point-set-topology

[stuck geyser]

also this is Rudin so i believe it best goes in #real-complex-analysis

[quartz horizon]

Sure..

Still, I worry it’ll just confuse you more

[stuck geyser]

If we introduce a different definition of "open" it'll be more confusing because now we have two definitions

[hollow geyser]

what we talking about in here?

[stuck geyser]

Intro chapter of rudin

How to prove (0,1) is open via the interior "every point is the center of a ball within the interval" definition

[hollow geyser]

Ah.

Who's needing the help? hyzae?

Oh my. Y'all been on this for a while I see.

[rigid path]

Hi I have a question

Let $f:X\to Y$ be a continuous function without any restriction on $X$ and $Y$. Is there a clever way to show that the diagonal function $id_X\Delta f$ is an homeomorphic embedding? I mean without proving that it is a bijection and bicontinuous

[gentle osprey]

Milo

[rigid path]

An homeomorphic embedding in the graph of f obviously

[umbral panther]

There is a map from X to XxY by the universal property of the product. It factors through the graph. There is a map from XxY to X by projection. Thus there are maps both ways between X and the graph

[hollow geyser]

I finally finished my Hausdorff chapter of my topo book.

I just need to make sure this last problem looks good.

And here's the Problem 11.8 I referenced

Am I applying it correctly? Does 11.10 look good?

[unreal stratus]

Yes

[unreal stratus]

Yes nice, though you can also do this directly from the diagonal being closed

Maybe I am too much of a fan of that fact aha

But nah yours is very fast

[hollow geyser]

Wow I'd have no idea how to do it from that

[unreal stratus]

Try to view it as the preimage of the diagonal under a continuous map

[hollow geyser]

C is the preimage?

[unreal stratus]

Yeah

[hollow geyser]

what do you mean by "the diagonal under a continuous map"?

[gritty widget]

preimage (of the diagonal) under a continuous map

[hollow geyser]

still lost

https://tenor.com/view/homer-simpson-the-simpsons-monkey-gif-15399757

Do I even know topology?

Okay so if $f :X\to Y$ is some continuous function (function and map mean the same thing)? Then I'm lost on what preimage of a diagonal means. Diagonal would be a subset of $X\times X$ and preimage of the map is in $X$ is it not? What am I not seeing?

[gentle osprey]

SWR

[gritty widget]

there's a diagonal in Y x Y too

you're trying to show that a certain subset of X is closed

your hint is to look at the preimage of a diagonal

[hollow geyser]

So what's the "diagonal under a continuous map"? Is it the set ${(f(x), f(x)) :x\in X}$ or am I way off?

[gentle osprey]

SWR

[gritty widget]

it's not "(preimage of) (the diagonal under a continuous map)", it's "preimage (of the diagonal) under a continuous map"

[gritty widget]

that's what i was trying to say here

[hollow geyser]

Okay. Let $f :X\to Y$ be our continuous function. Could you write what you mean in set notation?

[gentle osprey]

SWR

[gritty widget]

to take the preimage of the diagonal you're going to need a map to Y x Y (or X x X, but since you're given maps into Y, this seems unlikely to be the case, no?)

come up with a map into Y x Y

take the preimage of the diagonal

[hollow geyser]

So what? Like $f : \bR \to \bR^2$ s.t. $f(x)=(x, x^2)$ for example?

[unreal stratus]

The diagonal is a set

[gentle osprey]

SWR

[unreal stratus]

The preimage of a set

[gritty widget]

you're a set

[hollow geyser]

[gentle girder]

then you get all the points $x$ such that $x = x^2$ in the preimage of the diagonal

[gentle osprey]

polynomial of order 256

[gentle girder]

so the preimage of $\Delta$ is just ${ 0,1}$

[gentle osprey]

polynomial of order 256

[gentle girder]

we are literally treating the diagonal as a set here

and taking the preimage is just like taking the preimage on any other set

[hollow geyser]

Oh so if $f : X \to Y\times Y=(g(x), h(x))$, then "the preimage of the diagonal under the map $f$" is literally just $f^{-1}(\Delta _Y)={x\in X :g(x)=h(x)}$?

[gentle osprey]

SWR

[hollow geyser]

Okay I think I see it. $(f, g) : X\to Y\times Y$ is continuous, and $\Delta_Y$ is closed in $Y\times Y$, so $(f, g)^{-1}(\Delta_Y)$ is closed in $X$

[gentle osprey]

SWR

[hollow geyser]

That's a really cool way to prove it too. Sorry for the struggle, but I'm glad my eyes have been opened to it now.

[unreal stratus]

Perfect yes

There are lots of similar things to that

So for example, you can show that if A is a subspace of X, and if the inclusion i: A -> X has a left inverse r (i.e. ri = id) then A is closed

[hollow geyser]

The inclusion being the identity function restricted to A?

[unreal stratus]

Yeah

[hollow geyser]

left inverse meaning i is injective?

[unreal stratus]

Meaning there's (çontinuous) r: X -> A such that the composition r o i is the identity map on A

[hollow geyser]

ç

[unreal stratus]

Lol

[hollow geyser]

Oh I see

[unreal stratus]

One says A is a retract of X if this is true

[hollow geyser]

definitely never heard of a retract

but the name makes sense

[indigo frost]

I’m trying to show that the closure of the union of two sets is equal to the union of their respective closures

Is this proof fine? Or is there a mistake somewhere/step(s) that need more justification?

[hollow geyser]

How did you get x' != x? What if A=B={x}?

How did you get from step 1 to step 2 here?

[indigo frost]

The definition of limit point I’m using states that a point x is a limit point of a set A iff all its epsilon-neighborhoods intersect A at some point other than x

[indigo frost]

Apologies, I meant to write x’ in step 2 there. That follows from the definition of the union

this is what I meant

(step 3 is justified from the definition of the limit point)

[rigid path]

If I have an homeomorphism $f:X\to A\subseteq \mathbb{R}^k$ with k cardinal, $X$ Tychonoff and $A$ closed, what property are preserved in the restriction of $f|{F{\sigma}}$ to an arbitrary $F_{\sigma}$ subset of $X$? Can the restriction still be an homeomorphism?

[gentle osprey]

Milo

[rigid path]

Oh, I forgot to mention: can the restriction still be an homeomorphism to a closed subset of R^k

[formal tide]

It can't be a bijection if F_\sigma is not X, so it can't be a homeo

[rigid path]

What about $f|{F{\sigma}} :F_{\sigma}\to f(F_{\sigma})$

[gentle osprey]

Milo

[formal tide]

oh, in that sense any restriction of a homeo is a homeo

[tender halo]

what you are talking about is called a realcompact space

they have some interesting properties in regards to extending real valued functions on them, but re subspaces i only know that if you take an open F_\sigma set in them then the subspace is also realcompact

(obviously it will probably be isomorphic with a different A \subseteq \mathbb{R}^k than X was)

heres a full statement (its from engelking)

by a functionally open set he means a co-zero set

[rigid path]

Wow, I was trying to prove that ahahha

That is hereditary for f sigmas

[tender halo]

well now you know where to look if you want to cheat xd

[rigid path]

Yeah but the problem is that the encyclopedia of topolgy says that hereditary for f sigma implies hereditary for functionally open and for closed set

[rigid path]

So I was trying to prove that without this corollary

Sorry for my bad English

[patent path]

If $\Gamma_1$ is a finite index subgroup of a Fuchsian group $\Gamma_2$, then limit set of $\Gamma_2$ should be finite translates of the limit set of $\Gamma_1$ right?

[gentle osprey]

Symmetry enjoyer

[steel glen]

how come this doesnt allow us to say that the product of quotients is necessarily a quotient in Top?

specifically, why doesn't
[X/\sim_X \times Y/\text{id} \simeq (X \times Y)/\sim]
where
[(x,y)\sim(x',y') \iff x \sim_X x' \text{ and } y = y']
follow from that third point? are we missing some kind of structure in Top

ok done editing sorry

[gentle osprey]

maximo

[unreal stratus]

Sets

[steel glen]

oh

🤦

did you ever end up finding a counterexample?

[unreal stratus]

Oh I found some online yeah I can share

But yes I think ur question is still important in that like, why does it fail for Top when we have similar universal properties

[steel glen]

yeah im curious to see what makes locally compact important
maybe i should just look at the proof assuming locally compact and see how it is used

[steel glen]

if they're from MSE or MO i'll just look up and find them haha

[unreal stratus]

It'll presumably just be that you can't prove (A x B)/~ has the universal property without appaing to smth special about sets

Like if I give you maps into A/~ and B/~ then there is no obvious way to map into (A x B)/~

Like you have to pick representatives and show that you get smth well defined as sets

But like for top spaces you'd have to check that that map is actually continuous lol

[steel glen]

it would seem so obvious that the maps would be continuous but that's probably because every intuitive (to me) topological space is locally compact

thanks for pointing out the set thing though lol

[quartz horizon]

Q isn’t, for example

The key thing that makes this work for Set is that it’s Cartesian-closed

Which Top isn’t, (semi-)famously

[heady magnet]

This has been haunting my mind too

Lol

What I’m wondering is tho

What if we only take a quotient of A

So is (A/~) x B iso to (A x B)/~ (where we say elements in A x B are equiv iff their firts components are equiv in A and the second components are equal)

This example does a quotient both spaces in the product

Anyone know?

[unreal stratus]

No

[heady magnet]

Really ?

No way

[unreal stratus]

Indeed if f is a quotient map then f x id needn't be one

[heady magnet]

That’s so evil

Whaha

[unreal stratus]

I'll try to find a nicer counterexample lol

[heady magnet]

Interesting

[unreal stratus]

Well i think it's tricky idk

[heady magnet]

Yes please

Ping me if you do 😸

I am curious what could go wrong

[steel glen]

i was going to read the proof to see what kind of intuition we need about local compactness

that would probably help cook up a good counterexample

[heady magnet]

Sweet

[heady magnet]

If you do ping me 🙏

I had to take a break from it for a day it was driving me crazy lol

[unreal stratus]

from topology and groupoids

page 111

[steel glen]

could probably check "counterexamples in topology" too

im sure they'll have some good ones

[heady magnet]

Crazy

What an example lol

Awesome I am going to@thibk about this lol

[heady magnet]

Yea I should have done that immediately lol

I wonder what those suggestions were

Thanks guys interesting stuff I’ll be more careful from now on lol

[unreal stratus]

Say $G,H$ are abelian discrete groups and $G$ finite. Why is $\mathrm{Maps}(BG,BH) \simeq \mathrm{Hom}(G,H) \times BG$? Apparently the map $\mathrm{Maps}(BG,BH) \to BH$ given by evaluation at the basepoint gives you a homotopy equivalence on each component, and if I have this then I'm done, but I don't see why that should be the case

[gentle osprey]

Süßkartoffel

[formal tide]

In the shrinking wedge of circles the path that goes through each of the circles is continuous, yet is clearly "weird". Is there some property of it does singles it out as weird?

[unreal stratus]

do you mean like the hawaiian earring?

[formal tide]

yeah

[unreal stratus]

Hm I'm not sure in what sense it is weird other than by comparison to the normal wedge

hm

[formal tide]

not the space, but the path

[unreal stratus]

yeah that's what i mean

[formal tide]

oh mb

[unreal stratus]

Like it only seems weird to me when you compare it to paths in the bouquet of circles hm

One thing is that if you consider the classes (in π1) of maps which only go through finitely many then the group they generate doesn't contain the class of the weird path

[formal tide]

this doesn't really make sense but the "immediate successor" of 0 is clear in a path in the normal wedge, but it is not clear in the hawaiian earring

[hollow geyser]

@indigo frost this proof works, but it's a little long. You could do the whole thing in one go if you're familiar with just points of closure.

Hey @unreal stratus is there some clever way to prove this with diagonal mapping?

You seem to always have a diagonal map proof for every problem I've asked so far.

[gritty widget]

||G_f is the preimage of the diagonal under (x, y) -> (f(x), y)||

[hollow geyser]

This was my first thought, but then I have to prove the the diagonal is closed now yea?

[gritty widget]

sure

that's a classic exercise

[hollow geyser]

One I was not given. We went straight to showing all graphs are closed.

But perhaps showing just diagonal is easier.

[hollow geyser]

This look good?

"point of closure of $\Delta$", not $\overline{\Delta}$

[gentle osprey]

SWR

[gritty widget]

the "But X is hausdorff, so..." isn't clear

[lime sable]

yeah i agree, you suddenly wrap up what you have shown with the fact that U_1, U_2 were arbitrary

and then apply that to the disjoint opens you get from hausdorff

[gritty widget]

actually apply hausdorffness explicitly

[hollow geyser]

Is this not showing it?

Or should I state again at the end of the proof "but U1 and U2 were arbitrary"

[gritty widget]

you did not say they were disjoint

[hollow geyser]

That's because I'm not proving by contradiction

[gritty widget]

okay

[hollow geyser]

I'm proving that $(x, y)\in\overline{\Delta}\to(x, y)\in\Delta$\

[gentle osprey]

SWR

[lime sable]

hausdorff says if your points are distinct, then they have disjoint open neighborhoods. so when you start out with two points, they're either distinct or not. you're using LEM anyway

[hollow geyser]

LEM?

[lime sable]

law of excluded middle

[hollow geyser]

This is equivalent to saying if two points have no disjoint open neighborhoods, then they must be equal, is it not?

[gritty widget]

it is, you're just not making that clear in your proof

[hollow geyser]

ah. I didn't know it needed clarification

but for sure

Okay, well here's the rest of the proof. Just want to see if it looks good:

[gritty widget]

looks okay

continuity should be straightforward to prove right from the definition (or, you can appeal to universal properties if you don't want to write out open sets)

[hollow geyser]

I haven't done anything with continuity where the domain is a product set. I'm sure it's straightforward, I just haven't gotten to it yet.

[gritty widget]

it's the same as any other continuity proof but the domain happens to have the product topology

if you know what continuity is (you do), and you know what the product topology is (you do), then you shouldn't have any problems

[rigid path]

Anyone here knows the solution, or where to find it, to exercise 3.11.G of the Engelking?

[bitter pond]

is an open set of a open set of a topology also open ?

[formal tide]

yes

[bitter pond]

thank you!

[formal tide]

[coarse night]

Maybe try posting the problem

[rigid path]

I'm not at home anymore. However it says: show that if $f:X\to Y$ is a perfect open map from a realcompact space $X$ to a Tychonoff space $Y$, then $Y$ is realcompact.

[gentle osprey]

Milo

[bitter pond]

what about an open set of a closed set of a topology, is that set closed?

[limpid fern]

in the subspace topology?

[formal tide]

no, in the closed [0,2], the set [0,1) is open, but it's not closed nor open in R

[ebon galleon]

No, as Shiranai's example shows. You do know that if C is closed in X and A is closed in C (with the subspace topology) then A is closed in X. So it works if you match open with open or closed with closed.

[bitter pond]

thank you both!

[ocean viper]

Can k-theory be obtained using Ext?

Topological k-theory

The answer is no https://mathoverflow.net/questions/11421/is-higher-k-functor-the-derived-functor-of-k0

[formal tide]

Am I missing something? This space doesn't seem to be simply connected

[fading vale]

this is poorly worded but hes asking you to take those symmetries of the grid described there and apply them to the entire plane

[formal tide]

ohhhhhhhhhhhhhhhhhhhhhh

yeah my bad, it isn't even a covering space otherwise

thanks!

[jaunty summit]

im trying to prove that any homeomorphism from S^2 to itself of finite order with a fixed point must have at least 2 fixed points total

i think this is equivalent to the hairy ball theorem somehow but i cant figure it out

like the orbits of the points induce a combing somehow if such a homeomorphism existed

[umbral panther]

A general strategy is to think about the quotient. Away from the fixed points it is a covering space. And then you glue in the fixed points

(Actually you have to decompose the space according to all the different types of orbits. But start with Z/p, so it’s just fixed and free)

[jaunty summit]

wait this is just brouwer actually

[red yoke]

Brouwer gives one point

[umbral panther]

There is a (degree 1) self map of the sphere that has only one fixed point. Namely, translate R^2, which has no fixed point, and take the one point compactification

[tidal cedar]

The space of all of these should be SO(3) and thus RP^3 but I can't reason out why in my current state

[radiant walrus]

Is this necessary?

Like, I just did it like this\\
Since $M$ is bounded, then there is a $c > 0$ such that $||x-y|| \leq c$ for all $x,y \in M$. In particular, $||\alpha_j e_j|| \leq c \Rightarrow |\alpha_j| \leq \frac{c}{||e_j||}$ for each $j$

[gentle osprey]

Mari

[radiant walrus]

I didn't have to use the Lemma 2.4-1

[radiant walrus]

Okay, I think I get why. M is any subset, so aj ej might not be in M

[heady skiff]

how exactly does this question make sense? i'm letting $O$ be an open interval of $\mathbb{R}$, and I'm considering its preimage $d^{-1}(O)$. Now this is a subset of $X \times X$ - but i'm not sure how to show that it's open given that the definition of an open set in a metric space involves that metric and i don't know what the metric is supposed to be on $X \times X$

[gentle osprey]

okeyokay

[gritty widget]

does the phrase "product topology" mean anything to you

or have you only seen metric spaces

[heady skiff]

oh yeah oops i forgot about that

[gritty widget]

you can put a metric on X x X. one way to do it is to copy how the metric on R^2 is defined in terms of the metric on R

d((x_1, y_1), (x_2, y_2)) = sqrt( d(x_1, x_2)^2 + d(y_1, y_2)^2 )

you could also do max(d(x_1, x_2), d(y_1, y_2))

or any p-norm of (d(x_1, x_2), d(y_1, y_2))

a metric on X x X is not necessary to show continuity of d, though. all you need is the product topology

[heady skiff]

oh okay i think i got it using the product topology

thank you

nvm

i feel like i'm making this a lot harder than it should be

like i'm trying to find open balls such that the product of those open balls equals the preimage of some open interval of O under d

[brittle rapids]

hint: use ||triangle inequality||

[heady skiff]

yeah okay here's where i'm stuck on: i'm letting $O = (a, b) \subset \mathbb{R}$, and letting $(y_1, y_2) \in d^{-1}(O)$. then there exists $\epsilon > 0$ such that $(d(y_1, y_2) - \epsilon, d(y_1, y_2) + \epsilon) \subset (a, b)$. If $(x_1, x_2) \in B_{\frac{\epsilon + d(y_1, y_2)}{2}}(y_2)$, then $d(x_1, x_2) < \epsilon + d(y_1, y_2)$ by triangle inequality, but i'm having trouble getting the other side

[gentle osprey]

okeyokay

[brittle rapids]

reverse triangle inequality

[deep nova]

I mean I would just do it by triangle inequality and limits yeah instead of by definition

[heady skiff]

wdym

i don't know how you can show that if you already have the inequality d(x_1, x_2) < e + d(y_1, y_2)

[heady skiff]

wait

so the best approach to this question is just to find some open subsets $U_1, U_2 \subset X$ such that $d^{-1}(O) = U_1 \times U_2$ for any open interval $O \subset \mathbb{R}$ correct

[gentle osprey]

okeyokay

[brittle rapids]

no, this is probably not true

given two points p, q we wish to find radii r_p, r_q such that distances between pairs of points in B_{r_p}(p) and B_{r_q}(q) are close to d(p, q)

don't overcomplicate it

[heady skiff]

i see okay

yeah my brain is not working rn

[rigid path]

I was reading a proof, and during the argument the author wrote: given a subspace $F\subseteq Y$ of $Y$, enlarge the topology of $Y$ by making both $F$ and $Y\setminus F$ open. How can I show that this new family is still a topology on $Y$?

[gentle osprey]

Milo

[rigid path]

Of course the union and intersection of F and Y - F are the entire space and the empty set, so they must be open. But what about the intersction of an arbitrary open with F or F - Y? And what about the union of a family of sets which contains F XOR F - Y?

[heavy walrus]

is there any case where the continuous tangent vector field of an n-sphere has an Euler characteristic greater than n?

[knotty vine]

I cant speak for the intent of the author, but I suppose "enlarge" means we should generate all of those new opens, like for a subbase.

[rigid path]

Is it always possible?

[knotty vine]

Yes, look up subbases

[rigid path]

Ok, thank you

[quartz horizon]

Yep, so given any family of subsets of a set X, you can consider the topology generated by such a family

The intersection of all topologies containing the family

[rigid path]

Oh I see

[quartz horizon]

It’s very akin to the subgroup generated by a subset of a group G

[rigid path]

But there is another little problem, later in the proof he says that the new topological space is Tychonoff. (Y was tychonoff as well)

I studied that the axioms of separation are expansive properties for i less or equal to 2.5

So how can I justify that this new space is tychonoff?

[knotty vine]

I think only when n=0

[heavy walrus]

can you prove there are 2 sets separated by a continuous function?

[heavy walrus]

would the tangent vector field then be in dimension n-1?

[knotty vine]

Wait what did you mean by the continuous tangent vector field?

I thought you meant the tangent bundle

[heavy walrus]

if given a function so that every point p has a tangen of the n sphere

which can be expressed in a dimension = n-1

[knotty vine]

I dont understand

[heavy walrus]

womp

[knotty vine]

Sure thats a vector field of the 2-sphere

[heavy walrus]

I figure it can be defined in higher dimensions too

[knotty vine]

Yes

But what do you mean by (n-1)?

[heavy walrus]

and tangent itself is the tangent vector field of a 1 sphere

[heavy walrus]

the surface of an object can be expressed in 1 dimension lower

assuming all vectors were projected onto the flat plane

only accounting for the dimensions in which it is expressed

[knotty vine]

No, the picture you sent is a 2-sphere. It is a surface of dimenion 2

Anyway I think youre looking for the Poincaré–Hopf theorem?

[heavy walrus]

the boundry of a 3 ball

[median sand]

How is this a proof?

[heavy walrus]

the 3 ball is what I meant by "object"

[heavy walrus]

on an n-sphere if n<1 the would a tangent vector field become non-vanishing?

[knotty vine]

No sorry, I misread your question

[heavy walrus]

is there any definitive answer you could think of?

[brittle rapids]

i assume you know this already, but if a < b are distinct and a < t < b, then t not being in X implies that (-infty, t) and (t, infty) is a separation

[heavy walrus]

normal by urysohn's?

[brittle rapids]

did you mean to reply to someone else?

[heavy walrus]

no I'm just an idiot

[red yoke]

What's the Euler characteristic of a vector field

[heavy walrus]

how many points the magnitude is 0

[red yoke]

Like in Poincare-Hopf?

The Euler characteristic of Sn is 0 for odd n and 2 for even n

[heavy walrus]

If 0 is considered even the tangent vector field would be of -1 dimensions

[red yoke]

?

The Euler characteristic of S0 is 2 and the only tangent field has index 2

Not sure what you mean by -1 dimensions

[median sand]

I'm asking how this is a proof that intervals are connected, all he does is refer to 3.2.3. What you're saying is that non-intervals are disconnected, the converse of what I'm asking.

Proofs of the connectivity of intervals rely on the completeness of R (and are probably equivalent to it for all I know) and he doesn't use it in any discernible way here.

[hidden crag]

You implicitly use the characterization of completeness with sup and inf

I guess?

[median sand]

Yeah but that's the whole proof, lol

That's my point

I wasn't sure if I had missed something

[hidden crag]

I don’t see the problem then

[knotty vine]

I agree it's kinda weird. It is still left to prove that an open interval is connected. If "an open interval being connected" was considered an obvious fact then it's not such a stretch that "any interval being connected" is also considered obvious.

[median sand]

Exactly. Which is strange, since the book is very thorough otherwise.

[brittle jolt]

if there is a finite closed cover (or open cover) with empty intersection between any two elements, does the gluing lemma still hold?

[opaque scroll]

The gluing lemma still works if the closed sets don't intersect yes. And if the intersection is empty then two functions automatically agree on the intersection.

[paper wedge]

yo just a quick stupid question

the m-simplex [p1,...,pn] is homeomorphic to the cone [p1(omitted),p2,....,pn] x [0,1]/~ with vertex p1

by just gut sense i found the homeomorphim just sending the linear combination sum(t_ip_i) to (sum(t_ip_i),t_0) where t_0 is the coefficient of the ommited vertex

and it works

why cant i use any fixed t tho?

is it because to have p_1 to not be ommitted , you would have t_0 to be 1 which should be glued to have the vertex of the cone p_1?

cool

ty

[untold lily]

In the book, it says it is "trivial" to prove that a net $(f_i)$ that converges uniformly on compact sets to $f$ also converges to $f$ in the compact-open topology
while this intuitively feels clear, it took me a bit to prove it. Can someone check if this proof is accurate and/or suggest a better proof?

[gentle osprey]

lems

[untold lily]

Proof:

Let $f_i \to f$ uniformly on compact sets, and let $L(K,U)$ be an arbitrary neighborhood of $f$; i.e. $K$ is compact, $U$ is open and $f(K) \subset U$. As $f(K)$ is compact, $U^c$ is closed, and the two sets are disjoint, the distance $d(f(K), U^c)$ is greater than $0$. Now, let $\varepsilon = d(f(K), U^c)/2$ and consider the open set $V := \bigcup_{x \in K}B(f(x), \varepsilon)$. By the choice of epsilon, we have that $V \subset U$. Since the net converges uniformly on compact sets, there exists $\lambda_0$ such that $\lambda \geq \lambda_0$ implies $|f_\lambda(x) - f(x)| < \varepsilon$ for all $ x \in K$. It follows that $f_\lambda(K) \subset V \subset U$, whence we deduce $f_i \to f$ in the compact-open topology.

[gentle osprey]

lems

[stuck sleet]

Whats that theorem the characterizes topological spaces/toplogical manifolds/smooth manifolds/blahblah/ based on genus? I'm trying to remember a theorem that was something like "Everying [toplogical/smooth/analytic/SOMETHING] manifold is [homeo/diffeo/something-eo]morphic to to the genus-n shape that is a smooth connected sum of n tori"

[knotty vine]

https://en.wikipedia.org/wiki/Surface_(topology)#Classification_of_closed_surfaces ?

[lime sable]

classification of surfaces?

[stuck sleet]

@lime sable thats what i was looking for! Thanks

Really? The sphere is homeomorphic to a closed subset of R^@?

R^2*

my topology is a lil weak, but if the idea is to identify the edge with a point on the sphere, why doesn't this violate continuity? In the same way, why wouldnt a homeomorphism have to rip the sphere to turn it into, say, the closed unit disk?

[knotty vine]

What? There's no such homeomorphism

[stuck sleet]

ah nvm i was misinterpreting a picture on that page

[rustic lava]

You know those classic models of a "Klein Bottle"? Somebody said that they act (in terms of fluid) in the same way that theoretical Klein bottles would, I disagreed because it's not even in the same dimension but I'm not actually confident in my answer on that, any information about the difference in properties (physically, I know the difference in theoretical properties)

This is what I said, and I usually feel the idea of "99.9% of people have done less topology than me" but at the same time, I don't want to assume that I know more than other people 😅

[zinc siren]

I thought the reason why they wouldn't is because it's merely an immersion.

From the perspective of the abstract Klein bottle, you can go in a straight line along the place that the model says the surface intersects with itself without having any problems.

[rustic lava]

Yeah, I think the person just maybe didn't get that you can't truly "visualize" a real Klein bottle (except using xyzt space but I didn't really want to explain what that is)

[zinc siren]

I mean, the thing younger me didn't get was that it was merely an immersion. An explanation of "The abstract surface doesn't intersect itself" would've sufficed for me.

[rustic lava]

I didn't think it would suffice because they mentioned it was a non-orientable surface so I thought maybe they knew some stuff about it, but the more I read back the message, the more I think they might have just read the first sentence from Wikipedia on it

I'm not good at visualisations though, I mean like 90% of what I do is group theory now 😅

[zinc siren]

It would've sufficed for post "orientable" but pre "immersion" me

(i learned what "immersion" meant after learning what "orientable" meant for a differentiable manifold)

...

is it Differential Manifold, or Differentiable Manifold?

[unreal stratus]

2nd

sorry i am eepy

[zinc siren]

ah, it just looks weird right now for some reason.

I should nap

[rustic lava]

I learnt them both when I did an intro to Topology module so I think very similar times 😆

[radiant walrus]

How do you generally show a normed space is finite dimensional? Like, you would need to show it has a basis with a finite amount of elements, but I don't think you can do that when the dimension can be anything

[umbral panther]

There are infinite dimensional normed spaces, in case you are trying to prove there aren’t

The standard result is that a normed space is finite dimensional iff the unit disk is compact

[radiant walrus]

Yeah um

That's exactly what I'm trying to prove 🥲

Apparently this is supposed to help but I can't see how?

Like, I guess I could let Z = X, but then what would I set Y to be? And then how is that even going to do anything about it being finite dimensional?

It feels like a proof by contradiction works best but I can't seem to find a contradiction

Like, I guess I could set up a sequence of points in the unit ball where each one is linearly independent but then I don't know how that causes a contradiction

I mean, I can prove X is complete, if that helps

I just don't see how that lemma comes into play and how it can help

[brittle rapids]

if the sequence of points on the unit ball is spaced out enough, then it won't admit a convergent subsequence

[radiant walrus]

Wait a second...

How does this look?

[brittle rapids]

it looks good

[radiant walrus]

Suppose, by contradiction, that $X$ is not finite-dimensional. Let $x_1 \in M$ such that $||x_1|| = 1$. Since $x_1 \neq 0$, then ${x_1}$ is linearly independent. Let $X_1 = span{x_1}$. Since ${x_1}$ is a basis of $X_1$, then $X_1$ is finite-dimensional and therefore closed. Since $X$ is infinite-dimensional, then $X_1$ is a proper subset of $X$. By Lemma 2.5-4, iterating through each $n$, there's an $x_{n+1} \in X$ such that $||x_{n+1}|| = 1$ and $||x_{n+1} - x|| \geq \frac{1}{2}$, for all $x \in X_n$. Therefore, ${x_1, \hdots, x_n, x_{n+1}}$ is linearly independent, since there are no scalars $\alpha_1, \hdots, \alpha_n$ such that $x_{n+1} = \alpha_1 x_1 + \hdots + \alpha_n x_n$. Let $X_{n+1} = span{x_1, \hdots, x_n, x_{n+1}}$. Since $||x_{n+1}|| = 1$, then $x_{n+1} \in M$. Since ${x_1, \hdots, x_n, x_{n+1}}$ is a basis of $X_{n+1}$, then $X_{n+1}$ is finite-dimensional and therefore closed. Since $X$ is infinite-dimensional, then $X_{n+1}$ is a proper subset of $X$. We thus have that $(x_n) \in M$. However, for all $n,m$ with $n \neq m$, we have $||x_n-x_m|| = ||x_m - x_n|| \geq \frac{1}{2}$. Therefore, no subsequence of $(x_n)$ can be Cauchy, and since any convergent sequence is Cauchy, then $(x_n)$ has no convergent subsequence, contradicting that $M$ is compact. Therefore, $X$ is finite-dimensional.

[gentle osprey]

Mari

[radiant walrus]

When I read this, it just clicked for me. Idk why

[brittle rapids]

it's the entire idea

i remember struggling to prove this because i couldn't recall this definition of compactness

[radiant walrus]

Honestly, if I could go back to like, an hour ago, I would tell myself to just imagine an infinite-dimensional normed space and try to construct a sequence that doesn't have a convergent subsequence on the unit ball

Then I'd see why only a finite-dimensional one works

Trying to create geometric visualizations and come up with examples like that usually help so much to visualize the problem

[rigid path]

How can I prove that if $\mathcal{Z}$ is an ultrafilter on $X$ with the countable intersection property, then $\mathcal{Z}$ is closed under countable intersection?

[gentle osprey]

Milo

[kindred drum]

So, i need to give an example of the space $X$ and a subset $A \subset X$ such that $A$ is a weak retract of $X$, but not a retract. I am considering that the following example works, let $X$ be an unit disc at $(1,0)$ with radius $1$ and let $A$ be boundary circle without the point $(2,0)$, then if $r$ is a retraction it should fix $A$, but this would imply that it fixes $(2,0)$ and hence the whole boundary and so it can't be continuous. On the other way, consider $r(x,y)=(0,0)$, it obviously sends an entire disc into the point which is homotopically equivalent to $A$. Is this example correct?

[gentle osprey]

poTato

[unreal stratus]

Homotopy equivalent isn't enough for a weak retract right

Your answer seems to use a different definition of weak retract to what I find elsewhere

[kindred drum]

$ri$ should be homotopic to the identity on $A$, i think it is correct that for the punctured circle and it's point this works

[gentle osprey]

poTato

[unreal stratus]

Could you provide a definition

[kindred drum]

yep, one minute

[unreal stratus]

You omitted that it's a weak deformation retraction which is kinda important lol

[kindred drum]

oh, the second image is not the right one

[unreal stratus]

But okay

Oh

[coarse night]

inclusion of a point is always a retract, see if you get anything useful there

[kindred drum]

this is the right exercise

[unreal stratus]

okay sure

I don't think this is totally standard Terminology but yes ok

I agree with your answer yes

[coarse night]

there's an example in hatcher which should work

[kindred drum]

do you remember any keyword to search for it?

[coarse night]

I believe it's in the exercise of section 0 or 1

something that looks like a Infinite comb

also I believe the std terminology is weak homotopy equivalence

[kindred drum]

could not find it, but i believe my example is ccorrect

[unreal stratus]

No this is very different

[coarse night]

weak deformation retract? I was talking about that. I didn't see there's multiple def given

[unreal stratus]

You said weak homotopy equivalence

[kindred drum]

there's weak retract as well as weak deformation retract

[unreal stratus]

But weak def retract is also different

[coarse night]

Oh right

not same my bad

[unreal stratus]

A weak homotopy equivalence is a map induces isos on pi_i

[coarse night]

(all spaces are CW KeK)

[kindred drum]

one more question

strong deformation retract fixes the subset for any t

i thonk Mobius strip should work

woth $A$ considered as the "main loop"

[gentle osprey]

poTato

[kindred drum]

hence the boundary of Mobius strip

[coarse night]

Try this

[kindred drum]

i am not sure that for the last exercise this will work, strong and ordinary deformation retract are the same if you retract to the point

[coarse night]

Try picking your set A carefully

[kindred drum]

oh, okay

[coarse night]

Take the bottom line U some part of Y axis

[brittle rapids]

show that the collection of countable intersections forms a filter

then use maximality

[sullen nimbus]

how do i prove this
I am not able to make sense of the E intersection cl(U X/E) condition and why its significant

the sets E are not given open or closed specifically

[brittle rapids]

wlog the union of \mathcal{X} is just X. then in fact each E is open

[sullen nimbus]

oh so i had to work with relative topology?

[brittle rapids]

given a subspace S, and a subset A in S, the closure of A wrt S is contained in the global closure of A

so in the subspace you retain the disjointness

[sullen nimbus]

ah okay

[brittle rapids]

and in particular E is the complement

[sullen nimbus]

ah so thats why E intersection thing being empty was significant?

[sullen nimbus]

i thought this was related to pasting lemma after this , i can use that basically

[brittle rapids]

yes, afterwards it is that

[sullen nimbus]

so does that mean it generalizes to arbitrary top spaces?

[brittle rapids]

well the open statement of pasting lemma is the easy one

i think so

[sullen nimbus]

okay, Thank you

[signal kite]

can someone help with this? how do i get the convergent subsequence?

[opaque scroll]

Take x(i)_n to be a sequence in X.

Consider a subsequence [i in I0] such that x(i)_0 converges to x_0. Consider a subsequence of I0, [i in I1] such that x(i)_1 converges to x_1. Continue constructing In.

Construct a sequence J by picking the first element from I0, the second from I1, third from I2, etc. Then x(j)_n converges to x_n

[signal kite]

but how is this a subsequence of the original sequence? and why would it converge?

[opaque scroll]

It's a subsequence just by definition. To prove convergence you have to do a little more:

Consider eps > 0, consider n such that 1/2^n < eps/2. Consider N such that the distance between x(m)_k and x_k is less than eps/4 for all m > N in Ik and k <= n. Then
d(x(m), x) < eps/4 sum_k 1/2^k + sum[k>n] 1/2^k < eps/4 * 2 + 1/2^n < eps

(For any m in J larger than max(n, N))

[paper wedge]

would G(f_0(x),2t) and G(f_1(x),2t-1) work as a homotopy rel A ?

3.1

[distant hollow]

not sure where this question goes but I hope this is the right channel. Set
$$G(n) = \mathbb{Z}\Delta[n]/\mathbb{Z}\Lambda^0[n]$$
and
$$F(n) = \mathbb{Z} \Delta[n]$$
I know that
$$F(n) = \bigoplus_k G(k)^{\oplus {n \choose k}}$$
From this, how are retracts of $F(n)$ also direct sums of $G(n)$'s?

[gentle osprey]

havrebra

[distant hollow]

I am specifically looking at Charles Rezk's answer in https://mathoverflow.net/questions/195936/a-more-natural-proof-of-dold-kan and I am unsure how the above shows that every object in the Karoubi envelope of Z\Delta is a direct sum of objects G(n)

[umbral panther]

By Jordan Holder type arguments, two direct sum decompositions have a common refinement. But if End(G(n))=Z, then it has no summands

[distant hollow]

sorry I'm not sure I'm following. Could you explain a bit more?

[umbral panther]

An object in the Karoubi envelope is a direct summand of F(n), by definition. By this calculation F(n) has another direct sum decomposition as a sum of G(i). Composing the map for G(i) into F(n) to F(n) onto G(i) gives a idempotent endomorphism of G(i). It is either 1 or 0. Thus the given element of the Karoubi envelope is a sum of some subset of the G(i)s, the ones for which the endomorphism was 1

[distant hollow]

wait why is the given element of the envelope a sum of some subset of the G(i)'s?

[umbral panther]

Do you agree I have given you a recipe for which subset it is supposed to be?

[distant hollow]

right

okay right I get it

thanks a lot

[umbral panther]

Again, Jordan Holder is a very general technique. In fact, it is overkill in an additive category

A Karoubi envelope is an abstract setting, so you have to check that the intuitions can be actually made to work, but the intuition is to intersect filtrations

[worthy olive]

How good is prasolov book for self studying

[kindred drum]

Not sure about prasolov, i am currently using "a first course in algebraic topology" by kosniowski and it seems really good

although i think there is no chapter on homology

[hidden crag]

what

[unreal stratus]

Ig you could have a first course on pi_1

I had that

[hidden crag]

A first course entirely without homology sounds dreadful

[cedar pebble]

Straight to cohomology

[hidden crag]

[merry geode]

Would there be much difference

[cedar pebble]

a little bit yes

But it’s possible to teach it in this order

If you were forced to teach only either homology or cohomology for reasons of time constraints cohomology tends to be more useful

Although you can just as well teach them both in parallel

No reason to separate them so much pedagogically

[merry geode]

Indeed

..meanwhile I am still yet to properly read the cohomology part..

[slate bane]

I was wondering, it's known that if we have a topological space X, and Y a subset with the induced topology, then for a point y of Y, any neighbourhood basis N_y of y in X gives rise to a neighbourhood basis N'_y of y in Y: we take the family {U intersect Y for U in N_y}.
But are these the only possible neighbourhood bases of y in Y with the induced topology?
That is, if I suppose to start with a given neigh basis of y in Y, how do I show the existence of the corresponding neigh basis in X such that the above happens; or otherwise, how do I show a counterexample? I thought a little bit but couldn't come up with anything, even though it should be a simple question

Considering the nontrivial case when y is not in the internal part of Y

[granite slate]

Is it possible to have $$A \subset B \subset X, H_{n}(A) \cong H_{n}(B) \text{ for all n but } H_{n}(X, B) \neq H_{n}(X, A) \text{ for some n?}$$

[gentle osprey]

dompa

[tawdry valve]

yup, one example might be something like take X to be D^2 wedge S^1, A to be the boundary of D^2, and B to be S^1.

[granite slate]

A isnt a subspace of B though? their intersection is the point where the disk is attached to the circle

[tawdry valve]

ohh sorry I didn’t see that, hmm lemme think

[umbral panther]

If the inclusion induces the isomorphism, then the relative homologies will be isomorphic. But if it’s just an abstract isomorphism, you lose control

[granite slate]

That was my thinking too, but i can't really figure out subspaces with this property

perhaps i can use excision?

[tawdry valve]

maybe something like this would work for a counter example (when inclusion doesn’t induce the isomorphism)

I think H_1(X, B) should be trivial, but not H_1(X, A)

[umbral panther]

I always like examples with H_0
X = [0,1] cup {2}
A = {0,1}
B = X

[tawdry valve]

oh yeah that’s nice

I think my example is then the suspension of your example

[slate bane]

Does anyone have any suggestion?

[granite slate]

@umbral panther @tawdry valve thank you very much for the help, very intuitive examples

[granite slate]

Actually A and B dont have the same homology since A is path-connected and X isnt?

or is there some subtlety that i'm missing?

[unreal stratus]

A isn't path connected

it is {0,1}, not [0,1]

[granite slate]

what are the "unreachable" points?

or rather what two points cant be connected by a path in A?

[gritty widget]

there are only two points in A

A (given the discrete topology) is disconnected, so it certainly can't be path connected

[granite slate]

arent these intervals?

[gritty widget]

A is {0, 1}. the set containing only 0 and 1

note the curly brackets

[granite slate]

oh my god i'm so stupid

thanks

[paper wedge]

if i have two spaces homotopy equivalent

if i remove a point from each are they still homotopy equivalent

[grave solstice]

no

wait idk

ok

take R^2 and an isolated point far away from R^2

then this is homotopy equivalent to R^2 isn't it

[gritty widget]

No. R is homotopy equivalent to R². But after removing one point from R it becomes disconnected while doing the same to R² does not disconnect it

[grave solstice]

uhm maybe not because you have a path that is just the isolated point alone, I thought you wouldn't consider these

[merry geode]

Contractible space magic

[umbral panther]

Or R^1 and R^0

[knotty vine]

Or even [0,1] and [0,1] (which are clearly homotopy equivalent!)

[hexed steppe]

or for the failure of the converse, S^1 and S^0

[paper wedge]

ty everyone

[winged viper]

does the first stiefel-whitney class of a line bundle determine its isomorphism class?

[warm quiver]

yes, because H^1(X;Z/2) is just the Picard group of real line bundles over X

[winged viper]

sweet thanks

[sullen nimbus]

Wheres the fault in taking a Cauchy sequence in Y then taking it's inverse image then can't we say it's inverse image is also Cauchy as f^-1 is cts so it has a lim in X map that and claim the output of that to be lim of y

Ah idk if f^-1 is uniformly CTS right?

But f being a homeomorphisms still seems like it should be

Why would it fail if it did so

[hexed steppe]

that is the question, yes

[brittle rapids]

try to wildly distort cauchy sequences. hint: ||you don't need to go far beyond R||

[sullen nimbus]

I was initially trying to look in R I don't have several functions up my sleeve

Pls describe 'wildly distort cauchy sequence'

[brittle rapids]

homeomorphisms are not in general uniform. what's the best illustration for this?

[sullen nimbus]

Huh really!?

[hexed steppe]

why should it be true

[sullen nimbus]

Cause close things stay close up?

[hexed steppe]

but as you noted

you need some uniform control

which is much stronger than just continuity

[sullen nimbus]

Okay so I guess it's not true

Now to find a counterexample;-;

[sullen nimbus]

Is [0,infty) to [0,infty) f(x)=x^2 works?

I don't have any sequences on me such that xn-yn ->0 and x^2 n - y^2 n --> 0 but I think that x+y in (x+y)(x-y) should do sth 💀

I am actually quite bad with coming up with counterexamples

[brittle rapids]

no, no one is good at coming up with counterexamples

[hexed steppe]

maybe first just choose a continuous function that isnt uniformly continuous

[sullen nimbus]

So basically I need to close it so that it's domain is not comoact

But then again my mind goes blank when I think about it

[hexed steppe]

x^2 is not uniformly continuous

[sullen nimbus]

Oh cool

I am not interested in the proof should we just move on to the main topic

[hexed steppe]

wut

i think you need to understand the proof to come up with a counterexample lol

[sullen nimbus]

Oh 😩

[hexed steppe]

but its maybe not the best thing to work with

for this

[sullen nimbus]

I see

[hexed steppe]

idk just try things and see what happens

[sullen nimbus]

Isn't that like the most general advice

[hexed steppe]

never fails

[sullen nimbus]

Agree

[hexed steppe]

do you know any spaces that arent complete

[sullen nimbus]

Q

[hexed steppe]

ok that is complicated though

[sullen nimbus]

Any not closed set of R?

[hexed steppe]

there are examples that are much simpler topologically

yes

[sullen nimbus]

Uhhh

(-1,0)

[hexed steppe]

sure

can you find a homeo between that and something complete

[sullen nimbus]

I strech

It to fit the reals?

[hexed steppe]

yeah maybe try writing down a function though

it will be instructive

[sullen nimbus]

Uhhh 😭

[hexed steppe]

it might be a bit simpler to use (-1,1)

[sullen nimbus]

Okay

[hexed steppe]

though doesnt matter ultimately

[sullen nimbus]

x*-1 + (1)*(1-x)

Wait

Not

It's supposed to be the other way

😭

I can draw a Function

💀

I don't how to write it

It has asymptotes at x=-1 and x=1

[hexed steppe]

yes

you probably learned about a function like this in high school

not -1 and 1 but anyway

[sullen nimbus]

If you're pointing at 1/x then uhh

We probably need to modify that

By shifting it or sth

Okay tan x?

[hexed steppe]

sure

tan x is what i had in mind yeah

[sullen nimbus]

Man I had to go up on desmos to figure that out 💀

Idk what I'll do in exams

[hexed steppe]

anyway more importantly do you see why the vertical asymptotes are obstructing completeness

you should think about that

another approach to this problem is to try to just find a non-uniformly continuous function and strategically choose a cauchy sequence

so maybe think about what would happen if you did that with tanx

[merry geode]

(0, 1) being incomplete is such a tragedy

[sullen nimbus]

You should have known by now I am very bad(absolutely terrible) at strategies

But I'll try

Thank you

I'll come back if I still get no progress

[sullen nimbus]

Also here it's X is complete Y is ... So I need to take X as R and Y as (aymp1 , asymp2)

And consider tan^-1 ?

Also it has bdd derivative so it's uniformly CTS?

There's just tan is not cts 💀

Oh wait but I have restriction

[merry geode]

Restriction?

[sullen nimbus]

f is going from R to (..) right

f^-1 is from (...) To R

Aren't both cts

[merry geode]

Yes

[sullen nimbus]

So uhh tan^-1 works as a counter ex?

[merry geode]

,w plot tan^(-1) x

[gentle osprey]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by abstia

[merry geode]

Looks fairly continuous ime.

[sullen nimbus]

fairly cts ?💀

[merry geode]

It is continuous

Also uniformly continuous

[sullen nimbus]

It's differentiable

[merry geode]

Yeah

[sullen nimbus]

Yes

I guess this works 😩

[merry geode]

Are you sad that this happens to work?

[sullen nimbus]

I am sad that I have so much skill issue

Idk what will happen in the xeams

[merry geode]

Ahh

[sullen nimbus]

Anyways thanks again @hexed steppe

[merry geode]

It's okay, most others will struggle as well

[sullen nimbus]

[merry geode]

Solve a lot of exercises, then you could be confident!

[nocturne epoch]

I'm having a bit of trouble with a project I'm starting, so I've done my undergrad in math and I've completed a course in metric spaces, but now I want to do a project on a topic in topology, any suggestions of topics or how I can think of choosing one? I have a project superisor etc. but I want some ideas before seeing him, thx in advance.

[knotty vine]

Topology is incredibly broad, any particular directions you're interested in? Maybe algebraic topology or knot theory or manifolds or smooth manifolds or algebraic geometry or functional analysis or logic (heyting algebra/sheaves) or computer science (domain theory or persistent homology)?
If you've only done a course on metric spaces it will be good to read at least a text on point-set topology and how many definitions from metric spaces translate to topological spaces.

[granite slate]

What is the reason/intuition behind this? If q is a sequence of isomorphisms from H_n(A) to H_n(B), can't we construct the following diagram and isomorphisms fall out using five lemma? I suppose there is some issue with commutativity?

[umbral panther]

If the isomorphism is induced by inclusion, then it follows by the five lemma. If the isomorphism is arbitrary, you can’t build a map of long exact sequences. That is, you can’t make the diagram commute

[granite slate]

But why does inclusion yield commutativity?

[umbral panther]

That’s what it means to be a functor.

[last topaz]

The classification of surfaces is a fun one.

[granite slate]

Oh okay i see, so it's because they preserve composition?

[nocturne epoch]

Thank you so much for the response, I'm kind of knew to topology, I've only finished one book that was mostly focusing on metric spaces and I was very unsatisfied, I'll take this as a starting point and work from there, thanks so much for the great response.

[nocturne epoch]

I'll also look into this, appreciate the response thx!

[lament vortex]

trying to wrap my head around the formal definition of persistent homology
it's always defined as "a map in homology induced by an an inclusion between chain complexes" or something similar. but... what does this map actually do, and what does "induced" mean in this context?

[unreal stratus]

If you look at the definition of homology etc the point is that like

Any map X -> Y of spaces gives rise to a map C_*(X) -> C_*(Y) which gives rise to one H_*(X) -> H_*(Y)

How exactly that map is defined depends on what kind of homology you want

Cellular or simplicial etc

[opaque scroll]

You mean an inclusion of cell complexes or something?

The usual definition I've seen, is you take your data set and let X(r) be the union of balls of radius r around each point. Then the persistence homology is the homology of X(r) together with the maps from the inclusions X(r) -> X(r') when r' > r.

This is then usually restricted to only those r where the homology changes.

And then the idea is that the kernel of such a map is topological features that disapear, where's the cokernel are features being created. Then you're interested in features that persist

[heady skiff]

what's the easiest way to see that ${(x, y) \mid xy = 1} \subset \mathbb{R}^2$ is closed

[gentle osprey]

okeyokay

[gritty widget]

zero set of continuous function

[heady skiff]

zero set?

oh

preimage of {1} under multiplication

[gritty widget]

or level set sure

both work

[halcyon bone]

I'm actually just unsure where to go with this

😔

I think that I want to show that all the connected subsets of S containing p are open

and then the connected components of S must be open

[gritty widget]

this might not be true

[halcyon bone]

huh

how else do I show that the connected components are open

wait

I can just find a subset of them

that should be open

right?

wait then can't I just take the union of the neighborhoods V

[gritty widget]

take a point in one of them and show that you can find an open neighbourhood of it contained in the component containing that point

[halcyon bone]

or somethign like that

yeah and then that tells us the component is open

by the local criterion for openness

ok I don't know why I didn't see that earlier ty

[heady skiff]

how exactly does the deformation of the closed unit ball into the unit 2-sphere work again

is it just that the boundary is identified with the top of the sphere

like we pinch it

and then the interior is like the outside of the sphere

[fading vale]

If by closed unit ball you mean a disk then yeah

[merry geode]

Is it feasible to deform 3 dimensional ball into 2-sphere

[sly mural]

uhh I don't believe so - but a sphere with a point missing is homeomorphic to a disc

btw I would think of 3 dimensional ball and 2-sphere as being the same object

[hexed steppe]

?

[zinc siren]

No

Well

They aren't homeomorphic (Proof: We can look at the top homology group).

There is no deformation retract (Proof: Brouwer's fixed point theorem)

[merry geode]

I mean 2 dimensional ball and 2-sphere are not homomorphic, either.
(Not even homotopic iirc)

[zinc siren]

They aren't. Easy proof for this case looks at the homotopy group

2-ball is simply connected

Err, shit, nvm, 2sphere is too

[merry geode]

2-sphere is.. also simply connected. Yeah

[zinc siren]

Then look at homology.

Homology of ball is trivial because it's contractible

[merry geode]

This process delineates the deformation of closed unit disk into unit 2-sphere

[zinc siren]

Homology of sphere is ℤ in top dimension.

[merry geode]

Not “deformation retract”, ofc.

[zinc siren]

...how are you deforming it? what kind of deformation is allowed here

2-disk can be mapped to the top hemisphere

[merry geode]

Ah, maybe deformation is a poor choice of words

[zinc siren]

This construction was used in some exercise in Hatcher that proved something about a vector field on either the disk or sphere.

That's why I think of it

[merry geode]

You can identify the boundary into a point ye

[zinc siren]

It's just keeping the x,y coordinate but moving the z coordinate up

[zinc siren]

This is true.

[merry geode]

Would calling it 'deformation' be incorrect usage? Think I forgot the usual term for this..

[prime elbow]

I want an explicit description of closed sets in R(Standard topology) so is it union of (a, infinity) and (-infinity, b) ?

[naive trench]

Thats open

Closed sets in R are of the type [a,b] or (-inf,a] or [b,+inf) even points {a} are closed

[tidal cedar]

There's far more closed sets

[naive trench]

Just told him the form of few of them

[tidal cedar]

Cantor ternary set has no nontrivial closed interval [a, b] and is closed in R

[naive trench]

I know

But again just told him a few of them, the most common at least

[prime elbow]

Yes, then how can I write [0,1) as a union of closed sets and an intersection of open sets?

Because the intersection of the open set of R is the form of (a,b), right?

[naive trench]

[0,1+1/n] for example

[naive trench]

Intersection

Arbitrary intersection of closed is closed

[prime elbow]

Yes

[naive trench]

Oh you asked union

[prime elbow]

Yes

[naive trench]

Ig [0,(n+1)/n] works

[prime elbow]

First how I show that [a,b] is a closed set.

That means I want to show that union of (-infinity,a) and (b, infinity) is an open set in R.

[prime elbow]

But I want a finite union of closed set so how n+1/n goes to 1 in finite case

[naive trench]

Those are open and the union is open too

[prime elbow]

Yeah but how I show that (a, infinity) is open set

[naive trench]

When you said union I assumed countable at least my bad

[prime elbow]

I think a finite union of closed set

[naive trench]

You can express it as an arbitrary union of (a,b) with b>a

Like idk, union of (a,b+n) for n natural

[prime elbow]

Okay

[naive trench]

I think [0,1) cannot be expressed as a finite union of closed sets. Because it will implies that is closed and by definition is not. The complementary is (-inf,0)U[1,+inf) where [1,+inf) is closed not open

[naive trench]

As an union of closed must be countable like with n natural

[naive trench]

You can use this one

Since (n+1)/n<1

For the intersection of open ones mmmm

[prime elbow]

Okay thank you

For X is the union of elements of ∆, I want to show that ∆ is a basis for Ω.

I showed that ∆ is a basis but how I show that it is a basis for Ω?

[prime elbow]

Product topology has two basis?
For product topology on X×Y
First {U × V | U is open in X and V is open in Y}

Second {B × C | B belongs to the basis of X and C belongs to the basis of Y}

[knotty vine]

You write "the basis of X" but there is no unique basis. Otherwise, yes, prove it (every U x V can be written as a union of B x C's)!

[prime elbow]

Okay thank you

[feral copper]

Heyo; I understand that there are two versions of twisted (co-)homology: the one with systems of local groups, and the one with a reprensentation of the fundamental group. Steenrod describes for former; do you have a reference for the latter, and that explains how to get a system of local groups from a rep of the fundamental group?

[umbral panther]

My answer to all questions is
Bott and Tu

[unreal stratus]

i guess you mean local system of groups rather than system of local groups right?

[novel ember]

i cannot figure out a retract for this

my guess would be that f.g of this is Z*Z

[knotty vine]

Why? Isnt it just the circle?

[novel ember]

what

[knotty vine]

o wait nvm

[novel ember]

yeeah

we have a structure like this

[knotty vine]

Yeah thats a torus

[novel ember]

[knotty vine]

wth is that emoji

[umbral panther]

In particular the fundamental group is Z^2, not Z*Z
There are other configurations of a circle and a line where the fundamental group is Z*Z

[novel ember]

i know that f.g of a torus is indeed Z^2

i think im gonna want to figure out a way to separate this into U,V such that f.g of U,V is just Z, the intersection is Z, and have the relation be ab=ba?

because if we can do that we end up with <a,b|ab=ba> which is just Z^2

[knotty vine]

Thats not going to work

You need U,V with pi(U) = Z*Z, pi(V) = 0 and the intersection is Z

[novel ember]

[umbral panther]

I think it would not be that hard to write down a Z cover with fundamental group Z

[novel ember]

whats a Z cover

[umbral panther]

A cover with covering group Z. That is, a space X with a free action by Z whose quotient is your space

[feral copper]

Yes, sorry, I mean local system of groups indeed!
Thanks bw, I'll go check Bott&Tu then

Actually, no, I did mean system of local groups (Steenrod calls them this):

[umbral panther]

Archaic language

[feral copper]

[worthy olive]

What does a cohomology give? Like homology gives a set, and homotopy helps you find paths, what does cohomology give?

[tidal cedar]

Homology and cohomology and homotopy all output groups

Specifically n-indexed groups

The cohomology groups also have a product structure on them

You can package all the n-th cohomology groups, for all n

And you get a graded ring

[abstract saffron]

And this ring reveals structures which are non-trivial with homology or homotopy

[tidal cedar]

I'd also say homotopy groups don't really help you find paths

As much as tell you about the space you're looking at

I forget does cohomology ring actually reveal anything the homotopy groups don't for a nice space

[abstract saffron]

I don't remember either, but I definitely read about it

[worthy olive]

Thanks, the way I worded it was kinda weird

Like 2 homotopically equal paths

[feral copper]

Eilenberg Mac-Lane spaces: you know the homotopy groups of a K(G,1), for instance, but their cohomology is non-trivial (it is the group cohomology of G too)

[abstract saffron]

eyy, Matplotlib

long time no see

[tidal cedar]

This is still determined by the homotopy properties of BG

no?

[feral copper]

Heyo; indeed, I even lost the active role

[feral copper]

All higher homotopy groups are trivial; wasn't that the question?

You mentioned the groups, not the homotopy type of the space!

[tidal cedar]

That's true

[feral copper]

Because, at least in the CW case, it's roughly the same

[tidal cedar]

I tend to think of the two as being isomorphic

[feral copper]

(the homotopy type and the cell structure)

[tidal cedar]

Yeah I only really care about CW complexes haha

[feral copper]

However, there are CW complexes with all trivial groups but which are not weakly HE

So probably their cohomology would distinguish them

Probably

[umbral panther]

Huh?
A CW complex with all homotopy groups trivial is contractile
A CW complex with all homology trivial need not be contractile, but cohomology won’t help you

[feral copper]

Ah, yes, sorry, you're right; however, there are pairs (A,B) where pi_n(A)=pi_n(B) for all n but they are not WHE.
The special case of all pi_n(A)=0 indeed implies a WHE

Also, reading from Wikipedia, Whitehead+Hurewicz gives:

A continuous map f : X → Y between simply connected CW complexes that induces an isomorphism on all integral homology groups is a homotopy equivalence.

But again, having all (co)homology the same doesn't mean the isomorphism is geometric

[fickle elm]

I am starting to get confused. I think this system of local groups should be equivalent to locally constant sheaves, but when I try to read the original text, I come across Théorie axiomatique de la cohomologie by Henri, Carta. He says the sheaf cohomology is somehow different from Steenrod's local coefficients.

And why he says Steenrod's definition is plus particulier?

[feral copper]

I must say, I'm also very confused by this, and I couldn't find an easy self-contained modern exposition of this topic...

[umbral panther]

Local systems are sheaves, but they are only a special case (plus particular) of sheaves

[fickle elm]

Ah. I see now. Yeah my french is really terrible.

[chilly owl]

That being when the space deformation retracts onto the wedge sum S^1 v S^1 instead of S^1 x S^1 , right? You would have the line outside the circle for this to happen.

[fickle elm]

I find in Hatcher's algebraic topology chapter 3 appendix H: local coefficients. On page 331, he describes how to associate a group to each point of the space. Here Hatcher's "bundle of groups" is equivalent to Steenrod's system of local groups, and he shows how to obtain one from actions of the fundamental groups.

In this case it is not hard to see that this bundle of groups is just local system of groups using the etale spaces.

And to see that we have an equivalence of categories from local systems to representations of pi_1, there are a lot of literature.

e.g. search monodromy representation for local systems.

[feral copper]

Oh, thanks a lot!

[pseudo coral]

question

in math

is a distribution a generalization of a function

im trying to read this paper

https://math.mit.edu/~rbm/papers/deRham.11.May.2011.pdf

so the input valued are test functions i.e., real valued functions of compact support?

And I need help decoding Lemma 2 of this paper I am unsure what they mean by

$\Bbb{R}_y^k \times {0}_z$

[gentle osprey]

MyMathYourMath

[pseudo coral]

is that the y slice of R^k crossed with the zth axis

Im doing a project on de Rham cohomology and am learning it on my own so..

wasnt sure if this qualifies as alg top or diff geo

lol

[pseudo coral]

also, is this MOF worthy or MSE worthy for asking

[gritty widget]

MSE

[pseudo coral]

oof I asked MOF :/

[hexed steppe]

lol

[pseudo coral]

ill delete

and ask on MSE

[hexed steppe]

what are you even asking

“please tell me the definition of a distribution?”

[pseudo coral]

yeah some of the notation is throwing me off here

[hexed steppe]

surely you would already know the answer by now if you tried using google

[pseudo coral]

https://math.stackexchange.com/questions/4871881/a-remark-on-the-de-rham-theorem

I know this

but what is extendible distribution

[hexed steppe]

tldr

[pseudo coral]

I know a distribution is a generalised function

which acts on real valued functions of compact support

as apposed to just mere points

[hexed steppe]

never heard of this

[pseudo coral]

right?

[hexed steppe]

is it not defined there

[pseudo coral]

nope not defined

heres the paper

https://math.stackexchange.com/questions/4871881/a-remark-on-the-de-rham-theorem

sorry

[hexed steppe]

um distributions do not take points as input

Lemma 1: Any extendible distribution on the interior of an n−simplex S⊂Rn, i.e., any element of the dual of
{u∈C∞(Rn):supp u ⊂S},
is the restriction of a distribution on Rn with support in S.

[pseudo coral]

https://math.mit.edu/~rbm/papers/deRham.11.May.2011.pdf

[hexed steppe]

it literally

defines it

[pseudo coral]

dude

I just said

[hexed steppe]

in the same sentence

[pseudo coral]

as apposed to

taking inputs as points

[hexed steppe]

“as opposed to just”

[pseudo coral]

cause a regular function

inputs points

[hexed steppe]

implies that they take points, as well as functions, as input

[pseudo coral]

but distribution takes on functions as inputs

ahhh youre right the i.e. is the defn

well that clears that up

I still have questions

its in my MSE post

this papers a tough read with no exposiure to de Rham cohomology

im learning it on my own

[hexed steppe]

did you try reading a textbook

on de rham cohomology

[pseudo coral]

yes

I get the definitions and such

but like look at my question on lemma 2

the text doesnt answer these

like whats meant by that subset of rn

[hexed steppe]

i think you should learn more distribution theory

[pseudo coral]

is it what i wrote

do you have any references

[hexed steppe]

friedlander-joshi

just reading a small portion of it will clarify a lot

[pseudo coral]

ok thx

[hexed steppe]

like distributional derivatives

[pseudo coral]

thats what i need to learn more about ..

found pdf 🙂 thx!!

[hexed steppe]

a more comprehensive reference is hormander vol 1

the title is like “intro to pde” or something

[pseudo coral]

oh ive heard of this text

[hexed steppe]

but it is harder to read

[pseudo coral]

thanks man

[gritty widget]

link the paper in your question

[pseudo coral]

btw I hope I asked the question in the appropriate chanel ..

[gritty widget]

make your question as easy to answer as possible

[pseudo coral]

ok youre right @gritty widget

just did!

[umbral panther]

I think it’s a typo. He defines “extendable distribution” by saying “ie,” but his definition seems like it’s just the definition of a distribution. Extendible sounds like it should be the conclusion, not the hypothesis: an extendable distribution is the restriction of a distribution on a larger set; the function it defines on test functions on the small set extends to a function on all test functions

[fair idol]

You should read Conway's functional analysis texts there is a whole chapter on weak topologies. That's how you make sense of distributions I think.

I think rudin's functional analysis text actually has some exercises

[pseudo coral]

awesome thanks man!!!! I have both of those though I think i misplaced my grandpa rudin..

[pseudo coral]

ahhh its an extension theorem?

[pseudo coral]

oh wow for some strange reaspon (and I was like why would this text have it) but I thought yhou were referring to Conways Complex text

🤦‍♂️

I don’t have either 😢

Also is this question appropriate for this Chanel ?

It’s not diff geo is it

Or is it diff top

also, did some1 from here upvote me lol

on my MSE post for this question

[fair idol]

Yeah it's not appropriate for this channel you should go in advanced analysis I think

[quiet thorn]

when in doubt, go to foundations and have them point you to the right channel

[prime elbow]

I do not understand how I can relate the closure concept? If I have a set then how can I get closure of that set?

[pseudo coral]

its the set

union

limit points

@prime elbow

or

intersection of all closed sets containing the set

the latter is the definition

the first is a proposition or somthng

[gritty widget]

there are like a million ways to define the closure of a set

[pseudo coral]

true

but that imo at least, is the most pure way

the others saying u equal ur closure if ur closed n such are to be proven imo

but this is true some authors define it to be

set union limit points

or i should say space union

limit points of space

or just $\bar{X}:=\cap_{F_\alpha \supset X, F_\alpha \text{closed}}F_\alpha$

[gentle osprey]

MyMathYourMath

[pseudo coral]

$\alpha \in A$, $A$ arb indexing set

[gentle osprey]

MyMathYourMath

[pseudo coral]

and arb intersection of closed is closed so the closure is a closed set

[prime elbow]

Yes but I don't know how to compute a closure of given set

[pseudo coral]

find the limit points

union them with the whole space

i.e., the limit points that are not already in the set

if a set is closed

its its own closure

[prime elbow]

Okay thank you

[prime elbow]

I don't understand, sometimes to show one Topological space is different from another topological space, they take basis elements and sometimes open sets.

I am not sure about product topology and order topology. Is there any source from where I can learn?

[prime elbow]

Singleton set in R has no limit point, right?

So the singleton set is closed in R

[red yoke]

Yes

[red yoke]

Perhaps

[prime elbow]

Any lecture?

[prime elbow]

I want to prove that X has property that if all finite point sets are closed then for subset A of X, if x is a limit point of A then every neighborhood of x contains infinitely many points of A.

So let x be a limit point of A and assume there is a neighborhood of x U which contains finitely many points of A.

So by x is a limit point intersection of U and A-{x} has finitely many elements says {x1,......,xN}. And this set is open.

Now I use the result that if V is open and B is closed set then V-B is open set.

So U-{x1,......,xN} is open set containing x but this set has no other element of A except A, it contradicts that x is a limit point.
Hence our assumption was wrong.
Therefore every neighborhood of x contains infinitely many points of A.

Is it right?

[tiny obsidian]

"this set has no other elements of A except A" what do you mean by this (I assume it's a typo but want to make sure)

[prime elbow]

For x to be the limit point of A , every neighborhood of x contains an element of A which is not equal to x. So if x is the limit point of A then U-{x1,.....,xN} has a element of A except x but it has not therefore x is not the limit point.

Yes it is a typo. Thank you

In the Hausdorff space any singleton set has no limit point, right?

[tiny obsidian]

I think so

yes

You can go weaker than hausdorff to the same 'finite sets are closed' property (also known as T_1) and it would still be true

[prime elbow]

To prove the Subspace of Hausdorff space is Hausdorff space.

Let Y be a subset of X, then
let y1 and y2 belongs to Y then it also belongs to X so there exists U1 which contain y1 and U2 which contain y2 open sets in X such that they are disjoint.

So take intersection of U1 and Y , and intersection of U2 and Y so these are disjoint open sets in Y which contains y1 and y2, respectively.

Is it right?

[prime elbow]

Yes

[prime elbow]

There is a mistake, I used it except x no if x does not belong to A then intersection of U-{x1,........,xN} and A is an empty set , it also works but I assumed x is an element of A which is wrong.

[worthy olive]

Which book should ai use to self study primarily?

[prime elbow]

Show that X is Hausdorff if and only if the diagonal ∆ = {x × x | x ∈ X} is closed in X × X.

So first if X is Hausdorff space, then to show that ∆ is closed, I will show that any limit point of ∆ lies in ∆.

Let there is a limit point which is not in ∆ then it has form (x,y) where x≠y.

So if (x,y) is limit point then for all basis element U×V which contains (x,y) ( U is open in X and V is open in Y) has non-empty intersection with ∆.

But if x≠y then there exists U and V open sets which contains x and y respectively and they are disjoint so U×V contains (x,y) and intersection with ∆ is empty.

Therefore limit point of ∆ lies in ∆, thus ∆ is closed in X×X.

Now let ∆ is closed set.
For any x≠y, (x,y) does not belong to ∆ then (x,y) is not belong to closure of ∆.

If (x,y) does not belong to closure of ∆, that means there exists a basis element of X × X , U×V which contains (x,y) such that it is empty intersection with ∆ hence U contains x and V contains y is open set and they disjoint. Hence for x≠y, there is open set U which contains x and V which contain y and they are disjoint. Hence X is Hausdorff space.

Is it correct proof?

[prime elbow]

Upper limit topology on R has T_1 axiom how?

[drowsy iron]

I assume I have to do this by contradicting the connectedness of B but I can't find a way to do it. I don't even see how I'm supposed to use the hint.

[brittle rapids]

we want a partition that looks something like

      |
X\A  ∂A  A
      |

such that if B ∩ ∂A is empty, then both sides induce a decomposition of B

as is this isn't quite correct, but the corrected version is correct

[drowsy iron]

This helps, thanks

Ah I found a proof