point-set-topology

[unreal stratus]

Sure

[naive hare]

Easy looking at the presentation of the group

[unreal stratus]

"As a space" is a potential abuse tho lol

[naive hare]

🤔

[bright acorn]

You can construct a path connected CW-complex whose fundamental group is the tits group

[unreal stratus]

Unless you are being based and viewing spaces as infinity groupoids of which groups are a special case lol

[naive hare]

Yes

[bright acorn]

K(Tits,1)

[bright acorn]

homotopy hypothesis

[unreal stratus]

Now compute K_0(Tits)

[naive hare]

I think more in the context of honotopy type theory

Where it's more discrete

[unreal stratus]

Hi ryu

[naive hare]

Discrete constructive homotopy theory

S1:Type
base:S1
loop:base=base

One question

What's it called when you take a space then create a new space where the base point is the space and the path space is equal to the space equal to itself

X:Space
f:Space -> Space
base(Y):f(Y)
where
(base(Z)=base(Z))=(Y=Y)

[unreal stratus]

Basepoint the space?

[naive hare]

Yes

I couldn't find anything about it so I just called it bottle because you take a point from one Space to make a new Space.

It's more general than spaces

like you could have a Space K=S1+1
Then you have f(inl(base)) = S1

[coarse night]

hwllo

[coarse night]

btw can you actually do it lol

[naive hare]

$$\text{Bottle}:\prod_{A:\mathcal U}A\to\mathcal U$$

[gentle osprey]

s5

[naive hare]

$$\text{base}_A,a:\text{Bottle}_A,a$$

[gentle osprey]

s5

[coarse night]

[naive hare]

$$p_A,a:\left(a=_Aa\right)\to\left(\text{base}A,a={\text{Bottle}_A,a}\text{base}_A,a\right)$$

[gentle osprey]

s5

[naive hare]

Idk if this works

Tho I would just define it this way

$$\text{Bottle}A,a:\equiv\sum{x:A}|x=_Aa|$$

[gentle osprey]

s5

[unreal stratus]

idk alg k theory lol

[coarse night]

that's topological

K_0(K(Tit, 1))

[dusk heron]

In the book "Riemann Surfaces" by Simon Donaldson, I read the following (here \Sigma_g is an orientable surface of genus g):

How can I visualize this? My guess is that the picture corresponds to g=2, and that each ribbon pair corresponds to adding a handle on a sphere. But I am still having difficulty visualizing it.

[novel acorn]

the picture corresponds to g=4

[dusk heron]

Really? How would g=1 look then?

[novel acorn]

My horrible drawing lol

[dusk heron]

But then the boundary of has two components. How are you supposed to glue that together with the boundary circle of a disk?

Note that the boundary is connected in the figure I sent

[novel acorn]

or
I could be misunderstanding what a ribbon is in this case?

okay yeah I misunderstood

that is the case g=2

[dusk heron]

I am not exactly sure either, but I am pretty sure that in the figure I sent, you have a surface with boundary, and you glue it together with a closed disk, along the boundary

How, exactly, should I visualize the fact that this results in a two-holed torus? I suppose it would be enough to handle the case of g=1

so what I am really wondering is:

Why would that result in an ordinary torus if you were to glue together that and a closed disk, along their boundaries

[dusk heron]

I am guessing it is something like this, but I am still not completely convinced, seeing as a few of the curves get in the same places:

[quick bough]

idk if im being lost, but how does one show this

more like, idk how to use the assumption that U intersected with V is path connected

[dusk heron]

That's not true. Let $X=\mathbb{R}$ and let $U=(0,1)\cup(1,2)$ and $V=(1,2)\cup(2,3)$, so that $U\cap V=(1,2)$ is path-connected. Let $x=3/2$. Then $X'=X=\mathbb{R}$, and $U\cap X'=U=(1,2)\cup(2,3)$ is not path-connected

[gentle osprey]

gustavn64

[dusk heron]

Aha, wait, we assume that $X=U\cup V$...

[gentle osprey]

gustavn64

[quick bough]

yeah

[opaque scroll]

Hmm, we want to show U\capX' is path connected. It's enough to show that every point y has a path from x. We may assume y is not in the intersection. Let p be a path from x to y in X'. Then there is a smallest point t such that p(T) is not in the intersection for any T>t. Then I claim p(T) is not in V.

Thus p([t, 1]) is a path from p(t) to y in U\capX'

[dusk heron]

Why?

[quick bough]

why are assuming that y is not in the intersection?

[opaque scroll]

If y is in the intersection there is nothing to show

Because it's assumed to be path connected

[opaque scroll]

p([t, 1]) is a continuous function into the complement of U\capV

The image of a connected set is connected, so it's either contained in U or V

[quick bough]

im confused, aren't you supposed to take a point in the intersection and find a path from that point to x that lies entirely in the intersection

[opaque scroll]

Y is not in U\capV

It is in U\capX'

[quick bough]

oh, that's what you mean

ok, yeah

[opaque scroll]

There are still some details I haven't filled out, but that should be the basic structure

[quick bough]

why does such a smallest t exist?

[opaque scroll]

You can take infimums of real numbers

[quick bough]

okay, yeah, infimum works

[quick bough]

okay, i don't see how p(T) is not in V

there is a case that i can't seem to prove

[opaque scroll]

So let U' = U\V and V' = V\U. Then U' and V' are disjoint open sets. But the image p[t, 1] is contained in their union.

Thus that make it clear?

[quick bough]

and then what?

you say that p(1) is in U and then the whole image?

cuz i don't see why every point of the image would be either in U' or V'

cant it vary?

[opaque scroll]

The image of a connected set is connected

[quick bough]

oh

okay, you're right

do you even need to take such a smallest t then?

oh yeah, you do, nvm

[quick bough]

actually, can't you just use p([0,1]) is connected instead?

and how exactly do you conclude that it's entirely in U' then

because U' is not necessarily open

right?

[opaque scroll]

U' is open

[quick bough]

why

[opaque scroll]

Because the intersection of two open sets is open

[quick bough]

V is open, so the complement is closed

and U \ V is U intersected with the complement

[opaque scroll]

Right, sorry U' is open in the subspace topology on the complement of U\capV

[quick bough]

is it?

[opaque scroll]

Or if you prefer the preimage of U intersected with [t, 1] is disjoint from the preimage of V intersected with [t, 1] and they are both open

[quick bough]

open in [t, 1]

hm okay

i still don't rlly get where you are using the minimality of t

[opaque scroll]

That is actually the tricky part, which as I'm thinking of it more is trickier than I thought. You need to construct a path from x to p(t)

That's where the minimality was supposed to come in, but the argument I was thinking about doesn't seem to quite work.

[quick bough]

hmm

is there another way of proving this

[opaque scroll]

Idk, Im a bit stumped.

[opaque scroll]

Okay how about this:
Let t be the maximum for which t is not in U. Then t is contained in p^(-1)(V) which is open. Hence the is an h such that p^(t+h) is in V. But p(t+h) must be in U, thus is in the intersection.

Therefore you have a path from x to p(t+h)

[quick bough]

how does that help?

and i think you have to take the supremum, the maximum may not exist, idk if that would do something to the proof

because then t wouldn't be necessarily in p^{-1}(V)

and could you elaborate on the path from x to p(t+h)

[opaque scroll]

So the complement of U is closed. Hence it's preimage has a maximum

Any point not in U is in V, so this point is in the preimage of V.

The path from x to p(t+h) just comes from both being in U\capV, which is path connected

[vocal jay]

Hi, I am trying to prove the following statement. Along the way of attempting to prove this I have tried a range of ideas, one first of all I'd like to clarify (I'm a physics student doing maths in spare time so my foundations are not the strongest in proof) is this: A iff B, is this the same as notA iff notB? I did the truth tables for them and seemed to show they gave the same table, but a simple google search for this seems to show its not a common question for people to ask hence my doubt. I would like to clarify this before latexing up my answer which is based on this conjecture I had

[quick bough]

oh, i see

okay, but don't you have to get a path to an arbitrary y in U \cap X'

[opaque scroll]

How do you mean?

y was arbitrary

[quick bough]

im confused

[opaque scroll]

So p is a path from x to y

[hoary breach]

A iff B is equivalent to not A iff not B, A implies B is equivalent to not B implies not A (the contrapositive)

[quick bough]

yeah

but what about p(t+h) to y

why is that entirely in U \cap X'

[opaque scroll]

Which gives us a path from p(t+h) to y

[quick bough]

oh wait

nvm

by the maximality of t

right?

[opaque scroll]

Yes

[quick bough]

i see

seems right to me

[vocal jay]

thank you 🙂

[hoary breach]

you should still try to verify why that holds, maybe ask for help in #proofs-and-logic if you struggle with that part

[vocal jay]

A simple truth table proof is enough I think, which I have done. Was really confused why there are hardly no google results for not A iff not B equiv to A iff B but we move lol

[grizzled ibex]

If U inter V is path connected then it's contained= X', think about X \ (X') then, i think it'll work.

[quick bough]

i already thought about that tbh

what do i start with X \ (X')

[grizzled ibex]

im talking about the complement cause X' is contained in U union V and X \ X' too, but they are disjoint.

Its just a tip

[quick bough]

i mean its obviously in U union V because it is all of X

no?

[grizzled ibex]

yes its

but you should use that

[quick bough]

hm

[grizzled ibex]

if u started with X' doing U union V minus X' you'd end with a expression which has A intersection X \ X', union B intersection X \ X' which clearly is bad

but you can work with it anyway

[quick bough]

idk what you are trying to tell me

[quick bough]

with that

[grizzled ibex]

X \ X' tells you which points doesn't have a path with x

[quick bough]

yeah

all those points also don't lie in the intersection of U and V

but idrk what to do with that

[grizzled ibex]

(I changed the names for A,B instead of U,V) "A intersection X \ X' also tells you all points which aren't path connected, but those are not in the intersection of A and B, because the intersection is path connected

So they are points that belong only to A, and respectivly only to B.

[quick bough]

im not rlly getting your point

could you formalize it

like why does this hold: "A intersection X \ X' also tells you all points which aren't path connected"

[grizzled ibex]

Because X \ X' are points that aren't path connected, if it lies in the intersection of A, then those points (which are in A too) should share this propety

but like i said, those points belong only to A, not to B, because their intersection is path connected

likewise you can think of that to B

[grizzled ibex]

but you're forgetting another propety which relies on this and the fact that they are open, if im not mistaken

think a bit about it

[bright acorn]

hm

é o prelude que estou pensando que é?

-offtopic-

[vocal jay]

is this a valid proof for the forwards way?

[unreal stratus]

What is your definition of boundary and exterior

[vocal jay]

ill get it for you one second

[unreal stratus]

Np

[vocal jay]

Sorry im all over the place, just justifying the existence of neighbourhoods B_1 and B_2 of p, these are due to part a and b of this:

[unreal stratus]

Well you need to do more case work - it doesn't make sense to take the union of two neighbourhoods when we're only guaranteed that one of them exists

But yes I think negating both statements like you did is a good idea :)

I'd also write it out using words a bit more to help keep the ideas clearer

[vocal jay]

so i neednt take the union of them at all, just the fact that at least one of them is guaranteed to exist is the existence of the neighbourhood which satisfies the RHS of the statement?

[unreal stratus]

Yeah exactly

[vocal jay]

thats perfect, thank you for your help!

[unreal stratus]

Well also you wrote the negation of the 2nd statement slightly incorrectly in words but in symbols it is correct

Saying 'there exists a neighbourhood which doesn't contain a point in A or a point in X\A' sounds like you're saying the neighbourhood has empty intersection with A and empty intersection with X\A

I'd write it as like "There's a nbhd of p which is disjoint from A or X \ A" or "which doesn't intersect both A and X \ A"

[vocal jay]

I hate to admit I spent a good few hours today trying to negate it before i realised i should just represent it in symbols and do that. I have one more question, Since A is a subset of X, then each of the steps I have used are reversible and so the <= backwards direction also follows?

[unreal stratus]

Yeah I agree

[vocal jay]

Awesome

What do you use topology for? Are you a pure maths student?

[unreal stratus]

Are you asking ahout me personally or about what use topology has in general? xd

[vocal jay]

personally yeah

[unreal stratus]

Oh to me topology is sort of the end application, like the problems it studies are interesting in themselves

But there are cool interactions with geometry and algebra too

And yeah I'm a pure maths student

Hbu

[vocal jay]

im a theoretical physicist who just realised my uni hasnt taught me enough pure maths so that i can study general relativity or quantum mechanics properly so I'm teaching myself Topological manifolds and Differential geometry to fill in the gaps lol

[unreal stratus]

Ah noice

Yeah I used to be a physics student and that book you're reading was one of the things that started making me wanna do maths aha

But yes I hope you enjoy reading that

[vocal jay]

Its been a long time since i've done any pure maths so the theorem, proof lemma stuff is taking some getting used to. Very nice book so far even if I am 30 pages in haha

[unreal stratus]

Yeah it definitely takes time getting used to and can be painful initially for sure

Was for me lol

[vocal jay]

Do you have any intentions on using this sort of stuff for anything physics related?

[unreal stratus]

Not really, I'm potentially doing some GR next term and it'd be fun to read about stuff but not got any plans

Not really sure how physics interacts with topology other than through geometry or very complicated stuff xd

[vocal jay]

My current goal is working towards reading this: geometry topology and physics by Nakahara. But I envy you getting the opportunity to do pure maths, reading a book is nice but I wish my uni would just let me transfer onto maths, there are opportunities next year and I'll probably make the transition

[quick bough]

what happens to the mobius strip if we remove an interior point

it’s not homotopically equivalent to S^1 anymore, right?

[steel glen]

probably easier to see what happens if you take the construction from a square and gluing the top with the bottom backwards

then remove a point

[unreal stratus]

hope that goes welll :)

Yeah i think the geometric picture with the Möbius strip minus a point is like

You can imagine like drawing a lil line part way along the strip (away from the hole) and stretching the paper away so you just have the edges of the paper and that lil line

Now you can actually untwist the "wires" at the edges of the paper and just get a circle with that little line joining two points

[steel glen]

yeah. the idea is similar if you take the frame of a square and glue the top to the -bottom

[unreal stratus]

Yeah that's what I originally did lol cut and pasting

or (just a rephrasing) we can just kill the 2-cell and we're left with a graph which we can easily redraw

[steel glen]

nice

i wish i jumped at complexes more often

[thorny agate]

Stuck on the second part of the proposition, specifically showing that the map is open

Ok so

I mean all I have is the following

Let $U \subseteq Y \setminus A$ be open and $q$ the quotient $X \sqcup Y \to X \cup_f Y$. Then $q\mid_Y(U) = q(U) \subseteq X \cup_f Y$ is open if and only if $q^{-1}(q(U))$ is open in $X \sqcup U$ which is true if and only if $q^{-1}(q(U)) \cap X$ and $q^{-1}(q(U)) \cap Y$ are both open.

[gentle osprey]

spamakin

[thorny agate]

but I'm not sure how to show either

[steel glen]

aren't the equivalence classes in Y\A exactly the points in Y\A. you should be able to express any open set U in the disjoint union of X and Y as
(U n Y) u (U n Y^c) where the preimage of U n Y^c is empty, and U n Y is the intersection of open sets

[thorny agate]

aren't the equivalence classes in Y\A exactly the points in Y\A.
?

[steel glen]

the equivalence classes of Y \ A as a subspace of X u_f Y

i don't think i was clear enough
if you take some y in Y \ A, the image under the quotient map should just be y (or [y]), unless im missing something

[steel glen]

i think maybe i misunderstood. are you stuck on showing the map is an open map, or that the subspace is an open subspace?

the canonnical map should just act as a sort of identity map given what i said above, so it being a homeomorphism to its image (and so open) shouldn't be too far fetched

[thorny agate]

showing the map is an open map

[thorny agate]

oh because the only possibility is 1. in the list of possible relations

[steel glen]

do you know what the visual intuition is for this space?

[thorny agate]

not at all

[steel glen]

you're attaching the points of A to their image in X

[thorny agate]

I haven't found anything and this book does not have a picture

right right

[steel glen]

so it should make sense that q(Y\A), or q(U) for any open subset of Y\A should be "untouched" per se

[steel glen]

showing that amounts to this yeah

[thorny agate]

ok I think that clears things up

[steel glen]

if you ever do AT, this is a pretty important space

[thorny agate]

I plan to one day

but things like this, the mapping cylinder / cone

very unintuitive rn

[steel glen]

ye i feel that. at least you're getting going though, it's good

gl with the rest of your topology

[quick bough]

i dont seenot

see it*

doesn’t the point just basically “stay”
there

[cosmic socket]

One way to think about mapping cylinder for a map f:X ->Y is that it is a space that deformation retracts to Y, so replacing f by the inclusion X -> Mf is basically the same as remembering f. And furthermore this inclusion is well behaved (cofibration)

[steel glen]

well the idea is that you can deformation retract the 2-cell so as to just get the frame of the square

[steel glen]

it's what potato was saying earlier

if you don't know about cell complexes then don't worry about the cell thing

but essentially you can remove the interior points of the square and just attach the top to the bottom (flipped)

[trail charm]

the connected sum is an example of a disjoint union space, which helped me get some visual intuition for it

https://mathworld.wolfram.com/ConnectedSum.html

more generally, i like to picture the disjoint union X + Y as sort of combining X and Y into the same “universe” such that they’re disjoint, then you attach them by identifying points from one with their image in the other

here’s my attempt at drawing that idea

[quick bough]

i’ll need a picture for that haha, what does the circle with a little line joining two points look like

or more like i don’t get how to see that

[quick bough]

nvm, i think i see it

how does the homotopy look like is the question

[shadow charm]

Got this q in exam and ended up saying false but I’m having doubts: it asked if an n dimensional manifold with H1(M)=0 was necessarily oriented. I know this is true in the compact connected setting by PD, but I couldn’t think of a counterexample in the general case

[coarse night]

H_1 or H^1?

[shadow charm]

H_1

Mb

[coarse night]

yeah guessed

hint: Oriented double cover gives a subgroup of index 2

@shadow charm

[shadow charm]

Fuck okay yeah thought so

Darn

[coarse night]

[shadow charm]

It’s the one part of the course I didn’t really revise so it didn’t come to mind until after

[feral copper]

Wait, what about \pi_1 being a perfect group? (i.e. with trivial abelianization)
This only gives a subgroup of the fundamental group, not the H_1, right?

[coarse night]

if your M is not orientable you get a subgroup of index 2 if π_1, so abelianization can't be zero

i.e. π_1 is not perfect

[dusk heron]

I now see it! The picture I sent is clearly the same thing as the rightmost thing in the following image, which is of course just a punctured torus:

[feral copper]

The group G=SL(2,Z/5), which has order 6, has a 3-Sylow with index two. But G is perfect

Unless I'm missing something?

I think to prove the statement, it's better to say that M is orientable iff w_1(M) is zero, if you have this available

[dusk heron]

Doesn't SL(2,Z/5) have order 120?

[feral copper]

Really I don't know how to compute?
$#\mathrm{SL}(n,{\bf F}_q)=\frac{(q^n-1)\times(q^n-q)\times\cdots\times(q^n-q^{n-1})}{q-1}$

[gentle osprey]

matplotlib

[feral copper]

Yeah I forgot the last factor x')

[dusk heron]

Anyway, what is the subgroup of index 2?

[feral copper]

But I still don't see why having a subgroup of index two prevents from having trivial abelianization... Although that's #groups-rings-fields and not #point-set-topology anymore, and it's been a while since I last did groups xD

[unreal stratus]

Commutator is smallest normal subgroup with abelian quotient

If you have a subgroup of index 2 it is normal with abelian quotient

[feral copper]

Right... Sorry...

[coarse night]

@feral copper still have doubt?

[feral copper]

No no I was being a moron, it's obvious now

[coarse night]

[coarse night]

hmm this suggests that H^1 = 0 then it's also orientable

have to think thoughly

so w_1(M) \in H^1(M; Z/2) and by uct we have H^1(M; Z/2) = 0 as H_0 is torison free

ok idk UCT for H^1(M; Z) to H^1(M; Z/2), above statement is not correct

[feral copper]

Uct for cohomology uses Ext and not Tor

What is Ext(Z,Z/2)? Zero too I guess?

[shadow charm]

Yeah but Ext1 of an fg abelian groupll be it’s torsion subgroup

[feral copper]

Well whatever, I'm dumb when it comes to those things so I'll take that away

[shadow charm]

Isn’t Ext(Z,Z/2) just 0

[feral copper]

It is

But:

I'm dumb

[shadow charm]

Nah dw about it we all make mistakes

[shadow charm]

I mean H^1 = 0 should imply H_1 = 0 no?

(Over Z)

[coarse night]

not necessarily

[shadow charm]

Why not by UCT?

[coarse night]

H_1=Z/2

[shadow charm]

Right im a fool

[feral copper]

Like RP²?

This has H^1=0

[shadow charm]

Yeah lol

[coarse night]

H^n= F_n ⊕ T_{n-1}

[coarse night]

yes, Z projective. I didn't see the msg lol. IG H¹ also zero then

[novel acorn]

Ext^1?

[empty grove]

Ext without degree always refers to Ext^1

[novel acorn]

ah

[empty grove]

It was defined before the higher Exts

[tiny ridge]

the higher exts are irrelevant for fg abelian groups anyhow

[novel acorn]

Ext(A,B) is zero when A is projective or B is injective right

[tiny ridge]

yes

[novel acorn]

Goood
I am learning homological algebra well it seems

[opaque scroll]

This is even a characterizing property of being projective/injective. I.e. A is projective iff Ext(A, -) = 0.

[bright acorn]

If $X$ and $Y$ are topological spaces, what is the coarsest topology on the space of all functions between $X$ and $Y$ (denoted $Y^{X}$) such that the subspace of all continuous functions $\mathcal{C}(X,Y) \subseteq Y^{X}$ is closed?

[gentle osprey]

mistersystem

[bright acorn]

Can you explicitly characterize this topology in terms of a (sub)basis?

For some reason, I think the compact-open topology might be the one having this property.

[opaque scroll]

I mean you can define a topology where that set is the only closed one

Besides the empty and whole set

[coarse night]

coarsest would be { \empty, complement of continuous functions, all of Y^X}

that's not the desired answer i suppose

[bright acorn]

oh, ofc lol

[unreal stratus]

Seeing Y^X contrasted with C(X,Y) is making my brain melt

[coarse night]

so what are the restrictions on the topology

[coarse night]

lol true

[bright acorn]

i was trying to think of a characterization of the compact-open topology in terms of like

[coarse night]

(it's the same if you're working in Top)

[bright acorn]

it being the notion of convergence of functions with the least amount of requirements, but still retaining the property such that the limit of continuous functions is still continuous.

appearently

the compact-open topology is the coarsest topology making the evaluation map continuous

[coarse night]

there is one characterization for LCH spaces, you probably already know this. Z → Y^X cts iff X×Z →Y is

[bright acorn]

lmao

that's the notation my prof is using rn

[lavish geyser]

hi can someone prove this

i cant do caculus/topology its too rigorous

[coarse night]

don't then

[lavish geyser]

bruh

😭

[coarse night]

[unreal stratus]

||by definition||

[lavish geyser]

no this is metric space

in metric space closed set is when lim S=S

[bright acorn]

I was about to ask "what about spaces which are not locally compact tho?"

but fuck it

every space is locally compact

functional analysis is a hoax

[coarse night]

and Hausdroff

[quiet thorn]

You don't even need metric space structure to prove that statement

[bright acorn]

they probably defined closed sets in terms of sequences

for topological spaces, that only works for first-countable ones

[lavish geyser]

topology is too rigorous

who invented topology

[bright acorn]

while the "closed iff complement is open" is the most general definition

[coarse night]

are you sure about that

some cat theorist might come in

[opaque scroll]

Take a sequence in S converging to x in M\S. Use the existence of an open ball around x in M\S to get a contradiction

[unreal stratus]

oh lol defining in terms of sequences and stuff is interesting lol

[lavish geyser]

o

tysm

im so dumb

[queen prism]

just think more

[limber wren]

Suppose S is closed. Take x in M\S. If every open ball around x intersected S then x would lie in S (because x would be the limit point of a sequence in S, which is closed), so M\S must be open.

Conversely if M\S is open, and (x_i)->x is a sequence in S, then x in M \ S would imply some x_i lie in M \ S (a contradiction), so x \in S and thus S is closed.

[vocal jay]

Is this a valid proof: correction to last line A subset of B

Where my books definition of closure of A is:

[queen prism]

intersection of all closed sets containing A

[vocal jay]

Oh yeah

Lol

[balmy field]

Can't you just do this: The intersection contains A since it is the intersection of a bunch of sets that contain A, and it is contained in A since A is closed and hence in the intersection?

[unreal stratus]

Yes

[vocal jay]

Is there an easier way to prove the forward direction than this?

[unreal stratus]

,rotate cw

[unreal stratus]

Contraposition is a bit nicer.

[vocal jay]

hehe just finished doing that just now bro

hahaha

[unreal stratus]

Suppose there's a non-empty open set U contained in X \ A

Oh sorry lol

[vocal jay]

so many times in this book ive found contrapositive much easier

[vocal jay]

nooo you didnt know, but thanks for confirming

[unreal stratus]

Yeah I think the way i'd think about heuristically is like

Being given a concrete open set to work with is nicer than something about "every" thing

Often a useful thing to think about ig is how much data you are given to prove stuff with lol

[queen prism]

happens a lot in intro top I find (I’m still bad tho)

[unreal stratus]

Oh yeah and I think another bit of intuition for this is like

Often you will be proving statements about closed and open stuff

And it's easier to make sure that everything is in terms of open sets (or all about closed sets)

So for example you can change "every non-empty open subset of X contains a point of A" to "no proper closed subset of X contains A"

And I think it's obvious that "A is dense" is equivalent to that last thing

:)

[vocal jay]

Yes this was much easier to prove

Finally glad I made it past page 25 😂

[coarse night]

last line supposed to be N(x) ∩ A= empty

[lavish geyser]

i dont understand

what if a closed set doesnt have a preimage?

i mean what if there exists a point y in a closed set such that no point maps to y?

[gritty widget]

The empty set is closed

[lavish geyser]

o

ok thanks

[lavish geyser]

a path can self intersect right

so C \ {0} only has two paths up to homotopy

[balmy field]

Hint for this might be the pre image of the complement is the complement of the preimage

[lavish geyser]

and paths which "loop" around 0 are still homotopic to one of the two paths

[lavish geyser]

o

i mean for any two points

then theeres only 2 paths?

[vocal jay]

Yessir

[feral copper]

Non-orientable surfaces in a 4-manifold X deserve a special mention. Since they do not give rise to integral homology classes, ...
Hi! Could someone elaborate on this statement please? Does the author mean they are null-homologous, or that there's no such thing as the integral homology class of a NO surface? (Here, X is closed, connected, oriented)

[vocal jay]

Hehe I just did this one in my book, it’s a fun one. Have a think about how to construct a closed set for every open set

[limpid fern]

that's literally the definition of closed...

[vocal jay]

This works right?

[limpid fern]

yep

this looks like a lee exercise

are you using lee?

[vocal jay]

yessir

[umbral panther]

The latter
A homology class is a closed chain. A chain is a sum of oriented simplicies

[limber wren]

if f: X -> Y is a map and A is a subset of Y then f^{-1}(A) always exists, but it could be empty (if nothing in X maps into A)

[vocal jay]

Is this a valid proof?

sorry this one is better formatted

[coarse night]

in the "not continuous" part, you need to mention A, K are open

[vocal jay]

i did that in the previous "preface" didnt want to mention it too much, but yes they are open

otherwise is it okay?

[coarse night]

Mention open everywhere, otherwise sounds wrong

[vocal jay]

😄

[steel glen]

that’s not sufficient

what you showed is that there is an open set with open preimage, that doesn’t contradict f being not continuous

do you know what the subspace topology is?

[vocal jay]

oh yeah

is that the induced topology?

[steel glen]

i think so

what you should be doing is taking an arbitrary open set in the subspace A

and showing the preimage of the set under f|_A is open

given the definition of the subspace topology it should be straightforward

oh sorry i had this backwards

take an arbitrary open set in Y

and show the preimage is open in A

[vocal jay]

I see, I'll give it another go, and yes i realised it was meant to be the other way. Thanks for recognizing the fault in my proof 🙂

[balmy field]

Hint: pre image intersect A is open in the sub space topology

[vocal jay]

@steel glen hope you dont mind if i ping you, i think i got it. About to LaTex it up, do you reckon you could check it out for me?

[limpid fern]

if you post it other people can check it

[limber wren]

You could also just prove the map i_A: A -> X is continuous which is a bit easier, and then notice that f restricted to A is just f composed with i_A

[vocal jay]

I was thinking about this, but my book hasnt yet proved that the composition of two continuous maps is continuous so i tried to do it without that

It has proved that the identity map is continuous though ofc

[limber wren]

gotcha, you could probably prove it quick, it's actually pretty trivial to prove composition preserves continuity! But yeah, you can still prove it directly, just that proving i_A is continous is almost just repeating the definition of the subspace topology

[vocal jay]

it is the very next theorem to prove so I will prove the composition theorem and then reprove that result i just did to get a more concise one for my notes. Thanks a lot 🙂

[limber wren]

No worries! You're proof is equally good though, just an alternative way to think about it

[vocal jay]

Did you read john Lee's book for topological manifolds by any chance? That is where this question is from although it is obviously a very general theorem

I'm finding it takes me like a day atm to get through 2 pages, not that i'm spending all my time doing it but I feel very slow, then again I haven't done pure maths in years

[steel glen]

the choice of P for an open set leaves something to be desired haha

[limber wren]

lol that's normal, I started reading some algebraic geometry after taking a 6 year break from math after my phd (a completely new subject for me) and it's been about 1-2 pages a day and doing exercises

[vocal jay]

What was your phd in if you dont mind me asking

[tidal lynx]

probably in math if I had to guess

[limber wren]

Also for your proof, just for the sake of terseness, you could just write "since f_A^{-1}(U) = A \intersect f^{-1}(U) for any subset, f_A is continuous by definition of the subspace topology"

[limber wren]

I did it in analysis on compact matrix groups, basically doing integral calculus on matrix groups like U_n and SU_n

[vocal jay]

lmao i didnt even know you could do analysis on them

[limber wren]

algebra and representation theory surprisingly

[vocal jay]

I need to learn a bunch of representation theory so i can do particle physics in my 4th year i think

Ahhh either way so much i want to learn but not enough time

[limber wren]

yeah, doing analysis or probability on matrix groups almost always comes down to linear algebra and representation theory, because integrating over a group is like a projection operator in a way. If you don't have some sort of rigid structure like in group theory though, it's really hard to get practical formulas for integrals, the group structure is what actually enables you to come up with specific formulas you can calculate, even though they're really complicated

[vocal jay]

does this have anything to do with fibre bundles?

I remember watching this lecturer mention something about how bundles are something to do with a bigger space and integration can be done on them? Idk maybe i'm being silly

[limber wren]

not really, I mean the underlying abstract theory probably does, but when it comes to actually calculating integrals over groups, it's a lot of combinatorics, linear algebra, and representation theory. Actually one of the surprising results is that the most important structure related to calculating integrals over groups of matrices is the symmetric group, and the representation theory of the symmetric group is actually all combinatorics and drawing diagrams, there's almost no analysis involved

[vocal jay]

Lmao bet you enjoyed doing your art if there were that many diagrams, thanks for telling me about your PhD, it’s very useful getting an idea of what’s possible as someone who’s hoping to try and get one

[limber wren]

doing the diagrams was so much fun, I remember having to figure out how to make this diagram, it was really satisfying

[vocal jay]

what even is that phahaha#

[limber wren]

lol the diagram on the left is called the young graph, it's an infinite graph which connected diagrams differing by a single block, and it completely describes the representation theory of S_n for all n. The other stuff is weird and has to do with describing basis elements of each irreducible, lol it's weird stuff!

it took me a year of studying to understand how it all works and fits together

[vocal jay]

Didn’t really bother to latex it up but I think I got the right lines

Obviously I should have justified f inverse being continuous but this follows from Id on x being the composition of f and its inverse and we know Id on x is continuous

Actually I didn’t even need to do that ID_x is continuous by just setting the A in ID_A equal to X and since the continuity of ID_A has already been shown we get everything follows. Ahhh I’m so cluttered

[limber wren]

Why would U \subset A or A \subset X \ A for an arbitrary open set U?

[vocal jay]

Yeah I see, Ahhh idk I’m now imaging U could be an open set that included the boundary of A and X\A

I’m not too sure

Either way U is open so we can write it as the union of open sets, some of these may lie in A or in X\A, if they are in A then they correspond to a point in X, else they correspond to the empty set and hence ID_A is still continuous

?

I’ll write this up later or tomorrow

[limber wren]

it's more straightforward than that, i_A^{-1}(U) = A \intersect U, so it's basically a 1 line proof using the definition of the subspace topology

[feral copper]

Right thanks! Sorry for this silly question, but I thought I was thinking nonsense!

[queen prism]

so why is anime in the channel description here

[steel glen]

where else would they put it

[stable kite]

So, exp and the power functions share the property that they are a covering map on the punctured plane. Now I'm wondering whether this fact is true for any non-constant holomorphic function, that is, does the following actually hold:
If f is a non-constant holomorphic on domain U and A the discrete set of points where the derivative is zero, then f : U\A -> f(U\A) is a covering map.

I strongly assume that's wrong, but haven't found a counterexample yet

[stable kite]

Read about branched coverings on the Covering spaces article on Wikpedia and that one seems to suggest a variant of this is true for non-constant holomorphic maps between compact Riemann Surfaces, which appear to always be branched coverings

[novel talon]

Can someone clarify a bit a detail regarding complex projective line CP¹? I'm not sure if this belongs here, but it involves a quotient and a topology so at least it's close. I was playing around with it and defined the usual cover given by the sets U = {[x] | x_0 \ne 0} and V = {[x] | x_1 \ne 0} and then I found online that these are equivalent to something denoted by {[z : 1] | z \in C} = {w \ne 0} and {[1 : w] | w \in C} = {z \ne 0} and I'm not sure I understand how are these the same? These are supposedly called the homogeneous coordinates.

[gritty widget]

Can you not just translate a power function by a constant to get a counterexample?

[gritty widget]

Do you understand the definition of the complex projective line and the meaning of the homogeneous coordinates?

[stable kite]

How do you mean that? 🤔

[gritty widget]

e.g. f(z) = z^2 + 1

[novel talon]

I understand it as a quotient of S¹, but I don't understand these homogeneous coordinates

I saw that they are related to the different construction of the projective space where we fix a line at y = 1 I think and somehow normalize points on the plane to that line?

[gritty widget]

So homogeneous coordinates $[z_0 : z_1] $ define a set of equivalence classes, where $ [z_0 : z_1] ~ [\lambda z_0 : \lambda z_1] $ for all $ \lambda \in \mathbb{C} \setminus {0}$

If you play around with this you should be able to see that it describes the circle definition you're familiar with

[novel talon]

I can see this also the equivalence relation just relates points on the same line, but this [z : 1] and [1 : w] are weird to me. What is the deal with 1?

[opaque scroll]

[x:y] = [x/y : 1] whenever y is not 0

[novel talon]

I see they are of course on the same line. Is this common to normalize always the second entry to 1 if it's non-zero?

[novel acorn]

It's usually the first or last that you normalize

[opaque scroll]

Yeah, so you would usually think of the projective line as one copy of the normal line [z:1], and one point at infinity [1:0]

Exactly which entry you fix to 1 is not super important, since it's all symmetric

[quiet thorn]

With only 2 entries, you kind of just have a usual ratio so this is a pretty easy exercise

[novel talon]

Thanks for this. If it's not too much I would also like to know why on U \cap V we now have a relationship w = 1/z? If I think about this the intersection is just where neither w or z is zero so it should be isomorphic to C^*.

[opaque scroll]

When does [w:1] = [1:z]?

[novel talon]

Thanks(:

[abstract copper]

Is this the right place for questions about point-set topology? I'm working out of the book by Munkres

[balmy field]

Yes

[abstract copper]

I'm looking for some help for some help understanding this question (2.17.9).

Particularly the phrase "in the space X x Y". I'm aware that a topological space is a set along with its topology. So, in this case that would be {T, X x Y} (And I guess T is the product topology).

Where particularly is closure(A x B) and cl(A) x cl(B) living?

Are they in the subspace topology inherited from X x Y?

[empty grove]

They are both subsets of X × Y. You have to check that they are equal as subsets.

The closures of A and B are the closures in X and Y respectively

The closure of A × B is being taken in X × Y (under the product topology as you inferred)

As for how you are supposed to infer all of this from the problem statement alone, look at which spaces the 3 sets live in, and unless explicitly stated otherwise, always interpret closure of A to be its closure in the largest space mentioned in the problem that contains it

@abstract copper

[abstract copper]

So, in this case I'm considering cl(A x B) in X x Y and cl(A) in X and cl(B) in Y?

[empty grove]

Yes

[abstract copper]

Got it, thanks alot!

[lament steppe]

So Im trying to understand the back of the book solution to a problem im working on that is:
"If R is a compact region in a surface M, prove that R is a closed set of M"

Solution:

The notation with the U_{p,q} and U_{q,p} is something I dont understand.

Im not sure if this solution is essentially the same thing as a proof to the Hein-Borel theorm or if there is something new to learn here.

[opaque scroll]

For every pair of points p and q, you can find an open neighborhood U of p and V of q, such that U and V don't intersect. Here they are calling the neighborhoods U_p and U_p,q.

[lament steppe]

Im willing to accept that, but I guess then what is U_q,p?

[opaque scroll]

Seems to be a typo

[umbral panther]

It depends on p. You could call it V_p, but you can’t call it U_q

[opaque scroll]

U_p, q and U_q,p should be the same

[lament steppe]

Thanks all! I appreciate it

[vocal jay]

This question right here, i've proved the function is bijective

but showing a is continuous has proved very confusing for me

I understand both X and S^1 should have the induced topologies from their respective open ball topologies in R and C respectively, but then its the half open interval thats making it weird for me to try and describe any open interval in S^1

I have managed to do a construction where I can take any general open set in X and show the mapping of it under a produces an open set in S^1, but idk where to go from there

[unreal stratus]

Uh continuity is about the preimage not the image

[unreal stratus]

Also, this is false

Consider the image of [0,1/2)

[vocal jay]

But in the induced topology on S^1 this can be given by an intersection of S^1

nvm yeah you're right

[unreal stratus]

You can check that like the induced tpology on S^1 is just the topology induced by the metric

So the open sets are the intuitively open ones

[vocal jay]

I really don't know how to communicate or even write up my doubts with this question, That inclusion of 0 is annoying me

[unreal stratus]

Yeah the 0 is the annoying bit for sure

[vocal jay]

Let me draw a picture to show you my doubts

To me this needs an interval from X of the form [0,a) U (b,1)

To map onto it

[unreal stratus]

Yup sure

And [0,a) u (b,1) is open

so it's fine

[vocal jay]

so i argued that any sequence a_1, a_2,... a_n can be used to construct open intervals in X that will map to open intervals in S^1. Furthermore if you want to include that weird portion of S^1 demonstrated above, we would also need (a_1,a_2,) U (a_3,a_4) U...U [0,a) for some 0<a<=1

to me though it seems like there should be a simpler proof of continuity

[unreal stratus]

You just need to check the preimages of "small" open intervals in S^1

Like a basis of open sets in S^1

And you can write down the preimages of those hopefully without much difficulty

[vocal jay]

This theorem then?

I guess I am over complicating it yeah

[unreal stratus]

Yeah sure

[vocal jay]

If I can check the pre image of a tiny open set in S^1 then that will ofc have a pre image that is open I’ve showed that already. And then I just can use that in conjunction with the union of an arbitrary number of open sets is an open set and hence continuity of the map follows

[vocal jay]

yeah this gives it

[tidal lynx]

it’s kinda crazy to me how those hold but for a generic map f (not necessarily an inverse ) but intersections are only preserved when f is injective, or else one is a subset of the other

I guess the point is that inverses are well behaved in that they’re always injective

not crazy nvm

[queen prism]

preimages don’t work like images, that’s why

https://www.reddit.com/r/math/comments/9u00ku/are_there_intuitive_reasons_for_preimages_of/

[vocal jay]

Alright this seems to prove the continuity and fills in all my doubts I had regarding those weird half open sets in R but they are open in X

[calm ravine]

Hi, I struggle to prove that for a compact topological manifold the singular homology is of finite type (let say on Z).
If we have a differential manifold, I know how to conclude with the existence of a finite good cover. However this existence is related to a riemannian structure and doesn't pass to the topological case.

I have tried to use things like cellular approximation (in Hatcher) but it seems that an approximation of a compact manifold is not a finite CW-complex, so it is unclear that its homology is of finite type.

Finally I may have found this article on nlab
https://ncatlab.org/nlab/show/good+open+cover
with a theorem from osborne-stern which seems to imply with mayer vietoris that all homologies in degree <n-1 are finite. By adding the fact that the n-th homology is finite, it seems I can conclude that the n-1 homology is also finite. But this proof is very complicated.

So is there an easier argument I am missing ? Thanks in advance

[unreal stratus]

Hm well one way is that every compact manifold is homotopy equivalent to a finite cw complex

But here's an easier way

https://math.stackexchange.com/q/263064/932674

[gritty widget]

what are the accumulation point for [0,infty)

[calm ravine]

Oh I may have forgotten this theorem but it would definitely answer my question yes !

[limpid fern]

what do you think they are

recall the definition

[gritty widget]

is it empty ?

[unreal stratus]

Far from it

[gritty widget]

is it the set itself ?

[limpid fern]

Read the definition again

It should become clear

[gritty widget]

well i've read the definition

[limpid fern]

let x be a point in [0, infty)

is it an accumulation point?

[gritty widget]

yes

because every intersection excluding the x between ]x - epsilon,x +epsilon [ and A is different from 0

is it right?

oke thank you

how can i prove that (0,infty) is the interiors of [0,infty)

[quiet thorn]

I think you mean interior, and also review its definition

[gritty widget]

the proof goes like supposing that there is x in A and then applying the definition?

[limpid fern]

intuitively, The interior of a set is the largest open set contained in the set

so have a think

[gritty widget]

(0,infty)

[limpid fern]

Yeah it is true but again you're asking how to prove it

[gritty widget]

i don't know how to do that

[limpid fern]

what does your text use for the definition of interior

[gritty widget]

there exists delta such as (x-delta,x+delta) in A

[limpid fern]

this seems to be specific to R

[limpid fern]

so a point x is in the interior if this is true

I guess you know how to do it then

[gritty widget]

i don't

[limpid fern]

Just show double inclusion

show that the interior is contained by (0, infty)

and then (0,infty) is contained by the interior

[unreal stratus]

0 is also in the interior but yeah

[gritty widget]

is it

[unreal stratus]

Wait uhhh

I'm confused by the context ĺol

Interior as a subset of R?

Then no 0 isn't in it

[limpid fern]

Yeah it wasnt taken as the space itself

[gritty widget]

x is in the interior of A , then there exist delta such as (x-delta,x+delta) i A

from here i take delta = 1/n?

[limpid fern]

this is a real analysis textbook right

you just find the delta for each x then if you use this definition sure

[gritty widget]

the say that at infty 1/n=0 and x in A

[limpid fern]

Sorry I don't understand what you're saying

Are you asking me if it's correct to take delta = 1/n to prove that there is an open neighborhood for every x in (0, infty)?

[gritty widget]

yes

and then consider x-1/n as a sequence

[limpid fern]

Sounds like you're overcomplicating the approach?

there is an easier delta to take that doesn't require you to find a sequence

[gritty widget]

what is it?

[limpid fern]

That's for you to find out isn't it?

[gritty widget]

does it have anything to do with Archimedean property?

[limpid fern]

No actually

Your idea of using the archimedean property could work but seems very overkill here

[gritty widget]

i don't think i know the delta you're talking about

[broken nacelle]

[limpid fern]

let's suppose you drew a circle on your paper with a radius of r

what's an easy way to draw a smaller circle within that circle

[broken nacelle]

just take delta = x/2 or smth

hurb

[echo oyster]

I'm facing a bit problem while proving some space is connected.

Say R2 \ {(0,0)}
If it's not connected then it can be written as union of two disjoint open subsets, say A U B
Now I take a point in A say (1,0), then any nbd of (1,0) should lie entirely within A
After this I'm not sure how do I associate it with B

[unreal stratus]

I assume you already know about path connectedness - I'd recommend you consider that

Also, it's false that "any neighbourhood of (1,0) should lie entirely within A"

Indeed the entire space is a neighbourhood of (1,0)

[coarse night]

make it "connected nbd"

[echo oyster]

Yes with path connected it can be joined with a straight line if it doesn't go through origin else take another point not the origin and join with two lines

[echo oyster]

Its not clear to me how do I show in this approach, that if the set can be written as union of two disjoint open sets , then one of them must be empty

[coarse night]

maybe pick a point from the boundary of A

[craggy mauve]

Que: show that (0,(n-1)/n) is a open cover of (0,1)?
Ans: y_n=(n-1)/n is a increasing sequence which converges to 1. let x belongs to (0,1) there exist N such that x<y_N so x belongs to
(0,(N-1)/N) and since x is arbitrary hence proved
Is this correct?

[queen prism]

just justify why such an N exists

[craggy mauve]

Oh okay thanks !

[stable kite]

Is closed and nowhere dense equivalent to closed and discrete?

[queen prism]

consider the unit circle in R^2

[stable kite]

Ah that makes sense, of course. Thank you!

[obtuse meteor]

For this approach I would try to see why there can be no clopen sets besides the whole space and the empty set

This is equivalent and I think much easier

The key idea here in my mind is to exploit the idea of radius along with some “induction” on real numbers

Intuitively you could keep expanding any clopen set

Well radius is not quite right, but half planes is better

A more explicit hint: ||Let A be nonempty and clopen. Pick (x,y) in A. WLOG, x > 0, we will show all things in Right half plane are in A. Consider r > 0, R > 0 s.t the closed square box starting from x’ >= r and of lengths R on all sides lies in A. Expanding this in R,r is pretty easy….||

[hoary breach]

we know the fact that a collection of open sets is a basis if and only if given any open set and any point in the open set there is a basis member that contains the point and is itself contained in the said open set. Now, consider R^n with the euclidean topology. Given any open set in R^n and any point, we can find a ball of rational radius around a 'rational point' that contains the point x and is itself contained in the said open set. This gives us a countable topological basis for the euclidean topology in R^n. My question here is regarding the cardinality of the euclidean topology. By mapping each point in R^n to any open ball (which is just an open set w.r.t. euclidean metric) around it we get |R^n| \leq cardinality of the euclidean topology (call it T from now on). I think we can show that |T| = |R^n| = |R| = |2^aleph_nought| and for this we need an injection from T to R^n or from T to 2^(aleph_nought) and invoke cantor schroder bernstein to complete the proof. I don't think i'm getting anywhere right now, any hints?

[unreal stratus]

you're right about |R| <= |T| sure

[hoary breach]

i am trying to map open sets to points in R^n right now but i haven't been able to figure out the details to how the injection should look like

[empty grove]

Try enumerating your basis elements

And writing an open set as a union of basis elements

[hoary breach]

hmm so since we have a countable basis, we can write any open set as an at most countable union of basis elements but i can't figure out how to get separate points in R^n like this for the open sets

[empty grove]

The enumeration of the basis assigns each open subset of ℝ^n a subset of ℕ

[unreal stratus]

Just to check, if $X,Y$ are pointed simplicial sets, is their smash product just given by $(X \land Y)_n = \frac{X_n \times Y_n}{* \times Y_n \cup X_n \times *}$?

[gentle osprey]

potato

[unreal stratus]

where on the right the quotient is just of sets ofc

Wait no this may not make sense as the basepoint lies in X0 or Y0

[plain raven]

yes but it's conceivable there is a different definition that also works and is weak htpy equivalent

it makes sense. just phrase it categorically and you'll be fine

[unreal stratus]

Sure so like pullback of X -> pt <- Y ?

[plain raven]

it's more complicated than that.

that would be the ordinary product.

[empty grove]

Yes, when you go from a cat to its pointed version, the new monoidal product is given by the old one quotiented by the natural inclusion of the coproduct. All of these operations are taken pointwise for simplicial sets so this happens pointwise

[plain raven]

smash product is not monoidal i think. it doesn't have a unit?

[unreal stratus]

Okay sure rippies

[empty grove]

Going to the pointed version can be described as taking the under-cat of the terminal object

[umbral panther]

It might be better to say *_n rather than *, but *_n is a singleton

[empty grove]

Doesn't Δ⁰ ⊔ Δ⁰ work?

[plain raven]

ok yeah.

vool

i wrote this up a minute ago lemme see if i can find it.

ok here's my treatment of this

let C be a monoidal category with finite colimits which play nice with the monoidal product.

let I be a monoid in C

an object A is augmented by I if there are maps A->I and I->A so that I is a retract of A

here for smash products we take I=1
A morphism of augmented objects A->B should commute with the retraction maps.

Given (A, eta_A: I->A, epsilon_A : A->I) and (B, eta_B, epsilon_B)

write f for
(A\otimes I) +(I\otimes B) -> A\otimes B given by 1_A \otimes eta_B + eta_A\otimes 1_B

[hoary breach]

Let ${B}{k\in \mathbb{N}}$ be the countable topological basis for the euclidean topology $T$ on $\mathbb{R}^{n}$ and suppose $G$ is an open subset. Then, write $G$ as an at most countable union $G = \cup{a\in A}B_{a}$ where $A$ is the subset of $\mathbb{N}$ with its members the indices $a$ for which the basis member $B_{a}$ is contained in $G$ and forms a cover of $G$. This way, we form the mapping $T \to \mathscr{P}(\mathbb{N})$, $G \mapsto A\coloneqq {n\in \mathbb{N} \mid B_{n}\subseteq G}$. Is this the idea? I believe the injectivity of said mapping is trivial to verify and therefore this completes the proof?

[gentle osprey]

pikapikapikapikachu

[unreal stratus]

Yeah nice

Or you can phrase it as a surjection P(N) -> T which to me is slightly more natural but yeah

[empty grove]

Yes, just keep in mind the subtlety that the set A is not unique for a given G, so you begin the proof by fixing a choice of A for each G

[unreal stratus]

A is well-defined with that last definition right?

[empty grove]

Oh ok you are taking all basis elements contained in G

Yeah mb misread

Or rather didn't bother reading properly

[unreal stratus]

lol

kinda funny how looking up smash product in SSet is mostly giving me stuff about spectra

But that is what I am learning about it for

[hoary breach]

Hmm, thanks potato and Moldilocks1337

[hidden crag]

are you learning about simplicial homotopy theory potato

[empty grove]

Are you learning about spectra of simplicial sets?

Oh is this for Hovey Shipley Smith?

[unreal stratus]

I am learning about spectra and quasicategories basically

This in particular was for symmetric spectra

[empty grove]

But where in symmetric spectra do you need simplicial sets

Are you doing it all quasicategorically?

[plain raven]

write g for (A\otimes I) +(I\otimes B) -> I\otimes I -> I given by the epsilon maps and the multiplication of I

[unreal stratus]

Well Schwede develops a simplicial variant of symmetric spectra alongside the one based on Top

For convenience I think mostly, not read much of it yet lol

[plain raven]

write A\overline{\otimes} B for the pushout of f and g

[empty grove]

Ok yeah I think that's the version that the Hovey Shipley Smith paper does

[plain raven]

this is the reduced tensor product.

[unreal stratus]

Ah okay

[plain raven]

I'll draw a pic lol

[empty grove]

There is a more categorical way of defining this smash product btw, if you follow that you won't have to worry about what the symmetric spectra are made of

[empty grove]

[unreal stratus]

Which smash product do you mean?

Like on the stable infty cat of spectra

[empty grove]

The smash product of symmetric/orthogonal spectra. You can view it as an instance of Day convolution, which is a general recipe for producing closed symmetric monoidal products on categories of enriched presheaves given symmetric monoidal structures on their domains. You can view sequential/symmetric/orthogonal spectra as topological presheaves on certain categories but the cat for sequential ones doesn't have a symmetric product.

[empty grove]

On the model category level rather than only infinity cats

[plain raven]

this seems to give a reasonably workable definition of the smash product

[empty grove]

Oof everything you typed is now all scattered

[coarse night]

gravestone engravings

[unreal stratus]

Oh okay very nice I will look at that

[plain raven]

this also works for the reduced join

[unreal stratus]

I heard of that from the nlab article lol

[plain raven]

of pointed spaces

[empty grove]

Ye seems legit

Btw how do you take products of V-enriched categories? Is the monoidal product (where you tensor the hom objects) the product?

It doesn't seem like it should be the product

Because A ⊗ B may not be the product of A and B in V

Do you take direct products of hom sets instead?

[plain raven]

the objects in the product category are ordered pairs of objects (A, B)
Homs are tensor products of hom objects

hm wait. does that make sense? let me think.

[empty grove]

I find it surprising that that works as the product

If it does

[plain raven]

well i believe you may just not have cartesian product projections in the enriched setting.

[empty grove]

Yeah

[plain raven]

this necessitates the use of ends and coends rather than ordinary limits and colimits.

[empty grove]

Oh no I'm treating V-Cat as an ordinary cat

[plain raven]

i know but they're related

like Ab-cat has a terminal object

but that terminal object doesn't have the property that functors 1 -> C correspond to objects in C, i. e. constant functors

the terminal object is obviously the empty cartesian product so i am just pointing out that the weirdness of the product structure is related to the fact that limits behave poorly

[empty grove]

Why does the terminal object matter here though? To take the underlying ordinary cat of the Ab-cat Ab-Cat you should apply Ab(ℤ, -) right?

So shouldn't we look at maps from ℤ

I goofed

Though I still don't get what you were trying to say

[plain raven]

Let A and B be Abelian categories and F an additive functor between them. The limit of F should be a certain constant functor x: A -> B together with a cone x->F.
It is not obvious what a constant functor is in this setting though. in Cat you just say "it's a functor that factors uniquely through the terminal object". In Ab cat that isn't really a sensible approach.

[empty grove]

But that is a limit inside an Ab-cat

[plain raven]

if b is an object in B it is not clear what "the" constant functor from A to b is.

[empty grove]

I meant product of 2 separate Ab-cats

[plain raven]

Yes, I'm understanding you. I am bringing up an interesting and closely related fact, that when the monoidal product in V-Cat is a Cartesian product, ends and coends can be reduced to limits and colimits. It's somehow a special property of V-categories where V has a Cartesian monoidal product. for example V= CG-Haus

[empty grove]

ohh

So it isn't always true

[plain raven]

I don't think so, no. The natural monoidal structure on V-Cat isn't always Cartesian, I think.

For example in Ab-Cat i think you'd take the tensor product of the hom groups.

[empty grove]

That would make sense. I saw some notes which referred to that tensor product as the product and got confused.

I should read Kelley's book

[plain raven]

Hom((A, B), (C, D)) =Hom(A, C) \otimes Hom(B, D)

is what i think you would do

it's the cartesian product on objects lol

close enough

[empty grove]

Lol

[hoary breach]

(replace sigma with S)
Here's my idea at the proof for this theorem:

First step to this proof is to construct a 'nice' topology on $X$ that contains $S$.
Since $\mathscr{P}(X)$ is a topology on $X$ that contains $S$, we can set $T(S) = \bigcap_{\mathcal{A}\supseteq S}\mathcal{A}$, where the intersection is taken over all the topologies $\mathcal{A}$ on $X$ that contain $S$. It is trivial to verify that $T(S)$ is a topology on $X$ that contains $S$ itself.

Next, we will prove that this construction is the smallest topology on $X$ that contains $S$:

That $T(S)$ is the smallest topology that contains $S$ follows from the fact that if $E$ is another topology on $X$ that contains $S$ then it is included in the intersection used in the definition of $T(S)$ and therefore, it contains $T(S)$ itself.

Next we will show that the topology generated by $S$ coincides with the family of specified sets (call this set $K$):

Since $T(S)$ contains $S$ and it is a topology, it contains finite intersection of members of $S$ and arbitrary unions of these finite intersections. Therefore $T(S) \supseteq K$.

The proof should be complete once we show that $K$ is itself a topology on $X$ that contains $S$ (we can conclude due to minimality of $T(S)`$):

Since $K$ contains finite intersection of members of $S$, it contains members of $S$ themselves and therefore $K\supseteq S$. (I believe) All Members of $K$ are arbitrary unions of finite intersections of members of $S$. Finite intersection of arbitrary unions of finite intersections of members of $S$ is an arbitrary union of finite intersections of members of $S$ so $K$ is closed under finite intersections. Since arbitrary unions of arbitrary unions of finite intersections of members of $S$ are simply arbitrary unions of finite intersections of members of $S$, $K$ is closed under arbitrary unions.

Could someone verify this proof for me? Also, how do I go about proving for uniqueness?

[gentle osprey]

pikapikapikapikachu

[gritty widget]

ain't reading all that sorry for you or glad that happened

[brittle rapids]

if C and D are two smallest topologies, then C ⊆ D and D ⊆ C, such that C = D

i can't parse the generating part, but i think that's the general idea

like you can distribute things

[hoary breach]

oh that was it for the uniqueness? lol

[brittle rapids]

ya

[hoary breach]

ok, and do you think the overall idea in the proof is right?

[brittle rapids]

have you shown already that the intersection of topologies is a topology?

[hoary breach]

yep, trivial

[brittle rapids]

ok it's fine then i think

[hoary breach]

alright thanks

[brittle rapids]

btw the "set itself has this property, and intersection of subsets with this property have this property" thing happens a lot

and whenever that happens you always have this notion of a "smallest thing with this property that contains S"

[hoary breach]

the proof in the book was a bit too efficient for me to grasp lol hence the elaborate solution

[brittle rapids]

lol

this is a sketch of the proof ig

[hoary breach]

hmm, thankfully i was somewhat familiar with a similar idea regarding sigma algebras otherwise i was a goner

[brittle rapids]

indeed sigma algebras also have this property

[hidden crag]

This can be made rigorous with some logic stuff iirc

[unreal stratus]

I mean it's all small checks you could add details too if you wanted

[hoary breach]

huh?

[unreal stratus]

Wdym

[hoary breach]

what is timo referring to?

[brittle rapids]

if J is upwards-directed, and J -F-> Top is a functor, then there is a canonical map lim F -> ∏_{j∈J}F(j). is this always an embedding?

[novel acorn]

I mean the map exists in the case when J is discrete
But I don't think it's true generally

hm thinking about it now I'm maybe starting to doubt myself
but let me think about it

[brittle rapids]

i don't even know if it has to do with directedness

[novel acorn]

https://ncatlab.org/nlab/show/limit#ConstructionFromProductsAndEqualizers

nlab says if the product exists then lim F is a subobject of it

[brittle rapids]

oh nice

but subobject means injective here right

not necessarily an embedding

[novel acorn]

I believe in Top it means subspace

[opaque scroll]

In general when products exists the limit is an equilizer between such products. So in Top it should be a subspace of the product yes. Don't think you need upwards directed anywhere.

[plain raven]

it's monic

in top, it's a subspace

yes it's an embedding

[brittle rapids]

wait what

[plain raven]

sorry lol i know that was confusing

[brittle rapids]

embed [0, 1) into R^2 via t ↦ e^{2πit}

[plain raven]

the map is always a subspace embedding.

ok

[brittle rapids]

this is continuous injective but not an embedding

whoops i wrote 'embed'

[plain raven]

ok i thought you were talking about the limit of a categorical diagram

[brittle rapids]

oh, does that imply embedding

[plain raven]

yes

[brittle rapids]

h m m

why

i thought that as well, i think it probably works

[unreal stratus]

can't we construct the limit explicitly as a subspace of the product?

[brittle rapids]

but idk if the topology there is the one that's induced

oh wait it is

something about equalizers

[unreal stratus]

Yeah I mean explicitly here we have $S \subseteq \prod_{j \in J} X_j$ the set of sequence $(x_j){j \in J} \in \prod{j \in J}X_j$ coherent in the obvious sense and then any collection of maps into the $X_j$ induces a unique map into $\prod_{j \in J}X_j$, which lands in (corestricts to) $S$ if that collection of maps is `coherent'

[gentle osprey]

potato

[brittle rapids]

what's S

[unreal stratus]

This defines S

And I'm showing that it is the limit of the functor you were talking about

[brittle rapids]

i don't understand what's happening

what is a coherent sequence

[unreal stratus]

Oh I mean sequences compatible with all the maps between the X_j

[brittle rapids]

oh

[unreal stratus]

I mean really I'm taking a standard construction of the limit in Set

and saying that it still has the desired universal property in Top if we view it as a subspace of the product

[brittle rapids]

alright i'm convinced

i wouldn't mind the abstract nonsense explanation either

oh thank you btw

[plain raven]

yes

and all equalizers in top are subspace embeddings

[patent path]

When someone says the topology induced on $C([0,1],X)$ from $X$ what topology do they mean? The compact-open topology?

[gentle osprey]

strugglinggeometer

[patent path]

Related question, what do continuous maps from $[0,1] \to C([0,1],X)$ look like?

[gentle osprey]

strugglinggeometer

[empty grove]

Yes

[empty grove]

You can view a set function of that kind as a set function [0,1]² → X via currying. The former set function is continuous if and only if the latter is.

[patent path]

Thanks!

Are there any nice theorems about solutions of differential equations which have path homotopies between them?

[novel acorn]

Can't it also mean the weak topology?

[empty grove]

I thought C(X, Y) always means compact open unless stated otherwise

Though "topology induced by X" could mean that, idk

[woeful locust]

Hi ! I am a physics student and I am seeking some help to understand two model derived from the Hopf fibration.
So, the Hopf fibration is a space of homotopically nontrivial mappings (maybe I need to explanation already here xD) of spheres (S^3 -> S^2).
From this, we've build 1 model defined by an Hamilonian (see first picture)

The second model is similar but the two complex numbers z_up and z_down are changed as : z↑ = cos(kx)+isin(kx)sin(ky), z↓ = sin(kx)sin(kz )+i(−cos(2kx)+cos(ky)+cos(kz )−5/2).
In the article that describes this new model says : "This Hamiltonian resembles the ones introduced in [12] and [13] for one half of the Brillouin zone (k1 ≤ 0) with its point-reflected partner on the other half (k1 ≥ 0). An associated Pontryagin manifold of its ground state is shown in Fig. 4. Although the ground state map is not constant on the subtorus defined by k1 = 0, the figure shows that the Chern number of its restriction vanishes (the preimage for k1 = 0 is the empty set)"
And I think I really don't get that... I can provide the article in DM if needed 😄

[unreal stratus]

The Hopf fibration isn't a space, it's a map of spaces

[woeful locust]

Thank you for this clarification 😄

[late iron]

Is the orientable two-sheeted covering of a manifold M unique up to homeomorphism?

[feral copper]

Yes, even up to covering isomorphism, as it is given by a universal property
But note it's not necessarily the only non-trivial 2-fold covering

https://math.stackexchange.com/q/1478646/259363

[naive hare]

there are 2 possible semidirect products of the space of 2 points and the circle.

[unreal stratus]

What is a semidirect product of spaces?

[naive hare]

S1⋉2 has 2 solutions,
S1⋉2 = 2S1
S1⋉2 = S1

[naive hare]

I don't know

I'm coming from group theory

Also S1⋉S1 has 2 solutions as well.
S1⋉S1 = S1^2
S1⋉S1 = Klein bottle

[naive hare]

Pretty sure it's a fiber bundle

But I don't know the formal deffinition of a fiber bundle

Can you help?

[onyx raft]

Question from hatcher:

For every reflection A and B doesn't there exist a rotation R such that AR=B?

And since rotations have deg $1$ they are homotopic to id and so $A=A\circ id \sim AR=B$

[gentle osprey]

kaehler

[naive hare]

well a double reflection isn't a reflection

[marble anchor]

Weird question: the statement that a space is simply connected basically means that we can assign to each loop in the space some homotopy contracting the loop to a point. If we can do this, must there exist a continuous such assignment of homotopies to loops?

It's spooky to think that there might be a space that say, ZFC proves is simply connected, but where the homotopies contracting the loops aren't computable or something, and so the choices perhaps can't be made continuously.

[opaque scroll]

At least if you work with pointed spaces I guess the answer should be no. Since if I'm interpreting you correctly that would mean the loop space is contractible, which is not true in general.

Guess it's less restrictive if you consider free loop spaces, but my guess you still be no.

[marble anchor]

Hmm, my guess is no as well, but I'm a little surprised about the loop space becoming contractible... How do we see that?

You could get a continuous map from the loop space to the original space by taking the centres of contraction of the homotopies that are assigned to each loop...

But those are allowed to move around the original space, so long as they do so in a way which is continuous in the choice of loop

[opaque scroll]

For pointed spaces, the base point should always remain fixed. So each loop would be contracting to the base point. So it is a bit more restrictive

[marble anchor]

Oh, right, I didn't think about the pointedness properly

Hmm

Actually if you have a handy example of a pointed simply connected space whose (pointed) loop space isn't contractible, that would be interesting.

[opaque scroll]

S^2

In general $\pi_k(\Omega X) = \pi_{k+1}(X)$, so if the homotopy groups of $X$ doesnt vanish, then neitehr does those of the loop space

[gentle osprey]

jagr2808

[marble anchor]

Ah, good

Okay

(and, thinking about that example at a lower level, while you could get a continuous choice of homotopies for all the paths that don't go through the antipode of the base point for instance, it wouldn't have a continuous extension to those which do - there's no continuous way to make choices about which way around the sphere you'll pull the circle of great circles.)

[rapid lagoon]

Hi guys, I am rather confused on this problem

Specifically, on 3(f)

I do not understand what is meant by writing reals numbers as a union of non empty, disjoint open sets in S

Am I kind of "retro" contructing the reals? By first defining equivalence classes, then showing that its a field by defining the usual operations on the non-empty disjoing open sets in S?

[queen prism]

it means writing the set R as such a union

[rapid lagoon]

Oh wow

[queen prism]

the reals exist, but there’s a lot of ways to group them into sets

[rapid lagoon]

Thanks! I think this is easy, just take (-infty, a) U [a, infty) for any real number a

Both are open in S, both are nonempty and aredisjoint

thanks, btw

[urban zinc]

Let X be the set of positive integers regarded as points on the x-axis in R^2; let C'X denote the subspace of R^2 obtained by joining each (n,0) in X to (0,1) with a line segment. There is a continuous bijection from CX to C'X, but not a homeomorphism
Man, I get why this is true, but I hate this

[queen prism]

what is CX?

[urban zinc]

Cone over X (XxI/Xx{1})

[queen prism]

o

[coarse night]

you mean (1/n, 0)?

[gaunt linden]

(n,0) works.

Some of the line segments get rather flat slopes, but so what.

[urban zinc]

I mean (n,0)

[coarse night]

oh okay

[cosmic beacon]

Been reading through these notes on the stable homotopy category: https://people.math.binghamton.edu/malkiewich/stable.pdf

Can someone explain what the constructions these two bullet points are suggesting are?

[novel acorn]

wait

were you asking

[cosmic beacon]

I read those lol, I’m asking what the E*-algebra structure on E*(X) is

[novel acorn]

OK

I misunderstood ur question lol

[cosmic beacon]

Np lol

[cosmic beacon]

(E* is shorthand for the coefficient ring pi_*(E) = [S,E]_*)

I don’t see how to show ~E_*(Y) is a graded E*-algebra, and I don’t understand what the sentence “in particular, E*(X) is a graded E*-algebra” means (does it mean X is a space, not a spectra here?)

Furthermore how do I see that E*X is skew-commutative, using only the axiomatic construction of the stable homotopy category given in the notes up until this point?

[unreal stratus]

Given by the map X -> pt

[cosmic beacon]

Ahh thanks

[unreal stratus]

Wait ye that was OK ig

[cosmic beacon]

That makes sense

[cosmic beacon]

Ok two questions:

1. I understand given an unbased CW complex X and a ring spectra E we have a natural map

E*=E*(pt) —> E*(X)

induced by the map X —> pt. Explicitly though, what are the graded ring structures on E*(pt) and E*(X), and how do we know the arrow E* —> E*(X) is a morphism of graded rings and not just graded abelian groups?

Here’s my attempt at constructing the graded ring structure on E*(X), if you could tell me if it’s right: Let p,q >= 0, we want a bilinear map

E^p(X) x E^q(X) —> E^{p+q}(X).

Unraveling definitions, we want a map

[Σ∞X+,E]_{-p} x [Σ∞X+,E]_{-q} —> [Σ∞X+,E]_{-p-q}.

I can define a composition

[Σ∞X+,E]_{-p} x [Σ∞X+,E]_{-q} —> [Σ∞X+ ∧ Σ∞X+, E ∧ E]_{-p-q} —> [Σ∞X+ ∧ Σ∞X+, E]_{-p-q} —> [Σ∞X+,E]_{-p-q}

where the first arrow is induced by the monoidal (smash) product, the second arrow is induced by post-composition with the product E ∧ E —> E, and the last arrow is induced by a diagonal map

Σ∞X+ —> Σ∞X+ ∧ Σ∞X+

But I’m not sure about this since
a) is this composition actually bilinear? (in particular, does the smash product give an additive monoidal product on hoSpectra), and
b) does my supposed diagonal map necessarily exist, given the axiomatic construction of hoSpectra in those notes so far?

2. In Adams blue book in the construction of the ASS, it asks for a ring spectrum E and a spectrum Y such that “E_*(Y) is a projective pi_*E module”. How does the (unreduced) homology theory obtain an E*-module structure?

[hoary sphinx]

I have a question about the usefulness of generalizing metric spaces to topological spaces... What type of topology is useful somewhere else in math, which is not the standard topology of a metric space?

[novel acorn]

there are more than I could count lol

there's the product topology, the weak topology, the compact open topology, the cofinite topology the profinite topology etc.

all of these are extremely useful in their field of math

[hoary sphinx]

Maybe you could briefly explain one?

[unreal stratus]

For example many topological spaces occuring in algebra won't come from metrics I guess

[hoary sphinx]

Because when I think of topology I think of distances so it's not at all intuitive to me

[novel acorn]

Zariski topology my beloved

[novel acorn]

Topology is very far separated from distances

metric spaces are probably the most ideal possible topological space you could have

so they're not very representative

[unreal stratus]

Topological spaces are interesting spaces in their own right to many people and not having to carry around a metric gives a lot of freedom

Like, even when you're working with metric spaces, you often (presumably usually?) care about spaces merely being homeomorphic rather than isometric

And widening your scope to topological spaces allows you to do more constructions - as a small, example you can view a circle as just a closed interval with the endpoints stuck to one another. You wouldn't really be able to write that down as a metric space in the same way

[queen prism]

neighborhoods

[novel acorn]

like take the product topology \
Here we put a topology on the product of topological spaces $\prod_{\alpha \in A} X_\alpha$\
Now this product has a map ($\pi_\alpha: \prod_{\alpha\in A}X_\alpha \to X_\alpha$) that projects the product down to individual factors\
now the product topology on this new space is the topology consisting of unions of finite intersections of sets of the form $\pi^{-1}\alpha(U)$\ with $U \subset X\alpha$ open in $X_\alpha$

[gentle osprey]

ironyincarnate

[novel acorn]

essentially this is the coarsest topology such that the projection maps are all continuous

[queen prism]

i think even if you only work with metric spaces it can be easier to prove results using properties of open sets rather than metrics

[hoary sphinx]

Ok yeah this is interesting I see why you'd use that

[umbral panther]

The infinite product of intervals is called the Hilbert cube because it was originally discovered inside Hilbert space. [0,1]x[0,1/2]x[0,1/3]x…

So it is a metric space. But you want to talk about it topological properties like compactness and the categorical property of being a product

[hoary sphinx]

Would you say that in general useful topological spaces codify the concept of closeness (by closeness I mean two points being close together) or are there other useful topological spaces that codify some other structure?

[trail charm]

you can also define nets in topological spaces

which are analogous to sequences but have some neat properties which dont have analogous constructions in a metric space

e.g. universal nets (i think)

[pastel pier]

Yes, basically a topology endows a set with the notion of "neighborhoods" so that, in particular, it becomes meaningful to assign information that is "local" (through sheaves, essentially)

[hoary sphinx]

So for instance the topology of sets of the form (a,inf) what kind of structure induces?

Or is it just some pathological topology derived from the generality of the concept of a topological space?

[pastel pier]

What do you mean by (a,inf)?

[hoary sphinx]

Intervals

{x in R: x>a}

[brittle rapids]

how is that pathological

[pastel pier]

Since open sets need to be closed under finite intersections, one can obtain the open balls (a,b) from (a,inf) \cap (b,inf), so in the end it's the same topology as the one induced by the metric

[brittle rapids]

uh

[hoary sphinx]

What I mean is that it doesn't correspond to the concept of closeness one is used to

[brittle rapids]

yea well there are metrics you can put on R that also don't "correspond to the concept of closeness one is used to"

[hoary sphinx]

So I was wondering why it would be useful apart from being a curiosity

[hoary sphinx]

Yeah but that is just kind of redefining which things are close to one-another

But the concept of closeness is the same in a sense

[pastel pier]

The whole subject of Algebraic Geometry relies on a kind of topology (Zariski topology) that cannot be obtained as the topology defined by a metric

[gaunt linden]

Even if all we had were metric spaces, homeomorphisms between them are obviously important -- and it's natural to attempt to pin down exactly which structure it is that homeomorphisms preserve. Clearly they don't preserve distances. But once it turns out they do exactly preserve open sets (or closed sets, or interior/closure operators, or neighborhood filters), it's not a large step to wish to characterize stuff in terms of systems of open sets (etc) rather than in terms of distances, so one can be sure the stuff in question is preserved by homeomorphisms. For that one needs some axioms to start reasoning from, and pretty soon we're building a theory of topological spaces. The fact that it's not obvious how metric-space-like the axioms should be is evidenced by the plethora of different separation axioms that sometimes and sometimes not need to be assumed.

[plain quiver]

I can give some applied examples of non-metrizable topologies which are extremely useful, if you're still interested. Theoretical computer science is filled with them. But it also seems like your question has been answered

[abstract copper]

I'm having a hard time proving the product of two closures is closed in the bigger space

A subset X
B subset Y

I want to show that
cl(A) x cl(B)
is a closed set in space X x Y.

Clearly, cl(A) is closed in X by definition, and cl(B) is closed in Y by def..... So it makes sense that their product is closed. I'm sure I'm forgetting something obvious, but I'm at a loss on how to prove it...

[queen prism]

show that its complement is open? idk

[hidden crag]

Write cl(A) and cl(B) as complements of open sets

and try to go from there

[limber wren]

a natural question would be "is the product of the closures the closure of the product?"

[abstract copper]

I'm actually working on a sub-problem of proving this

[limber wren]

gotcha!

[limber wren]

For any sets A \subset X and B \subset Y you can write (A x B)^c = (A^c x Y) \union (X x B^c). If A and B are both closed, then this is the union of two open sets so (A x B)^c would be open.

and closures are closed obviously

[hidden crag]

that's what i was hinting at

[abstract copper]

^c denotes compliments?

[limber wren]

yah

[abstract copper]

(X x Y) - (A x B) =
((X - A) x Y) U (X x (Y - B))

(X - A) is open if A is closed
(Y - B) is open if B is closed
A and B are closed
So the union is open

Since (X x Y) - (A x B) is open,
(A x B) is closed

[limber wren]

exactly, assuming A and B are closed.

lol I realize using the same A and B as in your question is probably a bad idea, but basically the product of closed sets is closed is enough to prove what you need