomorphic to a circle

[swift sluice]

I've been stuck quite a while showing $\operatorname{int}{\mathbb R}(\operatorname{cl}{\mathbb{R}}F_N)=\emptyset$. It boils down to showing that there are no nonempty open sets $U\subseteq\mathbb{R}$ contained in every closed set $K\subseteq\mathbb{R}$ containing $F_N$. My best guess is that we should be able to pass from a nonempty $U$ to some nonempty open set $V$ which appears in the union defining $\operatorname{int}_{\Omega}(F_N)$, contradicting the assumption that this last set is empty.

[gentle osprey]

person2709505

[swift sluice]

I don't think the fact that this is taking place in R makes much difference. The other thing I was able to do is reformulate the condition that the interior of F_N in omega is empty as follows: for all U open in R, the set U cap Omega is either not contained in F_N or is empty.

I've also already shown that F_N = Omega cap (closure of F_N in R). I thought that maybe I could take interiors (in R) of both sides, distribute over the intersection, and make conclusions about the set we claim is empty that way, but of course the best we can do that way is show that this set is disjoint from Omega.

[rancid umbra]

the case when Omega is empty is trivial, so assume Omega is non-empty.

if you have an open subset U of R contained in cl(F_N) (closure in R), then U cap Omega is open in Omega and is contained in F_N, so U cap Omega must be empty. From here, assume for contradiction that U is non-empty.

U and cl(Omega) must have empty intersection. But U subset cl(F_N) subset cl(Omega), so after intersecting with U on all sides, we find that U is empty. contradiction

the only part left to show is that U and cl(Omega) have empty intersection

[rancid umbra]

to do this, assuming U is non-empty, put X = U union Omega. U and Omega are both non-empty, disjoint, open subsets of X (since X is open), which union to X, and so U and cl_X(Omega) have empty intersection (closure in X).
but cl_X(Omega) = X cap cl(Omega) = U cap cl(Omega).
therefore, U cap cl_X(Omega) = U cap cl(Omega) is empty

another way to argue that U and cl(Omega) have empty intersection is that U and Omega can be written as disjoint open intervals. wlog, assume that they are both connected, so that U and Omega are both open intervals.
if U and cl(Omega) have non-empty intersection, then U must intersect the boundary of Omega, and since cl(Omega) is a closed interval, U must intersect Omega, contradiction

[glossy wagon]

Hello, please how should I interpret this "considered as topological spaces with their Euclidean topologies"?

Since "subspace topology" was not covered yet by this point in the book, I'm guessing there's some other way?

If X and Y are not open

[opaque scroll]

Do you know how metric spaces are topological spaces?

[glossy wagon]

Yeah

[opaque scroll]

The euclidean topology is just the topology induced by the euclidean metric

[glossy wagon]

Ye but I mean if I take subsets then that would agree with the "general" subspace topology right

I guess that's what was meant thank you

[opaque scroll]

It gives the same topology as the subspace topology yes

[swift sluice]

Thanks for the detailed reply! This was very helpful

[glossy wagon]

Oh wait

This makes everything significantly easier

Holy that is a relief

[floral arrow]

(Lee's Topological Manifolds)

[floral arrow]

how do you see this

even assuming $$h(V \cap \partial N) \subseteq U \cap \partial M$$ don't see it

[gentle osprey]

The Royal Group

[floral arrow]

nvm I see it

[civic verge]

it is manifolds?

[floral arrow]

yeah topological manifolds

[civic verge]

what book are you using, I think I've seen this imagen

[rancid umbra]

this book is titled Lee’s Introduction to Topological Manifolds

[civic verge]

high level

[swift sluice]

I've shown the things mentioned in the hint, and now I'm trying to use it. I have demonstrated that [\operatorname{int}{\mathbb{R}}\left(\Omega\cap\bigcup_N\overline{F_N}\right)=\bigcup{\substack{U\subseteq\Omega\cap[\cup_N\overline {F_N}]\U\subseteq\mathbb{R}\text{ open}}}U\cap\Omega=\emptyset] and that [\operatorname{int}\Omega\left(\bigcup_NF_N\right)=\bigcup{\substack{U\mathbf{\cap\Omega}\subseteq\Omega\cap[\cup_N\overline{F_N}]\U\subseteq\mathbb{R}\text{ open}}}U\cap\Omega,] and the conditions for the sets that go in these unions tell me that the first set is contained in the second. But I need equality to establish the result.

[gentle osprey]

person2709505

[swift sluice]

Is the interior of a disjoint union the union of the interiors?

I'm specifically wondering about the case where we work in R if that makes things nicer

[gritty widget]

The interior is just the union of all open subsets contained in the set

If $P=X\coprod{Y}$, the interior is the union of all open subsets of $X\coprod{Y}$

[gentle osprey]

NAT Enthusiast

[swift sluice]

Not familiar with that notation 😦

Oh sorry it's just disjoint union

[gritty widget]

Idk if it’s the right one

yeah

might be the wrong symbol

[swift sluice]

$\sqcup$?

Yep

[gritty widget]

So to prove the interior of P is the union of the interiors of X and Y, you just have to prove every open subset of P is the disjoint union of an open subset of X and of Y(the other inclusion is clear, rhe union of the interiors is contained in the interior of P)

Or rather

That the union of all open subsets contained in P is just the union of all open subsets contained in X and Y

Same thing ig

[swift sluice]

It sounds quite obvious when you put it that way. Thanks!

[gritty widget]

Have you seen the proof of equivalence of the (various) definitions of interior?

[swift sluice]

I don't know if I've seen any other definitions apart from the union of all open subsets

[gritty widget]

Ah alright

[rancid umbra]

you are overcomplicating things. from what you proved in the hint,

$$\bigcup_{N\in\NN} F_N = \Omega \cap \bigcup_{N\in\NN}\overline{F}_N$$

taking interiors, we have
$$\text{int}\left(\bigcup_{N\in\NN}F_N\right) \subseteq\Omega \cap \text{int}\left(\bigcup_{N\in\NN}\overline{F}_N \right) = \emptyset$$
using the general fact that $\text{int}(A\cap B) \subseteq \text{int}(A)\cap\text{int}(B)$

[gentle osprey]

c squared

[pallid comet]

sorry, i just got back. can you explain how you got to $M \setminus {c} = \overline{C}$? thats the only part i cant get

[gentle osprey]

math eater

[rancid umbra]

hmmm. i may have concluded this a bit hastily… i’ll get back to you when i resolve it

[swift sluice]

Oh dear... that is quite simple. Thanks again!

[swift sluice]

Actually, how does this work? Is that the interior in Omega or R?

[thick aurora]

hello guys.
Let f:X->Y is a cont. function
if X is a simply connected space can we say that f(X) is also a simply connected space?

[gritty widget]

This is not true even for quotient maps

Hint:How would you get S^1 as a quotient?

[thick aurora]

I know very little about topology but was trying to understand a solution to a problem.
I think the solution includes topology but idk.
The author basically assumes that every point in a region is in the image of a function I was trying to understand why he assumed that.
Someone mentioned about connected spaces but idk.

[gritty widget]

The image of a connected space is connected, but the image of a simply connected space need not be simply connected

[thick aurora]

ok thank you bro ❤️

[rancid umbra]

so this was actually quite the headache to figure out.
we need to show two things in order to make this proof work:
(a.) The closure of M is M U {a,b}, and
(b.) M is connected.

By symmetry, the same will be true of N. To see why these facts are sufficient to complete the proof, fix a point c in M. Let C, D constitute a partition of X \ {a,c}. Either C or D must be a subset of M \ {c}, since otherwise, C and D would have non-empty intersection with N, and therefore separate N. WLOG, assume C is a subset of M \ {c}. Now, M \ {c} = C U (D n (M \ {c})) is a separation of M \ {c}, which shows that c is a cut point of M, exactly what we wanted to show

(a.) is used to prove that M is connected

here are outlines for (a.) and (b.):

(a.) M U {a,b} is closed in X since its complement, N, is open in X, so cl(M) must be a subset of M U {a,b}. cl(M) cannot be M, since otherwise, M would be a non-trivial clopen subset of X. Further, cl(M) cannot be M U {b}, since then N U {a} would be open in X \ {b}, and X \ {b} = (N U {a}) U M would be a separation of X \ {b}. Similar reasoning applies for M U {a}, and so cl(M) must be M U {a,b}. The same is true of N

(b.) If M is not connected, let C U D be a separation of M. Since M is open in X, then C and D are also open in X. Further, since cl(C) n D = empty = C n cl(D), then bd(C U D) = bd(C) U bd(D). It is not difficult to show that bd(M) = {a,b}, and so {a,b} = bd(C) U bd(D).

Now, in X \ {b}, one of bd(C) or bd(D) is empty. WLOG, assume C has empty boundary in X \ {b}. Now C is closed and open in X \ {b}, which means that X \ {b} is disconnected, a contradiction.

We conclude that M must be connected in X (the same is true of N, and so also demonstrates that X \ {a,b} has two components)

(if there are any mistakes or if you have any questions, let me know. there were a lot of hairy details here)

the hard part should be done now

https://www.ams.org/bookstore/pspdf/text-58-prev.pdf
please don’t ask me for a proof of the kline sphere characterization theorem. i will inevitably lose track of the little time i have today thinking about it lmao

after some more looking, you may find a more concise proof of this statement in willard’s general topology, the chapter on continua

[rancid umbra]

i wasn’t as clear as I needed to be: If $A$ is an open subset of a top. space $X$ and $B$ is a subset of $A$, then $\text{int}_A(B) = A \cap \text{int}(B)$. Here is a proof:

if $x \in \text{int}_A(B)$, then there is an open subset $U$ of $X$ containing $x$ such that $x \in U \cap A\subseteq B$. Since $A$ is open, then $U\cap A$ is an open subset of $B$, and so $x \in A \cap \text{int}(B)$. Conversely, if $x \in A \cap \text{int}(B)$, then there is some open set $U$ of $X$ with $x \in U \subseteq B$. Now, $x \in A \cap U$ and $A\cap U$ is an open subset of $B$ in the subspace topology on $A$, so $x \in \text{int}_A(B)$.

[gentle osprey]

c squared

[pallid comet]

what if Bd(C) = Bd(D) = {a, b}?

[pallid comet]

update: thanks! i proved the theorem on my own.

the idea is still to separate X minus a, b into M union N

for every pair of points m, n in M, N, either m must separate M union a, b or n must separate N union a, b

so if M union a, b has a non separating point that isnt a or b, N uniom a, b would be an arc

[swift sluice]

Thanks! I had come up with something much more convoluted so this is great

[chrome trout]

Do we know if the middle third cantor set contains any non-zero rational multiple of pi?

[tranquil cosmos]

I don't know for sure, however consider that none of the elements of the Cantor set are normal numbers in base 3 (they're those base 3 numbers than can be expressed with just 0s and 2s after the comma)

And pi and all its nonzero rational multiples are highly likely to be normal in base 3

[outer heron]

hi

[atomic island]

hey :) i'm struggling with a question.
let $A, B$ be connected sets in some topology X. assume $A\cap\overline{B}$ is non-empty. show that $A\cup B$ is connected.

[gentle osprey]

blutac

[atomic island]

(overline B = closure of B)

[prime elbow]

What did you try?

[prime elbow]

Try contradiction

[prime elbow]

Any hint?

[paper wedge]

the ordered topology is hausdorff

look at the compement

[prime elbow]

Yes it is

[prime elbow]

Complement is { x in X | f(x) > g(x) }

[paper wedge]

ye

[prime elbow]

How do I use the Hausdorff property and continuous property here?

[paper wedge]

u want to show that this complement is open

that is for every point in this complement you can find a nbd around it that is contained in this complement

pick some x such that f(x) > g(x)

use the hausdorff property to seperate f(x) and g(x)

u get two open sets say U_1 and U_2

can u continue

[prime elbow]

Actually I just saw the solution

But they didn't use the Hausdorff property

[paper wedge]

might work too

[paper wedge]

here u just take f^-1(U_1) intersec g^-1(U_2) and thats ur nbd

around x

[prime elbow]

Yes but then we have to show that this nbd contained in {x | g(x) < f(x) }.

[paper wedge]

yes

[limpid fern]

that's just using the properties of the order topology

actually if you draw a picture for this question it's not too bad

they probably just did something like

explicitly specifying the disjoint sets

I'm gonna guess they used (g(x), infty) and (-\infty, f(x)) and preimaged them with f and g respectively

[prime elbow]

Yes

[pallid comet]

is it true that in a $T_5$ paracompact topological space $X$, any open cover ${U_\alpha}{\alpha \in \lambda}$ has an associated closed cover ${A\alpha}{\alpha \in \lambda}$ for which $\forall \alpha \in \lambda, A\alpha \subset U_\alpha$ and $\bigcup_{\alpha \in \lambda} A_\alpha = X$?

[wise mist]

is there any particular reason(s) to go with one and not the other ?

its its fine as long as i know my definition

i like the idea of the wiki as how Tn becomes stronger and stronger

[gentle osprey]

math eater

[prime elbow]

What are the better ways to find the boundary points and interior points for E and F?

[pallid comet]

okay i just realized this is true so nevermind 😭

[silver ridge]

Can someone confirm my intuition here? We can consider the closed cover of R given by the singletons {x} for x in R, and then say f_x = 1 (the constant map) for x !=0, but f_0 = 0. Each f_x is continuous because they're constant, they agree on overlaps vacuously as there is no overlap but the function you get by gluing them all together is not continuous.

[warm kettle]

Looks good to me

[spark shale]

I came across a proof where someone claimed:\
If $x$ is a limit point of $A_1 \cup A_2$ then for every open set $U$ such that $x\in U$, there exists a point $p\in (A_1 \cup A_2)$ such that $p\in U$ and $p \neq x.$\
So far I see no problem but then they say:\
Because $p\in A_1 \cup A_2$ we have that $p\in A_1$ or $p\in A_2$ which is the same as saying that $x$ is a limit point of $A_1$ or $x$ is a limit point of $A_2.$\
Is this true? To me it seems like this is equivalent to saying that given two neighborhoods of $x$, say $B$ and $C,$ if $B$ intersects $A_1,$ $C$ necessarily intersects $A_1$ as well, which I don't think is true.

[gentle osprey]

Maxwell

[opaque scroll]

The statement is true, but I would agree there seems like a step is missing.

The idea is, imagine B doesn't intersect A1 and C doesn't intersect A2. Then BnC doesn't intersect either of them.

[plush folio]

The argument seems to use the proposition forall x, (P(x) or Q(x)) to conclude forall x, P(x) or forall x, Q(x), which isn't valid

[spark shale]

that makes sense, thank you for confirming :D

[spark shale]

are you saying that that proves the contrapositive of the statement "x is a limit point of A1 u A2 implies x is a limit point of A1 or x is a limit point of A2?"
if not I do not get what you mean by that

[opaque scroll]

Yes

[spark shale]

I see!
thank you

[tranquil cosmos]

Every open covering of a manifold of dimension n can be refined to a covering so that the intersection of n+1 different opens is zero, right?

[opaque scroll]

It can be refined such that the intersection of any n+2 is empty.

That's Lebesgue covering dimension

[tranquil cosmos]

n+2, sorry yeah

[hazy lotus]

All spaces here are metric spaces.

(8) seems intuitive enough, but I'm not sure where to start when proving it

[ruby delta]

notice that the straight line connecting x_k and x_k+1 is entirely inside B(x_k; ε)
also that there must exist a natrual number n such that ε/n < ε'

[hazy lotus]

To be honest, I still don't see it. I guess the main problem is that I know that I need to select additional points between x_k and x_k+1, but I don't know how to guarantee that the ε' balls around those points will actually stay within B(x_k; ε) U B(x_k+1; ε) (they have to because of part (i) in this book's definition of a chain)

[ruby delta]

small note, the points dont necessarily need to be in B(x_k; ε), but some B(x_k; ε') that is contained in E

[hazy lotus]

since ε'<ε, being in B(x_k; ε') means you're in B(x_k; ε)

[ruby delta]

actually, nvm, that's too confusing

just look back at the first post, lol

[hazy lotus]

are you saying to split the line xy into n different pieces?

I think there's still more to the problem because you have to show that each of those division points's e'-balls stay withing B(x_k;e) U B(x_k+1,e) (this is a requirement because of the weird definition of epsilon chain that is being used)

[ruby delta]

each B(ε/n) ball is guaranteed to be contained within at least one B(ε) ball, if you choose a large enough n - think about the triangle inequality

[hazy lotus]

that's the thing - we can't just choose a large enough n and shrink the balls down to an arbitrary size. The definition says that we need the ε'-balls to be contained in E. Not the shrunk down e/n balls.

[ruby delta]

ah, I see, good point - but I'm pretty sure the proof works regardless see below

[hazy lotus]

?

That would mean that the proof doesn't go through

how would it mean that the proof works if the proof doesn't work

[ruby delta]

you're right, I didn't think my response through - the approach I thought of ended up being kinda combinatoric, and Im trying to figure out if there's a better solution - not sure if there is one, actually but maybe someone else can find one:

(warning: full solution)
two cases - either ε' >= ε/2, or ε' < ε/2
||If ε' >= ε/2, then take the two ε' balls at x_n, x_n+1 and we are done.||
||Otherwise since ε' is still nonzero we can take a finite number of points along the line connecting x_n,x_n+1 such that the ε' balls along that line forms a chain. Suppose that y is one of such points.||
||Then suppose that p is some point inside B(y,ε'). by the triangle inequality we get that d(p,x_n) <= d(p,y) + d(y,x_n)||
||Equally we also get that d(p,x_n+1) <= d(p,y) + d(y,x_n+1). Now notice that as y is chosen as a point on the straight line connecting x_n and x_n+1, one of d(y,x_n+1) and d(y,x_n) must be less than ε/2||
||So min{d(p,x_n), d(p,x_n+1)} <= d(p,y) + min{d(y,x_n),d(y,x_n+1)} < ε' + ε/2 < ε/2 + ε/2 = ε and so p is in one of B(ε,x_n), B(ε,x_n+1), so every ball along the line is properly contained.||

[ruby delta]

let me know if the model solution is any better than mine

[hazy lotus]

There is no model solution. I also didn't completely parse your solution yet, but the first case looks wrong because if the e balls shrink to e' balls, there's no guarantee that the balls will contain the other centers.

[young stone]

the inequality guarantees this won't happen, draw a picture maybe?

[ruby delta]

True, I’ll have to fix that (will do later)