point-set-topology

[lusty trench]

(I love pulling these shenanigans when solving geometry problems, and don't want to manually and explicitly push data through obvious isomorphisms.)

[plush folio]

Hmm, that's too big brained for me

but you wrote (x_0, x_1, ..., x_n) -> (x_0, x_1/x_0, ..., x_n/x_0), did you mean (x_0, x_1, ..., x_n) -> (0, x_1/x_0, ..., x_n/x_0) ?

[lusty trench]

No.

The latter wouldn't be a homeomorphism.

[plush folio]

So we're first doing (x_0, x_1, ..., x_n) -> (x_0, x_1/x_0, ..., x_n/x_0) which is a homeomorphism from R^(n+1) to R^(n+1), since it's just a rescaling of some of the coordinates, then we do the homeomorphism to U

[lusty trench]

No, that isn't a self-homeomorphism of R^(n+1), because dividing by x_0 isn't allowed on H.

But it's a self-homeomorphism of R^(n+1) \ H.

[plush folio]

Ah yeah, that's what I meant

[lusty trench]

The projection to U isn't a homeomorphism. It's merely a continuous open map.

However, there's a homeomorphism R^(n+1) \ H -> U x (R \ {0}), which, followed by the projection to the factor U, recovers the original projection R^(n+1) \ H -> U.

Here, U carries the information about y_1, ..., y_n, where each y_i = x_i/x_0, and R \ {0} carries the information about x_0.

[plush folio]

That makes sense, but how do you show it is homeomorphic to R^n?

U x (R \ {0}) is homeomorphic to R^n x (R \ {0}) maybe?

[lusty trench]

R^(n+1) \ H itself can be rewritten as (R^n x R) \ (R^n x {0}) = R^n x (R \ {0}).

Argh, I need to draw diagrams.

Okay, first of all. Our homeomorphism (x_0, x_1, ..., x_n) -> (x_0, x_1/x_0, ..., x_n/x_0) means that we can assume x_0, y_1, ..., y_n, with y_i = x_i/x_0, are the new coordinates of R^(n+1) \ H.

Then the projection to U is (x_0, y_1, ..., y_n) -> [1 : y_1 : ... : y_n]. It can be written in two steps: (x_0, y_1, ..., y_n) -> (y_1, ..., y_n) -> [1 : y_1 : ... : y_n].

We want to show that the last step, (y_1, ..., y_n) -> [1 : y_1 : ... : y_n], is a homeomorphism. It's already a bijection. So we only need to show that it's continuous and open.

And for that we need to use the definition of the quotient topology of RP^n.

[plush folio]

So we want to show that the latter map, which is from R^n to RP^n is a homeomorphism?

[lusty trench]

No map R^n -> RP^n can be a homeomorphism, because RP^n is compact and R^n isn't.

[plush folio]

Oops, R^n to U

[lusty trench]

Yes.

We want to show that our map R^n -> U is a homeomorphism.

So let A be any subset of U. Its preimage in R^(n+1) \ H has the form { (x_0, y_1, ..., y_n) : [1 : y_1 : ... : y_n] \in A }. Because discarding the x_0 coordinate is a continuous open map, A is open if and only if { (y_1, ..., y_n) : [1 : y_1 : ... : y_n] \in A } is an open subset of R^n. And then I think we're done.

[plush folio]

Let V be open in R^n, then V is open in R^(n+1), and by the quotient topology V = f⁻¹(f(V)) is open so f(V) is also open

[lusty trench]

Wait, wait, what?

We're already using U for an open in RP^n.

[plush folio]

oops, let me call it V

[lusty trench]

Also, how can you say an open subset of R^n is also open in R^(n+1)? No matter how you embed R^n in R^(n+1), this is false.

[plush folio]

hmm, okay

[plush folio]

Do you mean preimage in R^n here? I don't understand this: { (x_0, y_1, ..., y_n) : [1 : y_1 : ... : y_n] \in A }

[lusty trench]

I said preimage in R^(n+1) \ H.

We have a projection R^(n+1) \ H -> U. So, for any subset of U, its preimage under this projection is a subset of R^(n+1) \ H.

[plush folio]

but why do you need to talk about R^(n+1) \ H when we investigate the map from R^n to U?

[lusty trench]

Because the topology of U is defined in terms of the topology of R^(n+1) \ H.

[plush folio]

I see. It's the preimage of the quotient map which defines RP^n, not the map we're investigating

[lusty trench]

Right.

Recall that U is the subset of RP^n consisting of the points of the form [1 : y_1 : ... : y_n], which I'll abbreviate [1:y].

Let A be a subset of R^n. Let B = {[1:y] such that y \in A}. Let C = {(x_0, y) such that x_0 \ne 0 and y \in A}. Then the following statements are equivalent:
a) A is an open subset of R^n.
b) B is an open subset of U.
c) C is an open subset of R^(n+1) \ H.

The equivalence between a) and c) follows from the fact that discarding the x_0 coordinate is a continuous open map.

The equivalence between b) and c) follows from the fact that the projection R^(n+1) \ H -> U is a continuous open map.

[plush folio]

Aha, that makes it easier to understand 👍

[lusty trench]

Great!

[plush folio]

Thanks for sticking with me 🙏 I feel I need a lot more practice, but I'm getting there

[lusty trench]

No problem.

[plush folio]

I'm thinking of starting on Topology: A Categorical Approach. I think the categorical approach might work better for my brain

[lusty trench]

IMO, at this level, category theory is best regarded as a tool for organizing ideas. But you first need to learn the ideas that you want to organize.

For example, now you know that, if H' is a hyperplane in R^(n+1) not passing through the origin, then
a) H' maps bijectively to its image in RP^n.
b) This bijection is actually a homeomorphism.
c) This lets you think of H' as an open subset of RP^n.
Would category theory have helped you with that?

[plush folio]

True, it's not a panacea

[untold lily]

pinging my question

[ebon galleon]

Say $K$ is an uncountable collection of $\alpha$ so that $X_\alpha$ is not trivial for any $\alpha \in K$; say each $X_\alpha$ has an open $\emptyset \neq U_\alpha \neq X_\alpha$. Consider the set ${\pi_k^{-1}(U_k) \mid k \in K}$, and suppose $\mathcal{B}_x$ were a countable neighborhood basis. Then, as you said, you can refine each $B \in \mathcal{B}_x$ to be an intersection of the form $\bigcap \pi_i^{-1}(V_i)$. Now you can just count how many $\pi_k^{-1}(U_k)$ can contain a fixed basic $\bigcap \pi_i^{-1}(V_i)$ and argue why that can't happen

[gentle osprey]

ryx!

[ebon galleon]

For lems

[untold lily]

ohh

thanks

I guess I was almost there, picking a nontrivial open set from each factor is the death blow

well, I guess the part I didn't think of was "refining" the nhood basis

[median sand]

Both of these are connected, but path-disconnected, like the sin(1/x) graph, right?

[red yoke]

Yes

[gritty widget]

Does anyone know how lattice structure, atom size and electric conductivity are related? I'm trying to find a network topology which has the best efficiency in transmission and I thought having a metallic lattice structure might help since conductivity is from the uniform structure

not sure if this is the right channel

[quiet thorn]

is this even the right server?

[opaque scroll]

It's not, but I'm not sure which server is the right one

[shy phoenix]

hey what's the best way to show the klein bottle and the projective plane wedge sum a circle are not homotopy equivalent. i computed their fundamental groups: for P2vS1 it's Z*Z2, and for K2 it's <b,c | b^2c^2=1>. I'm not really sure how to show these groups are not isomorphic. also, could computing their cohomology rings distinguish them?

[untold lily]

this is kicking my ass, idk where to even start (for either direction)
any tips appreciated

[outer harness]

Anyone know where I can find a list of common topologies that are used?

[tender halo]

Used where?

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

its a nice list split by categories

[languid patrol]

there is no condensed matter channel

[languid patrol]

Have you considered their second homotopy groups? That is easier

Equivalently: the universal covers are not homotopy equivalent

[languid patrol]

alternatively alternatively: the fundamental group of the Klein bottle is two generators a, b such that aba^-1 = b^{-1} so its infinite dihedral. Thus the infinite order elements are all in one abelian subgroup. On the other hand the fundamental group of the wedge has two non commuting infinite order elements.

[shy phoenix]

the universal cover stuff def cleared it up. i should probably review some group theory though

[zinc hearth]

I don't think this is the right server boss. Check #old-network

[pastel idol]

The physics server is likely to be more helpful. Find it linked in #old-network

[outer harness]

Thanks, I couldn’t find this for some reason

[rich cliff]

Ah so it's only appropriate to just say "the cohomology groups of X" if X is a real manifold?

[cedar pebble]

ordinary cohomology still depends on a choice of Abelian group A as coefficients, which produces A-modules for cohomology

if you know the cohomology with coefficients in Z you can recover the cohomology with coefficients in A for any Abelian group A by the universal coefficient theorem

but for example cohomology with coefficients in Q comes up almost as often as cohomology with coefficients in Z

[merry geode]

How do you recover cohomology of coeff in Z from those in Q

[rich cliff]

I think you have it backwards

at least what nGroupoid is saying

[merry geode]

Nah, I mean this way

[rich cliff]

Oh, I don't know then

[merry geode]

So you cannot go Q-coefficient -> Z-coefficient?

[zinc hearth]

You can’t really, taking it to Q kills torsion

[merry geode]

Gah

This might make me cry silently at some night

[zinc hearth]

Z is inital so u can only go out

But good news! It is very easy to get the cohomology with coefficients in the terminal object from cohomology with coefficients in Q!

[west brook]

Indeed
Just kill everything

[gritty widget]

Thanks for the info

[worthy fjord]

Does anyone else find it weird that we use the same notation for projections as we do for homotopic groups? Like the subbasis for the product topology on $X_1\times X_2$ is

$${\pi_1^{-1}(U_1)}\cup{\pi_2^{-1}(U_2)}$$

But then also the fundamental group of $X$ relative to $x_0$ us $\pi_1(X,x_0)$

[gentle osprey]

friendlyneigborhoodtopologydonut

[rancid umbra]

the forwards direction is a direct consequence of the Alexander subbase theorem

for the backwards direction, consider any open set in a topology which strictly contains the topology on X but is not open in X

[hidden crag]

Not really

[near sage]

Hello, I have a quick question
I'm aware that a space that is Hausdorff and Compact is Normal, but can I conclude that a space is compact if I have that it is Hausdorff and Normal?

[knotty vine]

Ive never had a problem with this, but the notation for projections is also not set in stone. If there's a clash, just use p_1 or something

[tender halo]

no

consider R

[near sage]

oof... I have a Hausdorff second countable space and I need to prove that it is compact how would that happen?

I have the exact space if needed too

[tender halo]

well whats the space in question

generally its proved...somehow

[near sage]

let X = N∪{∞ } where ∞ is a point outside N. On X we define a topology by making open sets be all subsets S⊆N and sets of the type N\T∪{∞ }, where T⊆N is finite

[tender halo]

this one is proved by definition of compactness

[near sage]

this one?

[tender halo]

yes

[grave solstice]

Is a subspace of a locally compact space always locally compact?

this is just true because compacts on subspaces come from compacts on the whole space, right

[tender halo]

no

a locally compact subspace of a locally compact space is always an intersection of a closed subspace and an open subspace

(and vice versa)

in particular, every locally compact space is an open subspace of a compact space

for a counterexample for your conjecture think uhh R^2 without a ray from 0 into the positive reals (excluding 0)

0 does not have a compact neigbourhood

[grave solstice]

Don't you want Hausdorff?

but uh also right

[tender halo]

uhh probably

i dont really consider non hausdorff spaces because im not a nerd

[grave solstice]

If Y subseteq X, and K is compact in X, then K cap Y is compact in Y, no?

wait I suppose its false

because K cap Y needn't be compact

[tender halo]

X = K = [0; 1], Y = (0; 1)

[grave solstice]

yeah

[alpine nest]

Some people include Hausdorffness in the definition of compactness, and those people are right.

[grave solstice]

disagree

[alpine nest]

I'm being facetious

[grave solstice]

In algebraic geometry people use quasicompact for compact and not necessarily hausdorff, which is a bit cringe imo

[alpine nest]

Well, I genuinely do hate non-Hausdorff spaces, but I accept that some poor souls do need them.

[tender halo]

im ok with non-hausdorffness but quasicompact is just better terminology

good things should have a short name and compact hausdorff is not a short name

[grave solstice]

compdorff

[tender halo]

CHaus is a better category name than Comp i guess

[grave solstice]

hauspact

[void wind]

let $(X, d)$ be a metric space
with $X \supseteq K={a}\cup{a_k: k\in \mathbb{N}}$, $(a_k){k\in \mathbb{N}}$ being an arbitrary sequence with elements in X that converges against a.
Is K compact? (I assume so?)\
If so, how do I prove it when given only the "finite subcovers" definition of compactness?
(in particular, we can't assume that we're working on $\mathbb{R}^n$, so Heine-Borel does not apply. We haven't proven that any other definition of compactness is equivalent.)

[gentle osprey]

Bob Goldham

[tender halo]

well you take the definition and apply it

there is like one step in the proof

[void wind]

why would any cover of any such K always have a finite subcover?

[tender halo]

take the element of the cover that includes a and think about what elements of K it doesnt cover

[void wind]

the first finitely many elements of a_k

[tender halo]

well, there you have it

[void wind]

because no matter how small we choose it to be, we always cover infinitely many elements because a_k converges

alright thanks

helped alot <3

[grave solstice]

how do you prove this tho

a hint would be appreciated

[tender halo]

which way?

[grave solstice]

the difficult one

i.e., locally compact --> intersection of open and closed

[tender halo]

prove that it is open in its closure

i.e. you can assume that the subspace is dense in the whole space

you dont need the big space to be locally compact btw

i.e. every locally compact subspace of a hausdorff space is an open subset of its closure

[grave solstice]

yeah, I just meant the subspace

[tender halo]

here is the solution

if you give up

or want to check yourself

[grave solstice]

I'll try in a bit

[storm bronze]

harsh

[gaunt linden]

I suppose it uses abstract nonsense a lot, too.

[tidal lynx]

Let $(M, d)$ be a metric space. Do we have
$$ A \subset M \text{ is discrete } \iff \inf_{\substack{x, y \in A \ x \neq y}} d(x, y) \text{ is positive} $$

[gentle osprey]

T Stepped

[tidal lynx]

Ok seems so

[ebon galleon]

Consider A = {1/n} in [0,1]. It's discrete and inf ( d(x,y) ) = 0.
The converse is true however.

[tidal lynx]

ahh crap

I really feel like there should be some similar equivalent statement

[unreal stratus]

Well really all you need is like

for each x, inf_{y != x} d(x,y) is positive

It's just that you can't take the inf over all x for the reason ryx said

[tidal lynx]

ok yea

that makes sense

[grave solstice]

how do you show that H is isometric to a nowhere dense subset of itself? Here H is the metric space of square-summable sequences of real numbers, with the metric d(x,y)=sqrt sum (x_k-y_k)^2

the most obvious isometries I can think of are reindexing all the sequences, but I don't think that works

[gaunt linden]

If you just shift each sequence a step to the right, don't you get a nowhere dense subset?

[tender halo]

maybe try inserting zeroes wherever

[gentle fjord]

Let X be an infinite set and τ a topology on X. If every infinite subset of X
is in τ, prove that τ is the discrete topology.

According to me the strategy for solving this should be to show, every singleton subset of X is in τ. That would involve creating two infinite subsets of X, with a as their only common element.

I have an intuitive idea, of how this can be done, but how should I explicitly write it. Any ideas will be appreciated

[grave solstice]

uh I think you are right

like every open set contains elements whose first element is nonzero

and the set you proposed is obviously closed

[gaunt linden]

Can you describe your intuitive idea in words, then we can look for a more precise statement together?

[gentle fjord]

Let A and B be two 'empty boxes' (for the time being), take an arbitrary element from X and put it in A, now if an element in X has not already been put in A then put it in B, keep on continuing this process forever.
This should give us two disjoint sets, one of them, should contain 'a'… put 'a' forcibly in the box that already did not contain it,
We re done

[heady skiff]

can somebody check my argument for a please? Let $U \subset \mathbb{R}$ be open. We show that $d^{-1}(U)$ is open. Given $(a, b) \in d^{-1}(U)$, choose $\delta > 0$ so that $B_{\delta}\bigl(d(a, b)\bigl) \subset U$. If $\epsilon = \frac{\delta}{2}$ and $(u, v) \in B_{\epsilon}(a) \times B_{\epsilon}(b)$, [d(u, v) \leq d(u, a) + d(a, b) + d(b, v) < d(a, b) + \delta] Moreover, [d(a, b) \leq d(a, u) + d(u, v) + d(v, b) < d(u, v) + \delta] so that $d(a, b) - \delta < d(u ,v)$. We conclude that $d(u, v) \in U$ so that $(u, v) \in B_{\epsilon}(a) \times B_{\epsilon}(b) \subset d^{-1}(U)$, so $U$ is open

[gentle osprey]

okeyokay

[rough arch]

Hello guys i tried the last 6 days for around 5 hours that day to tolle the most insane topology exercise Let (X,d) be a metric space : X has an dense and countable subset Y if and only if it has a countable basis B. And because it was so time consuming I couldn’t do the other exercises and since i need the points for the sheet insanely bad and I can’t just think rn because of the mental exhaustion this particular exercise put me through i can’t get the other ones so could you guys help me with the solution of this two exercises ?

Let X, Y be topological spaces and let T denote the coarsest topology on X × Y with the
Property: For all topological spaces Z, a map f : Z → X × Y is continuous if and only,
if πX ◦ f and πY ◦ f are continuous. Show that T is the product topology.

Let F : X → Y be a mapping between topological spaces. Show:
(a) Are U1, . . . ,Un open sets in X with U1 ∪ . . . ∪Un = X and if f |Ui is continuous for all i, then f
steadily.
(b) Are A1, . . . , An closed sets in X with A1 ∪ . . . ∪ An = X and f |Ai is continuous for all i,
then f is continuous.

[tender halo]

both of those are just "apply definition" exercises

[violet python]

same as what the above comment said. Just write out the definitions of everything in sight and you'll see you are basically done

you can do the same with the other exercises

[earnest ibex]

What does the boundary map $\partial:H_1(X,A) \to \widetilde{H}_{0}(A)$ represent? It doesn´t behave like a boundary operator in this case, right?

[gentle osprey]

ImHackingXD

[grave solstice]

Does a closed mapping preserve neighborhoods? In the sense that if U is a nhood of p then f(U) is an nhood of f(p) ?

[tender halo]

dont see a reason for it to be true

[grave solstice]

yeah I don't either

[heady skiff]

could I get a hint for a? I came up with something, but I'm not entirely sure if it works; I said choose $\epsilon > 0$ such that $|x_i - y_i| < 1 - \frac{1}{i + 1} < \epsilon$ for all $i$, so that $d(x_i, y_i) < 1$ for all $i$ but $\overline{d}(\overline{x}, \overline{y}) = \sup{\overline{d}(x_i, y_i) \mid i \in \mathbb{Z}^+} = \sup{d(x_i, y_i) \mid i \in \mathbb{Z}^+} = 1$, so in particular we don't necessarily have $U(\overline{x}, \epsilon) \subset B_{\rho}(\overline{x}, \epsilon)$

[gentle osprey]

okeyokay

[heady skiff]

i'm not sure if this works because I don't think we can choose e > 0 such that e < 1 and e satisfies this property

[tender halo]

as for a counterexample i guess take the quotient I -> S, does not preserve neighbourhood of the join point

or any quotient of a space by a closed set

[grave solstice]

well it could still work I think

but yeah I get what you are saying, thanks

[tender halo]

sure it can, just quotient it all into a point and it will preserve neighborhoods

[grave solstice]

I was doing this bit. I could show that if we are given f : X -->Y with the hypotheses and p in U subseteq Y, U open, then there exists a compact subset of X that is a nhood of each of the points in f^{-1}(p) and is contained in f^{-1}(U), but I'm not sure how to finish it

I believe that if U is a nhood of all points of f^{-1}(p) then f(U) is an nhood of p

when f is closed

?

Suppose f^{-1}(p) subseteq U with U open. Then f^{-1}(p) and U^c are disjoint, with U^c closed. So that p is not in f(U^c), which is closed. Then p in f(U^c)^c which is open and it is enough to show that f(U^c)^c subseteq f(U). This is true because f is surjective, so the image is the union of f(U^c) and f(U)

[fading vale]

It basically does. The boundary of your chain is a bunch of points and so the boundary operator sends a chain with boundary in A to the sum of the homology classes of each point in the boundary

It's just that that homology class only depends on the path component of each point

The fact that it's well defined is actually a little subtle though

(meaning that the epsilon map in the extended chain complex for reduced homology vanishes on it)

[earnest ibex]

Yeah, I did the computation of the boundary operator by definition and it's just $\partial[\alpha +C_1(A)]=[\partial c] \in \widetilde{H}_0(A)$

[gentle osprey]

ImHackingXD

[earnest ibex]

I need to read more about the 0th-reduced homology, I know that the classes of 0-th homology are the path components, but I'm not sure what a class of $\widetilde{H}_0$ represents

[gentle osprey]

ImHackingXD

[grave solstice]

In a Hausdorff topological group, can we describe closed sets in terms of limits of sequences (indexed by N)?

[jolly umbra]

i think most of the time u can just remove a Z from H0 to get H0~

[opaque scroll]

If you take (Z/2)^X with the product topology for some uncountable set X, then the elements with countable support should form a sequentially closed set.

[earnest ibex]

Yeah, but like, is there a geometric reason of what that does? Because the Z´s in H_0 correspond to path components, so removing a Z feels like ignoring one of the path components, but which one?

[opaque scroll]

The usual construction of reduced homology adds the summation map onto your (singular) chain complex. So singular homology basically has the added condition that all the path components "sum to 0"

Or said another way, when you think about 0-cycles in your space you only allow 0-chains with the same number of "positive" 0-simplexes as "negative" 0-simplexes.

[drowsy quest]

Guess who passed their Algebraic Topology class 😎😎

[thorny agate]

nice

[earnest ibex]

That makes sense, thank you!

[grave solstice]

So I think you have to verify Va, Vb and Vc

I could show Va, Vb but I'm thinking Vc doesn't necessarily hold

Let Y be some infinite set with the discrete topology (so that it isn't compact) and add a generic point 0, 0 not in Y. By this I mean that you form a new space, which I'll call Y', where the opens are of the form U cup {0} with U open in Y, and the empty set. {0} is obviously compact so we can consider Y' setminus {0}. If x in Y' setminus {0} then any nhood of x is not contained in Y' setminus {0}, because a nhood of x always contains 0

[grave solstice]

More generally, if V-c were possible, given a compact set L of Y you should be able to find another compact set K such that closure of L subseteq K, which doesn't hold in general

unless I'm missing something

mmh ok I'm assuming that L should both be compact and closed?

[near sage]

I think this is a topology question since it has the requirement for X to be a topological space so I'll ask my question here:

Let X be a topological space, G be an open subset of X and S, T⊆X such that Cl(S) = CL(T). Prove that Cl(S∩G) = Cl(T∩G)

Any idea how I should prove that? I tried using that Cl(G∩S) ⊆ Cl(G) ∩ CL(S) and then substituting Cl(S) for Cl(T) but I don't see how that is useful since its not always true that Cl(G) ∩ CL(T) ⊆ Cl(G∩T). Am I missing something?
I also tried to see if G ∩ Cl(S) ⊆ Cl(G∩Cl(S)) ⊆ Cl(S∩G) does something but it leads me to a dead end again and I think there is even less to do here

[red yoke]

A ∩ cl_X(B) = cl_A(A∩B) for any open A and subset B of X

[near sage]

cl_A means the closure of the set over A right?
Ill try to use this and will come ask again if I cant think of anything

[grave solstice]

Consider the inclusion Q-->R. Why should the induced map \beta Q-->\beta R of the Stone-Cech compactifications be onto?

I think it is enough to show that if R-->[0,1] is continuous and surjective then the induced map beta Q-->[0,1] is also surjective

[red yoke]

Yup

[near sage]

Alright I didnt manage to use that but I think I found sth so imma ask if it is correct
Cl(G∩T) ⊆ Cl(G) ∩ Cl(T) ⊆ G ∩ Cl(T) = G ∩ Cl(S) ⊆ Cl(G∩Cl(S)) ⊆ Cl(S∩G) => Cl(G∩T) ⊆ Cl(S∩G)
Cl(G∩S) ⊆ Cl(G) ∩ Cl(S) ⊆ G ∩ Cl(S) = G ∩ Cl(T) ⊆ Cl(G∩Cl(T)) ⊆ Cl(T∩G) => Cl(G∩S) ⊆ Cl(T∩G)
Therefore Cl(G∩S) = Cl(T∩G)
would that work?

[red yoke]

Why is cl(G) ∩ cl(T) ⊂ G ∩ cl(T)

[near sage]

I mean idk seems lie sth that should be true xd

[red yoke]

This fails for G = open unit disk in R², and T = cl(T) = x-axis

[near sage]

oh yeah... I was thinking about it the other way around...

[red yoke]

You just need to show that the image of Q → [0,1]^Set(Q, [0,1]) → [0,1]^Set(R, [0,1]) is dense in the image of R → [0,1]^Set(R, [0,1])

Alternatively, if you know the universal property of β, you can use the fact that free functors preserve colimits, hence epis

Then βA → βB must be surjective for any epi A → B

[near sage]

nope I cant think of anything to do with that... I tried going back to the definition of closure as an operator using interior and substraction but it just seems to overcomplicate things without changing anything

[grave solstice]

What you need is Hausdorff for this? Like if X-->Y is such that A-->Y is dense with A subseteq X dense, then X-->Y onto

[red yoke]

I don't think so?

[grave solstice]

well I don't see how it would follow from your claim then

[near sage]

is cl_A(A∩B) ⊆ cl_X(A∩B) true for any open A and any B subset of X?

[unreal stratus]

Should be true regardless of what A and B are

Indeed you can strengthen this

$\mathrm{cl}_A(B) = A \cap \mathrm{cl}_X(B)$ whenever $B \subseteq A \subseteq X$

[gentle osprey]

Crystalline Potato

[unreal stratus]

This is just by unwinding definitions

[near sage]

yeah but realized it doesnt do me any good for my question anyway even if its true

[unreal stratus]

Oh somebody said that lol

[grave solstice]

sanity check: Continuous functions Q-->[0,1] extend uniquely to continuous functions R-->[0,1] ?

[umbral panther]

No. Eg sign(x-sqrt(2))

[red yoke]

Do you mean βX being Hausdorff

[grave solstice]

no.

[grave solstice]

I meant to ask how is this sufficient

[umbral panther]

There is a hypothesis: absolute uniform continuity

[grave solstice]

By the way, I think the image would be the points $(f(q))_{f\c \R\to [0,1]}$ for $q\in\Q$

[gentle osprey]

croqueta3385

[red yoke]

This follows from βX and βY being compact Hausdorff

[grave solstice]

So if A is dense in X and X-->Y with the image of A in Y being dense in Y, with both X and Y compact+Hausdorff, then X-->Y onto?

[red yoke]

A map of CH spaces is epi iff it is surjective

[grave solstice]

ok thanks, that's what I was asking

[near sage]

wait if that is true does it mean it wont work for my problem since I have nowhere that either T ⊆ G or S ⊆ G??

[red yoke]

This still works

[grave solstice]

OK so this would follow from the fact that if r in R and f_1,...,f_n: R-->[0,1] are continuous then for all varepsilon we can find some q in Q such that |f_i(r)-f_i(q)|<varepsilon for each i. Which is immediate

I think

[red yoke]

Yea

[grave solstice]

So the beta functor is adjoint to the forgetful functor CompactHausdorff-->Top?

[red yoke]

Yea

You get unit/counit from freeness

Another hint is cl_X(B) = cl_X(cl_A(B)) when B ⊂ A ⊂ X

[near sage]

does cl_X(cl_A(A∩B) = cl_X(A∩B)? if it does can I do cl(S) = cl(T) => G∩cl(S) = G∩cl(T) => cl_G(G∩S) = cl_G(G∩T) => cl_X(cl_A(G∩S)) = cl_X(cl_A(G∩T)) => cl(G∩S) = cl(G∩T) ??

oh

[red yoke]

Sniped

[near sage]

so that should do it?

[red yoke]

Yea

But do you know how to prove the two lemmas

[near sage]

no idea about the second one but ill think a bit

[grave solstice]

I wasn't sure how much of what I wanted to prove followed from categorical considerations, once the universal property had been established. This kind of functors that arise from a universal property seem quite special, do they have a name?

[red yoke]

Adjunctions

Oh wait it's not unit/counit

It's uh

https://en.m.wikipedia.org/wiki/Adjoint_functors#Definition_via_universal_morphisms

[grave solstice]

yeah right, didn't remember the universal property definition of adjunction

Is there a criterion for X-->beta X to be injective?

[red yoke]

T3.5 apparently

Then it's an embedding

Uhh

[grave solstice]

Accoring to wikipedia the map X-->beta X to its image is a homeomorphism if and only if X is Tychonoff. But it could be injective without it being a homeomorphism with the image. Can you rule out this possibility?

[red yoke]

Hm what's the condition for points being separated by real functions

Cuz then eval_x and eval_y will be different

[grave solstice]

Dunno. There is this, which I think is what the Wikipedia article says

[red yoke]

But anyways it should be equivalent to this

Ah Urysohn spaces

Wait

What

The definitions for Urysohn space and completely Hausdorff space are swapped between wikipedia and proofwiki

[near sage]

btw unrelated question but does U = intU for every open set U? Id think it does but idk

[grave solstice]

yes, by triviality

[near sage]

ok thanks

[grave solstice]

I mean, just write down the definition

[alpine nest]

I think it can In fact be the definition of "open set"

[near sage]

it actually is... for a moment I was thinking that open set is any set that isnt closed... brainfarted

[red yoke]

Open/closed doors strike again

[grave solstice]

is the stone cech compactification of N homeomorphic to [0,1]^continuum, with the product topology?

wait

wait no no

[languid patrol]

Let me just say that you are not going to get an explicit description of B(N) as a space since even its behavior as a set depends on set theoretic axioms used to define N

so there is not going to be some simple answer where you relate it to a familiar space

[jolly umbra]

just to check, these two are homotopy equivalent right?

[grave solstice]

right, it will be ultrafilters on N

[untold lily]

trying to prove b) implies c) in part 3

I know I'm gonna need the results of 1 and 2

but like, how? do I take an arbitrary finitely inadequate family of open subsets and maximalize it? then they arent even necessarily open right

same for maximalizing the subbase

[red yoke]

Yes

You're stretching one node into two

[jolly umbra]

yeah also both are deformation retract of a disk minus 3 points

[untold lily]

another willard enjoyer 👀

[grave solstice]

If X and Y are discrete and X-->Y is injective, then beta X --> beta Y should be injective, right?

[rancid umbra]

what book is this from?

[rancid umbra]

it makes intuitive sense, at least, that adding open subsets generated by a subbase to that subbase generates the same topology.

it also happens that if both are finitely inadequate, then so is their union, since the stuff that you added is covered by unions of finite intersections of the subbase elements, at least, this is my hope

[untold lily]

willard - general topology

[untold lily]

I understand and agree with both of your pts, but I'm not sure how they are relevant

[red yoke]

All functors preserve split monos

[rancid umbra]

so, my working idea is the following:

let S be any finitely inadequate family of open subsets.
the topology generated by B (from the hypothesis of 2) and the topology generated by B U S are exactly the same.
B and S are both finitely inadequate, so i am hoping that the union is too

[grave solstice]

all functors preserve injections? 🤔

[red yoke]

Not all injections in Top are split

If it's split and an inclusion the left-inverse is called a retraction

[untold lily]

yeah the union would still be fin. inadequate, but we wan't to show that S is inadequate and I don't see how this line of reasoning takes us closer to showing that

and I'm certain that we must use the maximal finitely inadequate families in our proof

[grave solstice]

ah ok, got it

[untold lily]

I haven't used 1 and 2 anywhere else

[rancid umbra]

ah you're right; i kinda lost sight of the end goal

[untold lily]

I know that I must use 1 and 2 somewhere, this thm requires the axiom of choice since it proves tychonoff, and only 1 uses that so

[grave solstice]

I was thinking that continuous functions beta X -->[0,1] separate points on beta X, for an artbirary topological space X

[rancid umbra]

Willard - Tychonoff Theorem

[grave solstice]

I was thinking that if two points p,q in beta X are not separated then I can quotient beta X by identifying the two points to obtain another space, say Y. Then, Y is still compact and Hausdorff and clearly satisfies the Stone-Cech universal property

[grave solstice]

If X subseteq Y, as discrete spaces, is then beta X clopen in beta Y?

I was expecting this to be true, but I think it's false? beta X in beta Y consists of sequences of the form (a_f)_{f : Y-->[0,1]} but if f, g agree on X then so should a_f and a_g. But this property needn't be true if you consider the closure of beta X in beta Y

[gritty widget]

how ı find a Quotient space on topological space?

[unreal stratus]

It's not really clear what you mean

[jolly umbra]

let ~ be an equivalence relation on X, X/~ the set of equivalence classes, and q: X -> X/~ the quotient function
we say U ⊆ X/~ is open iff q^-1(U) is open in X
this defines a topology on X/~

kinda essential to know that to construct CW complexes lol

[gritty widget]

please give me a example

[unreal stratus]

Take the interval [0,1] and consider the equivalence relation which identifies 0 ~ 1 (and everything with itself etc)

Then the space [0,1]/~ can be identified with the circle; intuitively you've got a short line and joined the edges to each other to form a circle

[untold lily]

I wanted to say that I've finally been able to use this hint after proving the Alexander Subbase Theorem (funnily enough, the exercise that proves Alexander comes after this question, but whatever)

and also the other direction

thanks for that

[empty grove]

I'm not sure about openness, but βX is closed in βY because it's a compact subspace of a Hausdorff space

[heady skiff]

can somebody check my proof for b? Let any $\delta > 0$. Let $\overline{y} = (x_1, x_2 + \frac{\epsilon}{2}, x_3 + \frac{2\epsilon}{3}, \dots, x_n + \frac{(n - 1)\epsilon}{n})$. we consider the case where $\delta \geq 1$ and $\delta < 1$ separately. First suppose $\delta \geq 1$. If $\overline{z} = (x_1 + \epsilon, x_2 + \epsilon, \dots, x_n + \epsilon, \dots)$, then $\sup {\overline{d}(z_i, y_i)} = \sup {\frac{\epsilon}{i}} = \epsilon$, so that $\overline{z} \in B_{\delta}(\overline{y})$ but $\overline{z} \notin U(\overline{x}, \epsilon)$. Now suppose that $\delta < 1$. let $\overline{z} = (y_1, + \frac{\delta}{2}, y_2 + \frac{\delta}{2}, \dots, y_n + \frac{\delta}{2}, \dots)$. then $\overline{z} \in B_{\delta}(\overline{y})$ since $\sup {\overline{d}(z_i, y_i)} = \frac{\delta}{2}$ since $\frac{\delta}{2} < 1$. Moreover, $\overline{z} \notin U(\overline{x}, \epsilon)$; since $\frac{\epsilon}{n} \to 0$ as $n \to \infty$, we conclude that for $N$ large, $\frac{\delta}{2} > \frac{\epsilon}{N}$, so that [z_N = x_N + \frac{(N - 1) \epsilon}{N} + \frac{\delta}{2} > x_N + \epsilon] Hence $U(\overline{x}, \epsilon)$ is not open.

[gentle osprey]

okeyokay

[grave solstice]

Ah, right. Yeah, then it is open because the complement is beta (Y setminus X)

(beta respects arbitrary disjoint unions in general)

[untold lily]

has anyone heard of this guy? when I google urysohn's extension theorem only tietze shows up lol

[umbral panther]

Wikipedia says that it Urysohn’s name is sometimes included. Urysohn’s lemma is extremely close to the Tieze extension theorem

[jolly umbra]

i have heard of urysohn

i think of him as a point set topology guy

[untold lily]

sorry, my vocabulary can be a bit unusual

by "this guy" I was referring to the theorem

not urysohn

[untold lily]

I know that's why tietze shows up

[jolly umbra]

yeah i haven't heard of this theorem

[umbral panther]

Also read my first sentence. Or read the google hits you got

[untold lily]

?

again, the theorem in my screenshot is not tietze's extension theorem

that is something else

[jolly umbra]

the only extension theorem i know

is by caratheodory

[untold lily]

that theorem is almost as hard to prove as spelling caratheodory

[jolly umbra]

damn i butchered his name

[untold lily]

I don't blame you

maybe not even caratheodory would

[heady skiff]

why do we not need to show that V open in p(A) => q^{-1}(V) open in A?

we can't use the fact that p is a quotient map right since V open in p(A) doesn't imply that V is open in Y

can we just say that if $V$ is open in $p(A)$, then $V = p(A) \cap O$ for $O \subset Y$ open, so that $q^{-1}(V) = A \cap q^{-1}(O) = A \cap p^{-1}(O)$ which is open

[gentle osprey]

okeyokay

[heady skiff]

oh it's just that restricitng the domain and range of continuous functions preserves continuity

[rancid umbra]

just read through this and can confirm with you. what you are saying is just checking continuity; these maps are already continuous

[empty grove]

Ah ofc, cool

[heady skiff]

anyone?

[sonic tusk]

Im trying to understand open sets in a more intuitive sense but I just don't really get the conventional definition of open sets. "E is an open set if every element of E is an interior point"

From my understanding open sets are meant to generalize this idea of 'open intervals' in R^1 for example, or regions in R^2 where 'the boundaries are not there'

[gaunt linden]

Isn't that the same is you intuit "interior point" as "point (in the set) that is not on the boundary"?

[sonic tusk]

that's actually a good way of thinking about it, not too be pedantic but it wouldnt be consistent because then it would follow that a 'non-interior point' is a point that is on the boundary

[tender halo]

it is tietze's extension theorem

tietze proved the statement in question for metric spaces, and Uryhson extended it to normal spaces

[sonic tusk]

non-interior limit points are also in a way boundary points, and they also correspond to what we would call boundary points in R^n. Would a definition like "an open set is a set that does not contain non-interior limit points" make sense aswell?

[tender halo]

consider a singleton set, it is open by your definition

[earnest ibex]

Why is $\partial D^4 = \partial (D^2 \times D^2)$? We have that $\partial D^4= {(x,y,z,w)\mid x^2+y^2+z^2+w^2=1}$ and $D^2 \times D^2 = {(x,y,z,w) \mid x^2+y^2 \leq 1, z^2+w^2 \leq 1}$. Hence, I believe that $\partial(D^2 \times D^2) = {(x,y,z,w) \mid x^2+y^2 =1, z^2 + w^2=1 }$, but this is not the same as $\partial D^4$

[gentle osprey]

ImHackingXD

[merry geode]

Maybe you have more specific context where equality means different thing

[earnest ibex]

This is taken from Hatcher's Algebraic Topology, Example 1.24. He usually uses a different notation for when things are homeomorphic, which is why I think it is an equality

[merry geode]

In algtop, I think this could mean they are equal as a cycle.

Like, equal in H_3 homology.

[earnest ibex]

This is from chapter 1 of homotopy, homology is only introduced in chapter 2

[merry geode]

Can you elaborate on example 1.24? I do not have hatcher rn

[earnest ibex]

It's about torus knots, we want to compute pi_(R^3 \K), so we compute pi_1(S^3 \K), which is the same by Van Kampen

Then we want to show that S^3 \ K deformation retracts to a space X, and so this decomposition is made

[merry geode]

This basically is using homeomorphism

Abuse of notation, yeah, but not by much.

[earnest ibex]

hmmm okay, I had my doubts because of the common usage of $\cong$ for homeomorphisms and the second = being an actual equality, but I guess being homeomorphic changes nothing in the proof

[gentle osprey]

ImHackingXD

[earnest ibex]

Thanks for the help!

[merry geode]

Who knows, maybe a typo

No problem!

[gaunt linden]

Isn't is? What is your definition of "boundary"?

[unreal stratus]

They say "regarding", like you just pick a homeomorphism (there is a canonical one by regarding R^4 = R^2 x R^2) and treat the homeo as an equality to make it simpler

This is pretty common in maths for clarity and brevity

[uneven bronze]

I'm stuck with a basic claim. Consider n separable metric spaces Xj, and form their cartesian product X, with the product metric (the maximum of the metrics). Then each Xj contains a countable dense set Cj. How can I then show that the set of points in X whose jth coordinate is in Cj for all j is a countable dense subset of X? Seems intuitively clear, but I'm unsure how to actually show it.

Is the set that is claimed to be dense in X simply the cartesian product of the Cj?

[tender halo]

yeah

[uneven bronze]

ok, so being a cartesian product of countable sets, it is countable. But how do I show denseness?

[gritty widget]

X={1,3,5,9,11,18} and b equivalence relation on X
b={(a,b) 4|a-b}
X/b set of equivalence classes.
X/b={{1,5,9},{3,11},{18}}
Mean X/b quotient space in X
it is true?

[tender halo]

well i mean take the closure of it

should be the whole space

[uneven bronze]

yeah, indeed, thanks (using the identity that the closure of the cartesian product is the closure of the individual sets)

[heady skiff]

could I get a hint for showing that the $l^2$-topology is contained within the box topology? I let $B_{d'}(\overline{x}, \epsilon)$ be a basis element of the $l^2$-topology, and initially considered the open set $U \coloneqq \prod_{i \in \mathbb{Z}^+} U_i$ where $U_i = (x_i - \frac{1}{i}, x_i + \frac{1}{i})$, motivated by the $p$-series (for if $\overline{y} \in U$ then $(x_i - y_i)^2 < \frac{1}{i^2}$, so at least we know that the series $\sum_{i = 1}^\infty (x_i - y_i)^2$ is convergent). the problem of course being that we have no way of guaranteeing this series is less than $\epsilon$ - in fact if we bound by smaller neighborhoods $U_i$ of $x_i$, we can't really get a bound, the least being one since as $s \to \infty$, $\zeta(s) \to 1$

moreover diam$(x_i - \frac{1}{i^s}, x_i + \frac{1}{i^s}) \to 0$ as $s \to \infty$, so I question my abilities to find a small enough neighborhood $U_i$ of each $x_i$ as to make the sum less than $\epsilon^2$

[gentle osprey]

okeyokay

[heady skiff]

or maybe i'm just yappin

[red yoke]

Why not use another series that you can bound

[gentle osprey]

okeyokay

[red yoke]

I don't get what you mean by "if we use smaller neighbourhoods …, we can't really get a bound"

[heady skiff]

like my idea was that if we considered any neighborhood $U_{\delta} = (x_i - \delta, x_i + \delta)$ for any $\delta > 0$, then we could find $s > 0$ with $\frac{1}{i^s} < \delta$, so $V \coloneqq (x_i - \frac{1}{i^s}, x_i + \frac{1}{i^s}) \subset U_{\delta}$, so if $y_i \in U_{\delta}$, we might have $y_i \in V$, and we can repeat this process for all the $U_{\alpha}$ (given $\prod U_{\alpha}$) since they're all open intervals of $\mathbb{R}$ with positive radius

idk there's probably something severely wrong with my logic here lol

[gentle osprey]

okeyokay

[heady skiff]

like given any arbitrary $\overline{y} \in \prod U_{\alpha}$ there exists $s > 0$ with $|x_i - y_i| < \frac{1}{i^s}$ for all $i$ is what I'm worried about

[gentle osprey]

okeyokay

[red yoke]

But your s is chosen individually for each Uα

[heady skiff]

but couldn't we choose such a large s that it's true for every Ua

[red yoke]

Why is that possible

[heady skiff]

idk lol

it just seems intuitively plausible but i guess it can't be true

[red yoke]

If you have finitely many Uα then you can just take maximum

But when there's infinitely many you run into a problem

The choices can be unbounded

[heady skiff]

oh yeah i see

right eventually you might run into a neighborhood Ua such that (a - 1/i^s, a + 1/i^s) is not contained in Ua

i feel like i'm overcomplicating things lol

[red yoke]

Oh wait if 1/2^s < δ then 1/i^s < δ for larger i too

[red yoke]

Right should be that

[red yoke]

I think you just have to find a positive series with l²-norm less than ε

And take Uα to have those radii

[heady skiff]

so something like, find a series such that $\sum_{i = 1}^\infty M_i < \epsilon^2$ and then just take $U_i = (x_i - \sqrt{M_i}, x_i + \sqrt{M_i})$ (assuming $M_i > 0$ for all $i$)

[gentle osprey]

okeyokay

[heady skiff]

ay i'm very active again 😂

ugh i forgot everything about bounding series

crazy idea: what if we just considered the power series expansion for sin(x) near 0, since that's something we can bound by e^2

Let me know if this works: Let $B_{d'}(\overline{x}, \epsilon)$ for $\epsilon > 0$ and $\overline{x} \in X$. Since $\sin(x)$ is continuous, we have $\delta > 0$ such that if $|t| < \delta$, $|\sin(t)| < \epsilon^2$. Then $\sum_{i = 1}^\infty \biggl|\frac{t^{2i + 1}}{(2i + 1)!}\biggl| < \epsilon^2$. Let $M_i = \biggl|\frac{t^{2i + 1}}{(2i + 1)!}\biggl|$. For each $i$, let $U_i = (x_i - \sqrt{M_i}, x_i + \sqrt{M_i})$. If $\overline{y} \in \prod_{i} U_i$, $|x_i - y_i| < \sqrt{M_i}$ for all $i$, whence $(x_i - y_i)^2 < M_i$ for all $i$. Then [\sum_{i = 1}^\infty (x_i - y_i)^2 < \sum_{i = 1}^\infty M_i < \epsilon^2] so $\overline{y} \in B_{d'}(\overline{x}, \epsilon)$, as required.

[gentle osprey]

okeyokay

[heady skiff]

wait fuck i don't know if this works because we don't necessarily have $\sum_{i = 1}^\infty \biggl|\frac{t^{2i + 1}}{(2i + 1)!}\biggl| < \epsilon^2$ since that would invoke the triangle inequality intermediately

[gentle osprey]

okeyokay

[unreal stratus]

I don't think you shoul really need to do any explicit computations

[heady skiff]

yeah i guess i'll return to this at some point

[untold lily]

can someone explain how (a) in lemma 19.8 shows F is onto? (K_i, h_i) means that h_i is an embedding from X into K_i, K_i is compact Hausdorff and h_i(X) is dense (idk how standard this terminology is)

I guess, if a function maps a dense subset homemorphically to a dense subset, is it necessarily onto or is something else being used here

[rancid umbra]

whoops

[untold lily]

yes, I tried that and got a tautology like you did

lol

[rancid umbra]

i’ll think about this some more. got a bus ride in a few

any reason you’re going through this book? just for a class?

[untold lily]

I wanna improve my topology

also I like topology

summer plan is: go through a lot of this book, go through a differential geometry book, go through a complex analysis book, then starting grad school on september

[rancid umbra]

nice! i think that a lot of you’re questions are interesting since i am unfamiliar with those areas of topology as well.

good luck with grad school when you go!

[untold lily]

I really enjoy the exercises in this book, they are interesting

and thanks!

[untold lily]

forgot to mention algebra, man

[rancid umbra]

@untold lily
let $X$ and $Y$ be top spaces, $F:X\to Y$ continuous. If $X_1$ is dense and $F|_{X_1}:X_1\to Y$ is a homeomorphism onto its image, then $F$ is surjective. Here’s why:

let $y\in Y$. Since $F(X_1)$ is dense in $Y$, then there is a sequence of points $(x_n){n\in\mathbb{N}}$ in $X_1$ such that $$y = \lim{n\to \infty}F(x_n)$$
As $F$ restricted to $X_1$ is a homeomorphism onto its image and $(F(x_n)){n\in\mathbb{N}}$ converges to $y\in Y$, then by continuity of $F^{-1}$, we must have that $(x_n){n\in\mathbb{N}}$ converges to some $x\in X$. Now by continuity of $F$, we have $$F(x) = F\left(\lim_{n\to\infty}x_n\right)=\lim_{n\to\infty}F(x_n)=y$$
so $F$ is onto

[gentle osprey]

c squared

[untold lily]

you can't really use sequences here, but the argument stands with nets instead

the F^(-1) piece is key

[rancid umbra]

oh right, does compact hausdorff not imply first countable

[untold lily]

if it does, I haven't delved that deep into the book for it

[rancid umbra]

well, the function is continuous, so its sequentially continuous

you can get that one for free

[untold lily]

yeah but you can't guarantee a sequence such that y = lim F(x_n), can only guarantee a net

[rancid umbra]

are you worried about uniqueness of limits?

[untold lily]

no

[rancid umbra]

im not quite seeing why

[untold lily]

you are saying since F(X_1) is dense, there is a sequence in F(X_1) that converges to any y in Y

[rancid umbra]

how does that not just follow from definition of density

[untold lily]

that won't follow if sequences can't describe your space topologically

there are spaces where the only convergent sequences are eventually constant, I was going through it in the exercises actually

[rancid umbra]

ah okay

[untold lily]

in such a case, density simply cannot guarantee this

(I solved 1 and 2 but stuck on 3)

[rancid umbra]

ultimately constant is the same as eventually constant?

[untold lily]

yes, at least I hope so lol

only reasonable thing it could stand for

[rancid umbra]

do you mind posting the base definition of extremally disconnected as well? or is it condition b)?

[untold lily]

it's the first sentence

here are the other definitions drawn from

[rancid umbra]

where r u stuck in 3? some detail in the proof outline?

[untold lily]

tbh, I barely thought about it

just couldn't think of how to find the disjoint open subsets and the subsequence corresponding to it

and such that p is in the closure of all the U_k

hmm

[rancid umbra]

use the hausdorff property

[untold lily]

I guess you start with a nbhd of p

since x_n converges to p, eventually one x_k will land in this nbhd

keep it an open nbhd

[rancid umbra]

at each new value of the sequence, you can find a nbhd that doesn’t contain any of the previously seen values

[untold lily]

yes ofc, but that is not the issue

[rancid umbra]

by intersecting

[untold lily]

how do you guarantee p is in the closure

is the issue

and p can't be in any of the U_k for obvious reasons

[rancid umbra]

ah i didn’t even see that

[worthy fjord]

Hey, I'm learning algebraic topology, and apparently the inverse of ${[f]}$ in the fundamental group for some path $f$ is ${[\Bar{f}]}$ where $\Bar{f}(s)=f(1-s)$ and I just don't see how $f*\Bar{f}$ could be homotopic to $e_0$ (the constant path). Can someone please explain?

[gentle osprey]

friendlyneigborhoodtopologydonut

[worthy fjord]

Because I'm imagining this on the torus and it seems like it just can't be the case.

[untold lily]

why?

[rancid umbra]

you’re reversing the direction of travel. so you can just collapse the whole concatenated path to a point

[worthy fjord]

So if they're homotopic, then there's a continuous function $F:I\times I\rightarrow T$ such that $F(s,0)=(f*\Bar{f})(s),F(s,1)=e_0,F(0,t)=x_0,F(1,t)=x_0$, right?

So I'm thinking about a loop that goes all the way around the dougnut

and if you take that loop and connect it to a loop going the opposite direction, how can there be a continuous transformation from that loop to the constant one?

It seems like you would have to break the loop.

[untold lily]

no, what you are trying to do is homotope f to f bar,

[worthy fjord]

Why can you do that?

[untold lily]

what you need to do is homotope f * f bar to the constant path

[worthy fjord]

oh wait ur right sorry

[rancid umbra]

like, imagine walking to school and back. just shrink that path to a point

[untold lily]

f * f bar is traveling along a path and then traveling back along the same path

now imagine

instead of traveling the entire way, you travel halfway along f

and then return with f bar

[gentle osprey]

friendlyneigborhoodtopologydonut

[untold lily]

yes

[worthy fjord]

yeah but it seems like you would have to break the loop

[worthy fjord]

ok so for this

why can you do this?

you should have to travel the whole way and then the whole way back, no?

[untold lily]

no?

as long as you start and end at x_0, anything continuous is legal

[worthy fjord]

But isn't that the definition of $$? Because $(f\Bar{f})(s)=\begin{cases}
f(2s),s\in{[0,\frac{1}{2}]}\
\Bar{f}(2s-1),s\in{[\frac{1}{2},1]}
\end{cases}

[gentle osprey]

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

[sonic tusk]

is the following intuition for closed sets good? If there is a limit point then we know there must be infinitely many points in its immediate vicinity, a cluster or oasis of points in E if you will, if the limit point is not in E it can act as an entry point from outside to inside E and that’s why we call E closed if every limit point is in E

[untold lily]

yes, and?

[worthy fjord]

ok wait

[untold lily]

the point is, during the homotopy your path will look less and less like f * f bar and more like the constant path

[worthy fjord]

The notebook is coming out

[rancid umbra]

a new one?

[worthy fjord]

💀

[untold lily]

I don't really understand what you are saying, but since topological spaces can be finite I can say your intuition cannot make sense

[sonic tusk]

How could there be a finite closed set?

[rancid umbra]

{a} is closed in any non-empty hausdorff space

[sonic tusk]

Excuse me, I’m not familiar with all this topology mumbo jumbo

I’m just trying to learn analysis here 😞

[rancid umbra]

in R, any point set is closed

[swift fjord]

I have had news for you buddy

[untold lily]

if you are working exclusively in R, probably better to look at #real-complex-analysis

[worthy fjord]

oh, for analysis that's probably fine. But there are other channels for that #real-complex-analysis and #advanced-analysis

[sonic tusk]

Tbh I’m trying to understand it independently of R^n. But I am in the realm of metric spaces

afaik for metric spaces at least closed sets can’t be finite

[untold lily]

in metric spaces, every finite set is closed

that is as wrong as it can get

[sonic tusk]

you’re right I got confused

[worthy fjord]

So, if I'm understanding what you're saying properly, you're saying that if you cut the loop, so to speak, then bring both sides into the constant loop, that works as the path homotopy?

[worthy fjord]

Yeah, that intuition only works for intervals.

[untold lily]

you aren't cutting anything

first of all, you realize the loop in question is just one line

[worthy fjord]

yes

[untold lily]

and you travel to the end of the line, and then travel back to the beginning

so instead of traveling to the end

you travel to say, halfway point

and then go back

that is the idea

[worthy fjord]

But that doesn't consider the whole loop $f*\Bar{f}$.

[gentle osprey]

friendlyneigborhoodtopologydonut

[untold lily]

what do you mean?

[worthy fjord]

That's only part of the loop.

The loop goes all the way around.

[untold lily]

the "loop" is a line that goes from f(0) to f(1)

[worthy fjord]

yes.

[untold lily]

your new "loop" will be the same line from f(0) to f(1/2)

it's still a loop

it can be homotoped

I don't understand what the issue is

[worthy fjord]

wait wait wait no

[sonic tusk]

Sure, since finite sets don’t have limit points but for sets that do have limit points i think it makes sense no?

[worthy fjord]

The loop still goes from $f(0)$ to $f(1)$.

[gentle osprey]

friendlyneigborhoodtopologydonut

[untold lily]

I'm saying topological spaces themselves can consist of finitely elements

[worthy fjord]

also $f(0)=f(1)$

[gentle osprey]

friendlyneigborhoodtopologydonut

[worthy fjord]

because it's a loop

[untold lily]

no, unless you rebranded f to f * f bar

f(0) is not f(1)

[worthy fjord]

Well if $f(0)\neq f(1)$, then it's not a loop, no?

[gentle osprey]

friendlyneigborhoodtopologydonut

[untold lily]

or uh, wait this is a special case

okay in this instance yeah, but the same idea works even in paths that aren't loops

maybe that is why you are confused

[worthy fjord]

Okay can we restrict this to the fundamental group of the torus?

[sonic tusk]

? Sure but how does that relate

[worthy fjord]

I feel like something more concrete would help me understand.

[untold lily]

in this case, f * f bar would be a loop followed by the same loop backwards

[worthy fjord]

yes.

It's still a loop though, is it not?

[untold lily]

if instead, you traveled halfway along the loop and then went backwards

it's still a loop

[worthy fjord]

oh

I was misunderstanding what you meant.

[worthy fjord]

Here, you were just saying the definition of $f*\Bar{f}$, correct?

[gentle osprey]

friendlyneigborhoodtopologydonut

[untold lily]

yes

[worthy fjord]

okokok

I thought you were saying something else

But then, how is that homotopic to the constant loop?

[untold lily]

I just explained how, don't travel the entire loop and then go back

instead, travel some portion of the loop and then go back

and you still have a loop

anything that starts and ends at the same pt is a loop

[worthy fjord]

But then that wouldn't necessarily be homotopic to the original loop, right?

oh wait

you're right.

I see now

thank you.

[untold lily]

cool

[untold lily]

the "infinitely many pts" thing doesn't make sense in that instance, but every finite metric space has to be discrete anyway, so let's ignore that

I don't even know what definition of limit point you are working with

[sonic tusk]

Sure but if we ignore the case for finite metric spaces and only consider sets with limit points, would it make sense?

[sonic tusk]

does it matter though? They’re qualitatively equivalent anyways

[untold lily]

I guess

you can say that a limit point is infinitesimally close to E, and a set is closed if every point infinitesimally close to it must be in it to begin with

if I understood your point to begin with

[sonic tusk]

Yeah that’s definitely another way of looking at it but it’s harder see straight away how it nicely encapsulates this “closed” concept

[untold lily]

a set is closed iff you can't converge out of it
if you want to think of the space dynamically, sequences are the way to go

for intuition

[sonic tusk]

agreed

[rancid umbra]

how is this even working for discrete spaces? like, p would have to be in U_k.

[untold lily]

then the existence of such a U_k must follow from the assumption that there is a non constant sequence converging to p

such a U_k cannot actually exist, but the topology is tied to the convergent sequences of a space

[untold lily]

if not, it could also be that the hint is faulty of course, but I wouldn't be quick to claim that

[rancid umbra]

same

but it is a bit difficult to come up with these U_k’s

[untold lily]

yeah

I'm done with math for today but may come back to this tmrw

[rancid umbra]

why does this hint not break condition b) of extremally disconnected?
we are looking for disjoint open sets containing some specific points, but their closures all share at least one point in common?

[untold lily]

wait you are right, I should have noticed that way earlier

the hint is faulty then

[rancid umbra]

yea

i’ll start thinking of an alternative method

well

maybe some of the hint is viable

just thought i’d point it out

[untold lily]

okay, I looked it up and apparently you want p to not be in the closure

[rancid umbra]

interesting

[untold lily]

there is a prf of it in stackexchange, I only read the beginning of it

idk what the contradiction will be then

[rancid umbra]

ok cool thank you for the tip

[untold lily]

but yeah, construct a subsequence like that and see where it leads you

[near sage]

Hello again! I have another question I need help with

Let X be a topological space. A subset G of X is called regular closed when $G = \overline{IntG}$.
Prove that if $D={F_i}{i\in I}$, is a family of regular closed subsets of X then $\overline{\cup{i\in I}int\left(F_i\right)}$ is exact upper limit of D (considering $\subseteq$ ) and $\overline{int\left(\cap_{i\in I}F_i\right)}$ is the exact lower limit

[gentle osprey]

Hellofdeath

[north hedge]

what do you think about this nomenclature; Complex geometric ADG (ADG + hodge theory + algebraic geometry)

[unreal stratus]

What is ADG

[final grotto]

hey is every element of a topology open?

[zinc hearth]

the definition of open is that it belongs to the topology, so yes

[final grotto]

soo if I have a closed subset Y of a topology X and a subspace topology of Y then every closed subset of this subspace topology Y is also open right?

[unreal stratus]

No, but it seems you're conflating a few things

[gaunt linden]

Huh? Should some of those "closed" have been "open"?

[unreal stratus]

Remember a topological space is a pair of a set $X$ and a topology $\mathscr T$ on $X$. You don't talk about "closed sets of a topology", but rather closed subsets of the topological space $(X,\mathscr T)$ [which everyone writes as just $X$] or equivalently "complements [in $X$] of elements of $\mathscr T$".

[gentle osprey]

Crystalline Potato

[final grotto]

yes I meant subset not set my english is not so good sry

[unreal stratus]

I mean that's fine dw, I more meant because you said closed subset of a topology

It should be "closed subset of the (sub)space"

[final grotto]

closed element of a topology

[unreal stratus]

Sure, they are open (in the subspace) by definition

[final grotto]

okk I asked because I have to show that if Y is closed in X and X is normal then the subspace topology of Y is also normal but per definition is every closed subset of Y in the subspace topology open so yeah

thanks for answering my question ^^

[gritty widget]

what is product topology?

[patent path]

What's a natural metric to put on projective n space over reals? I'm thinking of taking representatives in S^n+1 and sort of looking at the minimum distance

[zinc hearth]

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

[last gale]

Any ideas on what co-dense is?

[ebon galleon]

Dense = it's closure is everything
Co-dense = its complement's closure is everything, or in other words, it has no interior

[last gale]

Thank you

[fading vale]

Is there some kind of naturality condition on the shriek/wrong way maps on K theory? As in if I have smooth embeddings X -> Y -> Z is the induced map on K theory from X to Z the composition of the ones from X to Y and Y to Z?

I've reduced this to verifying some annoying multiplicative identity on the thom classes of each of the three normal bundles after pulling them all back to K(Z) but I don't know how to verify that so I'm hoping somebody else already has

[umbral panther]

The Thom class depends on structure on the normal bundle. The easy case is if the normal bundles are complex. Then the Thom class is the exterior algebra. The exterior algebra turns sums into tensor product, which is what you need

[fading vale]

Sorry what do you mean sums?

Oh

Nvm I see

So you want to use that the normal bundle of X in Z pulled back to Z is equal to the pullback of the sum of the other two normal bundles?

Oh shoot I see

Yeah this makes sense, I was unsure about the even vs odd grading but it totally works out

The even part is the sum of tensor both even and tensor both odd, the odd part is sum of each even odd pair

Thank you bw

[gritty widget]

can you give me a example?

[zinc hearth]

I'll just be more explicit, The product topology is the topology on XxY consisting of arbitrary unions of sets of the form U x V with U open in X, V open in Y

so lets take {empty, {0,1}} and {{empty, {0,1}, {1}} to be our two topologies on X = {0, 1}, then the product topology would be generated by unioning the following sets:

\empty, {0,1}x{1} = {(0,1), (1,1)}, {0,1} x {0,1} = {(0,0), (0,1), (1,0), (1,1)}

[zinc hearth]

if memory serves correctly, there's an issue when you take arbitary products - but I don't remember it!

[tender halo]

the standatd topology on the plane is the product topology of R with itself

[rancid umbra]

the product topology is generated by a subbasis of preimages of projection functions

[tender halo]

in an arbitrary product you need the multiplicands of basis boxes to be the whole space almost everywhere

[zinc hearth]

ah. Like a direct sum. But backwards

actually no. Not backwards

[unreal stratus]

Yeah this is the nicest to me lol (preimages of opens though)

Like you want smallest such that projections are cts

So just do dat

[zinc hearth]

too much category theory

[unreal stratus]

Then only allowing for finite intersections means you don't get the full box topology

What cat theory does to a mf

[jolly umbra]

riemannian metric

[lusty trench]

Two natural candidates, which are probably the same anyway:

• Take the Fubini-Study metric on CP^n and restrict it to RP^n.
• Consider RP^n = S^n / G, where S^n is the unit sphere in R^(n+1) and G is the cyclic group of order 2. Since G acts on S^n by isometries, the metric on S^n arises from lifting a metric on RP^n. Take that metric on RP^n.

[sleek ravine]

I have a question related to handlebodies and 3-manifolds that I couldn't figure out.

So I consider my handlebodies with 3 handles as a solid 3-torus, and I glue my a copy of the handlebody with another copy of it by identifying their boundaries in the obvious way.

In the picture below, I have three disks A, B, C each along a handle, and another disk X that's between A and B.

When I glue my two copies of handlebody together, disks become S^2. How do I know the sphere that's from the disk X will contain the sphere of A and B in one of its sides (interior or exterior), but leaving the sphere of C on the other side? Or is the statement not true?

[red yoke]

Is there an interior / exterior

In the solid 3-fold torus you can go from one side to the other right

[sleek ravine]

Good question! My only intuition is that I can definitely divide S^3 into interior/exterior because S^3 is two copies of D^3 glued together identifying the boundary S^2. But I'm not sure for my picture.

[sleek ravine]

I think you may be right. Cut the handles open and I will recover my S^3, it does seem like by identifying the handles again I could pass from one side of it to the other. It's super weird I don't know what to make of it.

[oak smelt]

is the degree of a map: CP2 to CP2

always a square?

[tawdry valve]

yes! You can prove this by using cup products. The cohomology ring of CP2 is Z[a]/(a^3), where a is degree 2 (ie a cohomology class in H^2).

A map CP^2 -> CP^2 induces a homomorphism of cohomology rings going the other way, and the degree is the top multiplication. In our setup,

a^2 maps to (degree * a^2),

and since this is a ring homomorphism (ie the induced map respects cup products), since you have

a maps to k* a

for some k, you’ll get

a^2 maps to k^2 * a^2,

so, the degree is k^2

[oak smelt]

thanks!

[feral copper]

Heyo! Does anyone know of a simple proof of the fact that if W is a cobordism between Y and itself and f:Y->Y is a self-diffeomorphism of Y, then the manifold X=W/~ obtained by gluing ends via f has fundamental group the HNN extension of that of W?
I can't find anything, nothing; am I bad at Googling? I was kind of expecting a proof along the lines of that of SvK, but nope... Thanks!

A special case of that is the fundamental group of the mapping torus, but I'm interested in the general cobordism case

Hatcher speaks of graphs of groups, of Bass--Serre theory and of K(G,1) spaces, but I don't need that at all...

[misty grove]

One approach to proving it would be to look at the covering space associated with the universal cover of the cobordism W and then examining the action of the self-diffeomorphism f on this covering space. I dont think there is a quicky easy proof for this

[umbral panther]

That’s not true for a general cobordism. Take W the product cobordism and take the connected sum with S^n x S^1. This has fundamental group the free product with the original group

[feral copper]

Ah, I see! I guess (as usual, the problem I'm interested in is slightly more specific) I need to add the assumption that W is obtained from YxI only by adding 2-handles (I'm working in the 4-dim case, but whatever)

[umbral panther]

Yeah, SvK

[feral copper]

I think I have a cyclic covering by gluing an infinite number of copies of W at their ends; is that what you mean?

[umbral panther]

A handle attachment is like a cell attachment

[feral copper]

For SvK, I'm facing the issue that the intersection of U and V is not path-connected (then I can use the groupoid version, but I'd rather avoid that)

[feral copper]

The Wikipedia page for HNN extensions states:

HNN extensions describe the fundamental group of a self-glued space in the same way that free products with amalgamation do for two spaces X and Y glued along a connected common subspace, as in the Seifert-van Kampen theorem.
What's wrong then?

[umbral panther]

It is connected

Oh, sure, it’s fine. I thought you wanted the fundamental group of Y not W

U is a neighborhood of Y. V is the complement of Y

[feral copper]

U deformation retracts to Y, and V is a deformation retract of W
But U n V is disconnected

(using collar neighborhoods of the boundary components of W)

[umbral panther]

How about add a path?

[feral copper]

Also, now that I think about this statement, it doesn't matter, since you're changing W here. I'm willing to show that pi1(X) is an HNN extension of pi1(W)

So, disregard my previous claim and allow W to have 1-handles

[umbral panther]

If you add a path to V, you get the free variable. And the intersection is connected, identifying the two copies

[feral copper]

Unless W is simply-connected, I should get more in pi1(V) than just one free variable

I think adding this path means that pi1(V)=pi1(X), no?

(and I haven't said it yet, but thanks for helping me!)

[umbral panther]

V starts with W, and when you add the path you add the variable freely. The fundamental group of X is a quotient of that

[feral copper]

Ah ofc it's like wedging a circle

I think this outta work, thanks a lot!

[grave solstice]

Is a topological group totally disconnected iff 0-dimensional?

uhh wait no it's not true I think

[unreal stratus]

it's true assuming you have LCH spaces (so stuff like Lie groups would work since they're manifolds)

[grave solstice]

any LCH space is totally disconnected iff 0-dimensional, no?

[unreal stratus]

that's whati'm saying ye

i think you can do dumb things like put the indiscrete topology on a group to provide counterexamples in general

[grave solstice]

oh yeah I actually have to assume LCH

[unreal stratus]

like they're 0 dimensional but not totally disconnected unless they're a point

[opaque scroll]

Are we talking Lebesgue dimension here, or something else?

[unreal stratus]

oh yeah that was what i assumed

[grave solstice]

But that's not Hausdorff tho

0-dimensional as in topology

[tribal palm]

imagine doing dumb things haha

couldn’t be me

[unreal stratus]

lol

[grave solstice]

each point has a nhood basis consisting of clopen sets

[unreal stratus]

There are still different things ig but yeah ig for 0 it should coincide

[unreal stratus]

yeah but i didn't know if you were assuming Hausdorff tbf

[grave solstice]

yeah so zero-dimensional topological groups (assuming hausdorff) are necessarily totally disconnected

In general, 0-dimensional + T1 --> totally disconnected

[grave solstice]

this is pretty weird tbh

well I was wondering because Pontrjagin uses totally-disconnected and 0-dimensional for topological groups as synnonyms

[alpine nest]

Yeah, my first contact with the term "zero-dimensional" was in the context of compact metric spaces, so "totally disconnected" and "zero-dimensional" were treated as synonyms.

Took me a while before I realized they were not

[grave solstice]

I mean, what are even examples of Hausdorff+Totally disconnected that is not 0-dimensional?

Willard gives one example, but it's not simple

K setminus p is not 0-dimensional

Adding opens preserves total disconectedness and Hausdorffness. And given any topology on a group you can always add opens so that it becomes a topological group with that topology. So I'm thinking about considering product_{n=1}^\infty Q, which is Hausdorff, totally disconnected and zero-dimensional, and add some new open set containing the identity (like prod (-1,1)) and add more open sets. We would have to pray that in the process of adding more open sets prod (-1,1) never contains a clopen subset

[heady skiff]

if i've already shown a, doesn't c follow immediately? for example, if $U$ is open in the uniform topology and $V$ is open in the $l^2$-topology (both considered under the subspace topology given on $H$) then $U = O_1 \cap H$ for $O_1$ open in the uniform topology, $V = O_2 \cap H$ for $O_2$ open in the $l^2$-topology. But then since $O_1 \subseteq O_2$, $U = O_1 \cap H \subseteq O_2 \cap H = V$

[gentle osprey]

okeyokay

[heady skiff]

so the question is more or less asking where the product topology fits in the chain of inclusions

[rancid umbra]

it’s asking you to show containment relations among all four topologies for that specific space

so yes, you have the stuff from part a)
but it may be the case that some of the topologies are the same when your scope is just H

so there is still some work to be done @heady skiff

[sonic tusk]

I saw on math.se a guy asking that

(1) A closed set is any set that contains all of its limit points.

(2) A closed set is any set that contains all of its boundary points.

Does (1) imply (2), and vice versa, or is the implication only from (1) to (2)?

clearly the converse is false right? but somebody said it could be proven?

[rancid umbra]

the set of limit points can be written as a disjoint union of the interior points and the boundary points.

so if you contain the boundary, then you will contain all limit points, as any set contains its interior as well.

conversely, if you contain all limit points, then you will contain the boundary, as the boundary is a subset of the set of limit points

[sonic tusk]

the first part made absolutely no sense to me tbh, we can write the set of limit points as a disjoint union of the boundary and interior points how does it follow that containing all boundary points implies you will contain all the limit points what about the interior limit points?

[rancid umbra]

any set contains its interior

wait

sorry

[sonic tusk]

what about the singleton set

[rancid umbra]

i might be mistaken. those union to the closure of a set

[rancid umbra]

the actual interior depends on the topology, but its either empty or its not, both of which are contained within the set

[sonic tusk]

the neighborhood of the singleton set is always empty though no?

[rancid umbra]

not in the discrete topology

or if the space is just the singleton set

[sonic tusk]

say we have a metric space like R for example a singleton set would be empty though right?

[rancid umbra]

it holds in metric spaces too. but in R with the usual topology, no the interior of a singleton is empty

you definitely have the implication that boundary points are limit points

but the converse is not true

so, the idea is as follows, similar to what i said at first:

let S be a subset of a top space X. assume we know the identity cl(S) = S union bd(S)

(1) ==> (2) if S contains its limit points, then cl(S) = S and cl(S) = S union bd(S) = S, so bd(S) is contained within S

(2) ==> (1) if S contains its boundary, then cl(S) = S union bd(S) = S so S is closed, i.e., S contains its limit points

@sonic tusk

[sonic tusk]

thanks, i guess this makes sense but whats confusing me is that would R be a counterexample if you think of any closed region in R^n and remove a single interior point but I leave the boundary points alone then according to this that set is a closed set, or am I missing something here?

[rancid umbra]

a punctured closed disk is not closed. its boundary is the point you removed along with the circle bounding it, and the disk is the set of limit points of the punctured disk

[sonic tusk]

ok what about a shaded closed circle with centred at (0,0), (0,0) is an interior limit point and not a boundary point right? if I removed it, then it would be closed according to this?

or maybe Im confusing the definition of a boundary point

"A point is a boundary point of a set if and only if every neighborhood of. contains at least one point in the set and at least one point not in the set."

[rancid umbra]

yes it is an interior limit point and it is not a boundary point of this non-punctured region.

let D be the closed unit disk (so the open unit disk with the unit circle)

D - {(0,0)} is not a closed set. (0,0) is a boundary point (and thus a limit point) of D - {(0,0)}

the sets are changing when you remove points, so their interior, limit, and boundary points may also change

[sonic tusk]

umm (0,0) is still an interior point right? and it doesnt satisfy the def of a boundary point even after we removed it 🤔

[rancid umbra]

no, (0,0) is not an interior point of the punctured disk. you are correct though that it is not a boundary point. my mistake

no open subset of the punctured disk contains the point we removed

[sonic tusk]

soooo.. we have a counterexample of a set with all its boundary points but missing a limit point?

[rancid umbra]

gonna wait for somebody else to freshly explain this to u

i have butchered this

[sonic tusk]

no worries lol thanks for helping me out ❤️

[daring sparrow]

hey is someone available to take a look at my math help channel? It is related to klein bottles and it is in help 13. If you are interested in discussing please let me know!

[rancid umbra]

since the channel is dead atm and I had some time, let me try to properly address some of your points:

Does (1) imply (2), and vice versa, or is the implication only from (1) to (2)?
I have addressed this already with a proof that (1) and (2) are equivalent

what about the singleton set
let X be a top space and x an element of X.
it may be the case the {x} is open (in this case it is also closed), closed but not open, or neither; this all depends on the topology placed on X. int({x}) is either empty or {x} in any of these settings though

what about a closed circle centered at the origin...
this time, a more careful analysis:
let D be the open unit disk in the plane. cl(D) = D U C, where C is the unit circle (the boundary of D), is the region you were describing.

(0,0) is an interior point of cl(D) and a limit point of cl(D).
(0,0) is not an interior point of cl(D) - {(0,0)} because int(X) is a subset of X for any set X.

removing the origin from cl(D) makes cl(D) - {(0,0)} not closed, since (0,0) is a limit point which it does not contain.
Further, it is a boundary point: we have that bd(X) = cl(X) \ int(X). so

bd(cl(D) - {(0,0)}) = cl(cl(D) - {(0,0)}) \ int(cl(D) - {(0,0)})
= cl(D) \ (D - {(0,0)})
= C U {(0,0)}

[rancid umbra]

better to repost some of the discussion here tbh

[sonic tusk]

😎 nice

I see my error now, my misunderstanding was that I had though that the interior points need not neccesarly be a part of the set it is an interior point of. Thank you 🙂

[worthy olive]

Can someone please explain how you find actual homology and cohomology groups with Mayer vietoris, I can only see it as a method of finding induced mappings

[gritty widget]

every single algebraic topology textbook ever should have at least one example

wikipedia has like 4 examples

[earnest ibex]

Why is $f:\mathbb{C} \to [0, \infty[$, $f(z)=|z|$ a quotient map? I understand that $U$ open implies $f^{-1}(U)$ is open because norms are continuous, but I'm stuck as to how to show $f^{-1}(U)$ open implies $U$ open

[gentle osprey]

ImHackingXD

[rancid umbra]

take a point in U, say u. lets say f(z) = u. if u = 0, then z = 0, and there is an r > 0 such that D_r(z) is a subset of f^{-1}(U).
what can you say about the norm of each w in f^{-1}(U)?
does the same argument work if u > 0?
@earnest ibex

[earnest ibex]

I think I understand it now, for the case u=0, because f is surjective we get u belongs to f(D_r(0)) is a subset of U. Since f(D_r(0))=[0,r[, there is a neightborhood of 0 contained in U

for the case u>0, we get f(D_r(z))=]u-r,u+r[

[rancid umbra]

perfect!

[earnest ibex]

Thanks for the help!

[uneven bronze]

I'm stuck with a basic identity. Given I know $X\setminus S^\circ=\overline{X\setminus S}$, how can I show $$X\setminus\bar{S}=(X\setminus S)^{\circ}?$$ $X$ here is some topological space or metric space.

[gentle osprey]

Philip

[uneven bronze]

Also, I have seen some use the little circle on top of a set. Is this the same as the interior?

[novel acorn]

hint: take S = X \S in the upper equation

[rancid umbra]

yes

[heady skiff]

is it true that we have box > prod > l^2 > uniform on H? i was able to prove the inclusion prod > l^2 but I'm too lazy to type it out since the process was quite constructive

never mind my proof was in fact flawed

lol

[heady skiff]

why exactly is $g$ continuous here? i understand that one coordinate function is simply the projection $\pi_1$, however, I don't understand why the second coordinate function $f_2: X \to \mathbb{R}$ given by $(x, n) \mapsto \frac{x}{n}$ is continuous

[gentle osprey]

okeyokay

[sly mural]

this is a really dumb question and I should know the answer...
but if f: X-> Y is a homeo. and for subsets A, B of X if f(A) = f(B) then does that imply A=B?

i feel like the answer is obviously yes but I want to double check

yup it is obvious hahaha i thought i was going crazy

ty

[pseudo beacon]

if you equip the closed unit square in R^2 with the subspace topology, how do you prove that any neigbhorhood of a point on the edges is not homeomorphic to R^2? My understanding is that every point, including points on the edges are interior points under the subspace topology.

[unreal stratus]

Here's one argument. For any nbhd of a point on the edge, you can remove a point (on the edge) and keep it simply connected, which isn't the case for R^2

I'm unsure there is a way that doesn't use simple connectedness

[rancid umbra]

as manifolds

but uh

invariance of the boundary may seem a bit out of scope here

[unreal stratus]

Wdym

[rancid umbra]

homeomorphic top manifolds have homeomorphic bounaries

[unreal stratus]

Ye

[rancid umbra]

and any homeo of manifolds restricts to a homeo of the boundaries

but this seems overkill

this also follows immediately from invariance of domain

[merry geode]

Is there a notion generalizing boundary of open set and boundary of manifold

[brittle rapids]

consider the constant 0 sequence. let ε > 0. then there is a sequence which is nonzero for all its terms, whose summed squares converge to ε. but this sequence isn't contained in any product nbhd of 0

[brittle rapids]

g is the restriction of the map g': [0, 1] x [1, infty) -> [0, 1] x R defined by (x, y) |-> (x, x/y)

[rancid umbra]

what more do you want

[merry geode]

Ah, I mean unifying concept for the two

[unreal stratus]

Maybe I'm missing something, but is there any (co)homological way to check if a non-compact, connected n-manifold is orientable?

perhaps one thing is to use Poincaré and compactly supported cohomology / Borel-Moore homology to provide obstructions as you would in the compact case

[stray stratus]

Coffis Cup and Doughnut Exploration Group

[gritty widget]

Anyone know computational topology/topological deep learning?

[civic verge]

Hi guys could someone help me with this exercise I have done this but I don't know if I am on the right track

[gentle osprey]

Homology

[civic verge]

[lusty trench]

Recall that a bijection is a homeomorphism iff it's both continuous and open. Equivalently, a bijection is a homeomorphism iff both it and its inverse are open.

The forward implication is trivial, so you only need to prove the backward one, i.e., $f(int(M)) = int(f(M))$ implies $f$ is a homeomorphism.

[gentle osprey]

Eduardo León

[lusty trench]

Notice that $f(int(M))= int(f(M))$ already implies that $f$ is open. Because $f$ is a bijection, that's equivalent to $int(M) = f^{-1}(int(f(M)))$. That, in turn, is equivalent to $int(f^{-1}(N)) = f^{-1}(int(N))$, where $N = f(M)$. So $f^{-1}$ also satisfies $f^{-1}(int(N)) = int(f^{-1}(N))$. Therefore, $f^{-1}$ is also open, i.e., $f$ is continuous.

[gentle osprey]

Eduardo León

[worthy olive]

Can someone please explain how you find actual homology and cohomology groups with Mayer vietoris, I can only see it as a method of finding induced mappings

[opaque scroll]

Here's an example:

Let X be the circle, and let A be X without north pole, and B, be X without south pole. Then A=B=open interval, and AnB = disjoint union of two open intervals.

Since the interval is contractible, we know the homology of everything except X.

Then you have a sequence

0 -> H1(X) -> H0(AnB) -> H0(A)(+)H0(B) -> H0(X) -> 0

We have H0(AnB) = Z^2 = H0(A)(+)H0(B). And the map between them can be described by the matrix [1, 1; 1, 1]. So by computing kernel and cokernel we see that
H1(X) = H0(X) = Z

[worthy olive]

Thanks bro 😄

But how do you find the matrix mapping?

[opaque scroll]

You can do it completely explicitly.

H0(AnB) is generated by the a point from each component. And both of those end up in the same component when including into A, same for B.

[unreal stratus]

Often you don't even need to know the actual maps (or you only need a couple) since the form of the sequence tells you enough

For example in this case you could use the reduced Mayer Vietoris sequence and it'd just be 0 -> H_1(X) -> Z= tilde H_0(A cap B) -> 0

[uneven bronze]

I have an elementary question. Does every non-empty open set in R contain an interval (a,b)? I'm reading about the Cantor set being nowhere dense. The proof goes by contradiction; since the Cantor set is closed, we only need to show the interior is empty. Suppose the interior is non-empty. Then it "...contains some interval...". This is the part I'm doubting.

[rancid umbra]

yes, every non-empty open subset of R contains an interval
due to whatever definition of the usual topology on R you are using

[uneven bronze]

ah ok, thanks!

[median sand]

If X=A_1\cup\dots\cup A_n with A_j connected, can we conclude X has only finitely many components? This should be possible, but I only worked out the case where the A_j are closed.

[rancid umbra]

the number of connected components can only decrease when two of the sets are not separated
so there are at most n

[merry geode]

What is cohomology with group action on a space?

I wanna know how it is constructed and how you take the representatives

[median sand]

Nvm, I was doing something stupid.

[merry geode]

I mean equivariant cohomology here.

[naive trench]

Spooky stuff

I recommend you to learn category theory and sheaf theory

[merry geode]

Ah, I am quite familiar with both

[naive trench]

I dont know so much about cohomology but is what our professor told us for what is used all this abstract nonsense

[zinc hearth]

you start with your group G acting on a space X, then take cohomology of the space X x EG/(x, y) ~ (xg, g^-1y) with the natural free action of G on EG

[merry geode]

Ah, so that is the part X x_G EG

[zinc hearth]

yeah, it's a pullback of something you don't need to worry about

[merry geode]

Why does g act in different way?

The positioning reminds me of tensor product but

Ah wait nvm

[zinc hearth]

you're taking the orbit space of the (twisted) diagonal action of G on X x EG

I don't remember why preciesly I just know that it's better behaved

it's a composition of bisets. Definitely doesn't help understand it but I'm biset-pilled so I need to mention it at every possible opportunity

[gentle osprey]

Absta

[merry geode]

Now one big problem of mine is that I still don't know classifying spaces..

[zinc hearth]

oh classifying spaces are FANTASTIC

although maybe I'm not the best to exposit them... does "geometric realisation" mean anything to you?

[merry geode]

Yeah I did not learn what geometric realisation is

I can only guess that it is geometrically realizing some kind of e.g. moduli space

[zinc hearth]

it's geometrically realising a category 😹

idk what a moduli space is

[merry geode]

Ah, geometrized category sounds cool

Got some cobordism feel

[zinc hearth]

idk what that is either

the gist is, you generate some abstract simplical complex based on how the morphisms in the category compose, and then turn those into actual topological simplices

this is how you do homology/cohomology on categories and stuff like that

[merry geode]

Yeah, IIRC cobordism is similar idea, but it goes backwards to make category out of topological spaces

[zinc hearth]

for groups, you probably already know that H*(G;A) := H*(BG;A). BG is this geometric realisation of the single object groupoid corrisponding to G (i.e., a single object category with Aut(*) = G)

[merry geode]

(n+1)-dimensional manifold M between n-dimensional boundary manifolds N, N' becomes morphism M: N -> N', and then you get infinite nested(?) category etc.
But anyway

[merry geode]

Oh, neat

Actually I thought that was definition of group cohomology

[zinc hearth]

it is, hence ":=" not "="

[merry geode]

Ahh, yeah.

[zinc hearth]

I kinda went backwards starting with the more general object, but w/e. That's how I would construct the classifying space

[merry geode]

So what is EG?

Heard like EG -> BG is a G-bundle but..

[zinc hearth]

EG is the same thing but just with a different category

[opaque scroll]

Not really though, right.

Like the right hand side only make sense if A is a trivial G-module

[zinc hearth]

true, I should've put Z

[opaque scroll]

So really the definition of group cohomology should be
Ext_ZG(Z, A)
Which happens to equal H(BG; A) when A is a trivial module

[merry geode]

Hmmm

[merry geode]

ZG being the group ring of G over Z?

[opaque scroll]

Yes

[merry geode]

Thank you, that clears up a lot for me!

[zinc hearth]

I'm going to tentatively say it's the category with one object for each element of G, and one morphism between each pair

[merry geode]

Anyway, so what category does EG correspond to?

[zinc hearth]

iirc you think of this morphism g -> h "as" hg^-1

[merry geode]

Ahh

I guess with this category (geometrized), EG -> BG becomes G-bundle

[zinc hearth]

yur

easy to see too, you take the functor collapsing EG (the category) down to one object

iirc the geometric realsation of that functor is exactly the map in the G-bundle? but if I'm wrong I'm sure jagr will very quickly correct me

[unreal stratus]

yes this is one useful model

[unreal stratus]

and yes

[unreal stratus]

ye

a nice way to see the stuff is like

if you take a space EG (so contractible with a free action of G) then the singular cochains form an acyclic resolution of Z

[unreal stratus]

actually, are (-)^G-acyclic objects precisely those complexes with free G-actions?

feels like that should be the case from vibes

[languid patrol]

no

[unreal stratus]

Rip

Lol

[languid patrol]

0 \to Q \to Q \to 0

with trivial G action

[unreal stratus]

yeah sure lol

though is one direction true

I'll think

[languid patrol]

I think it's like

[unreal stratus]

oh wait yeah duh

i'm stupid

lol

you just get 0 after applying (-)^G i guess

Though I am now curious as to whether there is a nice criterion

[languid patrol]

i dunno, maybe for finitely generated modules one can say something

is G finite?

[unreal stratus]

i'm happy to assume that

[languid patrol]

if G is finite and R has |G| invertible then every finitely generated module is acyclic

[unreal stratus]

Yes

[languid patrol]

so I don't know if there is going to be a nice criterion

[unreal stratus]

Hm

Yeah good point

Thanks

I think I have mostly just worked with free resolutions for group cohomology lol

[languid patrol]

there is some interesting thing

[unreal stratus]

Well not done enough

[languid patrol]

like one can take $C^*R-Mod \to C^*R[G]-mod$ by base change

[gentle osprey]

Math_Discord_Final_Girl

[languid patrol]

and the reverse functor by fixed points

and this gives some kind of retraction

[unreal stratus]

I haven't thought about that construction interesting

Rn I am reviewing some hom alg for exams

Use oo-categories