point-set-topology

[candid hedge]

is it open?

And how is it closed by default i dont see that

[wispy veldt]

It certainly is open , you just have to prove it.

[candid hedge]

for any point in R, we can draw a ball contained in R that doesnt touch N

we're done

AH i see why

from the seq definition of closed okay

those sets are disjoint yes?

[wispy veldt]

Yeah , but that doesnt help you prove its closed

[candid hedge]

no it doesnt

i am just making sure

[wispy veldt]

In R\N but yes , any point you take in R\N you can find a open ball contained in R\N

[candid hedge]

yeah

its easy to write

[wispy veldt]

Or alternatively R\N is gonna be the union of open sets , namely (-infinity,0)U(0,1)U(1,2) etc (you can write it more neatly ofcourse) but same idea

[candid hedge]

yeah

[pseudo coral]

Question for a continuous 1-1 function from Hausdorff space into a non Hausdorff space can I consider the identity map from R with the standard topology to R with the trivial topology ?

[hidden crag]

yes

[pseudo coral]

The question asks if f: X -> Y is 1-1 and continuous and X is hausdorff is Y necessarily Hausdorff and I wanna say no

Cool thanks 🙏

And if Y is Hausdorff then if f is continuous and 1-1 it makes X hausdorff too, right? I feel I can prove this fact

[hidden crag]

yes

[pseudo coral]

Cool thanks 🙏

[hidden crag]

i'm pretty sure you only need injectivity for that

[pseudo coral]

Well you need that preimages of open sets need be open

[hidden crag]

that's continuity

[pseudo coral]

Yes

[hidden crag]

okay fair enough, injectivity and continuity

[pseudo coral]

Cool thanks 🙏

[candid hedge]

if A is a set in a normed vector space E , do we have that for x in E, D= { lx-yl, y in A} is an interval in R?

or at least a union of intervals

can i use the properties of sup and inf?

ofc not right?

and is this proof false

or is inf automatically in the closure of the set D

[unreal stratus]

Well the closure of A is the set of elements a distance 0 from A

[unreal stratus]

Hm no because you could just pick A arbitrarily to stop this, in say R

Like take R to be ur norned space, A any subset of (positive) reals and x = 0

Then D is just A

Which can be basically as interesting as you want

[candid hedge]

yeah

[candid hedge]

so is this false then?

the thing about finding a seq

[candid hedge]

Can i find a seq in D

that goes to d(x,A)

[unreal stratus]

Oh I mean you can find a sequence in there sure

[candid hedge]

why

[unreal stratus]

I just don't see how it links to being a union of intervals

[candid hedge]

what property

[unreal stratus]

Like just how infima work

Approximation property

[candid hedge]

we can always find a sequence going to infima

for any set of numbers

right?

Its just weird i dont remember this result passing anywhere

[unreal stratus]

Yeah it's approx property so like

If A is a set of eals with infimum a then for every d > 0 you can find some b in A with a <= b <= a+epsilon

[candid hedge]

yeah this ik

[unreal stratus]

Well

Or maybe signs weong

[candid hedge]

and for other sets its just the constant seq

[unreal stratus]

Well just pick one for each d=1/n

Ye

[candid hedge]

if we take the rationals bigger than square root of 2

[unreal stratus]

Yes

[candid hedge]

Ah ye nvm

ok i see then

[bitter smelt]

Does the one parameter family in the definition of the isotopy heegaard move have any restriction on the family? Like uh, it being an isotopy

[bitter smelt]

What is c_1(M) where M is a smooth manifold? Is this some spin-C thing?

Ex:

The abov doesn’t really matter for my Q but

There it is for full context

[rugged swan]

isn't it the 1st chern character ?

[unreal stratus]

ye that's what i'd assume

[bitter smelt]

Yeah that makes sense

[unreal stratus]

though if these are surfaces then c1 is just euler class right

equivalently

[rugged swan]

c1 is e2 yeah

imo

[unreal stratus]

Eh doesn't matter here cause we're already in the complex realm

[pseudo coral]

question

who here is familiar with nets

I have a solution verification for f is continuous iff convergent nets get mapped to convergent nets

[gentle osprey]

MyMathYourMath

MyMathYourMath

MyMathYourMath

MyMathYourMath

[pseudo coral]

is this ok for the forward direction.

[gentle osprey]

MyMathYourMath

MyMathYourMath

MyMathYourMath

[pseudo coral]

is all of this OK

[pseudo coral]

this is the statement

I have provided proofs for both directions

[little hemlock]

at best you've proved the same direction twice. you proved f => for all nets x_lambda -> x, f(x_\lambda) -> f(x) and the contrapositive of this statement

[pseudo coral]

for the reverse direction I proved the contrapositive supposing that convergent nets dont get sent to convergent netd and fins that f fails continuity

for the other direction I suppose f is continuous

i suppose theyre the same since theyre contrapositives but in an iff that would suffice, no>

?

[pseudo coral]

but I proved them correctly?

[pseudo coral]

i.e., proving two different versions of an iff would suffice

since its P iff Q you can prove

if P then Q

which i did

and for the if Q then P direction

youre supposing not P then showing not Q

this is a question of logic lol

[little hemlock]

i take issue with the reasoning here, but your direct proof of f cont. => for all nets x_lambda -> x, f(x_\lambda) -> f(x) is fine.

i.e., proving two different versions of an iff would suffice
you proved P => Q correctly. Then you proved not Q => not P (i think incorrectly)
but P => Q is logically equivalent to not Q => not P

You need to show Q => P or something equivalent like not P => not Q

[pseudo coral]

but thats logically equivalent to Q => P

[little hemlock]

not P => not Q is NOT equivalent to P => Q

[pseudo coral]

thats what I showed

no

Q then P

is what its equivalent to

which was the second one I was showing

[little hemlock]

for the second one, you attempted to show that
"if there exists a net $x_\lambda \to x$ such that $f(x_\lambda) \cancel\to f(x)$, then $f$ is not continuous"

right?

[pseudo coral]

no such that f(x_\lambda) DOESNT converge to f(x)

[gentle osprey]

kxrider

[little hemlock]

yea my bad, fixed

this is what you showed, right?

[pseudo coral]

yes

oh i see what youre getting at, i think lol

[little hemlock]

Let P be the statement that f is continuous and Q be the statement that f sends convergent nets to convergent nets.
Then the second statement you "proved" is not Q implies not P

this is equivalent to P => Q

which you already proved earlier

[pseudo coral]

youre right

I need to prove not P then not Q

[little hemlock]

yep

[pseudo coral]

or directly prove it

which would you recommend

[little hemlock]

you could try to prove Q => P, but I would recommend not P => not Q.
not P => not Q would involve constructing a convergent net which f sends to a non-converging net

[pseudo coral]

And not P is f not being continuous

[little hemlock]

ye

[pseudo coral]

so there exists and open set whose inverse is not open

[little hemlock]

right

[pseudo coral]

so how can I assure this net is in this one exsiting open set whose preimage isnt open

i think for this a straight forward prove would be easier

[gentle osprey]

MyMathYourMath

MyMathYourMath

[pseudo coral]

wait thats not true so far, is it

[little hemlock]

yea this follows from the definition of convergence

[pseudo coral]

ok so then

[gentle osprey]

MyMathYourMath

MyMathYourMath

MyMathYourMath

[pseudo coral]

does all of this work

[little hemlock]

x_lambda in V for all lambda >= lambda0 does not make V open

[pseudo coral]

it makes it a neighborhood of x

by definition of nets

if x_\lambda \in V and x \in V

[little hemlock]

wait, ur just assuming V is open

[little hemlock]

here

[pseudo coral]

then V is a neihghborhood of x

sorry

[little hemlock]

you need to show that f^-1(V) is open

[pseudo coral]

i meanti

f^-1(V)

[pseudo coral]

this should be

[gentle osprey]

MyMathYourMath

MyMathYourMath

[pseudo coral]

it was a typoclearly , how could x be in V if V is open set of f(x) lol

[gentle osprey]

MyMathYourMath

[pseudo coral]

@little hemlock

[little hemlock]

i don't see why f^-1(V) is open

[pseudo coral]

cause f(x_\lambda) \in V

for all \lambda \geq \lmabda_0

oh but thats given x_\lambda -> x

right?

[little hemlock]

sure, so x_lambda is in f^-1(V) for all lambda >= lambda0, but that doesn't make f^-1(V) open

[pseudo coral]

that makes it opeen!

by definition of nets

[gentle osprey]

MyMathYourMath

[pseudo coral]

f^{-1}(V) is an open set containing x since the x_\lambda are in it

for all \lambda >= \lambda_0

[little hemlock]

that's not how being open works. just because infinitely many points of, say, some convergent sequence lie in a set doesn't mean that set is open

[pseudo coral]

i thought that was how nets worked, if infinitly make values converge to a given point in a ball then that ball is an open ball about the given point

[little hemlock]

convergence of nets doesn't provide any sufficient conditions to prove that a set is open. It just says that if some net is eventually contained within every open set containing a point x, then that net converges to x

[pseudo coral]

youre right thats trhe definition

[little hemlock]

there is some intuition for why proving Q => P may be difficult.

[pseudo coral]

so I suppose f is discontinuous

so there exists some V in Y such that f^{-1}(V) is not open while V is open

so I need to construct a net that converges in X but that doesnt converge in Y?

[little hemlock]

yeah

[pseudo coral]

at first I dont see it cause in X the set isnt open but the net must converge somehow

but in Y the set is open

and the net must not converge

any hints on where to begin for this direection?

[little hemlock]

yea i think your net should be something indexed by something like set of open neighborhoods of a point x partial ordered by inclusion

also, the proof is sort of kind of analogous to the case of proving that a function between metric spaces is continuous iff f maps convergent sequences to convergent sequences.
it may be a good idea to study that as well

sequences work for metric spaces because each point has a countable neighborhood basis. In a general topological space, you might not have this, and this is why we need nets

[pseudo coral]

indexed bu directed set

[little hemlock]

yea, u should convince urself that this indexing set is directed

[pseudo coral]

let me think about this for a bit then

[little hemlock]

also probably should've said partial ordered by "reverse inclusion" since lambda > \lambda0 should imply that x_lambda lies in a smaller open nbhd of x than x_lambda0

[pseudo coral]

true

[pseudo coral]

can I say that since V is opeen in Y such that f^{-1}(V) is not open in X

then I can consturct a net in which

x_\lambda \not\in f^{-1}(V)

then f(x_\lambda) \not \in V

[little hemlock]

Yeah that’s the idea

[pseudo coral]

but where do I go from heree

[little hemlock]

so

since f^-1(V) is not open, there is a point x in f^-1(V) such that no open neighborhood of x is contained in f^-1(V)

(if this were not the case, you could write f^-1(V) as a union of open sets)

[pseudo coral]

so can we find an open neighborhood of x such that the x_\lambda converge to it there

[little hemlock]

you want a net converging to x such that no matter how "far out you go" there are points that lie outside of f^-1(V), and therefore points of {f(x_lambda)} that lie outside a nbhd of of f(x)

let Lambda = {open nbhds of x ordered by reverse inclusion}. This should be the index set u use to bulid your net

[pseudo coral]

so if $V \geq U$ then $x_V \subset U$

[gentle osprey]

MyMathYourMath

[pseudo coral]

something like that

[little hemlock]

Yes

This makes convergence of the net to x trivial

[pseudo coral]

i see now

so then f(x_\lambda) will be in f(U)

or f(x_V)

[little hemlock]

Right

To be clear though you want x_V to be in V as well to ensure convergence

[pseudo coral]

yes

[little hemlock]

is there a version of the fundamental groupoid for higher homotopy groups?

[pseudo coral]

hi

question for solution verification

[gentle osprey]

MyMathYourMath

MyMathYourMath

MyMathYourMath

MyMathYourMath

[gritty widget]

Yes.

Your proof is correct.

[gentle osprey]

MyMathYourMath

[pseudo coral]

ok sweet thanks!

[gritty widget]

Do you really need to make a new TeXit message for each logical step?

[pseudo coral]

lol sorry

[pseudo coral]

a new question

if convergent neets get mapped to convergent nets then f is continuous

solution verification

suppose convergent nets get mapped to convergent nets but f is discontinuous, then there exists some V in Y open such that f^-1(V) is not open in X (given f: X -> Y)

then one can construct a sequence where the directed set is neighborhoods of the element in the net ordered by reverse inclusion

[gentle osprey]

MyMathYourMath

MyMathYourMath

MyMathYourMath

MyMathYourMath

[pseudo coral]

it then follows that $(y_U)$ converges to $y$ in $X$

[gentle osprey]

MyMathYourMath

MyMathYourMath

[pseudo coral]

so T >= W implies T is contained in W

is that the trick? To let my indexing set be the set of neighborhoods of the nets point ordered by reverse inclusion?

the forward direction is so much easier!

[gritty widget]

Not really. It's just a bit harder to read than one continuous paragraph, personally.

It also makes it harder to tell when someone is done typing.

I'll try to phrase it less aggressively next time.

didn't seem aggressive

I didn't think it was either.

[pseudo coral]

Anyone can verify if I’m on right track for that proof

[cedar pebble]

there is a fundamental infinity groupoid which encodes the homotopy type (in particular all the homotopy groups) of a space

[grave egret]

Note sure where to post this, but does soemeone knows whether the ring of adeles is Polish?

[distant lichen]

I think it's Slovenian

[tidal cedar]

As in a Polish space?

[distant lichen]

though this is less of a variant and more of an extension

[gritty widget]

It's a joke I think

[gritty widget]

So ring of adeles is locally compact from what I read

And any locally compact separable metric space is Polish

So you're asking if ring of adeles is a separable metric space

The definition of restricted products is confusing me a little so I'll let you answer this yourself

So this is checking if finite adeles of a global field K are a separable metric space since A_K you multiply with finite amount of R or C

I'd try using Urysohn metrization theorem on it

Unless there already is some obvious metric on it

[rugged swan]

What do you mean by infinity groupoid of an higher group ?
In general, for a (oo,1)-topos T, you have a geometric morphism oo-Grpd->T, so a functor T->oo-Grpd which is the realization functor

[grave egret]

I figured it out, the product of all Z_p is a Polish open subgroup of countable index, from which it follows that the whole thing is Polish.

[gritty widget]

oh, so you are taking K = Q?

[grave egret]

Well in general the same should work with the product of O_K's, right?

[gritty widget]

I checked what the ring of adeles is just today. I have no idea

I don't study AG or whatever

So if you say it does, then you should know better

[grave egret]

Yep it still works for any K

[little hemlock]

Are you asking me?

[rugged swan]

yes

[little hemlock]

Idk anything about infinity groupoids. I was just wondering if there is a groupoid like the fundamental groupoid for pi_n, n>1

Or something similar in spirit to the fundamental groupoid

[rugged swan]

every space is an ∞-groupoid in homotopy theory

[empty grove]

Ye, the infinity groupoid that Zak mentions contains all the information of all the homotopy groups of the space. It is the infinity category in which the objects are points of the space, morphisms are paths, 2-cells (morphisms between morphisms) are path homotopies, 3-cells are homotopies between homotopies and so on. You can recover the nth homotopy group of the space with basepoint x as the automorphism group of the identity (n-1) cell on the identity (n-2) cell on the identity ... the identity 1 cell on x.

I don't think there's an ordinary groupoid thing that works well

[empty grove]

You can truncate (might not be the right term technically) this groupoid to an n-groupoid without losing information about the first n homotopy groups

[rugged swan]

yeah now the question is, if G is an ∞-group, how can you relate the structure of the ∞-groupoid |G| of G and the structure of G ?

my guess is that it should be somehow the same as the structure of G ?

[cedar pebble]

yes, this is basically what the fundamental n-groupoid is telling you (this is just the n-truncation of the fundamental infinity groupoid)

so the objects are points, morphisms are paths, 2-morphisms are homotopies between them, and so on; n-morphisms are n-homotopies between n-1 homotopies

[little hemlock]

huh okay interesting

[urban zinc]

why is a separable metric space called separable?

what does a countable dense subset have to do with separation

[urban zinc]

hm interesting

ty!

[gritty widget]

Hello! I'm new to topology, as I just started reading about it in my free time. I do, however, have two uncertainties regarding accumulation points,

The Wikipedia definition of an accumulation point states that x is an accumulation point of S if every neighborhood of x has at least one element of S (other than x itself, if it is the case).

It has, however, left me wondering whether or not it is unique for a given set or not. To me, it implies that every element of a set is itself an accumulation point of that set.

It also implies that if S is a subset of X, and x is an accumulation point of S, x must be the only element that is in X and not in S.

Both of these two implications of the accumulation point seem a bit odd to me, and I wanted to make sure if they are correct or not,

[gritty widget]

There are none, right ?

[unreal stratus]

No

[gritty widget]

Or, by neighborhood, you mean "centered neighborhood" (as I've seen a grade 11 book call it), the one where there are as many elements on the left as on the right (if we represent them as numbers on a line) ?

In that case, the accumulation points are 0 and 1.

Hmm, then what would an accumulation point of (0, 1) be ?

Because for every number that is in R but not in (0, 1), I can pick an interval that contains that number, and has no common elements with (0, 1).

[unreal stratus]

Perhaps the issue is yes you can pick such an interval, but not an open one containing 0

[gritty widget]

Yes - [-1, 0], for example.

[unreal stratus]

Aha

[gritty widget]

Oh

This changes everything.

So, then, the set of accumulation points of (0, 1) is [0, 1], right ?

I see.

My second concern about accumulation points is now gone too, knowing that the sets/intervals must be open.

Well, thanks everyone!

[devout sorrel]

if we have a metric space, and two sequences (x_n), (y_n) such that d(x_n, y_n) -> 0, and we have a subsequence (x_n_k) -> x, can anything be said about the existence of a converging subsequence of (y_n)?

[unreal stratus]

Indeed, I claim that $(y_{n_k})$ converges to $x$. Have a go!

[gentle osprey]

potato

[devout sorrel]

yeah thats what i guessed... hmm let me try proving that

yeah that was pretty straightforward through the triangle inequality. thanks

[unreal stratus]

np

[hazy lotus]

Hey all! I'm having trouble with the following homework problem:

What I've done:

I've shown that a and b are equivalent already (a->b is essentially taking {G \cap H: H \in \mathscr F} and it's an open filter finer than \mathscr F, so because F is an open ultrafilter, the two filters are actually equal and since G is contained in one, G must also be contained in the other. For b->a you suppose there is a strictly finer filter than F and you take G in the strictly finer filter such that G is not in the original filter F. The condition for b is satisfied because G is in the strictly finer filter and H \in \mathscr F is also in the strictly finer filter so they must intersect. This forces G in \mathscr F when we assumed G \not \in \mathscr F). I also managed to show that c->a by a similar argument to b->a. let \mathscr F \subsetneq \mathscr G and G \in \mathscr G - \mathscr F. Then X - \cl G \in F so X - \cl G \in \mathscr G. But then both G and X - \cl G are in the common filter \mathscr G, which is impossible since their intersection is the empty set.

What I've tried for a->c:

I thought this could be proven by using the fact that an open ultrafilter is a prime ultrafilter. If G \cup (X - \cl G) is in \mathscr F, then we could conclude that either G G \in \mathscr F or (X - \cl G) \in \mathscr F. However, I couldn't prove this and I ended up getting quite stuck.

Any help would be appreciated!

[hazy lotus]

whoops never mind, I got it - you do b->c by contrapositive of b. Then you can get that X- \cl G = \int(X - G)\in F because it contains the open set H \in \mathscr F found from b.

[devout sorrel]

A is dense in B and B is dense in C, how do you show that A is dense in C?

we have that cl_B(A) = B, cl_C(B) = C, but how do we get ot cl_C(A) = C?

[gritty widget]

cl_B(A) = cl_C(A) \cap B = B so B is a subset of cl_C(A)

Now take cl_C again

[acoustic dome]

Anyone here familiar with Bott & Tu? I am unable to show the map h defined on p136 sends \delta (H_d) to a D-something. I know those phi_i would be unique up to d-something but I don't know how to prove they form a D-something. Thank you!

[formal tide]

is this true? I haven't seen this version anywhere, any references?

for reference, this is the one I'm familiar with, an I can see it's a particular case of the former

[rugged swan]

yes it is

A should be path connected tho

I think

[formal tide]

Something about HoTT, not something that can be translated to point set topology as far as I can tell

I'm outside but it's from the HoTT book

[formal tide]

Can you elaborate on why? Can it be deduced from the other van Kampen I posted?

[rugged swan]

oh its in hott

then B⊔C is the homotopy pushout

in this case it is like

the pushout

C(C,A)←A×[0,1]→C(B,A)

you can take U = C(C,A) and B=C(B,A)

and it is the van kampen theorem

[plain raven]

maybe this is like, the homotopy pushout

oh zak just said this

[formal tide]

I see, I don't know them, I'll look them up later, ty

[rugged swan]

yeah the inclusion is in general cofibrant so that U ∪ V is the homotopy pushout of U←U ∩ V→V

[rugged swan]

btw C(X,A) for a morphism A→X is called the cone of this morphism

[formal tide]

ok I'm back, I did search it up but I didn't really find any good material on this so idk what you mean by this

i.e. what is C(C,A)?

[rugged swan]

the pushout of A×[0,1]←A→C

you won't have good material about it imo, the correct way to define homotopy pushout of topological space isn't really mine

it uses cofibrant replacement

it's a very confusing way to do it

tho you're in the HoTT setting, you should be able to show it only using HoTT

[formal tide]

I see, thanks

[opaque cloud]

consider the topological space [0, 1] with the standard euclidean topology

would {[0, 1/2], [1/2, 1]} be a cover of that space?

can the space itself be a cover of itself? since the every set is it's own subset?

[gritty widget]

Yes

Cover is anything that sums to your space

Doesn't depend on topology

You could say, it's not topology. But it is

[opaque cloud]

so then the space itself would not count as a cover?

[opaque cloud]

and refineing that cover is exactly what it sounds like? it's just a more finer collection of subsets as in there's more subsets now

[gritty widget]

The family containing it would

[opaque cloud]

like {X} would be a cover of X?

[gritty widget]

Refining a cover means making it smaller in some sense

It's not a partial order though

[gritty widget]

Precisely

[opaque cloud]

it's similar to making a coarse topology into a finer topology?

[gritty widget]

In topology we are often interested in various ways of refining a cover

[gritty widget]

Not quite

Taking subcover would be

Refining is something less rigid

More flexible

Because comparison of topologies is just comparison of (families of) sets

But for refinements you can have different sets in your new cover

[opaque cloud]

so the blue cover here would be a refinement of the red cover?

[gritty widget]

I could go on and on about different ways refinements are used in topology

Pretty wide topic

[opaque cloud]

very cool

I didnt expect it to be

[gritty widget]

The blue doesn't look like a cover but yeah

[opaque cloud]

lol let's just assume the union of all the blues would give you the original space

[broken nacelle]

Wait, you can type here without the advanced role?

[opaque cloud]

I guess that's what the bar role does

[broken nacelle]

Also, sup neam

[broken nacelle]

I thought it did the opposite

[opaque cloud]

nothing much, should be studying for my exams but instead im wondering stuff about point set topology

[broken nacelle]

[opaque cloud]

i see

[broken nacelle]

What are you using?

please say hatcher's notes

[opaque cloud]

my brain

Lol i'm not studying top, im just wondering about it

like im not doing any proofs rn

[broken nacelle]

Ah

[opaque cloud]

which brings me to what I was gonna ask blitz

Blitz you seem like you're pretty experienced with proofs in topology, what advice/tips do you have for a beginner doing topology proofs for the first time?

[opaque cloud]

but I will though! once I finish analysis and linear algebra

[gritty widget]

None

[broken nacelle]

Use a book

¯_(ツ)_/¯

Wtf

[opaque cloud]

Lol I meant when I'm using a book, what advice do they have

[broken nacelle]

I didn't send 4 shrugs

[gritty widget]

Oh wait yeah

[opaque cloud]

darQ spamming, banned

[gritty widget]

Think geometrically

[opaque cloud]

Like visualize stuff?

[gritty widget]

Like in R^n. Doesn't matter it's abstract

[opaque cloud]

mfw I need to think geometrically without a metric on my space (jk)

[broken nacelle]

You WILL learn at least a bit of pointset if you're doing analysis

[opaque cloud]

hm yeah okay

that's why im doing anal first 😌

[opaque cloud]

by "think geometrically, like in R^n. doesnt matter it's abstract" do you mean like this? Like when I need to understand a concept I just pick out an example in R^n?

[gritty widget]

Yeah. Usually helps a lot

[opaque cloud]

alright got it

[gritty widget]

Think about examples

Try to provide them

[opaque cloud]

also are topology proofs relatively easier compared to analysis and algebra proofs?

[gritty widget]

Not always

I mean it depends on the proof. Basic stuff is mostly straightforward

Proof of Urysohn lemma is definitely interesting and I wouldn't come up with the proof on the fly for example

Basic algebra is just generally easy

Analysis is a bit convoluted so, it's probably harder than basic topology

[opaque cloud]

welp, I better get to grinding all three then

thanks

[hidden crag]

I love that proof

[broken nacelle]

Based take

[obtuse meteor]

van Kampen approaches

[hidden crag]

[neat current]

how does this look
(first line should say closed in YxY, not closed in Y)

[vague brook]

how do you get phi((XxX)/R) = YxY/(phi(R)) @neat current

[neat current]

actually yeah that is incorrect reasoning

though you can argue by set inclusion that phi((XxX)/R) = YxY/ΔY

[ocean narwhal]

How easy something is probably depends on the person

[gritty widget]

I don't think I'm deviation from the rule in this case.

But people still might have different experiences ig

[pseudo coral]

Question 🙋‍♂️

[gritty widget]

The answer is two.

[pseudo coral]

Lol

I wish

Let X be locally compact space and let Y be its one point compactification . Characterize all continuous function. From X into R such that f has an extension to Y

So if we think of R as the space X (since R is locally compact) and it’s one point compactification be R U {\infty} can I safely assume@this is the one point compactification of R?

But how do you characterize such continuous functions f: X -> R which extend to Y

[gritty widget]

for R it should be functions vanishing at infinity

[pseudo coral]

Ok I had that feeling!!!

Cause you tackle on \infty

[gritty widget]

or well, "constant" at infinity I suppose

[pseudo coral]

So 1/z’s

Or 1/g(z) where g(z) blows up at i infinity

So 1/e^\infty works too ?

So for general X you want it to vanish at the point you tackle on?

[unreal stratus]

well you want it to approach some limit at that point but you can't talk about the point you tack on until you've tacked it on

[pseudo coral]

so f(x)=1/g(x) where g(x) \neq 0 and its limit is infinity as x appraoches inifity

for all x \in Y

[unreal stratus]

that isn't enough generality like

yours doesn't really allow for f vanishing anywhere

[gritty widget]

There must exist a point p such that if (a, b) contains p, then there always exists a compact set K such that f sends complement of K to (a, b)

pretty much by definition

and I'd say, it's good enough

[pseudo coral]

are you assuming here f:X -> R

[gritty widget]

well, of course

So this will be a sum of a constant function and a function "vanishing at infinity"

[pseudo coral]

so

f(x)=a+1/g(x) where a \in X and g(x) is nonzero and goes to infnity as x goes to infinity

a fixed

[gritty widget]

well, no

not all functions vanishing at infinity are inverses of some function

of course

[unreal stratus]

yeah idk why putting them in a specific form like this would be any more convenient than saying it approaches a limit at infinity

[gritty widget]

well, you could say it "approaches a limit at infinity" if you know what that means

[gritty widget]

so I'd still stick to this explanation

[unreal stratus]

oh i mean for R

[gritty widget]

since outside of some examples like separable metric spaces, we don't really know what that means unless we are advanced enough in topology

[unreal stratus]

but what the other person was saying was for the R -> R case ig

[gritty widget]

ig

[wise ruin]

Today my professor mentioned, as a theorem, that homology theories are isomorphic if there exists an isomorphism between all of the homology groups of a one-point space. I then asked if the theorem had a name or any keywords I could use to find it, and he just said to look up "Homology axiom".
Of course, he's talking about the ES axioms but I cannot find this particular theorem anywhere. Does it have a name or is it just a special case of a more general theorem?

[unreal stratus]

i believe it is proven in the paper by Eilenberg and Steenrod (and I'm pretty sure this only holds for the category of CW pairs), like i just have heard the theorem coming along with the axioms since they state the axioms and then prove uniqueness immediately after

[little hemlock]

this is also proposition 3.17 in hatcher

re: justAgift's question

[little hemlock]

anyone familiar with Mayer-Vietoris in operator/topological K-theory? I'm stuck trying to understand how the maps work when trying to compute simple examples like K*(S^1) for example

[little hemlock]

also, another really dumb K theory question: we have that $K_1(\bC) = 0$ in operator $K$-theory, so what is wrong with the computation
$K_1(\bC) = K_0(S\bC) = K_0(C_0(0,1)) = K^0((0,1)) = K^0({*}) = \bZ$?

[gentle osprey]

kxrider

[little hemlock]

Here I'm using that K_n(A) = K_0(S^nA), K^n(X) = K_n(C_0(X)), and contractibility of (0,1)

[silk ember]

Bit confused about a question. Trying to proof any two continues maps from [X, I] (X a topological space and I = [0, 1]) are homotopic. It seems like I can just take f, g two continues maps from X to I and define h(x, t) = (1-t)f(x)+tg(x) and since multiples and sums of continues maps are continues and h(x, 0) = f(x) and h(x, 1) = g(x) I have proven that that it is indeed a homotopy? I assume I am going wrong with continuity here so if someone could lead me in the right direction that would be awesome

[little hemlock]

What is [X, I]

[unreal stratus]

i assume you mean cts maps X -> I but yes this seems fine ^

A slightly more boring way to do this is to fix one cts map X -> I and show all of them are homotopic to that map. For example, every map is homotopic to the map sending everything to 0 (why?)

[silk ember]

Yes sorry, that is exactly it

And thanks!

[gritty widget]

It's because I is contractible

[gritty widget]

I saw [X, I] to mean homotopy classes from X to I before, so I guess it's a misunderstanding and they mean [X, I] consists of at most one element

Or it could mean continuous functions but that's usually denoted by C(X, I) or I^X

[unreal stratus]

Indeed

[candid hedge]

question... if there exists an r such that for all n, if lx-yl<r, lfn(x)-fn(y)l<e then f(x)-f(y)<e? with f the limit of fn

it makes sense right?

So if i am on a compact, can i say if the above property is verified, and i have simple convergence... then i have uniform convergence?

[unreal stratus]

Well <= in the conclusion

But yes

[candid hedge]

and the second property is verified? about uniform convergence of fn?

[unreal stratus]

Uh I'm a bit unsure what your hypotheses are exactly

does r depend on n, and do you mean uniform continuity or uniform convergence

[gritty widget]

No

[candid hedge]

why?

can you give me a counter example?

i am pretty convinced its true

there exists 1 r for alll fn

not just each fn is uniformly continuous

[gritty widget]

f_n(x) = x^n

[candid hedge]

this doesnt verify the property i specified

[unreal stratus]

(usually one would say satisfy not verify here ig)

[candid hedge]

each fn is uniformly continuous, not for the same r tho

sry

[gritty widget]

Oh wait I misunderstood you

https://math.stackexchange.com/questions/649588/equicontinuity-on-a-compact-metric-space-turns-pointwise-to-uniform-convergence

That's what you mean right?

[candid hedge]

wait

can you screenshot it?

[gritty widget]

Why

What you were asking is about equicontinuity, correct?

[candid hedge]

yeah ig

i didnt know that was its name

yeah yeah

So its true?

[gritty widget]

Its true

[bitter smelt]

What is a "positive generator" for homology?

If the group is Z, it's gonna be 1, but im not sure what this means in general

(Or rather, whatever homology class is identified with 1 under the isomorphism)

Does it not have meaning otherwise? Say if the group is Z + Z

Ok I guess it doesn't need to be what is identified with 1 if we let Z=<2,3> for example. Which I guess is suggestive of why we say a positive generator. Still my question remains for non Z groups

[bitter smelt]

And what does it mean for vector fields to be homotopic??

[winged badger]

a vector field is a section of the tangent bundle

two sections are homotopic if they are homotopic as continuous maps, and the homotopy stays in the set of sections

[bitter smelt]

Thank you, I could not find this anywhere

[vast estuary]

I understand the definition, but where do left and right uniformities come from?

I'm not able to prove uniform continuity

[gritty widget]

i don't know much about uniform spaces, but i think what's happening is that you're using the left/right multiplication maps to give G a uniform structure which makes all respective maps uniformly continuous

you need at least a uniform structure to even talk about uniform continuity

i found https://ncatlab.org/nlab/show/topological+group#uniform_structure

[vast estuary]

Thanks!

[gentle osprey]

Hausdorff

[gritty widget]

(g, h) to gh^{-1} is continuous as a map from G x G to G, so its restriction to a map H x H -> H is continuous

that map is continuous in the first place because it's the composition of (g, h) -> (g, h^{-1}) and then the group operation

[vast estuary]

ah yeah that makes sense

so that's where we need the product topology and the subspace topology to coincide

[gritty widget]

sure

yea

[vast estuary]

i'm sorta confused

[gentle osprey]

Hausdorff

[vast estuary]

This is what we have

[gentle osprey]

Hausdorff

Hausdorff

[gritty widget]

just take preimages

[vast estuary]

wym? like consider an open set X in H, and take psi^{-1}(X)?

[gritty widget]

yea

[gentle osprey]

Hausdorff

[vast estuary]

Is this correct?

(assuming X is open)

[gritty widget]

well, if you're taking preimages using the map in your big diagram, you should start with an open subset of G

if $V \subseteq G$ is open and $\phi \circ i_2 = i_1 \circ \psi$, then $$\psi^{-1}(V\cap H) = (i_1 \circ \psi)^{-1}(V) = (\phi \circ i_2)^{-1}(V) = (H \times H) \cap \phi^{-1}(V)$$

[gentle osprey]

TTeppa

[gritty widget]

so the preimage of an open subset of H under psi is an open subset of H x H because phi is continuous

(just using the definition of subspace topology)

so psi is continuous

[gritty widget]

Continuity doesn't depend on the image, in the sense that if $f:X\to Y$ is continuous and $f(X)\subseteq Z\subseteq Y$ then $f$ as a map $X\to Z$ is also continuous

[gentle osprey]

Blitz

[gritty widget]

so if you restrict the domain of a continuous map, then restrict the image, it's still continuous

not sure if you can see it somehow on a diagram but that's what you're using

restricting the domain can be done using inclusions

but you can't really compose a function with f to restrict the image

at least not that I know

[gritty widget]

it'd probably just end up being like the proof i did above

which is the "usual" proof but with fancy compositions of maps instead of just writing down the preimage lol

[gritty widget]

yeah

that's why I didn't bother writing it up

[vast estuary]

i've a couple of questions

how can we choose the basis {U_n} as stated?

i understand the choice of f_n's

not sure how to show bi-invariance of d and compatibility with G

[gritty widget]

as a metric space, G is first countable

[vast estuary]

bi-invariance is direct if it just means that d(x,y) = d(xh, yh) = d(hx, hy) for all h

[gritty widget]

moreover, intersections of neighbourhood basis at e is e, because well, it's Hausdorff for example

or even T1

[vast estuary]

hmm ok i don't understand this part

[gritty widget]

so like pick an element x different from e

there is open U containing e but not x

then U contains some U_n

[gritty widget]

we can do this because G is T_1

well it's even T_6

[vast estuary]

right, thanks!

[vast estuary]

is this correct

[gritty widget]

yeah looks correct

[midnight echo]

I'm trying to visualise what $S^1 \cup (\mathbb{R}^+ \times { 0 } )$ looks like, is it just a circle centred at 0 with a line on the positive y axis?

[gritty widget]

a ray but yeah

[gritty widget]

for compatibility with topology of G, you want to show they have the same open sets

[midnight echo]

Would the fundamental group of that be Z?

[gritty widget]

pretty sure S^1 is its deformation retract, so that the fundamental group is Z

[midnight echo]

Yeah that's what I was thinking. Thanks

[pliant atlas]

I was thinking about locally metrizable spaces that aren't metrizable, and I was wondering if the local metrics "agreed", is that enough to say the space is metrizable?

,tex I.e. if we have a space $X$ such that for all $x \in X$, there is some open neighborhood $U_x$ and a metric $d_x$ on $U_x$ with the following property:
(i) For all $x,y$, and all $a,b \in U_x \cap U_y, d_x(a,b)=d_y(a,b)$

[gritty widget]

the bot isn't working right now

[pliant atlas]

sad

let me paste into overleaf

[gritty widget]

you don't need to, it's clear what you mean from the plain text

[pliant atlas]

I think that if d_x is bounded on U_x we can maybe do something like the uniform metric? although im not sure

if there is some uniform bound we definitely can

actually maybe we can just shrink U_x by taking the ball of radius 1 then

hmm

if we define V_x by being the the ball centered at x with radius 1 in the d_x metric then we define d(a,b)=d_a(a,b) if b is in V_a and 1 otherwise, does that work?

[gritty widget]

I'd expect something like paracompactness to induce metrizability from local metrizability

like, there are examples of "manifolds-like" things that aren't metrizable

[pliant atlas]

this is true according to wikipedia

[pliant atlas]

wdym?

[gritty widget]

manifolds that aren't Hausdorff, say

like the line with double origin

but it is locally Euclidean

so you need some conditions to "patch it up"

the most natural is usually paracompactness

(by paracompact I mean paracompact Hausdorff btw)

[pliant atlas]

oh this maybe violates triangle inequality

if we take X=[0,0.5] and take standard metric and have all the U_x not equal to X

or the line with 2 origins as u pointed out

[gritty widget]

not sure if I ever saw a proof of this, but I think it might follow from Nagata-Smirnov metrization theorem somehow

We can take for each point a countable decreasing neighbourhood basis. Then we have a sequence of open covers U_n.
Now paracompactness should give us a sequence of open refinements V_n, each of them locally finite.
Now their union should be a sigma-locally finite basis, so by Nagata-Smirnov metrization theorem, this space needs to be metrizable

(paracompact Hausdorff implies normal)

now the converse is obvious from Stone's theorem that metric spaces are paracompact

if you know Nagata-Smirnov metrization theorem, then this should prove it

and few other facts

[gritty widget]

I was reading this proof recently, because I never encountered Bing's metrization theorem, and it uses a slightly stronger conclusion

the locally open refinement in Stone theorem can be in fact chosen to be sigma-discrete

[urban path]

random dumb question, but how did people know to define topology as the study of shapes without breaking it, i guess my main question is, how did we know to define topology as it is right now with these properties? Are there something special about these properties?

[gritty widget]

I don't have insight into the history of this, but my understanding is that, people were looking for basic axioms with which you can speak about things like continuity

[urban path]

oohhhh yea that's my initial thought too, just like how analysis started from the ground up

but without studying topology, why does "punching a hole" or "breaking it" seem so illegal

[gritty widget]

well its more like, algebraic topology I think

[urban path]

o shit im in the wrong channel

OOPS

[gritty widget]

lol? It's the topology/algebraic topology channel

[urban path]

my dumbass thought there was a diff channel for geometrical topology and algebraic topology

[gritty widget]

geometric topology is the study of manifolds so you're right on that one

well. It doesn't matter

I think #diff-geo-diff-top might focus more on the differentiable

[urban path]

dope tysm

[gritty widget]

I'm not redirecting you btw

[gritty widget]

it just does

[urban path]

LOL

[gritty widget]

for me, intuition comes first
then we make it formal, and explain it that way

[urban path]

right, it makes sense, but why, i was looking for a more mathematical motivation for why we defined it that way

[gritty widget]

topological space?

[urban path]

more like why it's illegal to punch holes or tear topological shapes

to transform one shape to another

[gritty widget]

because that wouldn't be continuous

kinda

[urban path]

oohHHh that's what you meant

makes sense now

[pliant atlas]

I don't think there was some mathematical motivation as to why people care about continuity

its just something that's sort of intrinsically interesting

what happens when you manipulate things in "smooth" ways (i'm using the word informally, not talking about differentiability)

[obtuse meteor]

There is in fact a lot of math motivation for studying topology outside of the intrinsic interest and I suspect the extrinsic motivation is why it became formalized

Topology inside of R^n becomes immediately useful in analysis of functions and understanding things of that sort (say even for physics)

Because a lot of properties of functions only care abt topological stuff and you can think of it purely in terms of that

Then you start studying more complicated things, like say RP^n (very concrete, studying lines through space) and you think about how to take what you learned about “closeness” and open sets in R^n and appropriately generalize

[silver umbra]

is the betti number b_2 of any manifold embedded in R^2 always 0?

[glass bison]

Can I do this with induction

or am I thinking about this wrong

I know there is continuous surjective map f: I to I^2 from the book section before this

I is unit interval betw

[unreal stratus]

Well not sure if this is the intended thing, but you can use induction (the hint!) to show you have a continuous surjection g: I -> I^n for arbitrarily large values of n

Once you have that, you can get all n (do you see why?)

[gritty widget]

Use that there is a surjective continuous map from the Cantor set to [0, 1]^n

Extend linearly

[glass bison]

Hmm

Had to get up for a bit will get back to yall once I try this

[glass bison]

we haven't really discusse the cantor set in class

[glass bison]

Well what I mean is

I see how to show I to I^4

since there is a surjective map from I to I^2 and a surjective map from I ^ 2 to I ^4

But what if its 3 or some other odd number, as in why is that the hint

it would seem to make more sense if you went: I to I^2 works, assume I to I^n works, does I to I^n+1 work

But thats not what the hint says which is making me doubt that

[unreal stratus]

Yup, that's why I'm saying this

Do you know a continuous surjection I^4 -> I^3?

(there are a few "easy" ones)

[glass bison]

I feel like I should know but I don't lol

some sort of projection?

[unreal stratus]

Exactly

[glass bison]

Alright

How do I know the projection I choose is continuous though

[gritty widget]

Basically

[unreal stratus]

Well the projection maps A x B -> A and A x B -> B are always continuous, essentially by definition of product topology

[glass bison]

oh

ok so that makes sense

induction for abitrarily large n, so we have a surjective map from I to I ^(2^k) for any k

then anything else has a projection

since there is some k

[gritty widget]

For any compact metric space X there exists a continuous surjection from Cantor set to X

[glass bison]

Yeah that seems like a cleaner proof but I think I'm intended to use the induction route here

with the hint

[unreal stratus]

Yeah exactly

:)

[glass bison]

Thanks y'all!

[unreal stratus]

np

[glass bison]

I may ask another question soon if y'all are free

I know there is a similar idea here in the hint

like I is continuous to I^2

and I assume there is some stitch in I^2 to make a quotient space to S^2

[glass bison]

My original idea was to make S^1

And then have an infinite series of functions, each containing an additional S^1 at an angle

But that doesn't really use the hint here.

[glass bison]

Current route I'm going is also showing surjective continuous map from S^2 to I via projection

[gritty widget]

note that S^2 is union of two sets homeomorphic with [0, 1]^2

surject [0, 1/3] onto one, [2/3, 1] onto the other

then extend

[pliant atlas]

you can also induct directly, if you have surjective continuous map from I to I^n then just take the nth coordinate to I^2 to get I^(n+1)

[candid hedge]

how to show that a subvector space is closed

without using the notion of ker of linear functions

id prefer a sequential proof if its possible

[mossy rivet]

from hatcher's - i try to understand the geometric motivation

LC are linear chains from the canonical simplex into some convex in a Euclidean space
$b_\lambda$ is just adding the baricenter

now I fail to understand what's the geometric meaning that hatcher talks about ( of T ), help would be greatly appriciated

[gentle osprey]

kingnaknik

[candid hedge]

show that this is connected with I an R interval

[gritty widget]

linear subspace of a vector space in general isn't closed

[candid hedge]

in finite dimensions only?

[gritty widget]

yeah. Finite-dimensional subspaces are closed, that's because they are complete

with respect to any given norm

[candid hedge]

Ah

well is there a sequential proof for finite dimension

[candid hedge]

also i am not sure how to show this seems complicated

[gritty widget]

it's just a triangle

[candid hedge]

yeah... but how to write that

[gritty widget]

it's convex so path connected

[candid hedge]

yeah

got it

[winged badger]

Does anyone know a proof of the following fact: a compact (hausdorf) group is totally disconnected iff the unit element is the intersection of all its compact open neighborhoods.

[gritty widget]

so if the quasi-component of e is trivial?

[winged badger]

Idk what is a quasi component

[gritty widget]

so, the meaning of totally disconnected differs from author to author, you see

what you're describing is a quasi-component of the unit element, basically

some authors define totally disconnected spaces as spaces for which all quasi-components are trivial

but here fortunately we have a compact space

[winged badger]

Oh I c

[gritty widget]

and for compact Hausdorff spaces it's known that quasi-components and components agree

[winged badger]

Yeah that

[gritty widget]

so yeah using homogenity of a group, the latter condition is equivalent to all quasi-components being trivial

and what you want is contained in this theorem I think I'll just screenshot to you

on wikipedia spaces with all components trivial are called totally disconnected, and spaces with all quasi-components trivial are called totally separated

[winged badger]

Thanks!

[winged badger]

Theres also this which I don't know a proof for: In a totally disconnected compact group, each neighbourhood of 1 contains an open normal subgroup.

[gritty widget]

the groups you are concerned with are called profinite groups and having a neighbourhood basis of the identity consisting of open normal subgroups is one of the characterizations of them
I'll let you know if I find any references

https://math.stackexchange.com/questions/55539/an-equivalent-definition-of-the-profinite-group?rq=1

@winged badger

[winged badger]

They don't have a proof there

But I solved myself

For any neighborhood U, there is a clopen neighborhood V in U

[gritty widget]

it's in the book in the comment

https://books.google.pl/books?id=iUSg040UoEsC&pg=PA3&source=gbs_toc_r&redir_esc=y#v=onepage&q&f=false

[winged badger]

Ic

Thanks anyway

[gritty widget]

lemma 1.1.1

[old oak]

why is it insisted that the space be compact?

[cedar pebble]

ehh I wouldn't call this a geometric motivation

when you're making a homotopy between two simplices \Delta^n you're trying to make some \Delta^n x I whose boundary is these two simplices

this subdivision is just a way to decompose \Delta^n x I into a union of simplices

[gritty widget]

https://www.jstor.org/stable/24893080
in here at the first page, there's a theorem that if we are in a complete metric space, and f_s are contractions, then such closed bounded set exists, is unique and compact.
My guess, is since because of this theorem, the spaces end up to be compact in practice, we assume it.

I know it's not the best explanation but I don't know more about the subject to be able to give you a better answer

[old oak]

hmm

could it be because this shows up in the context of fractals, and to look at the fractal dimension with respect to something like the box cover compactness is nice as it allows different covering numbers?

[mossy rivet]

And why do we want this one

Like - I feel like the algebraic definition makes much more sense - but unfortunately I need to include both in my lecture

[cedar pebble]

The algebraic definition does make more sense yeah

[cedar pebble]

You just have to choose one, the answer shouldn't really depend on this choice that much

[mossy rivet]

Ummmm

I'll think about it
Thanks!

[plain raven]

I came up with a categorical formulation of barycentric subdivision once

probably not helpful here

but if anybody wants to talk about it hmu lol

[shadow charm]

i asked a question in #probability-statistics but it kinda belongs here too

im wondering if its possible to generalize the notion of total variation metric, which is defined on finite spaces, to a total variation metric on a compact or maybe even locally compact topological space with its Borel algebra

by total variation metric i mean $$d_\infty(\mu,\nu)=\sup_{A\subset\Omega}|\mu(A)-\nu(A)|$$

[gentle osprey]

𝓛ittle ℕarwhal ✓

[shadow charm]

and now id be supping not over subsets of omega but over elements of the borel algebra

[gritty widget]

finite spaces?

don't you mean like, finite measures

[shadow charm]

finite probability spaces

and im talking about probability measures here so finite anyways

[gritty widget]

to make sense of this expression we need finite measures

and topology just means you can consider stuff like Radon measures

[shadow charm]

yeah but probability measures are always finite

wait am i being an idiot

you're right the sup is always defined for finite measures lmao

im not sure what i was thinking

[gritty widget]

yeah and I'm arguing that it's what it should be defined on

complex finite measures

[shadow charm]

yeah you're absolutely correct

[gritty widget]

and the topology doesn't bring anything to the table here

[shadow charm]

so i have notion of total variation distance for all probability spaces that's neat

[shadow charm]

yeah

i was just being a bit dumb

[gritty widget]

happens

[woeful coral]

For B3 does anyone know the general way to solve this

[gritty widget]

maybe, but first I'll have to break my neck

[woeful coral]

shoot sry

Assign to each positive real number a color: Either red or blue. Let D be the set of all distances d > 0 such that there are two points of the same color at a distance d apart. Recolor the positive reals such that the numbers that are in D are red and those who arent are now blue. If we reiterate this process will we always end up with all numbers red after a finite number of steps?

My claim is yes. But simply just because I haven't found any example that remotely behaves differently from every other example.

[gritty widget]

d is the set of distances d > 0 ?

[woeful coral]

D

D := { |x - y| | color(x) = color(y)}

this was on the putnam today and this question threw me in shambles

[gritty widget]

so we're having c:R_+ to {0, 1} and D_0 = {|x-y| : c(x) = c(y)}
Then say, putting c_1 = 1 on D_0^c and 0 on D_0
c_2 the same way, and so on

[woeful coral]

what is c_1 on D_0^c meaning?

[gritty widget]

I just thing naming things by colours is kinda dumb

like, we could name it c(x) = dog and c(y) = cat

but why

[woeful coral]

yeah so my way I relabeled it was by instead just 1 is red and -1 is blue

the farthest I got was noticing that if our set was a finite partition of R we could define equal parts of the set such that the leftsided numbers were 1 and the rightsided numbers were -1.

and then we would have an infinite loop

[gritty widget]

So if we're starting from some set $D_0$, we're basically taking $D_{n+1} = {|x-y| > 0 : x, y\in D_n,\ x, y\in D_n^c}$

[gentle osprey]

Blitz

[woeful coral]

Yes

[gritty widget]

what if we try D = dyadic rationals or something

Because if x, y are dyadic rationals, then |x-y| should also be a dyadic rational

I guess it doens't help much, huh

[woeful coral]

what are dyadic rationals?

nvm just googled it

After some thought I think this might hold, but it would be a pain to prove but my idea was that maybe we could prove that D_n eventually holds the form at which either it or its complement are finite.

then that would imply that the countably infinite set would just turn all red in the next step.

[gritty widget]

Denoting above by T(D), maybe it's useful to note that either D is a subset of D+T(D), or D^c is a subset of D^c+T(D)

[woeful coral]

My argument during the exam was like this.

And I said suppose by contradiction we never got this state. Then that would imply that D_n-1 had some unique property (i forget what I said) which would mean that we would get it all red for D_n-1

I think it was that if both were countably infinite then we are done. If both were finite then we could just use that to arrive at the Real numbers being finite which was a contradiction

[gritty widget]

Honestly this sounds like some sort of combinatorics problem

and I suspect the solution to be something kinda dumb

trivial when you see it sort of thing

[woeful coral]

Yeah thats generally how the putnam goes

[gritty widget]

If one of D, D^c is bounded then T(D) is the whole positive axis

so the only interesting case is when D and D^c are both unbounded

if D contains positive rationals, say, then we can tell that T(D) does so too

If D or D^c contains an interval of length J, then T(D) contains interval (0, J)

so then what I think would happen is that if x belongs to T(D), then (max(x-J, 0), x) would be contained in T^2(D)

and this would get longer and longer

but would it cover the whole space? Probably not if the gaps get larger and larger at infinity

but then the complement would contain arbitrary long intervals

so that can't really happen

so I think what I just established is that if D or D^c contain an interval, then T^n(D) is the whole real axis for some n

So the only real situation would be if both D and D^c are dense in the positive real axis

[woeful coral]

And jf D doesnt contain any intervals then we can use density to measure that any distance d will be in D after

[gritty widget]

can we?

I mean, it will be arbitrarily close to it for sure

[woeful coral]

Wait

NO

if you consider a point

Okay wait so suppose D^n doesnt contain any intervals

[gritty widget]

btw I was thinking about icicles in this argument

[woeful coral]

Then the nearest two points in D^n is seperated by a point in D_n^c

But also the same for D_n^c

So we apply D_n+1 and we still are missing these gaps that belonged in D_n^c

But then those gaps in D_n^c are of the same size in D_n so they will be mapped to the same D_n+1

And then also D_n+1^c has no intervals because D_n+1 has no intervals

Repeat.

No ending loop

[gritty widget]

let me think about what you're saying for a second

I think you're arguing that points in T(D)^c are symmetrical in the sense they are of the form |x-y| with x in D and y in D^c

[woeful coral]

yes kinda

[gritty widget]

well I'm not sure what you mean

[woeful coral]

let me draw it up

[gritty widget]

well. Another thing I'm noticing is that if we pick x in D and y in D^c, then we have a sort of symmetry going on in the interval (x, y)
In the sense that either z-x is in T(D) or y-z is in T(D)

[woeful coral]

yeah because z is forced to exist in either

[gritty widget]

they should be paying me for this

smh

[woeful coral]

lolol.

this question hurt me too much

i want to forget it

[gritty widget]

well, at least we seemingly pushed it a little

but maybe you don't even need that all

[woeful coral]

yeah you helped quite a bit a have a bit more insight now

[light flame]

you totally should

[hazy lotus]

Hey! I'm trying to prove this theorem from Willard about conditions for images of T3 spaces to be T2

[woeful coral]

Solution is out on ArtOfProblemSolvinv

Actually turns out there is no counterexample

[hazy lotus]

Everything makes sense except for the very last wentence when they conclude that U \times f^{-1}(W) can't meet A

I know that f(x_1) is outside of W so {x_1}\times f^{-1}(W) doesn't meet A, but it seems all bets are off for points in U that are not equal to x_1, as we can't use continuity to make a "closeness" argument without actually having an open set separating f(x_1) from W, which is what we're trying to prove

[candid hedge]

prove this equality

[gritty widget]

Only true in Hilbert spaces. You should stop posting functional analysis here

[candid hedge]

okay

should i go toreal complex analysis for norm questions?

[gritty widget]

#advanced-analysis

[candid hedge]

F a finite dimensions subspace of E(a hilbert space) .Show that for any a in E, there exists a b in F such that lla-bll= d(a,F)

[shadow charm]

They literally told you to move to another channel

[candid hedge]

this... this is topology no?

wtf why are my chapters in uni so messed up. okay sry

[shadow charm]

Okay fair enough it’s metric space stuff but if you’re talking about hilbert spaces it’s more worth asking in analysis channels

[candid hedge]

ok will do. ill move to advanced analysis

[hollow charm]

Let $Y = T^2 / M$ where $M$ is the meridian of a torus...

[gentle osprey]

mjachi

[hollow charm]

confused on notation -- particularly T^2 -- I have taken this to mean (Torus \times Torus) quotient (Meridian) but has felt fruitless

[hidden crag]

I've seen T^2 as just the classic torus

more generally T^n = S^1 x S^1 x ... n-times

[hollow charm]

mmm... it does make more sense

I'm trying to bounce between munkres, armstrong, and whatever is on the internet

I don't think that has helped. seen just T elsewhere, and "just words" even more commonly

[gritty widget]

yeah what Timo said, the 2 stands for 2-dimensional

[hollow charm]

ok.. thx

will play with that and see where it gets me

[hidden crag]

[serene ferry]

I need a quick sanity check:

[gentle osprey]

Volkenborn

Volkenborn

[candid hedge]

when is the closure of an open ball the closed ball?

[serene ferry]

What does this have to do with my question, I’m confused.

[gritty widget]

no, the entire space X is a neighborhood of y

assuming "O(y) subset Y" means "every element of O(y) is a subset of Y"

[serene ferry]

This is what I mean, yes.

[gritty widget]

it has nothing to do with your question

[gritty widget]

it's true in normed spaces at least. for general metric spaces i don't have a nice characterization off the top of my head

i'm sure if you try to prove that the closure of an open ball is the corresponding closed ball a nice characterization might show itself

[candid hedge]

sry it doesnt

[candid hedge]

how might i do that

let x be in A_

[gritty widget]

try to prove that the closure of an open ball is the corresponding closed ball and see if you can come up with any conditions that need to hold to make the proof work

[candid hedge]

B(x,e)inter B(0,1) is not the empty set

for all epsilon

then what

B(0,1) is the open ball

B'(0,1) the close ball

i am trying to prove it fro the ball A=B(0,1)

[serene ferry]

(Previously from #advanced-analysis) I'm trying to prove a problem about filters, here's the definition my book uses.

And here's the problem:

[gentle osprey]

Volkenborn

[serene ferry]

For the first direction, I am skeptical of my work because it seems too simple/and or wrong?

[gentle osprey]

Volkenborn

[serene ferry]

The reason I am asking about this is because I am skeptical about my assumption of how inclusion of filters work. If A<B and B is contained in the filter, does it also follow that A is in the filter?

[gritty widget]

according to your definition just take A empty and the answer is no

(you can have filters containing the empty set as an element, but then by the second condition such a filter is just going to be the power set)

(so those are boring and your definition excludes them)

[serene ferry]

But if A is nonempty does it still hold?

[gritty widget]

your proof here is correct, though. it is very simple

you didn't seem to use this false assumption anywhere

Y is an element of O(y) because it contains y and is open in X, so it's in F because that converges to y

[serene ferry]

What do you mean by "that" in this context, F or Y?

I'm assuming F because that is how convergence wrt the filter is defined, yeah?

[gritty widget]

yes

"Y converges to y" makes no sense

Y is not a filter; it's not even a set of subsets of X

in general

[serene ferry]

OK, so I guess I'll work on the second direction! Thanks again @gritty widget, that was very helpful once again.

[jagged sage]

Im trying to prove that for x_1,...,x_n distinct points in a Hausdorff space, that there exsits disjoint open U_1, ..., U_n where x_i in U_i. I tried proof by induction but dont know how to conclude in inductive step

any tips? not sure how to make sure neighborhood of x_n+1 point is disjoint

[hushed marlin]

Hey so I just got done with an undergrad topology course and I really struggled through it.

Is there a text anyone recommends that is good with a self study pace?

[plain raven]

The intersection of finitely many open sets is open

Does that help?

[serene ferry]

I've been working on the reverse direction for this, but I am skeptical of my thoughts. Assuming Y and O(y) are in the filter F (for some y in Y), then does taking the intersection of Y and O(y) finish that direction?

[gritty widget]

"the filter"?

the reverse direction assumes something holds for every filter satisfying some condition. which one did you have in mind?

[gentle osprey]

Volkenborn

[serene ferry]

I want to show that Y is open in X with the above assumptions @gritty widget.

The trouble I'm having is finding a way to show that Y is in the topology of X.

I was thinking of doing something with O(y), but I am not sure if I can do anything with it because Y is not assumed to be open for this direction.

[coral pivot]

hello topology friends so I need opinions on notation

cofibrant replacement, $QX$ or $\operatorname{Q}X$

[gentle osprey]

JohnDS

[coral pivot]

i see thats fair

right but I guess I will go with the former as its more flexible

well ok some of the things here feel a bit awkward i feel with just Q, like $Q(f)$ vs $\operatorname{Q}(f)$

[gentle osprey]

JohnDS

[coral pivot]

well maybe not that awkward

[winged badger]

Let $G$ be a profinite group. I'm trying to show that every continuous function $f: G\to A$ where $A$ is discrete factors through some finite quotient of $G$. This is equivalent to finding a normal open subgroup $U$, whose cosets tile each preimage $f^{-1}(a)$. Obviously each preimage is clopen, so there are only finitely many non empty ones, so I must prove the following claim: every clopen set is a union of cosets of some open normal subgroup $U$.
I know that each neighborhood of the origin contains an open normal subgroup, but idk how that helps for my claim. Would be glad for a hint

[gentle osprey]

Porphyrion

[pearl holly]

What John be teXing?

[little pulsar]

is it true that the boundary of an open set the same as the boundary of its complement?

[gritty widget]

It's true for any set

[hazy lotus]

It may be more difficult to prove depending on the definition of boundary you have. Typically it's cl A \cap cl (A^C) so the replacing A with it's complement gives you the same thing

[hot night]

Is the Hodge conjecture true for dimension 1?

[shadow charm]

Did you end up figuring this out? We spent some time on it in our math club but didn’t get farther than the conclusions you guys already made

[gritty widget]

Brayden said the solutions are out, so I guess they looked it up

[shadow charm]

Really? I tried to find it online but couldn’t

[gritty widget]

Yes, on ArtOfProblemSolving apparently

[shadow charm]

Ah okay found it

Answer is pretty disappointing

[gritty widget]

makes sense

[woeful coral]

I would have never gotten the answer

But now that i see it

…

[coarse night]

You have the answer?

[woeful coral]

Its on artofproblem solving

[high hill]

https://tenor.com/view/fujiwara-chika-fujiwara-chika-melting-blob-gif-22565166

[unreal stratus]

A triangulation of a space X is a simplicial complex with a homeomorphism to X

The point is you basically want to check that what you've drawn has an obvious simplicial complex structure (the space will be the same though)

So like each simplex is determined uniquely by its vertices etc

So in a sense we are just drawing a simplicial complex structure on our space lol

Yeah I mean it basically just is the same space like u needn't draw the map since it is identity

But you need to check it actually is a simplicial complex with the implied structure