point-set-topology

[odd flame]

holy shit moldi

long time no see

how is this maximal though? wont there always be a bigger D_i+1

[dry jolt]

I can't really tell what you mean through the letters 🙃 But to each edge in the cover you should label it either a or b corresponding to which edge it lies over in S1 v S1

I believe there should be three distinct covers in total

[woven sinew]

The whole plane will also be in your maximal totally ordered subset.

[agile chasm]

Hi guys, can someone tell me how can i show that the triangle inequality holds? For the second one its obvious but for the first one i dont know what i have to write down

[unreal stratus]

By 'the first one', do you mean the case where x and y are multiples of one another?

That is just the standard triangle inequality

So i'm not sure what you mean

[woven sinew]

If you want to prove the triangle inequality for the given d function, it seems like you are missing some cases. Take three points x, y, and z in your space. You have first the two cases x = lambda*y or x != lambda * y. In the first case (x = lambda*y), you can then have either y = mu*z for some mu or no such mu exists, so you see that there are more cases.

[agile chasm]

Yeah, I have to Show that all thre equations hold if i understand the task corectly. In the last equation for the second case its true because d(x, y) < = d(x, y) + some constant is allways larger than the left side.

Can i say for the first case i subbtrac the same amount z as i add so the worst case would be |lambday-y|=|lambday-y|

[hazy lotus]

Hey all, I'm having trouble with this problem from Willard about attaching spaces:

So far, I know that you can prove this by showing that the following is a homeomorphism:

The only problem is that it seems extremely messy to try to prove that h is continuous. You would have take an open ball in the subspace of S^1 and then you would have to take inverse images of them under multiple possible cases.

After you work out what the different preimages are in different cases, it you then have to work with open sets in decomposition spaces to decide whether the preimage is open or not.

I was wondering if there was a cleaner way to do this.

Another problem with doing it this way is that it's actually not easy to compute the preimage of an open ball in S^1 directly and exactly, even if you know intuitively what the preimages look like and that they're open.

[carmine plover]

I'm trying to show that the orthogonal group is closed in R^(n^2), that is, the set of AA^T \neq I open in R^(n^2)
Is someone online who could help me? A lot of literature assumes it's trivial, but I fail to see its apparent triviality

[gritty widget]

AA^T = I is a system of equations for continuous functions of the entries of A, so it defines a closed set.

[unreal stratus]

So for example it is the preimage of the zero matrix under the map A to AA^T - I

Lol

Sniped ig

[gritty widget]

This

[carmine plover]

ohhh, thank you!!

both of you!

[stable kite]

I've been looking at this exercise for some time now but I just can't figure out how the linear extension of phi to a map from S_1(X) is supposed to work. I'd like to understand that to be able to start the actual exercise. 😓

[lunar yoke]

So S_1(X) should be the free abelian group on the simplices Delta^1 -> X

so if you specify where these basis elements should be sent into another abelian group, this automatically linearly extends to a map from S_1(X) into the other abelian group

this works just like for vector spaces

concretely, an element $\sum_{i=0}^n a_i \sigma_i$ will then be sent to $\sum_{i=0}^n a_i f(\sigma_i)$ where $f$ specifies where you send each 1-simplex

[gentle osprey]

Phil

[lunar yoke]

the a_i are just integer coefficients here

and every element in S_1(X) is of the form of such a linear combination of simplices

[stable kite]

Ohh wait that's exactly what I tried to make sense of - but somehow I haven't seen so far how the right sum actually makes sense as an element of the abelianized fundamental group. 🤔

[lunar yoke]

well its an abelian group

[stable kite]

Like, the left sum can be an infinite formal linear combination, can't it?

[lunar yoke]

no

only finite

we're doing algebra

[stable kite]

Huh

[lunar yoke]

in vector spaces you also only take finite linear combinations of vectors

unless you're doing functional analysis or smth

[stable kite]

Wait, I thought free abelian group over some index set means each element corresponds to an assignment of some integer to each generator

[lunar yoke]

so if you prefer to use the path concatenation notation we could send $2\sigma - \tau$ to $\phi(\sigma) * \phi(\sigma) * \overline{\phi(\tau)}$ where the overline is again the inverse path

[gentle osprey]

Phil

[lunar yoke]

just you're technically dealing with equivalences classes of homotopy classes of paths since you took the abelianization of pi_1, so we often just use +

[lunar yoke]

yes, but only finitely many assignments are nonzero

the free abelian group on a set X, denoted Z[X], consists of all finite linear combinations with integer coefficients of elements of X

[stable kite]

I legitimately didn't remember that fact... That makes everything a lot easier, wow.

[lunar yoke]

so $\mathbb{Z}[X] = {\sum_{i=1}^n a_i x_i \mid n \in \mathbb{N}, a_i \in \mathbb{Z}, x_i \in X,\ i=1,\dots,n}$

[gentle osprey]

Phil

[lunar yoke]

lmao how did you do all the homological algebra questions on the previous sheets then

[stable kite]

Uhh I haven't managed to solve many questions these last weeks, probably rusty algebra knowledge had its contribution

[lunar yoke]

haha i feel you, when i took the course last year I also had 0 homological algebra knowledge before

you'll learn a lot if you stick to it though 🙂

[stable kite]

That's very motivating, haha. Thank you kindly, and have a nice rest of the day! ^^

[zinc siren]

the idea being that even with an infinite basis set, you still have manageable elements.

because there's a metric fuck ton of singular simplices (or maybe you're doing simplicial homology, in which case there's still often a fuck ton of simplices, though maybe not as infinitely many).

[gritty widget]

what do you call this process of taking a shape like a torus and turning it into a 2d depiction? like this

also my bad if this isnt rly topology

ah yes ty!

[odd flame]

vague question but why is path connectedness important

atm im thinking of it as stronger connectedness that has some nice properties for ordered sets but idk

[gritty widget]

Path connectedness is quite important in algebraic topology, where you deal with paths and their homotopies a lot.

e.g. the fundamental groups at every point of a path connected space are all isomorphic.

Path connectedness is also usually easier to prove.

At least in my experience.

It's not really directly related to your question, but it might be worth noting that "connected" and "locally path connected" together imply path connectedness.

[gritty widget]

if I were given some abstract topological space, then it'd probably wouldn't be as easy

I was planning on clarifying.

I think arc-connectedness is more important

[hidden crag]

depends on the area of interest i guess

[gritty widget]

because knowing you can order some piece of your space that connects two points, is pretty nice

[odd flame]

point set topology class

just following munkres

[hidden crag]

another example is that fibrations with path connected base spaces induce sequences of homotopy groups

[gritty widget]

ah. Then let me give you an adequate answer.
Path connectedness is important because it might appear in your exam

why would it be a problem

you can do cool things with arcs that paths don't allow you

no

[odd flame]

[lunar yoke]

besides the other answers, it is often used to somehow extend a local property to a global property

its a pretty standard but powerful proof technique

so e.g. if you have some locally constant function into say the reals on some connected space, it already has to be constant

oh sorry i guess this is the more general phenomenon of connectedness instead of pathconnectedness, but in lots of spaces we care about thats the same thing, like cw complexes

[lunar yoke]

so the general way to go about this is to take some nonempty subset and show it is both open and closed, then by connectedness it must already be the whole space

for the example above, lets say we have f : X -> R with X connected. I pick some x in X and consider the subspace U = {y in X | f(y) = f(x)}. This is nonempty since x in U. It is open since f is locally constant, and the same argument shows its complement is open: Indeed, say y is not in U. then since f is locally constant y has an open nbhd where f has the same value as y, i.e. not the same one as x, so this whole open nbhd lies in the complement of U, showing that U is closed. Hence by connectedness of X we have U = X and so f is constant

[lunar yoke]

this does not need the base space to be path connected though

you just need to pick basepoints

[odd flame]

this helps!! thank you phil

unrelated but how does the lebsgue # connect to uniform continuity

[odd flame]

i know this is late but did you meant that this is B right? not the maximal set it's asking for

yes you did nvm

[odd flame]

will bump this tho bc im not sure how to put those bunch of sets between Di and Di+1 that mold mentioned

[peak crystal]

does there exist open map from closed unit disk in R^2 to R^3

[alpine bolt]

Is it possible to have a (not necessarily continuous) bijection from a space that is not compact to a space that is compact?

[viral atlas]

You can certainly have one by cardinality arguments alone

R is in bijection with [0,1]

[hidden crag]

The proof for it that I saw was restricted to path connected base spaces

Aah that was only used to have it end on 0

Nvm then

[odd flame]

bump to this timo 👉 👈

[gritty widget]

B is already totally ordered. Use Zorn's lemon to extend it

[odd flame]

hrmmm

im not sure i see how

from what i understand the lemon says that an "infinite process" must stop right

so do we just say that by zorn's lemon we take the maximal D_n and we're done?

[gritty widget]

Take a chain of totally ordered sets containing B

It has an upper bound

Hence we can use Zorn lemma to conjure up a maximal one

And that's mathemagic

[odd flame]

in the most literal sense

i'll be back this is a lil icky for some reason

[odd flame]

shouldn't it be a collection of subsets though?

oh i guess the collection of every maximal subset for every possible chain?

maybe im overthinknig this

[gritty widget]

No, definitely mathemagic

[odd flame]

[gritty widget]

[odd flame]

for real though, a subset of A has to be a collection of subsets no

[gritty widget]

Chain of totally ordered sets containing B

The totally ordered sets are subfamilies of A

[odd flame]

[empty grove]

No need Zorn

You can explicitly find sets in between the integer radius disks lol

There's a super obvious choice of sets to put in between

And then you show that you can't add anything more to that collection

[odd flame]

squares?

though idk if that fits "between"

i dont think it does

just...more disks?

[empty grove]

Ye lmao

Non integer radius disks

The set of all disks centered at 0 is a totally ordered set

And is maximal too

Last thing needs proof but is easy

[odd flame]

,ti

[gentle osprey]

The current time for stμ₂dying is 04:20 AM (EST) on Mon, 14/11/2022.

[odd flame]

nothing is easy right now

[gritty widget]

That's false

[odd flame]

gottem tho

(it's the we*d number) ((please laugh))

[empty grove]

Wait really?

[gritty widget]

Yeah. Like open disk for instance sits inbetween them

[empty grove]

Ah

[gritty widget]

And even if we allow open disks it's not enough

[empty grove]

Nvm use Zorn

[gritty widget]

What you mean is to use polar coordinates ig

[odd flame]

so density means it doesnt work right?

[empty grove]

What density

[gritty widget]

Density is a random thing to say

[odd flame]

nvm then

[empty grove]

[gritty widget]

You can still use Moldi's idea, just take polar coordinates and order lexographically

Let the sets be open and closed intervals with endpoint an element of R^2

So it'll be open, closed balls, and things inbetween

And whole R^2, empty set, {0}

[odd flame]

ig my main doubt now is that it says "find a maximal totally ordered subset..." not to show that one exists

actually ig the chain is that

[gritty widget]

Imo that just means to show it exists

shrug

[odd flame]

on a different note now, any pointers here

just hints though please

[gritty widget]

This is called Alexandroff extension of your topo space

a) holds iff your space is locally compact Hausdorff

b) always holds

c) I'd have to think about it more

[small jackal]

For countably infinte metric spaces, can we not take the product metric to just be the minimum of all the individual metrics? I mean will that not work because the number of such metrics are infinite?

[gritty widget]

Countably infinite product?

[odd flame]

i was thinking this one to be false

[small jackal]

yes

[gritty widget]

It is. Q is not locally compact

[odd flame]

why is that iff true then

i haven't seen that before sorry

[gritty widget]

This might give you box topology or product topology.

And probably few different topologies as well

[small jackal]

Ah so the metric is valid

[gritty widget]

Well maybe not box topology

[small jackal]

I was basically reading this textbook and it mentions

this for countably infinite product

[gritty widget]

Yeah. It's standard

I mentioned it before to you

Remember?

[gritty widget]

Prove b) first, and assume X^+ is Hausdorff to obtain contradiction

X is its open subspace

[small jackal]

Yessss

I was wondering if choosing the max or min would make sense for the product metric

[gritty widget]

Yeah. I was thinking of sup d_i

[small jackal]

Ah right

[gritty widget]

Wait. You wanted min anyway

My bad

But yeah. Here they use min(1, x) to make them into bounded metrics

Is all there is to it

[small jackal]

Right but would min(d1, d2,...) be valid?

[gritty widget]

No because, the inf can be = 0 for non-zero points

And it often will be

[small jackal]

Didn't get you

Oh right

and if it's 0, it'd mean the two points are the same which they might not be

max still works right?

[woven sinew]

For (c), try $\mathbb{Q}^+ = A\cup B$ (where the $A$, $B$ are open and disjoint), and see if you have any results about compact subsets of the rational numbers.

[gentle osprey]

TimeTravellerOne

[odd flame]

i think it's connected

[gritty widget]

If A contains infinity, then B is contained in a compact set of Q

Can a non-empty open set of Q have compact closure?

[odd flame]

sorry to divert a bit but some googling pointed out that proving that a cpt subset of Q has empty interior is useful

i dont really see how it's even true tbh but is it necessary?

[odd flame]

ok i showed that it was compact but idk about a) still

[dry jolt]

what happens if both points are in Q

[odd flame]

Q is hausdorff so that's fine no?

blitz said it was false before though so im definitely missing something

[dry jolt]

right, the issue arises when you consider consider a point in Q and +infty

if the space were hausdorff, then every point in Q would have a compact neighborhood. Do you see why that statement is true?

[odd flame]

they'd be closed in R since it'd contain all Q in the interval as well as irrationals no?

or is it better to say that we could a finite subcover for any open covering of a subset of Q

though idk if that's worded well/true

[dry jolt]

well the point is that Q isn't locally compact, which is why the space isn't Hausdorff

in my earlier message I meant do you see why the statement "if the space were hausdorff, then every point in Q would have a compact neighborhood" is true

[odd flame]

ahhh nvm

i dont full see that so brb

im gonna take a break from that one - am i right to think that Q^+ is connected though?

wait if a space X isn't hausdorff it has to be connected right?

based walter listening to tyler

[dry jolt]

this definitely isn't true but i'm too sleepy to think of a counterexample rn

[odd flame]

same

,ti

[gentle osprey]

The current time for stμ₂dying is 08:28 AM (EST) on Mon, 14/11/2022.

[odd flame]

i definitely slept

[dry jolt]

but yes, I believe it's true that Q^+ is connected

[odd flame]

pf by "i said so"

[dry jolt]

yeah i mean it's not trivial to prove but I think if you start by observing that a separation of your space would yield a clopen compact set in Q then you can use some properties of these sets in Q to get a contradiction

[odd flame]

that just leaves first countability

google said no

[woven sinew]

It's not true - as a counter-example, take the disjoint union of two non-Hausdorff spaces.

[small jackal]

Do uncountably infinite product of metric spaces make sense? If so, how? I cannot find out an instance where it does.

[gritty widget]

Yes

But it won't be a metric space

Any uncountable product of non-trivial metric spaces is not metrizable

[small jackal]

Ah

How can I formally show that

I tried finding some reference online, couldn't

[gritty widget]

Try to see if some topological property that metrizable spaces have fails.

[small jackal]

the metric which is imposed needs to have a finite value for all pairs of points?

And this would fail in the case of uncountably infinite spaces

[cerulean oriole]

You're trying to show that the product (as a topological space) is not metrizable — ie, is not the topology arising from any metric.
In particular, you need to show it can't come from a metric regardless of what the metric is. So you don't know how the hypothetical metric is related (if at all) to the metrics on the spaces you started with.

[cerulean oriole]

So a property whose statement depends on the metric would be more difficult to show to be true since you don't know what the metric is.

[cerulean oriole]

This hint is to find a topological property — ie, a property that only depends on the topology (e.g. connectedness, compactness, Hausdorff, …). That way, you can check whether it is true (hopefully) straightforwardly, since you know what the topology has to be (the product topology).
If it's a property that the product doesn't have but you know any topology of a metric space has, then you're done.

[gritty widget]

first countable

note: it can be separable

[wise ruin]

nLab defines simplicial homology in terms of S_n ⋅ A = Z[S_n] ⊗ A, for an abelian group A and n-simplices S_n. Namely, it is the chain homology of the complex C.(S⋅A)=C.(S,A). How does this generalize the typical way simplicial homology is introduced (just using boundaries of n-chains with no extra ab groups involved), and what benefit does this generalization provide?

[cedar pebble]

I mean this generalization is necessary because a lot of natural geometric constructions produce classes in H_k(X,A) with A some Abelian group different from Z.

On the other hand the universal coefficient theorem says you can recover these H_k(X,A)'s from H_k(X,Z).

[zinc siren]

good example is the relationship between de rham cohomology and singular cohomology actually uses the singular cohomology over the reals instead of the integers

[pseudo coral]

how do I show C is not a covering space of C \ {0,1}

do I suppose there is a covering map from C to C minus 0,1

and find some contradiction

@ me with a response 🙂

[gritty widget]

Analytically speaking? Since if you care about homeomorphism or even diffeomorphism, the claim is true.

I'd figure you meant the first one since you also posted this in the real/complex analysis channel.

[pseudo coral]

Yes analytically speaking

[little hemlock]

here psi is an element of Delta^1(X; Z)

im having trouble making the connection between the values psi takes on edges of 2-simplices and curves crossing the 1-skeleton transversely

[unique wolf]

Guys please tell me where I am going wrong. I have to prove that excluded point topology in R is lindelof. However take this cover of say (0,1) where excluded point is 2 for our topology.Clearly (0,1) is open and every point is open in this topology so I collect all singletons in (0,1) as my open covers for this interval calling it O={x such that x is a point in (0,1)}. Now this cover is uncountable since (0,1) is uncountable but if you remove even one cover from this set then you cant cover (0,1) since you have lost a point from (0,1). So how this space lindelof where am I making a mistake

[little hemlock]

you need an open cover of the whole space. Not just a subset

for example all your argument shows is that R with the excluded point topology is not hereditarily lindelof

[unique wolf]

Thanks for pointing that out .

[little hemlock]

why is the cohomology ring different for even dimensional spheres? Having trouble finding the explanation in hatcher.

[little hemlock]

how does one see the underlined part?

[lunar yoke]

He probably just uses the künneth isomorphism inductively for the S^1 components

[undone ridge]

very basic question:

I don't see how d(x,y)=d(x_1,y_1) implies xR_dx_1 and yR_dy_1 (illustration above).

(matter of fact, I just woke up so maybe I'm just missing the obvious here)

Btw I get the point of basically quotienting a semi-metric space X with the relation x ~ y if d(x,y)=0, and getting a metric space X/~ out of it, but it's the way it's being phrased that confuses me.

[gritty widget]

Semi-metric can have multiple meanings

[undone ridge]

here it means a metric where the condition of $d(x,y)=0 \implies x=y$ isn't required.

[gentle osprey]

anøk

[gritty widget]

Trivial then

[undone ridge]

(it's also called a pseudometric)

[undone ridge]

yeah, it was more of an "english problem" that anything else lol

[peak crystal]

Is there any?

[undone ridge]

mmmmm (considering that the discs D² and D³ are centered at the origin) if you choose like, an open ball of radius r in R² centered at (x,y), and map it to an open ball of radius r centered at (x,y,0), does this work as an open map? I mean, the topology of D² is generated by open balls right

[little hemlock]

where does this map send the closed unit disk though?

[undone ridge]

it sends D² to the unit disc on the xy plane no?

[little hemlock]

choose like, an open ball of radius r in R² centered at (x,y), and map it to an open sphere of radius r centered at (x,y,0)
also i think you must mean open ball in R3, not "open sphere"

[undone ridge]

oh my bad

[little hemlock]

but then the map isn't open

like the image of D^2 should be an open set in R3

[undone ridge]

oh ok ok

[little hemlock]

i think if we were working with the open unit disk, something like ur idea should work tho

[undone ridge]

No wait wait wait

We send D² to D³

The closed unit disc D² goes to D³

[undone ridge]

@little hemlock

[undone ridge]

It actually sends to the whole space

[little hemlock]

D^3, the open unit ball in R3?

[undone ridge]

Closed unit disc in R3

[little hemlock]

thats not open in R3

[undone ridge]

But it is in the D³ with the subspace topology no?

The open map is supposed to be from D² to D³ right?

[little hemlock]

yeah, but i believe they asked for an open map cl(D^2) -> R3

[undone ridge]

Oh ok

[peak crystal]

This is the question

If there is such map then D map to closed set right in this case there is not continuous open map

So in this case we need to find an discontinuous open map.

If we send to R^3 then how to show that it is open map?

Or if we can show that there is local homeomorphism

Yes this is

[little hemlock]

not sure what u are trying to say here. like, what points are supposed to map to what points?

[peak crystal]

Correct

[little hemlock]

also why do you think D has to map to closed set?

[peak crystal]

Under continuous map compact set map to compact then it is closed

[little hemlock]

ohh true how could i forget

i think that's all there is to your problem. they didn't ask for an open discontinuous map

[peak crystal]

But open map can be discontinuous so in that case how to approach

[little hemlock]

idk the context of this problem, but generally when we're talking about topology, and someone asks about a map, we're generally implicitly talking about continuous maps.

[peak crystal]

Okay then, thanks alot.

Is there any argument to show that R^2 and R^3 are not homeomorphic? Except dimensions argument

[little hemlock]

yeah, the theorem is called "invariance of domain" but it requires algebraic topology

[peak crystal]

In point set topology?

For that we need algebraic topology. Same for open ball in R^2 and open ball in R^3?

[little hemlock]

hmm, the point-set methods that show R is not homeomorphic to R^n would not work. I'm not aware of a more elementary method that could show R^2 and R^3 are not homeomorphic

Same for open ball in R^2 and open ball in R^3
yea invariance of domain also applies here

[peak crystal]

Okay, Thank you.

[undone ridge]

but an open map need not be continuous (if the condition of continuity was required, then the solution would easily follow from the fact that D is compact and R^3 is not)

Or we could solve it by looking at the fundamental groups of these spaces. (I still think tho that, even if the condition of continuity isn't assumed, we can prove that such an open map can't exist by looking at Int(D))

@little hemlock

[undone ridge]

Wait I have an idea

[little hemlock]

but an open map need not be continuous
i agree. my point is that it is very likely "continuous open map" is what was implicitly being asked for. A common abuse of language in topology.
solution would easily follow from the fact that D is compact and R^3 is not
the failure when the open map is continuous does not have to do with D being compact but R3 not being compact per se. Its because D is compact, but (nonempty) open subsets of R3 are not compact

[wise ruin]

In order to compute the simplicial homology of S^2, we must give it a delta-complex structure. The typical structure is its fundamental polygon with an edge through its diagonal. However, as far as I can tell, delta-complex structures have no notion of identified n-simplices, so such a structure must be the quotient space of a bona fide delta-complex structure. So, in order to complete the computation rigorously, must we first prove some properties of the simplicial homology of a quotient space?
I think I might also be a bit confused since in order to have the simplicial homology on a quotient space, we must have have a delta-complex structure on that quotient space, which is exactly the original problem. Where might I be confused?

[coarse night]

you can think of S² having a the simplicial structure

• 1 0-cell
• 0 1-cell
• 1 2-cell attach via the boundary going to the 0-cell

which will give you the simplicial complex
... →0 →ℤ →0 →ℤ →0

[wise ruin]

Sure, but that's a cell-structure. I'm asking about computing it via a delta-structure.

[coarse night]

Ok I'm no longer sure what the question is

[wise ruin]

Which part of my question is confusing?

[coarse night]

the description I gave is a simplicial structure on S²

[unreal stratus]

Sounds like ur giving it a cw structure

[coarse night]

it's also a CW structure yes

you can verify rigorously actually,
send Δ⁰ →0
Δ² ∖ ∂Δ² → S²\0 (by the identification map)

see that the maps restricted to the interior is injective indeed and that restricted to the boundary is another delta complex.
also it's the weak topology which is by def of CW cpx. So yes it's also a delta-cpx structure

[tawdry valve]

this is not a delta complex structure, at least how Hatcher defines it. If f:Δ^2-> S^2 is the map of the 2-cell, you need the restriction of f to its boundary to be maps of 1-cells in your delta complex structure, but this does not work because a restriction g:I->S^2 of f will not be injective in the interior (since they are constantly mapping to the point which is your 0-cell)

[tawdry valve]

I think an easier way to see it is that, though we prefer drawing pictures of the unidentified quotient spaces, when we label the edges/faces, we really mean maps to the quotient space.

For S^2, rather than thinking of the picture being about maps f: Δ^n-> square, you should instead post compose with the quotient map π. Then, you have honest maps from the simplex to the space S^2. You should then verify that your collection of maps assemble into a delta complex structure of S^2 (ie check the 3 things on Hatcher page 103)

[gritty widget]

Does anyone have tips for like visualizing one point compactifications?

For example for (0, 1) x R as a subspace of R^2, I don't really know how to picuture it (although I feel like it will look like a sphere)

[winged viper]

one way is maybe to see that (0, 1) x R is homeomorphic to R^2, which is homeomorphic to a 2-sphere minus any point (by stereographic projection)

then the one point compactification is just filling in that point

maybe a good way to think of it is to try to mold your space so that it looks like a compact space minus a point

for example you can think of (0, 1) u (2, 3) as the space given by two circles joined at a single point (looks like the number 8) minus the joining point

so the one point compactification is just the 8

[gritty widget]

oh yeah good point

[supple heart]

How is an interval [a,b] closed??

Doesnt every point the interval have a neighbourhood there?

[hidden crag]

What about a and b

[supple heart]

Doesnt specify

It says its closed cuz the complement is open

That makes sense and all but I dont see how it is closed

Does a not have a nbd in there?

Or will it go to the "left" and out of the interval?

[hidden crag]

Yes

[supple heart]

Ahhhh

That makes sense

[gentle osprey]

justAwork

[wise ruin]

EDIT: [SOLVED]

I guess a shorter version of this question is How do you represent a topological space by a simplicial set so that both simplicial homologies (that of the delta-complex structure of the space and that of the simplicial set) coincide?

[agile chasm]

Hi im trying to solve this problem, im not quite sure if im on the right path. Im also not sure what im trying to do. Sure i need to get the coefficients but im kinda lost her. Can someone help?

[gritty widget]

honestly this fits better in #advanced-analysis , but let me answer it anyway
You're wrong about the part in the red, it's supposed to be an integral

$$\sqrt{\int_0^h (ax+b)^2\mathrm{d}x}$$

[agile chasm]

i see integral over the whole domain

[gritty widget]

yes

sqrt(a^2h^3/3 + abh^2 + b^2h)

[agile chasm]

shouldnt it bee (ax+b)^2 * dx and not x?

i dont get were the x comes from

[gritty widget]

yes because I messed up my LaTeX

[gentle osprey]

Blitz

[agile chasm]

thx

[agile chasm]

@gritty widget additional stupid question but what do i do next how do i get the c? Do i just compare the coefficients because all three equations seems similar?

[gritty widget]

well yeah, you should try to vary them and find an optimal constant

[agile chasm]

ok thanks again

[jagged sage]

How can I prove that for two quotient maps q1: X to Y1 and q2: X to Y2 that if q1(x)=q1(x’) iff q2(x)=q2(x’) then Y1 and Y2 homeomorphic

Or at least intuition/structure of proof

[gritty widget]

the properties of quotient maps tell us that q_1q_2^-1 and q_2q_1^-1 are well-defined and continuous, and clearly inverses of each other

hence Y_1, Y_2 are homeomorphic

[gentle osprey]

loganb

[lunar yoke]

I think the space is homotopy equivalent to the genus 2 torus

and iirc the pi1 of that is <a,b,c,d | [a,b][c,d] > where the brackets denote the commutator

[gritty widget]

i dont think its abelian...

[lunar yoke]

that group is not abelian

[gritty widget]

pi1 (T^2) = pi1(S^1 x S^1) = pi1(S^1) x pi1(S^1) = Z x Z is abelian no?

[lunar yoke]

that is the genus 1 torus

[gritty widget]

oh wait genus 2 torus i misread

yeah ty my bad

hm i dont see the homotopy

[lunar yoke]

you fatten up the s^1 v s^1

to basically the "interior" of the genus 2 torus

and then you collapse the R^3 - that interior to the surface

[gritty widget]

ahh okay

ill draw some pictures and see if i can make some sense of that

thank you!

[lunar yoke]

wait sorry this is probably wrong

the same "argument" would get you that R^3 - S^1 retracts to the torus, but this should be S^2 v S^1 instead

doing this solely through imagination is probably not the best idea ^^

so yeah maybe your initial guess of S^2 v S^1 v S^1 makes more sense

[gritty widget]

because i thought like draw a sphere around the two circles and then retract onto it plus two different diameters of S^2

[lunar yoke]

yeah this is a nice image for S^1

[gritty widget]

and then bring the points where the diameters intersect S^2 together

yeah ive seen that

i wasnt sure if it works with two

[lunar yoke]

i think so. if you start like this where the red part is a "hole"

then you push it upwards and downwards you should just get S^2 v S^1 v S^1

[gritty widget]

what do you mean by push it upwards and downwards?

[lunar yoke]

the filled in space above and below the red part deformation retracts towards the sphere

[gritty widget]

ahh

[lunar yoke]

or towards the two holes

its hard to describe haha

[lunar yoke]

and my drawing skills cannot compare to this

[gritty widget]

so is S^2 union a cross section missing two points

where is S^1 v S^1 coming from

[lunar yoke]

what do you mean by cross section

result should look like this

[gritty widget]

hm

[lunar yoke]

and then you take the two lines and move start and endpoint together

so what i mean is you can fix the bottom point of the line where it is attached to the sphere, and the top point you can move along the surface towards the bottom point until they coincide, giving you a loop

finally you can move the two attaching points of the circles together if you want

[gritty widget]

i see how to go from the last one to S^2 v (S^1 v S^1)

ah and i guess the middle step is drawing diameters through the infinity sign and pushing the inside of the circle towards it and outside the circles away from it ?

[lunar yoke]

well the diameters are kind of there already

the sphere in my drawing is supposed to be filled in except the red part

like in the drawing for just S^1

[lunar yoke]

you do the same thing as here

[lunar yoke]

so technically everything here except the two diameters should be red but then you couldn't see anything so i didnt draw it

[jagged sage]

How does q2 and q1 being inverses of each other makes Y1 and Y2 homeomorphic

[gritty widget]

A map is a homeomorphism iff it's continuous with continuous inverse

this is definition of homeomorphism, no?

also I never said q_1, q_2 are inverses of each other

that wouldn't quite make sense

[jagged sage]

Yeah, thought when you said clearly inverses of each other you meant q1 q2. Is the homeomorphism you’re referencing the q1q2^-1?

[gritty widget]

yes

[fallow shale]

is the wedge sum of homotopies bilinear? f v (g+h))=f v g + f v h? Writing out definitions it seems like it, but it isn't written anywhere....

[summer sequoia]

how do i go about this please help If ( A \subseteq \R ) is closed then ( A^c ) is open.

[gentle osprey]

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

[swift mountain]

Let x in A^c then if every neighbourhood of x intersected A then x would be a limit point of A and hence in A since A is closed, but x in A^c so there is some neighbourhood that doesn't intersect A, this is true for all x in A^c so A^c is open

@tight oasis

[sage trail]

That's... the definition of a closed set

[neat stag]

some authors define closed and open sets in terms of limit points and interior points rather than closed sets as complements as open sets

[sage trail]

Oh, I did not know this depended on the author. I thought the general definition for a closed set was "its complementary is open" because it's the only definition that you can build from scratch in general topological spaces. Or at least that's what I believe.

[neat stag]

Yeah, but in the context of metric spaces (like in real analysis) it's often defined in terms of that

[gritty widget]

Well there's another good definition in topological spaces

A set A is closed iff for every x, if every neighbourhood U of x intersects A, then x is in A

[sage trail]

Oh yes you are right, that's equivalent

Also, if we define the notion of limit, we actually do not need closed sets right ? If we say that a sequence converges to x if the sequence enters once and for all every neighbourhood of x, we can define the adherence of any set and define a closed set as equal to its adherence ?

[opaque cloud]

and it's a diffeomorphism if both it and it's inverse are C^k/C^inf?

[gritty widget]

yes

to be more precise, a diffeomorphism is a map f such that both f and f^-1 are differentiable
and a C^k-differomorphism is a map f such that both f and f^-1 are C^k maps
but people differ in conventions here and I think I saw before the usage of the word, that two manifolds are diffeomorphic if there is a C^infinity map with C^infinity inverse between them

but to keep it short it's like homeomorphism but with differential structure (like a differentiable manifold structure)

but this is not topology

[wise ruin]

I'm back with another misconception.
In order to compute the simplicial homology of S^2, I'm using the delta-complex structure in the image.
Since partial_2(F)=-partial_2(G), the image of partial_2 is Z<a+b+c>. Now, since there is only one 0-simplex, the kernel of partial_1 is Z^3. Hence, H_1(S^2)=Z^2.
This is clearly wrong but I'm not sure where my reasoning fails.

[coarse night]

there are actually 3 distinct vertex for your structure
both a leaves
a enters b leaves
both b enters

[wise ruin]

Ah, so the above image (without c) is not equivalent to the fundamental polygon of S^2?

[coarse night]

No it actually is

but the boundary maps you are getting are not the correct one

Let me redraw the picture for you

@wise ruin

[wise ruin]

So, I see how it works when you give it three distinct vertices, however, I don’t understand why identifying all of the vertices gives you the wrong boundary map. It’s still a delta-complex structure of S^2, no? And thus should give the same boundary map

[coarse night]

Because under your identification the vertices are not even the same

[wise ruin]

Huh? In my drawing, a, b, and c are just three distinct loops based at v

[coarse night]

These vertices are inherently different you can’t say the are same

One is starting of A and another is ending of a

I mean at least identifying them won’t give you S2

Try cutting a paper and see why

[wise ruin]

Ok, yeah so this goes back to this question. Earlier you said yes but if it is in fact no then I understand.

[coarse night]

Well I meant except the vertices, rest is fine

Mb

[wise ruin]

Well I appreciate the help

[coarse night]

It’s like folding a Taco

[autumn dagger]

https://images-ext-2.discordapp.net/external/DUyEh3esFgKjBQJlePPOtCsnWYLnx9gn46pbQQCUL3g/https/i.gyazo.com/thumb/1200/7e89630f875f4983cc86af6e9a67d8ce-png.jpg

https://images-ext-1.discordapp.net/external/YzMvdxZmk6Jk9ReHeDSduAQSpyFguM-GFDct7mQKfOM/https/i.gyazo.com/thumb/1200/9a1a7c9d6b82ed047bf7af358ed472be-png.jpg

Hey guys, these are two conflicting definitions for a basis, the second conition in wikipedia uses subseteq, but munkres uses proper subset, so I'm just wondering which one of these is correct? Thanks c:

[shadow charm]

c does not denote proper subset

Proper subset is $\subsetneq$

[gentle osprey]

𝓛ittle ℕarwhal ✓

[autumn dagger]

so they are the same

💀

[shadow charm]

Though to be fair a few people will use $\subset$ for proper subset but it’s ambiguous

[gentle osprey]

𝓛ittle ℕarwhal ✓

[shadow charm]

This is more standard notation

[shadow charm]

Yup

[autumn dagger]

thanks <

[gritty widget]

not a few

a lot of people

[shadow charm]

Bad people

[gritty widget]

bad people use $\subset$ to mean subset

[gentle osprey]

Blitz

[shadow charm]

That too

[unreal stratus]

don't most people use that to mean subset lol

Virtually every time i've seen \subset it just means subset

[gritty widget]

definitely not

[unreal stratus]

i wonder what you'd get if you polled mathematicians lol

[rough cedar]

\subset => proper subset
\subseteq => proper subset or equal => improper subset

my 2 cents

[gritty widget]

1 < 1 am I right

[gentle osprey]

Blitz

[empty grove]

Imagine writing subseteq every time you mean subset

🤢

[pearl holly]

Imagine seeing moldi online

[devout sorrel]

how do you show that if you take away a finite set from R^2, its still connected?

[gritty widget]

When you go for a walk, you tend to go around the construction on the road, right?

[pearl holly]

No.

[gritty widget]

https://en.wikipedia.org/wiki/Umarell This you?

[devout sorrel]

unless i happen to have a topo exam coming that week

[gritty widget]

Path-connectedness feels more intuitive here. That's what I'm getting at.

[pearl holly]

Omg how did you know???

[gritty widget]

It's arc-connected

[devout sorrel]

hmm

[gritty widget]

And you can take away any set of cardinality < continuum

Normally for R^2 you would consider a linear path between x, y

Do a little wiggle.

:catwiggle:

There are continuum many piecewise linear paths from x to y, disjoint apart from the endpoints

Try justifying it

[devout sorrel]

i dont get how you show that one of the non-linear paths does not intersect with any of the removed points

[gritty widget]

Compare cardinalities

If all intersected we would have a injection from a set of size continuum to a set of lesser size

{Family of continuum many disjoint paths} to {removed points} and the map injective

[devout sorrel]

hmm right

[gritty widget]

How have I never seen a solution like this before?

That's nice.

It's standard you have seen it for sure

[opaque cloud]

ah makes sense

[opaque cloud]

[devout sorrel]

how do you show that if f : X -> Y satisfies the property f(cl(A)) subseteq cl(f(A)) for any subset A of X, then f is continuous?

[unreal stratus]

You're made a typo, since what you wrote always holds

[devout sorrel]

yeah sorry

corrected

[unreal stratus]

Cool yeah, so what you want to do I'd take a closed subset C of Y and pick an appropriate subset A of X which will show you f^-1 C is closed

That's more or less the only thing you can do, but hopefully if I put it like that it becomes more clear aha

[waxen swan]

A continuous function is one where the pull back of closed sets are closed sets.

Take any closed set S in Y. Look at its pull back T. Consider U = cl(T). f(U) = cl(S) = S. Therefore T = U. Therefore T is closed and we are done.

[acoustic dome]

Hi guys how could I show the converse, $F(U_{\alpha_0\cdots\alpha_p})\cong F(P)$ for all $P \subseteq U_{\alpha_0\cdots\alpha_p}$ (the def of locally constant presheaf)?

[gentle osprey]

shiburin

[opaque cloud]

what's the difference between an immersion and an embedding?

[gritty widget]

Immersions are only locally embeddings.

[rain ether]

an immersion is a local embedding

Klein's bottle can be immersed but not embedded in 3-dimensional space

[magic geyser]

immersions dont have to be injective. even an injective immersion doesnt have to be an embedding bc it doesnt have to be a homeomorphism onto its image

[velvet breach]

It's not injective and surjective....
m right??

actually I'm learning group theory for first time, so it's pretty confusing for me

needs someone help to solve this question

[gritty widget]

#point-set-topology?

[velvet breach]

anyone please

[gaunt linden]

Rakko's point (if a bit tersely made) was that the question off-topic for this channel. Group theory belongs in #groups-rings-fields.

[onyx dust]

Can someone recommend a good resource for homology groups for a student pursuing an Advanced Mathematical Physics Course?

[coarse night]

topological viewpoint?

[winged viper]

If you’re just looking to learn the basics of homology, the last chapter of Lee Topological Manifolds is pretty good. If you are looking for something more comprehensive, Hatcher chapter 2 is okay

[coarse night]

got an interesting problem from point set topology, if anyone wants to try-
Show that there cannot exist a continuous bijection from lR to lR^2. (Not asking for homeomorphism btw).
(Generalize this to show there is no cts bijection from R^n to R^m when n and m are not the same, though it requires knowledge of homology)

[silk ember]

Is X = [0, 1] x {a, b} with euclidean topology on [0,1] and trivial topology on {a, b} compact?

[hidden crag]

The product of two compact spaces is compact

[silk ember]

Oh yes you're right. Didn't think about looking at them individually lol. Thanks

[silk ember]

Wait to be clear. {a, b} is compact because it contains finite many points so that is compact and [0, 1] is compact because it is closed and bounded in R right?

[hidden crag]

Yes

[silk ember]

Awesome thanks for confirming

[unreal stratus]

In fact one way to prove being closed and bounded is equivalent to being compact for subsets of R^n is to show [0,1] is compact directly

[gritty widget]

yeah. Any finite space is compact automatically

no matter the topology of {a, b}

that's why we don't need to ask what topology you mean

[magic geyser]

for R to R^2 what happens when you delete a point from R?

[gritty widget]

the proof doesn't require knowledge of homology tbh

but some theorems like the Brouwer fixed point theorem are probably needed

the proof I am thinking about is probably essentially the same though

[junior adder]

I need help on 2a). Specifically, I think I've proved most parts that it's an embedding, but I haven't been able to prove that the inverse is continuous (or that f is an open/closed mapping). Any help would be appreciated

[magic geyser]

whats an elementary way to show deleting an S^n homeomorph from R^m doesnt disconnect R^m for n<m-1?

smoothly imbedded i can just use transversality

[gritty widget]

Suppose f(x_m) converges to f(x). Then d(x_m, a_n) converges to d(x, a_n) for all n.
d(x, x_m) <= d(x_m, a_n) + d(a_n, x). Choose n so that x is close to a_n

[junior adder]

How does that show that the inverse is continuous? sorry, just having a hard time following

[gritty widget]

What does it mean for the inverse to be continuous?

It means that if f(x_m) converges to f(x), then x_m converges to x

[junior adder]

ohhh ok

[magic geyser]

or non-elementary lol

[junior adder]

can you explain why f(x_m) converging to f(x) implies d(x_m, a_n) converges to d(x, a_n) for all n? is there any specific metric you're using for convergence on [0, 1]^N? I'm thinking of the metric max_{n \in N}{d(a, x_n)}/n but i'm not sure. Thanks!

[gritty widget]

definition of Hilbert cube

it has product topology, the vector f(x_m) converging to f(x) in the Hilbert cube means that each coordinate of f(x_m) converges to each coordinate of f(x)

no need for a particular metric

[sage trail]

Hello
If (X,O) and (Y,O') are topological spaces and f : X -> Y is a function between the two, can we say that f is an homeomorphism if and only if it's an isomorphism between the monoids (O, intersection) and (O', intersection) and between the monoids (O, union) and (O', union) ?

[junior adder]

Sorry to keep using this but now I'm struggling with this component. I've proven s <= d(x, y) but I don't know how to approach the direction d(x, y) <= s

[gritty widget]

Take a_n convering to x. Then |d(x, a_n) - d(y, a_n)| converges to d(x, y).

[junior adder]

do you mind if i ask how the second part follows from a_n converging to x? Thanks and sorry for the inconvenience.

[gritty widget]

From continuity of the metric

[junior adder]

Alright, I get that part now! Thanks! And I'm assuming the limit as n goes to infinity of |d(x, a_n) - d(y, a_n)| is supposed to be the supremum? But how do we know that? Is it a monotone increasing sequence somehow?

[gritty widget]

We aren't taking n to infinity

And what you do is, d(x, y) is a limit of things <= s

So it's <= s

Maybe I wrote it wrong but I didn't mean that a_n converges to x

We take some sequence b_m consisting of elements of the form a_n which converges to x

[junior adder]

oh alright I see now

[cursive flume]

Given I solved the exercise, how do I conclude thus...?

[unreal stratus]

You can characterise topologies in terms of which sets r closed

[cursive flume]

Knowing closure of a set how does this help

[unreal stratus]

And we know whether a set is closes by comparing it to its closure

[cursive flume]

I can define topo in terms of closed sets right,but not by closure

Ah, you mean i take any subset,and if its not its closure,its not in the topology?

[unreal stratus]

Well

Its complement isn't

[gritty widget]

This is saying that first countable spaces are Frechet-Urysohn

[cursive flume]

Yes this is where I encountered it

I am familiar with Frechet Urysohn spaces

But idk why first countable spaces are characterized by sequences

[gritty widget]

https://dantopology.wordpress.com/2010/06/21/sequential-spaces-i/

Look at the proof that first countable implies Frechet-Urysohn

[cursive flume]

This?

[coarse night]

where's the contradiction

[coarse night]

tbh yes, all you need is Sn and Sm aren't homeomorphic

[magic geyser]

where you able to do it using that?

[coarse night]

?

[magic geyser]

you said " all you need is Sn and Sm aren't homeomorphic" which suggests you managed to prove it

[coarse night]

well yes

[magic geyser]

how did you prove it?

[coarse night]

it's the same trick you use for R to R² case

were you able to sole R to R²?

[magic geyser]

yeah you just delete a point

and look at connectedness

[coarse night]

no that's not gonna work

[magic geyser]

ohh yeah ur right i cant get that images are open

so how do you do it?

[coarse night]

you want a hint or the solution itself

[magic geyser]

sure give me a hint

and then the solution depending on whether i can immediately do it from the hint 🙂

[coarse night]

ok so we can show the any [a,b] cannot be homeomorphic to a square [c,d]×[e,f] in R² but we won't know if there is an interval that maps to a square

can you show there must be one such nbd

[magic geyser]

ur asking can i show theres a space filling curve?

[coarse night]

use ||Baire category||

[magic geyser]

im aware there are space filling curves

[coarse night]

that's not one-one

but you can guarantee it'll be

at some nbd

[magic geyser]

ur saying you can make a space filling curve bijective?

maybe can you just tell me how you did it?

[coarse night]

ok 1 moment

[magic geyser]

sorry to make you type. i thought it would be short

[coarse night]

,tex say $f$ is a continuous bijection from $\mbb{R}$ to $\mbb{R}^2$. Then

$$\cup_{n \geq 1} f[-n, n] = \mbb{R}^2$$
But from Baire category, there is one such $n$ s.t. it contains an open ball $B(x, r)$. But now $f$ restricted to $[-n, n]$ is a homeomorphism and $\overline{B}(x, r/2) \subseteq f[-n, n]$. Now look at the preimage of the closed ball, the preimage must be of the form $[a,b]$ as this is a homeo and compact connected. But we know that cannot happen. $\qed$

[coarse night]

I'm on my phone that's why it's taking so long

why it didn't even render

[magic geyser]

ah ok BCT nice

thank you

[coarse night]

render it mentally now

[magic geyser]

i got it

[junior adder]

alright I have one final question. Albeit this may require a pretty comprehensive answer so I don't know if I expect one? I need to prove that the sorgenfrey is completely regular as defined here. It'd be easy if I could just say that the projection of a closed set in the Sorgenfrey plane is closed (in either coordinate) because then I could use the fact that the Sorgenfrey line is normal and therefore completely regular to build a function, but I don't know if I can say the projection is a closed map in this instance. For example, the Sorgenfrey line isn't compact so I can't use that theorem that states that if Y is compact then p_1: X x Y to X is a closed map

[little hemlock]

i think u might be overthinking a bit. Like, let x in R_l and B be a closed set not containing x. Then x has an interval [a,b) disjoint from B and you can just define your function separately on the pieces (-\infty, b), [a,b), and [b, infty)

[gritty widget]

Yes

[opaque cloud]

not topology but fyi you can also do \bR^2 instead of \mbb{R}^2

[coarse night]

yes

[gritty widget]

uh. Totally a preference. Let's not go this far to comment on someones LaTeX

[opaque cloud]

No I meant, it's easier to type since they mentioned they were on their phone (even on a PC i use \b since it's less characters )

[gritty widget]

as kxrider mentioned, [a, b) is clopen

[coarse night]

well I can also do $\R$

[gentle osprey]

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

[coarse night]

ok nvm I thought I had the macro setup

🗿

[gritty widget]

this means that you can define a function which is say, 1 on [a, b) and 0 on the complement

and it'll be continuous

[opaque cloud]

does the lebesgue covering dimension work for any kind of topology? or does it have to be like hausdroff or something for it to work

[gritty widget]

any

and for the Sorgenfrey plane, since product of clopen sets is clopen, we can use the same procedure

[feral copper]

Hi! Can you help me spot what's wrong in the following proof please?

Assume CP² is smoothly embedded in some Rⁿ, n⩾5. Then, because H₄(Rⁿ)=0, we see that C² bounds some 5-manifold. But because the signature of CP² is non-zero, it cannot be cobordant zero, so there cannot be an embedding CP²⊂Rⁿ.
(because obviously, by Whitney embedding, this fails for n=8)

[feral copper]

Oh, right...

How would you go proving that CP² does not embed in R^5 then? x')

Wait, no, your argument does not work, because H_0(R^n)=Z here, with the point as a generator

No, no, what I'm saying is also wrong... 🤦‍♂️

[pastel linden]

If you could embed CP² in R^5, then treating the tangent bundle as a real vector bundle over CP², then the even stiefel whitney classes vanish and the odd classes 2i + 1 are 3Ci mod 2 (so w(CP^2) = 1 + a + a² if a generates H²(CP², Z/2)), and the embedding gives that TCP^2 direct sum NCP^2 is trivial. So by the Whitney product theorem w_i(TCP^2) is the convolution inverse of w_i(NCP^2). (1+a+a²)^-1 = 1 + a, and deg a = 2, so CP² can only be embedded at least in R^6 = R^(2*2 + 2). I think that argument makes sense?

[feral copper]

By means of SW classes I agree that you get this bound indeed! However, I'm supposed not to use them and resort to cobordisms... ^^"

My guess is that I have to use the fact that the signature is an isomorphism from the group of cobordism classes (with disjoint union) to the integers, but I don't know how x)

[odd flame]

is a countable product of spaces necessarily second countable?

if theyre all second countable then obviously yeah

i mean arbitrary spaces

let's say non-empty

[gritty widget]

Uh?

Of course not?

[odd flame]

ok figures i saw a solution for this problem that starts by saying "suppose X is second countable"

wait maybe im reading it incorrectly

does it mean a countable number of dense subsets or dense subsets that are countable

[tardy meadow]

am I stupid or could you just take a space not second countable for the first space then a singleton {0} for the rest or something

[pastel linden]

Dense countable subset

that's what a separable space is

[odd flame]

so it's saying X is a product of separable spaces

fuck the terminology though

bc this isnt separable in the sense of disjoint open sets

[tardy meadow]

separable always means that but it's a bit confusing that a partition into open sets is often called a separation

[odd flame]

any hints on doing it though pls

obvious start is to take the product of dense subsets of each X_i but i doubt that works

[gritty widget]

It won't work for cardinality reasons. The set you get won't be countable.

[odd flame]

pain!

at least i was right about being wrong

brb then

[gritty widget]

But it's a good first guess. Maybe you can modify it to make it countable but still dense.

[odd flame]

ok but does this depend on the topology that X has?

the question seems to be pretty vague about that so idk

[gritty widget]

[hidden crag]

X has the product topology i suppose

[gritty widget]

Products only have one important topology, despite what Munkres might have you believe.

The box topology is genuinely useless.

[odd flame]

[gritty widget]

The box topology literally only exists to motivate the definition of the product topology.

[hidden crag]

lol

[odd flame]

well that's fun

what abt uniform topology?

[gritty widget]

I don't know what that is. Isn't the problem stated in terms of arbitrary topological spaces?

[odd flame]

yeah probably an unrelated consideration, nvm

back to the Q tho, showing it for a product of two spaces should be enough for it follow for countable products by induction right

[gritty widget]

No.

That would only prove it for finite products.

[odd flame]

agh

[gritty widget]

Ordinary induction only proves that some property holds for each natural number.

[odd flame]

ok im gonna guess it has to be a subset of the product of all dense subsets

you said make it countable, so all points in that product of dense subset with some fixed index? that's kinda arbitrary though

[gritty widget]

I can't understand what you mean. Can you write it in mathematical notation?

The bot is dead, so consider drawing it in paint or using something like mathb.in.

[odd flame]

im guessing that the dense subset of X_1 x X_2 x ... is a subset of D_1 x D_2 x ... , where each D_i is dense in X_i

[gritty widget]

That part was clear.

The second part wasn't.

[odd flame]

well you said to make it countable

changing it now: but i was trying to do that by taking all the points in D (will use this for the product of dense subsets) with all but countably many indices fixed

that probably still isnt clear

[gritty widget]

You're taking a countable product, so "with all but countably many indices fixed" is not going to change anything. You just get the original product.

[odd flame]

*finitely many then

[gritty widget]

Sounds better. Is this countable? Dense?

[odd flame]

sorry im kinda just poking around here, lol

[gritty widget]

It smells right.

You can try to prove if it works or not.

[unreal stratus]

False

Claim: sum 1+1+1+... is finite

Proof: if n is finite so is n+1, and 1 is finite

XD

[marble socket]

🙈

[unreal stratus]

Monke

[marble socket]

[rain stratus]

Jesus how do you not go mad after understanding the relations with Physics?

[unreal stratus]

Which relations

[gritty widget]

Relevant.

[rain stratus]

that middle point was truly crazy, I love that I picked up stoicism before going to this side man

and catholicism

I would've gone nuts without God and my admiration of stoics

but so uhm does anyone know if Allen Hatcher's book will have an updated print any soon or should I get the 200X ver?

[marble socket]

nobody i know likes hatcher :p

so i don't plan on reading it anytime soon

moldi once told me to read jp may

[unreal stratus]

ngl i prefer spanier or tom dieck cause it's like

may's concise without saying lots of stuff is easy etc

[marble socket]

oh

[rain stratus]

this information is more useful to me depending on who 'you know', you think you can describe their beliefs a bit?

[rain stratus]

so is Hatcher a bit like Munkres in that it is talkative? I prefer books like Rudin, it makes finding what I have to remember easier

[unreal stratus]

yes Hatcher is very talkative

at least by most people's lights

[rain stratus]

thanks for the heads up, I hate it now 🙂

[unreal stratus]

lmao

[marble socket]

lol

but not just about talkative

a lot of things it does are weird

a prof i know once said something along the lines of 'hatcher pretends that he knows a lot more than what he actually does'

and i was laughing real bad

there was another prof who my friend once reached out to asking about some serre spectral sequence thingy and again his response was to ignore whatever hatcher is doing

at my level, i think i find it annoying that hatcher delays defining categories so much and while doing homology theory sticks to Z for no reason at the start >.<

if the proofs are same, i would rather just keep the coefficient ring to be any R

there are probably other things that people find annoying about it

[trim saffron]

Rudin has a topology book??!

I am just starting to read Munkre's.

It is 🤌🏻

[gritty widget]

Where did they say Rudin has a topology book?

As far as I can tell from looking online, Rudin has never written a book on purely topology. They're all analysis.

[trim saffron]

They didn't say that but I just was guessing.

Also, Rudin's wife was a topologist.

And the proof of Nikiel's Conjecture was devoted in loving memory of her.

[gritty widget]

yes. She worked on Dowker spaces among others

[trim saffron]

lol cats

[gritty widget]

I assumed "Rudin" referred to Walter Rudin as it commonly does.

I didn't know his wife was a mathematician. That's very nice.

[trim saffron]

Yep

[gritty widget]

well. Maybe worked is a wrong word. She came up with a first example of Dowker space

[trim saffron]

And what's a dower space?

[gritty widget]

Dowker space is a normal space such that its product with unit interval is not normal

[marble socket]

weird >.<

[gritty widget]

a space is what surrounds us, darkness and endless freezing embrace of space

[trim saffron]

😆

[marble socket]

you just gave me some existential crisis >.<

[unreal stratus]

I feel like i have only ever stumbled upon papers by his wife lol

[gritty widget]

That's what happens when you spend all your time writing books.

[trim saffron]

RIP W. Rudin

[unreal stratus]

lol

[trim saffron]

sylvain cappell
jalal shatah

Anyone heard of these gents?

[gritty widget]

Well, if I wanted to give you an existential crisis I'd talk about all the different ways in which humanity could eventually but inevitably die
And even very atoms in our bodies will eventually die, because of the ever expanding nature of our universe

[trim saffron]

Oh and my favorites

Very cool mathematicians

Eddie Woo
Blackpen Redpen

lol

[rain stratus]

No, which kind of is the point. Topology doesn't make sense for me unless it's with Functional Analysis or Measure Theory

[marble socket]

lol your definition of space still sounds a lot more scary to me 🙈

maybe because the different ways i would die will happen a lot after i die normally :p

but the thing about darkness and freezing embrace felt like the space has trapped me and i can't escape even if i wanted to >.<

[opaque cloud]

I've been numb to existential crisis for a while

[unreal stratus]

Mood

[haughty cypress]

Hello, would anyone mind sharing a good resource for self research on the topic of Fourier transform please?

[odd flame]

can i get another nudge on this pls i think i got close before but idk

[odd flame]

actually fuck that, what is this shit

is C(I, R) standard notation?

[gritty widget]

Yes.

Maybe not the curly C, but C(X, Y) is common notation for the set of continuous functions from X to Y.

[odd flame]

ohhhh

im not sure i see how that's a subspace of R^I though

or does R^I denote functions from I to R

[gritty widget]

It does.

[void tapir]

i thought the homotopy was called the deformation retract?

[gritty widget]

I think it should be retraction

But we are arguing semantics tbh

[woven sinew]

Yeah, there's a small misuse of terminology. Sometimes people use it about the homotopy, sometimes about the function r.

And if I can bring out an archaic reference, mathworld uses it about the set A: https://mathworld.wolfram.com/DeformationRetract.html

People will know what you mean, no matter which one you use.

[void tapir]

yeah i know, i just want to be precise with my terminology

[void tapir]

yeah fair enough, thanks

[woven sinew]

Good, I too am a fan of precise terminology. 😎

[gritty widget]

I still wouldn't call it a deformation retract

[odd flame]

bump to starting this, not sure the hint is veery enlightening

[gritty widget]

but a deformation retraction

[gritty widget]

use Stone-Weierstrass theorem

[odd flame]

not sure it's presented in munkres? but i'll check

[unreal stratus]

Well you can just use the fact continuous functions on C[0,1] r uniformly continuous

To approximate it using piecewise linear stuff

[gritty widget]

What the hint doesn't say, but it should, is that values of those functions should also be rational

[unreal stratus]

Well rational endpoints ig

[gritty widget]

so that they are defined by giving a finite set of pairs of rationals

[unreal stratus]

So maybe means the nodes are in Q^2

Idk

Ye

[gritty widget]

it's written badly either way

idc

[unreal stratus]

Ye

[gritty widget]

But yeah, this is like, you consider piece-wise linear function given say $f(k/n) = f_k\in \mathbb{Q}$ defined by $f(k/n+t/n) = (1-t)f_k+tf_{k+1}$, $t\in [0, 1]$

[gentle osprey]

Blitz

[gritty widget]

You want to take some continuous g and use uniform continuity to find delta > 0 such that |g(x)-g(y)| <= epsilon for |x-y| <= delta

then take n small enough so that 1/n < delta

choose f_k to be close enough to g(k/n)

this will prove it

[odd flame]

ok im sorry but i dont see this lol

[odd flame]

starting here, that's an odd way to define a function?

[unreal stratus]

You're just saying f(k/n) = f_k and extend linearly

[candid hedge]

the interior of the set of functions that verify : f(x)->0 when x goes to +infinity and f in the space mentioned later

how to even think of this

[unreal stratus]

As a subset of what space

[candid hedge]

the space of continuous functions from [0,+inf] in R

[unreal stratus]

Oof

Is that meant to be +inf)

But yeah

[candid hedge]

yeah

[unreal stratus]

Okay well suppose f is in that subspace. Is f+epsilon in that subspace for any epsilon?

[candid hedge]

yeah

no

[unreal stratus]

No

Ye

So what does that say

About the interior

[candid hedge]

the interior is empty?

[unreal stratus]

Indeed

[candid hedge]

okay if we took the ones where the limit exists

whats the interior of that

i guess everything

[unreal stratus]

I mean have a think like you can do a similar trick

[unreal stratus]

no

[candid hedge]

okay i am gonna think

[unreal stratus]

Can you add smth small to get smth not in the subspace?

[candid hedge]

no

[unreal stratus]

Ye you can

Have a think owo

[candid hedge]

owo?

[unreal stratus]

Lol

Oopsies

[candid hedge]

is that a face? xd

[unreal stratus]

Indeed

[candid hedge]

very nice

mmmm

i mean... any e i am gonna add isnt going to cause my function to diverge to inf, so my only hope is oscillation... but i dont see how

Ah

i see how

[unreal stratus]

Yes this is good

What is the easiest way to add a small oscillation xd

[candid hedge]

the periodic function 0 on most of the domain with an e every once in a while

[unreal stratus]

Yeah sure

I'd just add epsilon sin(x)

For small epsilon

[candid hedge]

yeah

i see

[unreal stratus]

But it works either way

[candid hedge]

okay so empty as well?

[unreal stratus]

Well actually yours isn't continuous

But yes

[candid hedge]

Ah fuck

right

[unreal stratus]

Another way to see it is uh

Well like

If it contained an open ball then since it's a vector space you'd get like

Everything

[candid hedge]

Ah right

[unreal stratus]

I think lol

[candid hedge]

no thats very true

thanks a lot

[unreal stratus]

Nppp

[candid hedge]

if f is a vector space and A is not in f then d(A,f)>0 right?

[odd flame]

can you elaborate on this step pls

[gritty widget]

can you show me some work

[odd flame]

ahhh im handwriting this hw, not latex'ing feel free to hate

im confused by most of it though but i can at least semi follow until that part

[gritty widget]

well, did you try, say, |f_k - g(k/n) | <= epsilon

[odd flame]

ok my first simple confusion is what f_k looks like

[gritty widget]

random rationals

don't worry about it

we don't need to know what they look like

we need to know they exist

and that we have

[odd flame]

ok i mostly got that last one but im now stuck trying to prove that a regular lindelof space is normal

[coarse night]

Prolly a dumb question but I don’t see how we are getting coker i*

[little hemlock]

ker(B_n-1* --> H^n(C; G)) = im(Z_n-1* --> B_n-1*) and coker i*_n-1 = B_n-1*/im( i*_n-1) = B_n-1*/im(Z_n-1* --> B_n-1*)

[bitter smelt]

What’s the idea behind this remark

Nvm, found a SE post

How am I meant to see this as S^2

Text suggests to think of S^2 as R^2 with point at infinity

And the two circles on the left are identified, along with the two circles at the right. So alphas become circles

But how could S^2 - alphas possibly be connected

Any (nondegenerate) closed curve in S^2 will disconnect it

I guess I’m confused on what the handlebodies are meant to be

And how this identification results in S^2

[void tapir]

i agree

retracts are sets and retractions are maps

[coarse night]

got it

[candid hedge]

if A = B then closure of A = closure of B ... no?

A and B are 2 sets

if x is in A_, then there is a seq xn in A that goes to x, and xn is in B too, so the proof is done

yeah nvm

[gritty widget]

you're saying that A = cl(A)

[candid hedge]

no?

how?

A_ is the closure of A

[gritty widget]

you should be the one answering this question

[candid hedge]

I cant

how am i say A_=A if i am saying if A=B then clos(A)=clos(B)

[gritty widget]

listen

when you don't write something correctly, then it's entirely all your fault if people misunderstand you

[candid hedge]

yeah i see that

what did you misunderstand

[gritty widget]

it won't help if you just try to correct yourself based on my advice

you should do that actively

[candid hedge]

yeah but i dont see my mistake

[gritty widget]

were we talking about any mistake?

[candid hedge]

fine if were talking in general i agree

[gritty widget]

you have some thoughts going on in your head
and why would I correct what you write if to know what's in your head I need to know what you write in the first place

and if you write it incorrectly then it's up to you to correct yourself
there's only so much that people can guess

[candid hedge]

I am asking if A=B implies A_=B__

I am not sure what was lost in translation

[gritty widget]

it does

[candid hedge]

but now its obvious

great

thanks

ive been stuck at this for an hour now

if 1/p+1/q=1

[gritty widget]

closure is a function so you're saying if x = y then f(x) = f(y) really, has nothing to do with topology

[candid hedge]

then this inequality hold

[candid hedge]

Ah

[gritty widget]

this again is called Young inequality and has nothing to do with topology

[candid hedge]

It's in my topology chapter

its called holder norm

But ok

[gritty widget]

it has nothing to do with topology

[candid hedge]

i see

nvm then :l

[unborn lotus]

@gritty widget can you not be annoying the entire time you’re trying to help someone

[pearl holly]

metal, this has nothing to do with topology nor anime :/

[unborn lotus]

chrew

[unborn lotus]

there was nothing unclear about this btw dont feel gaslit

[gritty widget]

I did a mistake, my bad

But I'm not "annoying the entire time". It only happens occassionally and I do try this not to happen so get off of me

[candid hedge]

it's alright, Blitz is very helpful and this was a bad quest anw :/

A = {n : n natural} and B = {n+1/n : n natural, n > 1}... how to show those 2 sets are closed

[wispy veldt]

In R i assume?
What definition of "closed" are you working with?
The easy approach is to prove that the set contains all its limit points vacously since we are working in R

But the better approach is to prove the compliment is open.

[candid hedge]

yeah

any definition

you can use the seq one

or that the compliment is open

i am not sure how to move forward

there isnt really any limit points

so we're left with compliment ig?

[wispy veldt]

Well since there is no limit points you kinda get its closed by default cause the set contains the empty set

But the critical approach is to indeed use compliments , can you guess how the compliment of N will look like in R? If you find that out the rest is easy.

Once you figure out the compliment of A in R and prove its open the idea for B is very similar , do give it a try! And ask if you get stuck again

[candid hedge]

the compliment is R without those points