point-set-topology

[little hemlock]

The singletons are just the points which aren’t glued to anything

For example, to make a circle, you can start with [0,1] and glue it’s endpoints together

You’ll have a partition like {{x} : x in (0,1)} \cup {{0,1}}

This cylinder construction you’ve given is basically this, except you are forming a circle on each fiber of the square

[jagged sage]

Sure, intuition makes sense. For the circle example, how do you conclude a circle is being formed just from the {{x} : x in…

[little hemlock]

There is a natural map [0,1] -> S^1 given by e^{2pi i t } which is constant on each element of the partition of the interval.

So it induces a continuous map on the quotient space

And this will be a homeomorphism from the quotient we constructed to the circle

[jagged sage]

Hm ok if we needed to construct a partition of [0,1] x [0,1] to form something complicated for me like Möbius strip how would I try

[little hemlock]

Can you guess where (0,0) should be glued? What about (0, 1)?

[jagged sage]

Prolly same location as in cylinder but how would I twist one side

[odd flame]

when are these all "useful"

[little hemlock]

It can’t be the same location as the cylinder. Otherwise you’ll just get the cylinder

Glue (0,0) to (1,1)

Glue (0,1) to (1,0)

[gritty widget]

Last one is useful to say that the product is a metric space

First two are useful in functional analysis

[little hemlock]

Think about what to do for the rest of the point on the edges

[little hemlock]

The first two are induced by norms on finite dim spaces. Therefore you can use whichever you want whenever it’s convenient

[odd flame]

yeah i realized that munkres says soon afterwards that the topology they induce is equivalent to product topology

[gritty widget]

Even though the norms are equivalent it's still useful to distinguish them. Just not for topology reasons.

[odd flame]

im assuming that the definition of convergence is basically the same as one you might find in an analysis course?

like for all points in the sequence, there exists an N s.t for all n > N, subsequent x_n will be contained within some "radius"

or i guess since we dont necessarily have a metric of the space something something nbhd of x_n

[gritty widget]

A sequence (x_n) is said to converge to a point x if, for every open neighborhood U of x, there is an N such that x_n is in U for all n > N.

I stress "a" because, in a non-Hausdorff space, a sequence could converge to two limits. The classic example is (1/n) in the real line with two origins. A more contrived example is any indiscrete topology, wherein every sequence converges to every point.

Since you were unsure what the definition of convergence should be, you should try to prove that convergent sequences in Hausdorff spaces have unique limits.

The converse is false, though. There are non-Hausdorff spaces in which every convergent sequence has a unique limit, such as the cocountable topology on an uncountable set.

(I googled that one.)

The moral of the story is that all spaces are Hausdorff.

[ocean narwhal]

*all spaces are sequential

[unreal stratus]

All spaces are path connected compact hausdorff

[gritty widget]

I'm not topology brained enough to know what that means, and I don't want to be.

All of my spaces are smooth manifolds.

[unreal stratus]

Based

[gritty widget]

LCH BTFO.

[unreal stratus]

All my spaces are (up to homotopy) finite cw complexes

[odd flame]

me neither and i have an exam in t-minus 12 hours

[gritty widget]

Good luck.

What topics are on it?

[odd flame]

sections 19 - 26 of munkres

product metric quotient topology, connectedness and compactness

doesnt sounds like a lot when i say it like that but

[gritty widget]

Trying to look back at my old general topology materials and see if there are any good questions that weren't pulled right from Munkers.

... most of them were.

[odd flame]

im just reviewing material for now, in about two hours i'll switch to reviewing problems/examples

appreciate it tho but i'll def be posting goober questions in here

[gritty widget]

The optional stuff on topological groups might be fun, and you could tackle it since you're posting rings in the algebra channel.

[odd flame]

yeah i wanna do that in my free time this weekend

im literally taking topology and algebra this semester so it's only natural lol

[gritty widget]

Here's one I remember being tricky. Prove that every infinite Hausdorff space has an infinite discrete subspace.

I think the idea is to do it for metric spaces (easy/easier) and then generalize.

Lee's book Introduction to Topological Manifolds also has nice problems.

It's really more of an introduction to topology through manifolds than a specialized course on manifolds.

[opaque cloud]

differential topology?

oh wait top manifolds not smooth manifolds

[gritty widget]

Topological manifolds. The second book in that trilogy is the differential topology one.

[opaque cloud]

nice

so yk how for a manifold to be smooth, the transition functions between its overlapping charts must be C^inf or at least C^k? I've seen some people define them as "locally diffeomorphic to R^n" as well

are those two things equivalent?

is it because like composition of differentiable functions is also differentiable?

[odd flame]

what is the significance of uniform convergence

oh i guess uniform limit thm looks important

[unreal stratus]

uniform convergence is rly important for many theorems and is a natural way to talk about functions converging

[opaque cloud]

okay do I guess that's true

does that extend to complex manifolds as well?

like instead of transition maps being holomorphic

"locally (holomorphic)homeomorphic between M and C^n"? like the homeomorphisms being holomorphic

[odd flame]

sorry completely unrelated to my last question but i'll go back to that soon - how is a quotient map stronger than contuity?

isnt this first condition just saying that a homeomorphism is a quotient map?

this is weird, so a quotient topology depends on the topology of a space that maps to a set

i guess that makes sense if its a quotient similar to how we think of quotients in algebra

[void tapir]

maps aren't necessarily surjective, quotient maps are

[void tapir]

no, it implies an injective quotient map would be a homeomorphism

first property there doesn't say anything about homeomorphisms

[odd flame]

[trail plaza]

For the lower limit topology, if you have a function $f: \mathbb{R}{ho} \to \mathbb{R}{ho}$ given by $f(x) = -x$, is the function continuous? person I'm working w/ says no cause w/ an example like $[0, 1)$, you'd get $(-1, 0]$, but I'm not entirely convinced that's true

[gentle osprey]

JesusChristSusan

[woven sinew]

The function x -> -x is indeed not continuous, when you put the lower limit topology on both the domain and codomain. Why? You have found an open set whose preimage is not open.

[odd flame]

how often do maps from compact spaces to hausdorff spaces come up

and how relevant is that consequence that its a closed map

[unreal stratus]

A lot of spaces you work with in point set will end up being compact and or hausdorff, so the answer is probably 'very often'

[unreal stratus]

25 is pretty important in particular

[gritty widget]

Only all the time

[gritty widget]

I object, 24 is equally important because it says in particular that it's a quotient map

[unreal stratus]

Oh I mean like it's an important consequence of 24

[gritty widget]

26 looks interesting, haven't seen it before

Wait no. That just says if f is perfect then preimages of compact sets are compact

So I've seen it before

Just didn't realize

I forgot all those proper/perfect map stuff. Never used it

[odd flame]

quotient maps still befuddle me, how does 24 show that

[gritty widget]

Continuous + open/closed implies quotient map

[odd flame]

silly question but what does an open set of the quotient topology look like

it depends on quotient maps right

so it must in turn also depend on the topology of X in p: X -> Y

[hidden crag]

it's the final topology wrt. to the quotient map

[odd flame]

wdym final

[hidden crag]

the finest topology such that the quotient map is continuous

(i called it "final topology" because this is a concept that can be generalised for families of functions into some fixed space and is called like that)

this gives you that U is open in the quotient space iff. p^-1(U) is open in the parent space where p is the quotient map

https://en.wikipedia.org/wiki/Final_topology#:~:text=In general topology and related,makes all those functions continuous.

in case my explanation is too chaotic

[odd flame]

mildly related - what is the significance of a map being closed/open

[gritty widget]

We are doing topology

[odd flame]

[gritty widget]

Closed open sets are important

Maps which preserve those are important

[odd flame]

ok i'll rephrase - why is an open/closed map not necessarily cont.

[gritty widget]

Because there are counter-examples to that

Take any bijective continuous function and inverse it for example

are closed/open maps which aren't continuous important? Probably not

[gaunt linden]

The identity map from R-with-the-usual-topology to R-with-the-discrete-topology is open and closed, but not continuous.

[gritty widget]

yeah, two topologies on the same set, one strictly finer than the other

[jagged sage]

How do I prove the real projective plane RP^2 is compact. It’s defined as the quotient space obtained by taking the antipodal points of S2 but other than that not sure

[woven sinew]

As a hint, note that the real projective plane is the image of S^2 under some map, and that S^2 is compact.

[unreal stratus]

is that even a hint lol that is the proof xd

well i suppose you haven't given the map

[woven sinew]

Nor the theorem used to conclude that RP^2 is compact. But yes. 😆

[jagged sage]

I have proved before that continuous image of compact space compact. For the map, I’m not sure but could it be mapping opposite points of the 2 sphere to each other

[unreal stratus]

the image of that map is S^2

Use the fact that RP^2 = S^2/~ for some equivalence relation ~. In general, what is the most clear map from A to A/~?

[lusty trench]

Is there any easy description of the generator of the first nonzero homotopy group of the complex Stiefel manifolds $V_{n-q}(\mathbb C^n)$? (Which Milnor and Stasheff, citing Steenrod, assert is the $(2q+1)$-th one.)

[gentle osprey]

Automorphism

[lusty trench]

I am reading Steenrod as well, and he gives an explicit description of the generator of the first nonzero homotopy group of the real Stiefel manifolds. But nothing similar for the complex ones.

[cedar pebble]

V_k(C^n)=U(n)/U(n-k), now just use the long exact sequence of homotopy groups

[lusty trench]

Thanks! What I want is the fact that U(k) acts trivially on the homotopy groups of V_k(C^n). That is just a consequence of the fact that U(k) is connected, right?

[cedar pebble]

I believe so yes

but yeah you're just doing the same thing you do with real Stiefel manifolds replacing orthogonal groups with unitary groups

the homotopy groups of unitary groups are easier than for orthogonal groups

so the answer ends up being a bit simpler 🙂

[lusty trench]

For orthogonal groups, O(k) does not act trivially on the homotopy groups of V_k(R^n). And that sometimes causes a twist in the bundles of homotopy groups of V_k(E), for E -> X a real vector bundle.

Anyway, thanks!

[small hemlock]

Prove that if $U_i \subset X_i$ is open, then $\prod_{i=1}^{n}U_i$ is open in $\prod_{i=1}^{n}X_i$. To prove this, do I just pick up an element in $\prod_{i=1}^{n}U_i$ , define the basis element it's in, and show that it's contained in a basis element for $\prod_{i=1}^{n}X_i$?

[gentle osprey]

J.Ross

[small hemlock]

I don't think I'm following this stuff very clearly lol

[unreal stratus]

how are you defining the product topology

(because under some definitions this is literally by definition)

[small hemlock]

We've only defined it for finite cartesian products for now, defining the product topology on $\prod_{i=1}^{n}X_i$ by the basis $B_{P} ={\prod_{i=1}^{n}b_i : b_i \in B_i}}$.

[gentle osprey]

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

[unreal stratus]

Oh is this like given bases of the X_i

Huh haven't seen someone define it like that since that is basis-dependent lol seems to just complicate it

[small hemlock]

Oh yes, sorry, each X_i has a respective basis B_i

He's using his own text, which is extremely unfortunate, but I have to roll with it

[unreal stratus]

Ah okay

[small hemlock]

For X_1,...,X_n

I can show the proof of it being a basis if it's any consolation

[gritty widget]

Then you need to prove that the topology is independent of the basis chosen

[small hemlock]

He stated this which I suppose makes sense

The "next exercise" is in part asked above

[gritty widget]

Take $x\in \prod_i U_i$ then $x_i\in U_i$ so $x_i\in B_i\subseteq U_i$ for some $B_i\in \mathcal{B}_i$ so $x\in \prod_i B_i\subseteq \prod_i U_i$ so such products form a basis for product topology

[gentle osprey]

Blitz

[small hemlock]

Ahhh okay that makes sense

[candid hedge]

can someone give me an example of an open dense set

i cant find any

[gritty widget]

R - {0} in R.

[candid hedge]

in R if there is any

[unreal stratus]

R in R

[candid hedge]

other than the set itself

[unreal stratus]

but fr R minus finitely many points is a nice example ^

[candid hedge]

Ok so we can only remove points

[gritty widget]

R^2 minus a line.

R^3 minus a plane.

etc.

All good examples.

[candid hedge]

but those arent open

right?

[gritty widget]

They are.

[unreal stratus]

They are, since the plane/line etc are closed subsets of the space

[candid hedge]

Ah right i was thinking of the line not being open

okay got itt

the intersection of infinitely many open dense sets is also dense and open?

[unreal stratus]

I guess if you want you can do something like R minus loads of copies of the cantor set and stuff like that lo

[unreal stratus]

Not in general, but for countably many this is Baire's theorem

[candid hedge]

Yessssss

how do i prove that

[gritty widget]

Careful. The intersection may not be open.

It will be dense, though.

[unreal stratus]

Okay sorry good point i didn't see that lol

[candid hedge]

so it can not be open?

even if its dense?

i know the intesections of infinitely many open sets can not be open, but is it the same when theyre dense

[gritty widget]

Find an example.

[candid hedge]

from the examples youve given me it doesnt worj

even if i remove infinitely many itll always stay open

[gritty widget]

Intersect R - {1/n} for n ranging over the positive integers. These are all open and dense. Is their intersection open?

Dense?

[candid hedge]

it seems open and dense to me :/

[gritty widget]

Why do you think it is open?

[candid hedge]

we're just removing points

and a set of points is always a closed set

so its compliment is open

[gritty widget]

"A set of points is always a closed set" is completely false.

[candid hedge]

it needs to contain its limit 0 okay

[gritty widget]

So you agree that the set I wrote down is not open?

[candid hedge]

so {1/n} for n going to infinity isnt closed? :/ i am a bit lost now

[gritty widget]

It is not closed.

[candid hedge]

is there an intuitive sense for this?

[gritty widget]

0 is a limit point but it is not in the set.

[candid hedge]

or just sequential definition?

okay yeah

yeah yeah its good

[candid hedge]

yep

but its surely dense

nice

got it

thank you

[gritty widget]

Yeah because R\Q is such set

[candid hedge]

yeah :(

indeed

[gritty widget]

Yeah. To see this enumerate Q

What an irrational example.

You can remove any other countable set from R too

Baire space is best space

Who doesn't love zero-dimensional Polish spaces

Bear space.

You just walked into the wrong bar.

[jagged sage]

Trying to describe quotient space D^n/S^{n-1} generally. I know Sn-1 is boundary of Dn so I try to visualize case for n=2. D2 is the unit disk in R2 right and I collapse boundary to a single point

Do I just end up with unit circle in R2

[lunar yoke]

no, you get the 2-sphere

and this generalizes to D^n/S^{n-1} = S^n

[jagged sage]

The 2 sphere is unit sphere in R3 right

[lunar yoke]

yes

[jagged sage]

And D2 is unit disk in R2?

[lunar yoke]

yes

[jagged sage]

How does collapsing boundary of disk end up with S2

[lunar yoke]

see if this answer helps

https://math.stackexchange.com/questions/3729399/dn-sn-1-cong-sn

[jagged sage]

Yeah image is helpful. How does the bottom half of the circle get constructed

[lunar yoke]

ok i found this animation

https://www.youtube.com/watch?v=IVkPGGC_2R8

[jagged sage]

Thanks makes way more sense. When you “collapse a set to a point” the point is arbitrary right? Like in this example, bottom semi sphere is formed by considering collapsing of boundary to point below and above the disk for example

And quotient space contains set of these collapsed arbitrary points?

[lunar yoke]

i dont really understand your question

the points in the quotient space are equivalence classes

the whole set you collapse is mapped to a single equivalence class, i.e. to a single point in the quotient space

[jagged sage]

If I have V_n nonempty closed subset of compact space X with V_{n+1} subset Vn for all n in N, how do i show that if intersection of V_n for all n in N is subset of U for open U, V_n subset U for some n

I have proven the intersection is nonempty but not sure if it's useful here

[glass bison]

Yo could someone check if my proof here is sufficient

[wispy veldt]

Suppose otherwise , that V_n intersect U^c is non-empty for all n , conclude.

[jagged sage]

What is the contradiction? V_n intersect U^c should be non empty, no?

[wispy veldt]

If thats the case for all n
Then that would contradict the assumption " intersection of V_n for all n is a subset of U "

can you prove why

[jagged sage]

Not sure but maybe something to do with X\U closed

[wispy veldt]

To show some set A is not a subset of a open set U its enough to show A intersect U^c is not phi \

Notice if V_no intersect U^c is not phi
then so is intersect Vn for n<=no

Do you see where we are going with this?

[jagged sage]

Right, I agree and everything makes sense. But I don't see where the contradiction will be. If the intersection of all Vn with U^c is nonempty, we can use earlier result and say that U^c has same characteristics as V_n I think

[wispy veldt]

Dont know what you mean by "U^c has same charachteristics as V_n "

The contradiction is || if V_n is not a subset of U for all n , then V_n intersect U^c is not phi for all n , thus the infinite intersection of Vn has to intersect U^c (because otherwise there is a V_n that is a subset of U), this proves the intersection of Vn for all n is not a subset of U which is against our assumption ||

[jagged sage]

I see I was being foolish and thinking wrong about hypothesis. Thanks for help

[gloomy bridge]

Question about concept of continuity
So it's been a while since my undergrad top course, but I randomly got in the mood to work through some basic stuff.

[gloomy bridge]

The short version I recall for what makes a function $f$ continuous is preimage of open is open" which I think means Given some function $f: X \to Y$, $f^-1(Y)= {U \in \tau_X: f(U)=V \in \tau_Y }$. So the preimage of the codomain $Y$ is the set of all open sets of $X$ that also map to open sets of $Y$."
\
\
The intuition I had at the time (and still do) was that what makes this definition capture the notion of continuity is that things that are close" in $Y$ must have come from things that are close" in $X$. And if that's NOT the case, then that means that there are some things that are close" in $X$ that got split" while traveling on their way to $Y$.

[gentle osprey]

Vin

[gloomy bridge]

Alrighty, so with this intuition in mind, I'm confused by the function $f: [0, 2\pi) \to S^1$ where $S^1$ is the unit circle in $\mathbb{R}^2$ and $f(\phi)=(cos\phi, sin\phi)$. The claim I've come across is that $f$ is continuous, but it's hard for me to see how this is true. That is, what's close" in $[0,2\pi)$ that is split" apart in $S^1$?

[gentle osprey]

Vin

[gloomy bridge]

My gut is that I might be incorrectly believing that the space $[0,2\pi)$ has the usual topology of $(a,b) \in [0,2\pi)$ where $a,b\in \mathbb{R}$ and $0 \leq a < b < 2\pi$.

[gentle osprey]

Vin

[gritty widget]

What do you know already? Are cosine/sine continuous? How are they defined?

For me... it's kind of obvious

This is literally exp(it)

Which is continuous because exp and multiplication by i are continuous

Or, looking differently, it's continuous because the functions cosine and sine are continuous

We have different equivalent notions of continuity and why would you work with just one, anyway

The preimage definition is good because it's comfortable in lots of arguments, they become purely set-theoretical

[gloomy bridge]

So from a metric space sense, sure $sin\phi$ and $cos\phi$ is continuous (ie, open balls and stuff). That's not what I'm after, though

[gritty widget]

That's the same

You prove it once and leave it

[gloomy bridge]

What's I'm trying to grok is which sets of $[0,2\pi)$ don't stay inside of another open set in $S^1$

[gentle osprey]

Vin

[gloomy bridge]

again, I think there's something wrong with my assumption that an open set in $[0,2\pi)$ looks like $(a,b) = { x \in [0,2\pi): a < x < b}$

[gentle osprey]

Vin

Vin

[gloomy bridge]

Oh, I think I found it by writing through this. Just need some confirmation that my logic (and more importantly, foundational understanding) doesn't have holes:

[gloomy bridge]

Take the coordinate point $(1,0) \in Y = S^1$ and a neighborhood around it $V$ that lies on the unit circle $S^1$ and is arbitrarily close to $(1,0)$ (can't remember how to easily parameterize the interval I'm imagining, but oh well). Then there is no open set in $X = [0,2\pi)$ that maps to $V$ since you need to be on both sides of the coordinate point $(1,0)$ on the unit circle $Y=S^1$.
\
\
For instance, if I grab some small open set $U_1 \in X = [0, 2\pi)$ that's super close to $2\pi$, then the image of $U_1$ under $f$ never grabs the coordinate point $(1,0) \in Y = S^1$. On the other end, if I try using some really small $U_2$ that's close to 0, I run into the same issue. The open set $U_2$ is either not close enough to the lower bound 0 (so it misses the coordinate coordinate point $(1,0)$) or it oversteps like $U_2 = (0 - \epsilon, 0 + \epsilon)$ and falls outside of our space $X= [0,2\pi)$ making it actually not an open set

[gentle osprey]

Vin

[gloomy bridge]

In any case, I should probably stop procrastinating on my day job by rehashing topology stuff for fun. I'll check back later to see if someone can confirm/reject my reasoning or intuitions. (Miss my days in uni when there was guaranteed feedback of thinking through this stuff; maybe one day I'll try to utilize Adga or Coq to just check my logic)

[little hemlock]

This is \emph{not} the topology on $[0,2\pi)$. A set $U$ is open in $[0, 2\pi)$ if $U = [0,2\pi) \cap V$ for some open set $V$ in $\bR$. For example $[0,1) \cup (2, 2\pi)$ is open in $[0,2\pi)$ since $$[0,1) \cup (2, 2\pi) = [0,2\pi) \cap ((-\infty, 1) \cup (2, \infty))$$
So with this in mind, you can think about how to construct an open set of $[0,2\pi)$ whose image is contained in an open nbhd of $(1,0) \in S^1$.

[gentle osprey]

kxrider

[little hemlock]

Also it seems that your "intuition" that $f$ shouldn't be continuous is actually the intuition for why $f^{-1}$ is not continuous. This is because $f^{-1}$ will take a small connected nbhd of $(1,0)$ and send it to some disconnected set in $[0,2\pi)$.

[gentle osprey]

kxrider

[little hemlock]

going to bed, but hope that helps clear things up somewhat

[urban zinc]

I'm confused, where do we use the fact that A_alpha is path-connected?

it doesn't seem like we need it..

[gritty widget]

It isn't used, but it's nice to assume path-connectedness when you're doing SVK .

[urban zinc]

oh okay

what's svk?

[gritty widget]

I got the capitalization wrong. Seifert-Van Kampen.

[urban zinc]

ahhh

alright ty

have a good day! :)

[gritty widget]

That is what Hatcher's doing, right?

The assumption is there because you want the fundamental groups at each point to be isomorphic.

[urban zinc]

haven't gotten to van kampen's thm yet

this is in the proof that the fundamental group of S^n = 0 if n ≥ 2

[gritty widget]

This lemma is used in the proof of the theorem as well.

[urban zinc]

ahh okay

[gritty widget]

The lemma basically says that the map in van Kampen's theorem is surjective.

[gritty widget]

Some googling suggests that the path-connectedness assumption is there only to be able to go from the fundamental groupoid version to the fundamental group version.

This is outside my expertise, though.

[unreal stratus]

Yeah I don't think you should need anything to do with path-connectedness of the $A_\alpha$, cause really all we care about is the path components of the $A_\alpha$ containing $x_0$ and that the intersection of those corresponding path components is also path connected. But it doesn't really matter since we could just redefine the $A_\alpha$ to be those components anyway

[gentle osprey]

potato

[unreal stratus]

Yeah in this lemma all you care about is a given basepoint but to be able to move between different basepoints you need those additional path connectivity assumptions

which hatcher just seems to be doing anyway but that's chill

[gritty widget]

That's what I thought. Nice.

[urban zinc]

ahhh alright

ty!

[tawdry valve]

I'm wondering if for y in N subset V, when do we have H_n(V, V-N) -> H_n(V, V-y) induced by inclusion is an isomorphism. I would want something like N deformation retracts onto y, but it's not working out for me yet.
here's what I've tried so far: look at the LES's of the pairs (V, V-N) and (V, V-y). We get a chain map between them via i_*, so I would like to get isomorphisms of H_n(V, V-N) and H_n(V, V-y) by applying the 5-lemma

[gentle osprey]

Joseph

[tawdry valve]

but I don't know if i_* is an isomorphism of H_*(V-N) -> H_*(V-y). Is there a homotopy inverse to the inclusion, given that N deformation retract onto y, or do I need more assumptions?

I guess this will work in manifolds and stuff just by doing a radial projection sorta thing

[gentle osprey]

Nobody

[unreal stratus]

does simplicial homotopy theory get more intuitive or do I just have to think of everything syntactically? lol

Like I'm working through stuff on Kan complexes and homotopy groups etc and I see the analogy to topological spaces but can't really visualise anything (probably unsurprisingly)

[unreal stratus]

Of course, sure thank you :)

[plain raven]

potato have you computed a good handful of geometric realizations

[median sand]

Is it true that if a subset of R^n is discrete, then its intersection with every compact set must be finite (and vice versa)? I suppose it would be enough to show that the closure of a discrete subset is discrete, but that doesn't seem to be the case (e.g. take {1/n:n>=1}).

[gritty widget]

Your first question is yes no, because of your example

[coral pivot]

that aint discrete

theres a limit point at 0 right

[gritty widget]

That's discrete

There's nothing preventing discrete sets to not be closed

[coral pivot]

yeah sorry my bad

[gritty widget]

If your set is closed and discrete, then intersection with any compact subset of R^n is finite

If a subspace of R^n is closed and intersections with compact sets are finite, then because it's a K-space, it must be discrete

[unreal stratus]

Not yet

Only rly for simplicial complexes

[median sand]

What's a K-space?

[gritty widget]

Space which topology is determined by topologies on compact sets

[jagged sage]

Trying to prove RP^2 compact by using def of RP2 as quotient space obtained by taking antipodal points of S2. If I take open cover, union U_i of RP^2, union f^{-1}(U_i) will be open cover of S2 and S2 compact so we have finite subcover, ig union from i=1 to n f^{-1}(U_i) finite. So i have union from i=1 to n U_i is finite and union of open sets in RP^2 but how do I ensure it actually covers

[unreal stratus]

Note if $S^2 = \bigcup_{i=1}^{n} f^{-1}(U_i)$, then since $f$ is surjective we have $\mathbb{RP}^2 = f(S^2) = f(\bigcup_{i=1}^{n} f^{-1}(U_I)) = \bigcup_{i=1}^{n} f(f(^{-1}(U_i)) = \bigcup_{i=1}^{n} U_i$, where i've used the fact that if $f$ is a surjection then $f(f(^{-1}(U)) = U$

[gentle osprey]

potato

[unreal stratus]

In fact, this proves more generally that the image of a compact space under a continuous map is compact and since quotient maps are surjective, in fact any quotient of a compact space is again compact - which is nice to know

e.g. Mobius strip, Klein bottle and so forth are all compact, being quotients of a square

[jagged sage]

Clean explanation thank u

[rancid sand]

I'd like to understand my error on this question

Where I got no right point

Maybe one was true

[hidden crag]

you might want to give translations of the question and the statements

[graceful abyss]

I think I know where you're from looking at the website lol
What course is that?

[unreal stratus]

dope np

[jagged sage]

If X is infinite set with cofinite topology and A subseteq X finite, what is cl(A)

[jagged sage]

thought cl(a)= empty since only two closed sets are X and empty but it not correct

would make sense if cl(A)=A but idk why empty set wrong

[little hemlock]

its the cofinite topology. doesn't that mean finite subsets are closed?

[jagged sage]

Yes

[little hemlock]

so its not true that the only closed sets are X and the empty set

[gritty widget]

The closure of a closed set is itself...

[jagged sage]

Yeah that is dumb so X not closed

[little hemlock]

X is closed

X is the only closed infinite set

[jagged sage]

Ok yes empty set open and complement

I’ve realized I was thinking A was open

Sorry for bad question thanks for help

[little hemlock]

its all good npnp

[paper wedge]

yo

A U Bd(A) is closed

[wispy veldt]

What are some example's of a sequential space that is not Frechet-Urysohn ?

[paper wedge]

proof: let x be in the complement --> x is an exterior point of A --> there exists a nbd of x subset of X-A --> N intersect A is phi

what then

this is over a metric space

B_e intersect A is phi --> B_e/2 intersect (AUbd(A)) is also phi

so B_e/2 is a subset of A U bd(A) all complement

so its open

done?

[gritty widget]

https://dantopology.wordpress.com/2010/06/21/sequential-spaces-i/
See examples. Note that Frechet-Urysohn spaces are precisely the hereditarily sequential spaces

[paper wedge]

anyone

[unreal stratus]

It's a bit hard to read

(Why do you say is phi instead of is empty 😭)

[paper wedge]

my bad

[gritty widget]

And it's a mess

[unreal stratus]

Also, how are you defining the boundary

Under some definitions, this is a triviality for example

[paper wedge]

points that have nbds that intersect with both A and A complement

[unreal stratus]

Okay sure

[wispy veldt]

alright ile give it a look and come back with any questions , thank you

[summer sequoia]

hm question, can there be a set with both cluster/limit points and isolated points?

[swift fjord]

sure

take for example the set [0,1]\cup {3} in R

3 is an isolated point clearly

but [0,1] is closed and in particular contains all its infinitely many limit points

[summer sequoia]

oh cool thanks

would this be one

${.9,.99,.999,...} \cup {100}$

[gentle osprey]

imnotrachel

[summer sequoia]

oh why are there no brackets

[swift fjord]

you need to do \{\}

this has a limit point that's not in the set

but 100 is an isolated point

[gritty widget]

every point is isolated, in fact

[summer sequoia]

oh i see so it would need to be in like an interval such as what you put [0,1]

[swift fjord]

true

what are you trying to achieve imnotrachel

[summer sequoia]

im trying to understand a new concept im learning in real analysis

just got introduced to topology

that was just a question my professor left us to wonder about till next time

he also asked if its possible for a set to be an open set whose complement is not closed

which im not sure what that means entirely

[gritty widget]

one of the definitions of closed sets is complements of open sets

it might be trickier to see depending on your definitions, but the answer is no

it's impossible

[summer sequoia]

so its impossible because the complement of an open set is closed

[swift fjord]

well, in R this is not obvious a priori

and needs to be proven

[tidal cedar]

does anyone know if the two complexes in Dolbeaut cohomology are chain homotopic up to a shift

[tidal cedar]

as in the ones given by partial and partial bar

[pearl holly]

this must be super trivial, but I really do not see how the "long equation" involving pullbacks proves that the formal group law is x + y + xy (oops I cropped away the last part, but the last equation says f(x, y) = x + y + xy)

the formal group law is given by the image of the "Thom class" under this map, the induced map characterizing the tensor product of two line bundles, but I don't even see how last expression in the "long equation" in the prev. post lives in the image

I should just go to sleep

also wait, is E^*(point) the same as pi_*(E)?

[unreal stratus]

So I've read that if X is some space then we have a weak homotopy equivalence |Sing(X)| -> X. Is this a nice way to show that cw approximations exist and can be defined functorially? because otherwise i'm surprised i'd not seen it earlier lol

maybe it's just the initial hurdle of defining simplicial sets etc which is the issue

[plain raven]

yeah this is proved in may's book on simplicial homotopy theory

[unreal stratus]

Dope

I'm enjoying this stuff hehe

[small hemlock]

This was in my topology homework but I think it's really easy which is, obviously, making me overthink so I just want to check. "Let x be a point in a metric space X, and suppose $r > r_0 > 0$. Show that $B(x,r_0) \subseteq B(x,r)$". I just had as proof: Let $y \in B(x,r_0)$. Then $d(x,y) < r_0$. But $d(x,y) < r_0 < r$, so $d(x,y) < r$, and $y \in B(x,r)$. Eh?

[gentle osprey]

J.Ross

[gritty widget]

This is fine.

[small hemlock]

Swick

[gloomy bridge]

Ah, thank you! I knew something was wrong...

Amazing, thank you for this 🙏

[small hemlock]

"A subset U of a metric space (X, d) is open if and only if for every x \in U, there is some B(x,r) \subseteq U, r>0". Don't both implications follow directly from the metric topology having the set of all balls centered at points in X with r>0 as a basis?

[gritty widget]

Since you were assigned an exercise like the one above, I think you're expected to provide a little bit more detail. But yes, that is the idea.

I would normally say that this is pretty much immediate, but you were assigned a similarly immediate exercise above, so there's probably some expectation of giving all the details.

[small hemlock]

Oh most definitely, I just wanted to make sure I had a decent intuitive idea first. Thank you for the help! I'll be sure to right it our formally as to why it follows so directly

[supple heart]

Which definition for topological spaces is most commonly used?

The one with open sets or the one with neighbourhoods?

Ik theyre equivalent but the open set one is easier to remember

[pearl holly]

@empty grove sorry for ping but you did stuff like this before right?

[unreal stratus]

not sure what you mean

[gritty widget]

you can be given a basis of neighbourhoods for each point

[unreal stratus]

Yeah it's just vague as stands imo

[gritty widget]

not really. Anyway the open sets definition is clearly used more often

[unreal stratus]

Well I've never seen someone define what a topological space is (in general terms) in terms of neighbourhood bases

only specific examples

[gritty widget]

I think Hausdorff used to define topological spaces in this way, historically
but I haven't seen such definitions myself either

it's archaic

[supple heart]

Allright thanks

I still dont fully understand how open sets work. What if the point we are considering the neighbourhood of is at the "margin" of the space? Wouldnt the spherical neighbourhood with epsilon radius "go outside" of the space?

I mean a part of it

Or epsilon has to be in the subset?

[gritty widget]

You're asking why something like, say, [0, 1/2) is open in [0, 1]

right

there's no problem here

the neighbourhood of 0 in [0, 1] is not the same as neighbourhood of 0 in R

[supple heart]

Something like that, yeah. But what of the neighbourhood of 1 in [0;1]?
Cant the neighbor of 1 go over the 1] so like (1; something)?

Thats what I dont understand

[gritty widget]

same thing

If you're working in a topological space [0, 1], then you forget about everything that R is

3? Doesn't exist for you

well not literally but what I mean is you restrict your view to [0, 1] alone

you can't take subsets outside of your space of interest and claim they are open

[supple heart]

And every point in [0,1] has a neighbour there?

[gritty widget]

where

[supple heart]

In that interval

[gritty widget]

well yes, multiple neighbourhoods

[supple heart]

In [0,1]

[gritty widget]

a whole bunch of them

[supple heart]

Ah I see now. Well my confusion them stems from the fact that the book didnt give any example of set

I imagined it worked a bit different then what you tell me

[hasty echo]

Hi everyone, i have a question that i cant seem to get anywhere with

F:R2 -> R
f(x,y) = x+y^2
Show f is an open map

This should be easy but im stuck

I want to prove its a quotient map by showing its a continuous and open map.
Continuous is simple. But to show its an open map i couldn't

[gritty widget]

$$f((a, b)\times A) = \bigcup_{y\in A} (a, b)+y^2$$ is open as union of open sets

[gentle osprey]

Blitz

[gritty widget]

so f is open

[hasty echo]

Oh

[gritty widget]

yeah. Images have this useful property that $f(\bigcup_i A_i) = \bigcup_i f(A_i)$, which is why it's enough to check it for the basis

[gentle osprey]

Blitz

[gritty widget]

(saying this for completeness sake)

[supple heart]

Would (0,1) be closed then?

[gritty widget]

in what space

[supple heart]

Idk, R?

[gritty widget]

no

[supple heart]

In what space would it be closed?

If its disjoint?

[gritty widget]

what's the point of this question

[supple heart]

I cant think of a closed set, but thats probably because im thinking of the space R^n?

[gritty widget]

so you just want an example of a closed set?

[supple heart]

Yep

[gritty widget]

take any singleton in R^n

[hasty echo]

Blitz are you a professor?

[supple heart]

Facepalm ensues

[grizzled ibex]

you can define it with filters too

bourbaki did it if i'm not mistaken

[hasty echo]

Im asking because im doing graduate in math, and i still feel inadequate

[supple heart]

I thought that if the empty set is open then a singleton would be open

[hasty echo]

Shit is hard

[supple heart]

For some reason

...the empty set is open right?

[grizzled ibex]

empty set is open by vacuity/definition

[supple heart]

Yeah

Thanks

[grizzled ibex]

i like introduction to topology based on filters

[supple heart]

I have this book topology by h .schubert

But he gives noo examples he only did for metric spaces

[grizzled ibex]

do you think i can get it online for free?

hmm

take cofinite topology or product topology

it could be good abstract examples

[supple heart]

Ill see what I can scour up

[grizzled ibex]

take zariski topology too

another very abstract example

[supple heart]

This is the most abstract subject Ive ever studied. The error in my thought was that I saw open sets as subspaces

And I did a lot of mental headcanon

[grizzled ibex]

I see

trying to get a grasp of the intuition behind it isn't always good

i understand you

if you associate it with "weak" things like spaces you will lose some of the general intuition sometimes

try to study topology with filters

[supple heart]

Wdym with filters?

[grizzled ibex]

1 sec

https://en.wikipedia.org/wiki/Filter_(set_theory)#:~:text=A%20filter%20on%20a%20set%20may%20be%20thought%20of%20as,a%20filter%20is%20an%20ideal.

tf is wrong with this link

any topology expert here?

[supple heart]

Well, Id catch up on general topology before spicing it up with filters or any sort of variation

[grizzled ibex]

If you find a good book.

Which im not sure which one to recommend to you

filter could give you a better grasp of topology

[supple heart]

Its fine ill figure it out

[grizzled ibex]

than any metric space

idk if alot of people agree with me

[supple heart]

But do you see how that could cause me confusion when going back to normal topology?

[grizzled ibex]

Yes and no..

with a good book

or a good teacher

maybe you can do the "parallel thinking"

@wispy veldt do you agree with me?

[wispy veldt]

Do you know of a good reference for filters in topology @grizzled ibex? ive been really interested in studying it.

[grizzled ibex]

Bourbaki probably

"Bourbaki General Topology"

that's probably your way

@supple heart https://math.stackexchange.com/questions/2925070/book-reference-for-topology-can-i-avoid-filters

[unreal stratus]

lol

depending on how much you wanna know, i found a nice handout on ultrafilters & tychonoff a while back

oh yeah, this https://ericmoorhouse.org/handouts/ultrafilters.pdf, short but goes up to like proving tychonoff and introducing stone-cech compactification etc

[grizzled ibex]

just like the dude in math.stackexchange said, filters are very powerful.

[unreal stratus]

ig it depends on what you wanna do right

[supple heart]

Allr

[grizzled ibex]

just give it a try

[grizzled ibex]

but in your opinion, what do you think?

[unreal stratus]

wdym

[wispy veldt]

The connection between ultrafilters and various theorems has been a fascinating subject to me recently, i was just about to ask for references on its relation with tychonof lol

[grizzled ibex]

is it really necessary? or just very useful in specific topics?

[unreal stratus]

Oh

From what I can tell they are only useful if you have a specialised interested in general topology lol

[supple heart]

Oh

[unreal stratus]

Like I've never used them in algebraic topology for anything and when I took my first topology class I was practically dissuaded from learning about them as opposed to smth else

[grizzled ibex]

makes sense

[gritty widget]

In functional analysis nets are indispensable and filters are just cumbersome

[unreal stratus]

But they are nice to understand stuff like Tychonoff for sure and I'm sure they have plenty of applications in general topology lol like blitz will know better

[grizzled ibex]

what do u mean?

with cumbersome

[gritty widget]

I know few applications of filters but it doesn't strike me as the miraculous thing that people are trying to paint it as here

[hasty echo]

Is this sound enough?

[grizzled ibex]

people argue that both nets and filters are good

[wispy veldt]

What are some advantages nets have over filters? ( in the case of topology ofcourse)

is it just easier to work with ( which i assume so from what you said)

[gritty widget]

neither is easier to work with

both are easier in some circumstances while cumbersome in others

[grizzled ibex]

now i understand your opinion

[wispy veldt]

fair enough

[gritty widget]

no

also it's hard to read if it's not in latex

what you want to take is something like the origin and point (0, 1)

if you just take some point close to the origin, that won't work

[tidal lynx]

can someone explain the counterexample to me

if max|f - g| = 0 then f(x) - g(x) = 0 for all x, so f(x) = g(x) for all x, and so they are the same function

they say that it is enough for f, g to intersect at exactly one point, but then max|f - g| is clearly not 0 right?

[gritty widget]

The argument in the picture doesn't make much sense. As you say, d(f, g) = 0 does imply f = g.

[gritty widget]

it doesn't define metric because it's not well-defined

[tidal lynx]

oh like if f diverges and g = 0

[gritty widget]

it's bonkers, whoever wrote it is wrong

[wary terrace]

Hi, I I’m currently trying to teach myself abstract algebra and wanted to learn a little bit of topology for fun. I’m completely new to topology and know basically nothing about it, but I wanted to learn a little bit of the basics. Computation wise, where should I start?

[old violet]

Or Topology without tears if you prefer something relatively easier to read: https://www.topologywithouttears.net

[old violet]

or Topology Now! https://www.amazon.com/Topology-Now-Robert-Messer/dp/1470447819 if you want to learn a bit of what "modern" topology research is about. Quoting from one of my profs: It gives you some taste of what part of topology topologists are interested in. Although I personally don't like the writing of it..

[little hemlock]

Or topology: a categorical approach 🤓

[wary terrace]

Thanks for the suggestions 🙂

[jagged sage]

How do I prove union A_i for i in I connected if {A_i | i in I} is indexed subfamily of path connected subspaces with each Ai Aj nondisjoint

intuitively makes sense dont know how to rigorously prove. presumably find path connection for union but not sure how

[gritty widget]

Path connectedness is stronger than mere connectedness. You only have connectedness, so you can't use paths to prove this.

To do it rigorously, start from a separation of the union and get a contradiction.

[jagged sage]

I've just realized i might have formulated question wrong. I am trying to prove the analogue of this question for path connected spaces where the original was connected. I would guess that this means A_i is indexed subfamily of path connected subspaces instead of connected

any suggestions with this in mind

[gritty widget]

Start with two points and try to connect them with a path.

Just start to prove it as you normally would and see if you can smell anything.

[jagged sage]

It seems obvious but idk take x,y in union, then x,y in some U_x and U_y respectively and since U_x, U_y path connected there is probably some way to generate path connection for this union u_x and u_y

[gritty widget]

There is. Think about it.

Remember your assumption on the intersections of these sets.

Draw a picture if it helps.

[gritty widget]

In case you haven't gotten it yet, ||there is a point z in the intersection of U_x and U_y. Since U_x is path connected, you can find a path from x to z. Similarly a path from z to y. Concatenate them.||

[jagged sage]

Yeah i had got it thanks for help

[empty grove]

Nope, can't really follow it

[ocean narwhal]

topology >computation wise

[pearl holly]

Okay, thanks for taking a look tho

moment

[gritty widget]

since you're new I suggest reading about the topology of metric spaces first

I didn't read it myself but I found this book: Topology of Metric Spaces by Kumaresan

[pearl holly]

Lol nlab has an introduction to point set https://ncatlab.org/nlab/show/topology

I’m not saying I’m recommending it btw, I just find it funny

Okay it’s not too much point set in there

https://ncatlab.org/nlab/show/Introduction+to+Topology+--+1

This one has a lot more lol

[viral atlas]

I think that topology of metric spaces is a better starting point as well

[frigid patrol]

Can we phrase hilbert theorem 90 in terms of some classifying space being simply connected?

[rugged swan]

What

What is your intuition

Are you talking about the theorem on elements of relative norm 1 ?

[frigid patrol]

Yes

H90 says

Group cohomology can be interpreted as the cohomology groups of some Eilenberg-Maclane space

So my intuition is that H90 is equivalent to saying some space is simply connected, but what space is the correct one to consider here?

[rugged swan]

H1 = 0 doesn't imply simply connecteness

Somehow you should be able to write H^1(Gal,L) = [Gal(L/K),K(1,L*)]

[tidal plume]

Hey, we just learned about cw complexes in my alg top class and im kinda struggling with the intuition behind it, specifically when you "are allowed" to attach n-cells to your space. The exercise im currently working on requires you to construct the solid torus as a cw complex, i know that you can construct a regular torus by first attaching the endpoints of two 1-cells to a single point and then you can somehow attach a 2-cell to this construction to get your torus, therefore i am tempted to say that we can create a solid torus by attaching the boundary of a 3-cell to this 2-cell so we "fill the inside", but i know this is not the way to do it because the euler characteristic of the solid torus is 0. Im wondering if there is some easy way to see in which cases you can/cannot attach an n-cell, for me even the regular torus is a construction i don't get

[golden gust]

you can only add an n-cell by gluing it to the (n-1)-skeleton. with your torus example, you're not allowed to fill the inside with one 3-cell because the two "ends" of the 3-cell would need to glue to each other

[tawdry valve]

just to clarify, you have to glue the n-cell to the (n-1) skeleton by giving a gluing/attaching map from the boundary of the n-cell (an n-1 sphere) to the (n-1) skeleton.

Then, the new space you get is by putting the n-cell down and doing the gluing. In particular, you’ll get that the interior of the n-cell is still homeomorphic before the gluing, it is just the open n-disc. For the solid torus, the volume inside is homotopy equivalent to the circle (you can deformation retract the open solid torus onto its inner circle). If your construction were correct, the open 3-disk and the circle would be homotopy equivalent, which is false.

with regards to seeing how the (hollow) torus is a CW complex

we should talk about how we attach the 2-cell to fill in the skin of the torus

attaching the 2-cell amounts to giving a map from the boundary of the 2-cell (a circle) to the 1-skeleton. Maps from a circle are just loops, and we can write them down by giving a sequence of paths that starts and ends at the same place

in our case we attach the 2-cell via the attaching map which goes first along a, then b, then a backwards, then b backwards

[lunar yoke]

maybe as an aside, but this picture helped me when i was learning about this

(this represents one possible cw structure of the torus)

[tawdry valve]

ooh that’s a really nice picture

[tidal plume]

Okay, I've seen the notation aba^-1b^-1 being mentioned before for attaching the 2-cell to the 1-skeleton, but because I only saw the picture of the square I didn't really understand why it would look like a torus but now I see why a loop like that is needed, thank you. As for the solid torus, if I were to attach another 2-cell to the 1-skeleton by attaching it only to b then the "inside" of the 2-skeleton you have would basically be a cylinder, right? Then you can attach a 3-cell to it, if I understand correctly

[tawdry valve]

yeah that’s right!

[tidal plume]

Awesome, thanks alot!!

[pseudo coral]

question

is [0,1) open in [0,1] as it can be written as (-1,1) \cap [0,1]

[gritty widget]

You are correct.

[pseudo coral]

and can someone give me some example problems on subspace topology and when a given space is open in a subspace or a link to exercises

[pseudo coral]

cool thanks! preparing for a topology midterm and coming up with examples on the fly lol

[gritty widget]

Munkres probably has what you're looking for in its second chapter.

[pseudo coral]

ive donee those lol but thanks 🙂

[gritty widget]

Hold on. I know a good list of general topology problems.

Let me find it.

[pseudo coral]

so to get things straight

okie doke

[gritty widget]

http://www.math.toronto.edu/ivan/mat327/docs/biglist.pdf

[pseudo coral]

A is open in a subspace Y of X if A can be written as Y intersect open sets from X?

[gritty widget]

You might find what you want here.

[gritty widget]

The open sets of Y are defined as the intersections of Y and open sets from X, so yes.

[pseudo coral]

thanks!!!!

sweet thanks!

and thanks for the link!

[gritty widget]

To clarify: if $A \subset Y \subset X$, then $A$ is said to be open in $Y$ if there is a $U$ open in $X$ with $A = U\cap Y$.

[gentle osprey]

rakko

[gritty widget]

Just the definition of the subspace topology.

[pseudo coral]

makes total sense, thanks!

tysm!!!!

[pseudo coral]

question

is [-1,0) NOT open in [-2,2] as theres no (a,b) such that (a,b) \cap [-2,2] = [-1,0)

[gritty widget]

answer

kind of? You should be checking it for all open sets, not just a basis

[pseudo coral]

but theeres no open set such that its intersection with [-2,2] is equal to [-1,0)

[steel glen]

does the lower limit topology have any particular uses

[gritty widget]

probably only as an example

[steel glen]

this feels extremely trivial, isn't that just the definition of a basis for a topology?

never mind, i guess the intersection property isn't immediately apparent

[gritty widget]

yes?

it's one of the definitions

[potent sky]

I ran across an exercise that requires i show something is a homotopy but im really rusty on this. $h_t(e^{i\theta}) = (e^{i\theta}, e^{i (\theta + t\varepsilon)})$ as a homotopy from $i(e^{i \theta})= (e^{i \theta},e^{i \theta})$ to $f(e^{i\theta}) = (e^{i \theta}, e^{i (\theta + \varepsilon)})$ where these are maps from $S^1$ into $S^1 \times S^1$

[gentle osprey]

*-algebra

[potent sky]

is the easiest way to show this thing is continuous really to look at the preimage of open sets in the torus?

maybe i can say im composing continuous functions?

\theta + t\varepsilon into f?

but it's not really a composition is it..

hm

like i am not sure how to "write down on open set of the torus and take the preimage to show it's open"

oh i guess it will have the same preimage as f and i really anyways

[gritty widget]

can someone share some geometrical insight on why the fundamental group of the torus is abelian. I mean I understand algebraically why it is, but it doesnt feel geometrically that the product of two equiv classes of independent loops should commute

or at least it is hard to believe that the loop which first goes around T^2 vertically and then horizontally is homotopic to doing this in the opposite order

[potent sky]

my question is also abotu the torus 🙂

if you want to help me whilst you wait

[gritty widget]

ill take a look 🙂

[potent sky]

im also a little confused how to take a derivative of f in the circle

do i do like $df_{e^{i\theta}} = \begin{pmatrix} 1 \ e^{i\varepsilon}\end{pmatrix}$

[gentle osprey]

*-algebra

[potent sky]

looks weird

i guess that works

[gritty widget]

you just show that h_0 = i and h_1 = f and h_t is continuous, but this is clear as each component function is continuous.. (e^i*theta is constant wrt t so it is cts and e^i(theta + e \epsilon) is cts by basic analysis)

that you can look at the continuity of each component comes from the definition of product topology

[potent sky]

i thought it was a mistake for f(x,y) to show that f(x, -) and f(-, y) are continuous separately?

Also i feel like $f(e^{i \theta}) = (e^{i \theta}, e^{i (\theta + \varepsilon)})$ should intersect with ${ (e^{i \theta}, e^{i \theta})}$ but i cant show it

[gentle osprey]

*-algebra

[potent sky]

eps is bounded between 0 and 2pi

but these have to intersect right? i can see the picture in my head

there should be 2 points

but i can find them

[gritty widget]

i think the best way to go is using by the universal property for product topology, h is cts iff both canonical projections are cts

so it suffices to show the two maps pi_1(h_t(e^itheta)) = e^itheta = id is cts ... done and that pi_2(h_t(e^itheta)) = e^i(theta + tepsilon) is cts, here you can take preimages or use composition of cts fns

[potent sky]

ok sure

you agree that those two subsets should intersect btw right

these channels are so dead with rakko

btw

@gritty widget https://www.youtube.com/watch?v=nLcr-DWVEto

this might answer your question though

[gritty widget]

cant you just solve the system e^itheta = e^itheta' and e^i(theta + epsilon) = e^i(theta' + epsilon)?

[gritty widget]

thank you!

[potent sky]

i figured stackexchange would have the algebra and youtube would have the geometry

thanks logan

i dont think that system works

one is the diagonal of the torus

the other is not

so it's literally $e^{i \theta}= e^{i \tau}$ and $e^{i \theta} = e^{i(\tau + \varepsilon)}$

[gentle osprey]

*-algebra

[potent sky]

you get that $e^{i \varepsilon} = 1$ but you can't have that because $0 < \varepsilon < 2\pi$

[gentle osprey]

*-algebra

[gritty widget]

then the sets cant intersect it seems

[potent sky]

these both live on the torus

well im being asked to find the number of points they intersect

lol

so i feel like they must

i guess they really dont..

[glass bison]

Hi all, I dont really understand how the last line of the first paragraph of this proof can be concluded

Haven't we only shown that there is an open set in the component C, not that C itself is open?

[gaunt linden]

Note that x was arbitrary. It's a general fact that if a set contains a neighborhood of each of its points, then it is open.

[glass bison]

Ah ok, I thought about it being the case for every x, but I thought the author would say something along those lines so I got confused

Thank you!

[summer sequoia]

how do i prove Two closed sets whose intersection is not closed.

i believe this is impossible

and that the intersection is closed, but i do not know where to start the proof

[gritty widget]

But the intersection of any collection of closed sets is closed.

To prove it, well, you must have a definition of "closed set" to start from.

[summer sequoia]

a set contains all of its limit points

right now, all i can think of is imagining a number line and intersecting two closed sets

hm i think i have an idea

[summer sequoia]

Impossible. Let sets $A,B \subseteq {R}$ then the intersection of the sets would be $A \cap B$ where $\exists x \in A,B$ therefore proving that the intersection is closed contains all of its limit points

[gentle osprey]

imnotrachel

[gritty widget]

This doesn't prove anything.

There isn't even a discernable logical step here.

I don't even know how to comment on this so I'll instead just say how you actually should prove this.

Start with two closed sets. Consider a limit point of their intersection. Show that the limit point is in each of the sets (and therefore in their intersection).

That's just for the case of two closed sets. The intersection of any arbitrary collection of closed sets will be closed again, but I think it would be good for you to start from the simple two sets case.

[summer sequoia]

oh okay i see, ill try again with this information thank you

[warm hedge]

today first time i read about nets and i think i cant understand them

are all nets sequences ? or are all sequences nets ? or neither of these is correct ? 😛

[gritty widget]

nope

example of a net in $\mathbb{R}$ would be something like $x_t = t$ for $t\in [0, 1)$ for example.

[gentle osprey]

Blitz

[gritty widget]

all sequences are nets though

[warm hedge]

so nets can be uncountable ?

[gritty widget]

yep, they can be of any cardinality

[warm hedge]

oh i see

[gritty widget]

they just generalize all notions of convergence, because sequential convergence is too weak to capture them

it's not as hard as it seems

[warm hedge]

also i saw an example like that but t is the rationals in (0,1) in that case couldnt we replace the net with a sequence ?

[gritty widget]

depends on what you mean

replace, then the question is, for what purpose

[warm hedge]

the example was that this net with the {1} isnt compact set but all convergent sequences with their limits are compact sets

[gritty widget]

that's true, this is a one peculiar property of sequences that nets don't share

other than that, they are pretty much the same

[warm hedge]

but couldnt we replace the above net with a convergent sequence that would have limit 1 and then they would be the same set ?

[gritty widget]

In what sense?

I'm not sure what you mean by replace here

[warm hedge]

actually i will try to think about it more and i will come back 😛

[gritty widget]

You can't order rationals in (0, 1) in such a way, that it forms a sequence, and the sequence is convergent

[warm hedge]

i see

thank you

[gritty widget]

np

[devout sorrel]

if {A_n} is a finite collection of sets, how do you show that
int(\bigcup_n A_n) \subset \bigcup_n int(A_n)?

im aware this does not hold in the infinite case

[unreal stratus]

Usually you make life easier for yourself and say like

Suppose we have an open set contained in the LHS (when you remove the int I mean). Can you show it is contained in the rhs?

Or actually there are other similar ways

[devout sorrel]

hmm this is that theorem about finitely many open sets' intersection being open right?

[unreal stratus]

Usually a definition / pretty much a definition and here we have unions anyway so I'm not sure what you mean

[devout sorrel]

yeah i guess so

[unreal stratus]

did you mean to do intersections

[devout sorrel]

oh right sorry

[unreal stratus]

Like as it stands I'm pretty sure this always holds

[devout sorrel]

$int(\bigcap_n A_n) \subset \bigcap_n int(A_n)$

[gentle osprey]

CoolShot

[unreal stratus]

Sure

So there are a couple of ways to do it so like

One way is to take any open $U$ contained in $\bigcap_n A_n$ and show that it is contained in the RHS

[gentle osprey]

potato

[unreal stratus]

Another way is to show that $\text{int}(\bigcap_n A_n) \subset \text{int}(A_n)$ for each $n$ and in fact this is easier

[gentle osprey]

potato

[unreal stratus]

Have a go at one of those

[devout sorrel]

hmm let me try the second one

[gritty widget]

this holds for arbitrary amount of sets

[gritty widget]

and it just follows from this which is the most optimal way to show this imo

[devout sorrel]

does it?

take {[-1/n, 1/n] : n in N}

then it doesnt work righjt

[gritty widget]

it works

in both cases it's just {0}

for finite amount of sets we have the equality

[devout sorrel]

oh

[devout sorrel]

ok yeah youre right i was able to prove subset in the general case

[pseudo coral]

I have a question regarding my solution verification

Let R_sup have the topology of all intervals of the form (-\infty,a) a \in R union with empty set and R itself

Let f: R_sup -> R_standard be continuous then f is constant

So what I did is suppose f isn’t constant then f(x) \neq f(y) so by Hausdorff ness of R_standard there exists disjoint open neighborhoods of f(x), f(y)

Then their pre images are open by continuity and disjoint

But their pre images under R_sup are NON disjoint

Forcing f to be constant

Does this work

@ me with an answer if you’re helping me 🙂

[coarse night]

Try to justify why preimages won’t be disjoint, otherwise looks fine

[pseudo coral]

Thanks 🙏 @coarse night

[jagged sage]

For any topological space X with f:X to S^n, f continuous and not surjective, how do I show f null homotopic

[gritty widget]

This is effectively a map from X to R^n

and R^n is contractible

[jagged sage]

just learned homotopy def today, and I just looked up contractible which seems like quick solution to problem. is there a way to find explicit homotopy between f and constant function

also what is relevance of non surjective in your contractible explanation

[gritty widget]

it's so that we can treat f as a map into R^n

You might want to clarify that it is because S^n minus a point is homeomorphic to R^n.

Why would I? You already said that

lol

[jagged sage]

Lol that makes sense

Other than that, any tips on how to explicitly make homotopy between constant and f? want to try even if for practice

[gritty widget]

But R^n being contractible basically means that something like H(t, x) = tx maps the identity functions to the zero function

and here after the identification with R^n we can also just do H(t, x) = tf(x)

and this is a null-homotopy

[jagged sage]

Ok seems super simple when explained thanks for help

[summer sequoia]

im confused at the end. when we choose an epsilon > 0 why can we then say that x is in complement of A

oh nvm its a contradiction therefore x is in A complement

im brain dead

[little hemlock]

what is BWOC supposed to mean? "By way of contradiction" or something? haven't seen that abbreviation before

[summer sequoia]

oh yes my prof does that

actually i do have a question

why exactly does it provide a contradiction

[little hemlock]

because if x is a limit point of A^c, any open set containing x should intersect A^c. But V_eps(x) lies inside A, so it doesn't intersect A^c, contradicting x being a limit point of A^c

[elfin topaz]

if $A^c$ has a limit point $x$, then the open neighbourhood $V(x,\epsilon)$ contains points of $A^c$. Therefore $x$ can not be an interior point of $A$ and, since $A$ is open, $x\notin A$. Finally $x\in A^c$ and $A^c$ is closed because contains all limit points ($x$ is arbitrary).

[gentle osprey]

José Rosales

[summer sequoia]

ohh okay that is clear. thank you guys

[analog yarrow]

hi! how would i be able to calculate the genus of the Grötzsch graph?

[cedar pebble]

here is some sage code that does this:

G = graphs.GrotzschGraph()
G = Graph(G, sparse=False)
bt = sage.graphs.genus.simple_connected_genus_backtracker(G._backend.c_graph()[0])
bt.genus()```

The genus is 1.

In general computing graph genus is pretty bad, even this sage implementation hangs very badly on reasonably small graphs. But in this case we're okay and we only need to attach one handle to the plane to embed this graph. You should end up with an embedding something like this

[neat current]

how does my proof look

[gentle osprey]

Nobody

[wispy veldt]

You also dont need all this C buisness , simply being in B and boundary of A implies A intersect B is none empty by definition of boundary.

[gritty widget]

I am trying to compute the fundamental group of R^3 \ {S^1 V S^1}, i can do it using van kampen and I see it is Z free product Z, but I cant see what it retracts to or and algebraic way to compute

For example another one which i was able to do was fundamental group of $R^3 \setminus {L_1 \cup L_2 }$ where $L_1$ and $L_2$ are disjoint lines: for this I computed $\pi_1(R^3 \setminus {L_1 \cup L_2 }) = \pi_1(R \times (R^2 \setminus {p ,q}) ) = \pi_1(R) \times \pi_1(R^2 \setminus {p,q}) = 1 \times \pi_1(S^1 V S^1) = \pi_1(S^1) * \pi_1(S^1) = Z * Z$

[neat current]

good point, originally I had a thought somewhat along these lines but for some reason I couldn't formulate it properly

thanks

[gentle osprey]

loganb

[gritty widget]

i dont see why they become lines? why not two rays?

[hazy lotus]

I'm having trouble proving some things I need for homework:

1. I have to show that S^1/~ is homeomorphic to S^1 where ~ identifies antipodal points. This was mostly fine.

I used the continuous function f=z^2 from S^1->S^1 and noticed that since the function takes the same value on antipodal points, you get the unique continuous function h: S^1/~ -> S^1 satisfying f = h o pi where pi is the canonical surjection.

If you let g: S^1->S^1/~ be the multivalued square root function, you can see that h and g are inverses as their compositions are identity maps.

However, I'm having trouble showing that g is actually continuous.

2. Let X be the union of the lines y=0 and y=1 in R^2 with the induced topology. Define ~ on X as follows: (x_1, y_1) ~(x_2, y_2) iff x_1=x_2 =/= 0. Is X /~ homeomorphic to X?

I had a bit more trouble deciding on an approach here. X is clearly not connected, and I believe that the decomposition space is connected, but I had trouble actually showing this. If you try a cut point argument, you can say that every point of X is a cut point, and intuitively, no point of X/~ is a cut point, but that seems just as difficult as showing X/~ is connected in the first place. If you use continuous functions to a 2 point discrete space, it just feels a bit unwieldy and I'm just not as familiar with how that interacts with decomposition spaces.

3. If f:[0,1]->S^1 is defined by f = e^{2pi i t}, then f is a closed mapping. It seems intuitive, but I couldn't really make any traction on actually proving it.

Any advice would be appreciated! Thanks!

[unreal stratus]

For the first bit, a general thing is that if f: X -> Y is a quotient map and ~ the relation on X given by x ~ y iff f(x) = f(y), then f factors through to give an homeorphism X/~ -> Y

And your map z -> z^2 is automatically a quotient map (think why!)

As for the second, I think you should try checking path connectedness (often it's more easy to prove this than connectedness)

[hazy lotus]

This should be because it's open and continuous (easy to prove because it suffices to check that z->z^2 is open for open intervals). This does bypass a lot of the work I ended up having to do, but just out of curiosity, do you happen to know any sources that prove that the set-valued square root function is continuous?

[unreal stratus]

Uhh I mean I would show continuity by saying it is the inverse of the other one lol

But there is a general theorem which shows ur map is a closed map

so you don't even need to do that check

The same theorem works for 3

Like cts map from compact to hausdorff is closed

[hazy lotus]

I don't think I ever would have thought to use this.

[unreal stratus]

It is really useful for quotient spaces

Especially since many examples you get will be at least one of the two lol

since it means often you can just write down a map and immediately know it is a quotient map

[hazy lotus]

This also definitely seems like a good way to go about it - for two points a,b \in X, you get two cases based on whether or not one of the points is (0,1).

In the first case you let the path be f:t->[(a+t(b-a),0)] and in the second let the path be g:t->[(a+t(b-a),1)]

To show that these paths are continuous you fix some t_0\in[0,1] and take an arbitrary open set U containing f(t_0) or g(t_0). the elements of f(t_0) (or g(t_0)) both have to be interior points of the union of elements of U in the underlying space X, so you can find open balls entirely contained in the union of elements of U centered at the elements of f(t_0) (or g(t_0)). Say the radius of those balls is d. Then by choosing a ball of radius d about t_0, it seems you can prove continuity.

It's still a bit of a hassle to work with arbitrary open sets of a decomposition space, but I think I can get around it now.

Let me know if there are any holes with my argument or simplifications that can be made, and thanks for your thorough help!

[unreal stratus]

yeahh I think you can also define paths by like

Well, you can define paths in X and then compose with the quotient map to get some paths

Then you can glue them together to get more paths (where you may not have been able to glue paths before)

If you do smth like this then you don't have to explicitly talk about open sets and stuff ^

I'm slightly unsure what your proof is doing but perhaps I'm just not thinkign about it properly lol

[wise ruin]

In class, we were taught a method of representing connected sums of surfaces as attaching (twisted/untwisted) ribbons to a disk.
Is there a name for some formalization of this? I understand it's in the realm of Morse theory, but searching online for that doesn't get me very close to this specific process.

[wise ruin]

Sure, so in this context, you can show that T#RP2 is homeomorphic to K#RP2 by attaching a twisted ribbon to the second diagram and "moving" where things are attached so it looks like the first diagram.

[analog yarrow]

thank you!

[little pulsar]

hiii, I'm wondering if anyone could help me with this maybe? I understand that X excludes from R the reciprocals of integers, and I tried mapping some open sets in X to X/~ but so far no luck. Any hints?

[gentle osprey]

Nobody

[gentle osprey]

Nobody

[little pulsar]

update: now I'm just stuck at proving X/~ is not Hausdorff. Could anyone please give me suggestions to proceed?

[gentle osprey]

Nobody

[little pulsar]

Could you take a look at my question by any chance? 🥺

[small jackal]

Hi guys, a very silly question but I am writing a term paper and am kind of forgetting what was special about this metric and it's name as well (was it Frobenius metric?)

It's basically for the countable product of metric spaces with pn as each individual metric

[gritty widget]

Nothing, just one of many ways to give metric to the countable product of metric spaces

The only important bit is that the topology generated from it is product topology

[small jackal]

Ah alright

I am probably overthinking this bit

[gritty widget]

maybe you don't know what product topology entails?

[small jackal]

Ah yeah, I am still exploring that part

That's what the term paper is about

[gritty widget]

just to be safe, it basically means that the convergence in this metric is the point-wise convergence along all the coordinates

[small jackal]

Because I counldn't find any good resource on those and how they're important

[gritty widget]

I wouldn't say they're important, using x/(1+x) is one of the canonical ways to make bounded metrics

but you could just as well use min(x, 1) for that

the idea is, make metrics (uniformly) bounded, say by 1, sum them up with coefficients whose series converges

and it should always give you the same thing

in terms of topology/convergence

[small jackal]

Right, I kind of understand what you're trying to say

Thanks!

[gritty widget]

np

[swift fjord]

Frechet is how I know it

[jagged sage]

If I have X compact Hausdorff with f:X to X continuous and X_n+1 = f(X_n) for all n >/ 0, doesnt intersection from n=0 to infty of X_n nonempty follow by definition of X_n

[gritty widget]

yes

from finite intersection property

X_n is decreasing, and the members of the sequence are all closed and non-empty

assuming X isn't empty in the first place

[jagged sage]

thanks that makes sense. I'm also trying to show f(A)=A for what I guess is nonempty subset of X. To do this, first i need to show A_n+1 subset A_n then it gives set of A_n has FIP. From here, how do i show A subseteq f(A) and f(A) subseteq A? If this is right approach

[gritty widget]

Are A_n iterates of A

[jagged sage]

I just let A_1=f(X) and A_{n+1}=f(A_n) similar to X

[gritty widget]

Is A their intersection

[jagged sage]

yes

[gritty widget]

inclusion of f(A) in A is obvious for sure

from commutativity of composition of f with powers of itself

[jagged sage]

sure. I'm not sure for this other inclusion tho

[gritty widget]

I feel like I saw something similar before. We might need to assume that X is metrizable, not sure

Let's take $x\in A$, then $x = f(x_n)$ for some $x_n\in f^n(X)$. Let's take a convergent subnet $y_\alpha$ of $x_n$ converging to $y$. Then $x = f(y)$. For any $n$, we can find $\beta_n$ such that $y_\alpha \in f^n(X)$ for all $\alpha \geq \beta_n$. Then taking limits, $y\in f^n(X)$ for all $n$. So $y\in A$.

[gentle osprey]

Blitz

[coarse night]

Do you need Hausdroff for this ?

[gritty widget]

yes, for the image to be closed

[coarse night]

Actually yes you do otherwise regularity would suffice?

[gritty widget]

as in regular spaces?

[coarse night]

Yes, it’s just something minor nvm

[gritty widget]

this only uses countable compactness of X, but we still needed compact Hausdorff to show that the intersection is non-empty ig

[gritty widget]

If you just assume X to be countably compact, then you still get that f(A) = A, just not that it's non-empty

[jagged sage]

Never heard of subnet, but just looked it up and not sure if I can use this. Is there alternative way to formulate without nets or subnet

[gritty widget]

you can assume X is metrizable if that bothers you

and take a subsequence instead

[jagged sage]

Ok ig I'll use that instead thanks for help

[jagged sage]

In this question I'm asked to consider equivalence relation defined by x~y iff x-y integer and justify what the resulting quotient space is. It is clearly R/Z which I know is homeomorphic to S1 and can show that but idk what exactly question refers to

[gritty widget]

I can't understand you. What exactly are you bothered by?

You seem to have answered the question just fine.

[jagged sage]

I don't know what the question is asking. It asks me to justify what the resulting quotient space is. What is there to show to prove quotient space is in fact R/Z?

[gritty widget]

It is R/Z by definition. The question is rather open-ended, but it probably wants you to prove that R/Z is homeomorphic to S^1.

There is nothing you have to do to prove that it is R/Z.

[jagged sage]

Ok thanks, thats what I was thinking just wasn't sure

[little pulsar]

Hiii, I'm wondering if anyone could help me with the last question? We haven't covered enough to prove this, so I guess I only need to state what's the intuition, but I find it hard to visualize this... is this like a cone?

[little hemlock]

just apply -1 ~ 1 first. can u see what you get then?

[little pulsar]

a sphere?

[little hemlock]

like a 2-sphere? nope.
-1 ~ 1 is like pinching two points together on the circle

[little pulsar]

yeah okay

kind of hard to visualize that but okay...is it like a fortune cookie?

[little hemlock]

hmm i don't think so. i think ur overthinking

[little pulsar]

ah

[little hemlock]

ur only identifying two points together on the circle

[little pulsar]

could you just tell me what it is🥺

[little pulsar]

oh....

[little hemlock]

then ill let u think about what also identifying i ~ -i comes out to

[little pulsar]

so how would I describe this thing..

[little hemlock]

qualitatively its a figure-eight kind of shape. Its generally called "a wedge of two circles"

the wedge sum is just a kind of operation that takes two spaces and identifies them at a single point

[little pulsar]

so something like this?

doesn't look very pretty

[little hemlock]

are you trying to draw 2 dimensional spheres or something?

not sure what im looking at tbh

[little pulsar]

no I'm just trying to do what you did again

[little hemlock]

the bottom-middle one looks kind of like what i drew but its a bit cut off. Not sure what the 4 other drawing are tho

[little pulsar]

how would you draw it?(sorry I'm not asking you to do my homework I just really can't picture it)

[little hemlock]

do u mean the 1 ~ -1 and i ~ -i picture?

[little pulsar]

yeah

[little hemlock]

the final picture is the desired space

it would be a wedge of four circles

i just drew the intermediate steps to get across the intuition that you are just "pinching" these points together

[little pulsar]

ummm but why are you associating i, -i with 1 and -1 the same point?

[little hemlock]

oh good question.

because i did it wrong

my bad

[little pulsar]

does this look right then?(I'm just moving i and -i together)

[little hemlock]

yeah, its kinda hard to draw

but its the disjoint union of two circles which are identified at two distinct points

[little pulsar]

oh yes I think I'll use this description

[little hemlock]

like taking the figure eight shape and folding it onto itself

[little pulsar]

sounds good

[unreal stratus]

yeah if you take two disjoint circles and glue one to the other at, say, i and -i as in your picture then there's a map from S^1 into that space given by traversing one circle then the other which should factor through your given quotient to give a homeomorphism

[little pulsar]

sounds reasonable! thank uuuuu for both of ur help 🙂

[unreal stratus]

Go sleep jk

Lmao

OK I am sleep now gn

Gl

[gritty widget]

excuse me I am new to topology

why {open set} x {open set} is also open in product topological spaces... is it definition or what?

and I found the definition of a "basis" for a topology in Munkres is somehow different from the def in a topology book published later... is there something wrong?

[dry jolt]

I think it's easier to just think about what a degree 2 cover must look like

If you think of S1 v S1 as a graph with 1 vertex and two oriented edges, then a degree 2 cover ought to have 2 vertices and 4 oriented edges

wait no, that's not how it works

hmm maybe I forgot how this goes

[dry jolt]

Ok I think this is how it works

So if we label the edges of S1 v S1 with a and b

Then in a degree 2 cover we have two vertices and four edges. For the covering map to be a local homeomorphism, each vertex in the cover should "look" like the vertex in the base space, meaning it should have an edge corresponding to a going into it, an edge corresponding to a going out of it, and similarly for b. Note that the edge going in and the edge going out can be the same edge or it can be distinct

This gives you constraints on what the cover can look like and it's not too difficult to explicitly draw the graphs and see which ones are distinct

I'll think about how one might recover this from index two subgroups and ping you if I come up with something

[odd flame]

any pointers on starting this

[empty grove]

Can you visualise B

It's gonna be all the integer radius disks centred at 0

You want a larger totally ordered set, so just put a bunch of sets in between each D_i and D_(i+1), till you cannot put any more sets in between