point-set-topology

[pseudo coral]

silly question

but for an open covering that doesnt cover (0,1) does this work

(0+1/n,(n+1)/n) as n ranges over positive integers

or 1-(n+1)/n for the upper bound

[gritty widget]

an open cover that... doesn't cover it? why not just take an open set that's not (0, 1)

[pseudo coral]

i mean that doesnt have finite subcover

sorry

poor wording

[gritty widget]

if you want to work with open covers whose elements are open subsets of, say, R, it works

[pseudo coral]

yes

standard open sets

[gritty widget]

if you intersect with (0, 1) you get (1/n, 1) and that's the standard example

which is simpler and still works

[pseudo coral]

ive seen that but i was trying to come up with an interval with n's in the upper and lower bound

[unreal stratus]

well like

(1/n, 1- 1/n) then

which i imgine is what you were shooting for

Okay, so there is an obvious typo (should be RP not R at one point) but how is it easy to see this is a surjection (lol accidentally clipped that off)?

It feels a bit weird when we don't have a definite description of what the map actually is; maybe I'm missing something.

[pastel linden]

Bott-Tu defines a locally constant presheaf as a presheaf where all the restriction morphisms are isomorphisms (and a constant sheaf when the morphisms are the identity). nLab defines a locally constant presheaf as a presheaf over a space X with an open cover that is constant when restricted to each element of the open cover. are these two equivalent?
Bott-Tu implies that resX,U: F(X) -> F(U) is an isomorphism, so F is isomorphic to the constant presheaf F(X) over each U since resU,V resX,U = resX,V for any open subset V of U, but i'm not sure how the other definition implies that each morphism is an iso

For any subset V of U, an element of the constant cover, we have resU,V is an iso since it's the constant sheaf, but I'm not sure how this implies that every restriction morphism is an iso, at least using presheaves. So is this only equivalent for sheaves?

[cedar pebble]

hmm I'm not sure that this is the definition of locally constant sheaf...

[cedar pebble]

why should F(X)->F(U) be an isomorphism? What if U is disconnected?

[pastel linden]

I see where my confusion is now

they specifically define locally constant and constant presheaves on a good cover

[cedar pebble]

Aha

[pseudo marsh]

What’s a good topology book

[tardy meadow]

point-set i would go with munkres

[gritty widget]

I wouldn't

I'd read Dugundji

also #book-recommendations is kinda more relevant

munkres is a snoozer

i feel like a lot of people (possibly including myself) just don't know any better though

For metric space topology there is Kumaresan which people online recommend

which I think people should start with metric space topology anyway

after that I'd go for some more advanced book, maybe from point set topology, but not really Munkres (like Dugundji)

but you can just go straight to the point and read Munkres of course

[neat current]

what exactly is wrong with munkres?

[tardy meadow]

I had also used a book by a Sunderland or Sutherland that was quite good

idk how well-known it is

"Introduction to Metric and Topological Spaces" by Sutherland it was. think i mainly used it for exercises

[unreal stratus]

i have only heard about that book from oxford/cambridge and never heard it mentioned elsewhere

[tardy meadow]

it was on the suggested reading list for my topology course

[unreal stratus]

Yeah nice

[tardy meadow]

looking back it's a bit basic in terms of how far it goes though (what you'd want from a first course and pretty much nothing more)

[untold lily]

I started reading dugundji recently, want to vouch for it

it is very damn nice so far

[coarse night]

ya dugundji is nice

[vivid ingot]

Problem: show that the version of the topologist’s sine curve shown here is connected.

Progress: Not far. I started trying to show that any nonempty clopen set C in (we’ll call it) S is all of C by showing that each x in R has an image lying in S. That seemed a bit much after writing it out and not conducive to using clopenness at all.

I am now trying to suppose per contra that there is some clopen set strictly between Ø and S.

My problem with just letting A and B form a separation of S is I don’t know how to whittle it down to (0,0) being the “separation point”.

Is this an analytical proof? I saw one online using an infimum of x-values along the right half, but it was a different version of the topologist’s sine curve.

[gritty widget]

closure of a connected set is connected

[vivid ingot]

What would I do with the limit points after that?

Aren’t the limit points S, plus the line segment from (0,1) to (0,-1)?

Important to my original method, or to using Blitz’s?

Oh, I see. Yes, that was in the statement of the original problem.

Good lord, the one on Stack exchange translates to this version almost immediately.

I’m pretty sure I can translate this precisely. Thank you both and sorry for not reading more carefully.

[serene ferry]

I've been reading Pedersen's book Analysis Now, and I wanted to double check that I'm understanding the wording of one of his proofs.

That's his proof, but I thought the wording was slightly confusing, so I tried to reword it a bit in my notes.

Is this equivalent to the above proof?

[plain raven]

This screenshot does not explain what sigma is

[serene ferry]

Sorry, it’s based on the definition of continuity given by Pedersen - it means the topology of Y.

[plain raven]

It means the topology of Y?

What is O(f(x)) then?

I assumed it meant the set of open neighborhoods of O(f(x)) but then the intersection would be redundant, as obviously the set of open neighborhoods of f(x) is a subset of the topology of Y.

[gaunt linden]

The argument looks like it might make sense if O(y) is the set of all neighborhoods of y (though where the letter O comes from is then a bit of a mystery).

[hidden crag]

oll neighborhoods

[coral pivot]

nice, pederson is my fav book.

[pseudo ocean]

[swift fjord]

I think they use the definition of nbhd as "set containing an open set containing the point"

[gritty widget]

Open?

[plain raven]

I think troposphere's point is that if a neighborhood is not open then why call the set of neighborhoods O(f(x))?

imo it's a more natural definition for continuity at a given point

which is what he's talking about here

A function is continuous at a point if the preimage of every neighborhood of f(x) is a neighborhood of x

[plain raven]

because the preimage of an open neighborhood is not necessarily an open neighborhood if the function is only continuous at that point

[chrome ridge]

Is triangulation equivalent to finding a delta-complex structure on the space?

[unreal stratus]

Usually it would mean a simplicial complex structure

[plain raven]

I would say it's a matter of context. "simplicial complex" is itself sometimes used in a more generic sense

[prisma arrow]

is there a name for this.
the collection of open sets that have boundary A

[void gazelle]

Hi, guys, why such B always exists?

[gritty widget]

you could take a chart at p contained in U and let B be an open ball strictly contained in that chart centered at p, for example

[void gazelle]

oh, I see, regular ball

Thank you!

[untold lily]

can someone explain why Ex.4 is a counterexample here?

E^1 = real line

[gaunt linden]

{0} cup {1} is a subset of E1 whose connected components {0} and {1} are not open. Yet E1 is locally connected.

[untold lily]

oh god, I was so focused on whether 0 cup 1 is locally connected in the subspace topology and it made no sense

thanks that clears it up

[stone cipher]

Let $(M,d)$ be a metric space and let $A\subset M$. I already know that $A$ is compact iff $A$ is sequentially compact. How to prove that $A\subset M$ is relatively sequentially compact iff $A$ is relatively compact. In other words, how do I prove that for every sequence in $A$ there is a subsequence converging in $X$ iff the closure of $A$ is sequentially compact?

[gentle osprey]

RaD0N

[stone cipher]

ok. I found it

[lean cedar]

I am currently working with cohomolgy theories in the sense of Eilenberg and Steenrod. It is a standard result that any two cohomology theories satisfying the Eilenberg-Steenrod axioms gives the same result on CW-complexes. Any recommendations on a short/consise ressource that proves this? What I have seen so far is lengthy and has quite big leaps in logic. I also thought about it myself and hoped we can give a more "top down" proof: Somehow a CW complex is always a direct limit (in Top) of its k-skeleta. Now I recall (not sure) that cohomolgies behave nicely with filtered colimits (in particular sequential ones). Thus it should essentially suffice to prove that any two such cohomologies agree on k-skeleta for any k. And these are somehow finite disjoint unions of disks modulo gluing -- so again a colimit. Hopefully this behaves nicely with cohomology again. What do you think?

[lean cedar]

Hmm, I can see this is related, but I cant quite make the connection to CW-complexes! This seems very general!

[lean cedar]

Ah I see

But that is not really concise and uses a lot of machinery

I see the point, but its not really what I am aftee

[warm hedge]

I am reading about triangulations. I understand the idea that we are putting triangles on a surface with some properties but i dont know how to actually do it. For example it says that we cant make a triangulation on the sphere with less than 4 triangles. How can i show that ?

like lets say i can make a triangulation with 3 triangles, how i will try to create it to show a contradiction ?

[lean cedar]

Thanks, I will check it out

[warm hedge]

i tried to create one and then i show that the Euler characteristic cant be 2 (which is for the sphere), but is this a correct way, i dont know if this is enough to convince me or not 😂 like what if i could create a different triangulation which would give me 2 for Euler's characteristic

btw sorry if i am interupting a conversation.

[plain raven]

Suppose you have a triangulation of the sphere. You have a set of vertices, K0 a set of edges K1 (which is a set of unordered pairs of vertices) and a set of faces K2 (which is a set of unordered triples of vertices). Any face in K2 has all of its edges in K1.

I think you might be able to just work out what all the spaces are with 3 or fewer faces and classify them up to homeomorphism. There are not that many of them. None of them are a sphere.

For example if there are 3 faces and each face has 3 vertices then all in all that's 9 vertices. But it could be fewer than that if the faces have vertices in common.

Does that sound reasonable? You are classifying small simplicial complexes up to isomorphism.

In the case of the sphere you can discard complexes that are not connected or simply connected, and you can discard any space which is contractible.

[warm hedge]

yeah i think it makes sense

thank you

[prisma arrow]

how do you know if a complex is contractible without looking at homotopy

[plain raven]

You don't know this. I was taking it for granted that we understand that a sphere is not contractible, however this is difficult to prove.

Anyway, given any simplicial complex with fewer than four faces and a small number of edges/vertices I am confident I can give you a simple proof that this is not homeomorphic to a 2-sphere.

So the problem can be attacked on a case by case basis

[prisma arrow]

okay got it

[gaunt linden]

Even before considering the Euler characteristic, a triangulation of any closed surface with an odd number of triangles would need to have a half-integral number of edges, which is absurd.
This excludes "1 triangle" and "3 triangles". There is a "triangulation" with 2 triangles, though: draw a great circle on the sphere, and declare three points on it to be "vertices" where two triangles meet. That gives you a triangulation with two 180°-180°-180° spherical triangles. If you want to exclude that (say, because it complicates notation and bookkeeping that a face is not given uniquely by its set of vertices), you'll need to build an exception for it into your notion of "triangulation".

[plain raven]

Yeah this is where the right choice of definition of triangulation can make your life easier

What troposphere describes is a triangulation by a delta complex but not a triangulation by an abstract simplicial complex

[warm hedge]

yeah like for my triangulation there are some specific properties like every edge must belong to only 2 triangles and for every 2 different triangles their intersection will be the empty set or one common vertice or one common edge. So if i understand correct the triangulation with 2 triangles their interection will give 3 vertices so its out of my triangulations

[zinc siren]

is what you were doing an abstract simplicial complex? it looked like just a simplicial complex to me.

(do note that i don't know barely any algebraic topology, so i likely will be not all that coherent)

[gaunt linden]

In an "abstract simplical complex", you forget everything about each simplex except its set of vertices. This is usually enough to reconstruct the original space up to homeomorphism when the complex is abstracted from an actual geometric simplical complex -- except in degenerate cases like mine where the two triangles become indistinguishable when we're throwing away their interior points. You can always avoid this situation by subdividing sufficiently many of the simplices.

[plain raven]

It was just a simplicial complex yeah but I was using the term "abstract simplicial complex" to be more specific about what exactly I meant. I have seen the term "simplicial complex" used to mean "delta complex" or "geometric realization of a simplicial set" but the term "abstract simplicial complex" always refers to a specific definition

[obtuse meteor]

Barycentric subdivision

[plain raven]

yo faye

I came up with a gigabrain definition of barycentric subdivision at one point lmao

Do you want to hear it

Ok "I came up with" I kinda stole it from Kan

First I define a notion of the "subdivided join" of two simplicial sets, if X and Y are simplicial sets then the subdivided join is constructed by taking the product diagram
X <- X x Y -> Y
and taking the homotopy colimit, so you replace those projection maps by their mapping cylinders.

Now you check that this 'subdivided join' defines a monoidal product on SSet. Let's call it \ast.

Now define the following functor sub : \Delta -> SSet: it sends [0] to a point 1, it sends [n] to the iterated tensor product 1 \ast 1 \ast ... \ast 1, n+1 times. You can check that this is the barycentric subdivision of the n-simplex.

the left Kan extension of this functor sub along the Yoneda embedding defines an endofunctor SSet -> SSet which gives the barycentric subdivision of an arbitrary simplicial set by gluing it out of the barycentric sudivisions of its simplices

[empty grove]

Neat

[obtuse meteor]

it's ok that's what I mean when I say "I came up with" too

also ew

[gritty widget]

is there a topology (R^d, O) such that there exists a p with open neigbour B in O but is not an openneigbour in the standard topology for Rd if so can u give me example.

[unreal stratus]

like most topologies other than the standard one should work right?

e.g. you can take the discrete topology

[gritty widget]

ohh yeah the question is kinda dumb

[unreal stratus]

or even uh like

take any set which isn't open in R^d and then consider the topology it generates

[gritty widget]

thanks, im just speedrunning through maths to get an overview so my understanding is very limited

[unreal stratus]

ye dw

[pseudo coral]

or consider the lower limit topology on R^d as the half open intervals aree open in this topology but not in standard

or like potato mentioned the discrte topology gives you the same thing as all subsets are open in there

[unreal stratus]

it's probably worth pointing out though that if you induce a topology from any norm then you'll just retain the standard one

[plain raven]

What's the easiest definition of a weak equivalence between simplicial sets which are not necessarily Kan complexes and doesn't involve passing to the geometric realization

like the definitions I know are:
f : X -> Y is a weak equivalence if it induces isos on homotopy groups (but the homotopy groups are hard to define if X, Y are not Kan complexes)

f : X -> Y is a weak equivalence if |f| : |X| ->|Y| is a weak equivalence (and |X| and |Y| are CW complexes so this happens iff |f| is a homotopy equivalence)
f : X -> Y is a weak equivalence iff Ex^\infty X -> Ex^\infty Y is a weak equivalence

[lunar yoke]

Another one is that f is a weak equivalence iff precomposition induces a bijection on homotopy classes f* : [Y,K] → [X,K] for all Kan complexes K.

[plain raven]

ooohhh nice

https://ncatlab.org/nlab/show/simplicial+weak+equivalence

i found this page after posting which has a good list

[lunar yoke]

Since we are then looking at maps from cofibrant to fibrant, it does not matter whether we take left or right homotopy and wrt to which cylinder or path objects. So generically we would look at homotopies X x Delta^1 → K

[plain raven]

gotcha

[lunar yoke]

Neat

[compact sonnet]

I have an argument that I'm not sure how to formulate rigorously:
I claim that infinite unions of intervals of type [-a,a] are either going to be equal to:

1. (-k,k) (such as for all a < k)
2. [-k,k] (such as for all a <= k)
3. R ( such as for all a)

It seems clear those are the only possibilities, but I'm not sure how to prove that rigorously.

[shadow charm]

Take the sup over the a’s in your union

[compact sonnet]

So I can say that if a sup exists, then the union isn't R , but how do I know there isn't some crazy other possibility other than 1) or 2)

[rancid umbra]

it is a symmetric, (path)connected subset of R, so its a symmetric interval, and the only symmetric intervals are of the form 1., 2., or 3.

[compact sonnet]

I feel like I'm being pendantic but how do we prove that?

[compact sonnet]

I feel like bringing in path connectedness shouldn't be necessary for a question so close to basic principles though. Maybe I'm wrong about this?

[rancid umbra]

its not necessary, just a fast way to do this problem since (path)connected subsets and intervals are all the same thing in R

[rancid umbra]

he means either the sup is in the union or its not, in which case you can get arbitrarily close to the sup using elements from the union (this follows from the characteristic property of the supremum in R)

[compact sonnet]

So we need the completeness axiom for this?

[rancid umbra]

no, you are studying all cases: the union has no supremum in R, the union has a supremum which is in the set, or the union has a supremum which is not in the set. these exhaust all cases involving the supremum and the cases only rely on the order structure of R (since we are considering supremums)

[compact sonnet]

I see, that makes sense. Thank you!

[rancid umbra]

try not to get bogged down with the specifics of the proof. if you just have the definition of a symmetric interval J, you should be able to deduce that it is either all of R, [-sup J, sup J] or (-sup J, sup J) with similar case work

J is symmetric if whenever x in J, -x in J.
J is an interval if whenever we have x,y in J and any real number z between x and y, that implies z must be in J

[compact sonnet]

In this particular case I really care about the specifics -- I'm going back to the subject decades after college and I'm going through textbooks really wanting to understand it fully from first principles, as opposed to taking a class that has timelines.

[plain raven]

Imo you do need the completeness axiom to say that there is a sup

[rancid umbra]

i disagree. any subset of a poset either has a supremum or it doesnt. if it does, its either in the subset or its not

oh good point, thanks

[untold lily]

Isn't the completeness axiom something literally built in to the real numbers anyway? You do need it but it's not a big deal that you do

[untold lily]

that's true

[quiet mirage]

hey anoyone there?

how to start off Knot theory?

[gritty widget]

you should start off with some algebraic topology knowledge

[quiet mirage]

Like what are the required pre requisites from alg top?

[supple sable]

Well as the name suggests it is algebraic, so you need some knowledge of abstract algebra first. Group theory and such. A second part that would be helpful is to have a good foundation in real/complex analysis and proof writing. Then I think you are good to start with algebraic topology

As for knot theory I don't know, haven't studied it

It sounds really cool tho 🙂

[thorn citrus]

Thanks. I'll buy Colin Adams' Knot book too.

I didn't ask the question but recently I needed that topic to understand Wilson loop deeper.

[plain raven]

It's not important to have a background in complex analysis to study algebraic topology

[quiet mirage]

Hmm aight thanks

[gritty widget]

in a metric space, if p in B=B_r(q) then B is an open neighbour of p?

cuz it is in the top. right?

[hidden crag]

yes

[gritty widget]

what is the topological space (R, P(Q)) called?

i think it maybe a good example to use to study topologies and train my intuition

[plain raven]

What

What is (R, P(Q))

Are you talking about $\mathbb{R}$ with the powerset of $\mathbb{Q}$ as the topology?

[gentle osprey]

diligentClerk

[plain raven]

The powerset of $\mathbb{Q}$ does not contain $\mathbb{R}$

[gentle osprey]

diligentClerk

[gritty widget]

yeah

[gritty widget]

i mean every element of a powerset of Q is a subset of Q and there fore R and (trivially) their unions and intersections are again in P(Q) so it is a topology on R right?

i was wondering if it had a name

or maybe sth like top. space (M, P(N)) where N is dense in M

if it had a name

[hidden crag]

no

a topology needs to contain the base space as open set

[untold lily]

why would R be in P(Q)?

[hidden crag]

this will generally not be a topology for the same reason

[gritty widget]

ah forgot the very first condition

[untold lily]

I mean you could just add R to P(Q) and make it a topology but at a first glance this topology doesn't seem to possess any interesting properties

[gritty widget]

its hard to visualize

[untold lily]

is it?

[gritty widget]

i mean i cant

[west spindle]

having a little bit of a hard time with this homology theory problem:

Find the homology groups of the union of all k-dimensional faces of the n-simplex.

getting a bit lost in the details, and it isn't really helping that i have had to skip a lot of class and essentially self-study due to Circumstances™️.

so far i have written down that the i'th chain group of this thing is isomorphic to Z^(n+1 choose i+1), which is just... a combinatorial thing

i cannot even write down the boundary maps without getting into a huge mess.

<@&286206848099549185> long shot, but maybe?

ping me if replying

[west spindle]

(resolved)

[gritty widget]

How do you know if this is a continious map for it to be bundle

[unreal stratus]

Is that Schuller

[gritty widget]

yeah i like his lecture style tho there are only 2 courses on yt

[lean cedar]

How does one prove that the (Cech-)cohomology of the warsaw circle $W$ is that of $S^1$? I have seen an argument for homology that uses the fact that $W$ is the infinite intersection $\bigcap_{n\in \mathbb N} W_n$ where $W_n = W \cup R_n$ where $R_n$ is a montonoically decreasing collection of rectangles covering the singularties of $W$. Then $W = \lim_{\leftarrow } W_n$ and $H_\bullet(\lim_\leftarrow)=\lim_\leftarrow H_\bullet$ gives the result as all $W_n \simeq S^1$ in a compatible/natural way. Does a similar argument hold for $H^\bullet$?

[gentle osprey]

MrMonday

[gritty widget]

my brain is fuken fried, if (M, O) is a manifold with dim d, and there exists p in M such that {p} is in O, what does it imply

[unreal stratus]

Ye tho he is v concise so make sure to read up more if you wanna learn topology properly

[gritty widget]

any book recommendations?

best if it has a video lecture based on the book

[gritty widget]

O?

this notation isn't universal

[wicked yew]

how do I go about civ)

[gritty widget]

R is complete and C is ______

[wicked yew]

oh yes i forgot about that proposition

thanks

[gritty widget]

compact metric spaces are complete in any metric

[pseudo coral]

are these sub n's supposed to be sub k's?

[gritty widget]

looks like it

[pseudo coral]

cool i edited it lol

[pseudo coral]

also i dont understand why

U_m \subset U_n for all m \geq n

is that just basic set theory

by the way the U_n are defined

[pseudo coral]

is this because. compact in metric is seqwuentially compact thus any sequence has convergent subseq thus so does any cauchy sequence and is thus complete

[gritty widget]

yes

[gritty widget]

Cauchy sequences either converge or have no limit points

but in a compact space any sequence has a limit point

[pseudo coral]

compact metric space you mean? I thought sequentially compact and compact were same in metric spaces only

cause you could give a topology on a 3 point set in which no sequence converges, I wanna say

or there is a sequence that isnt convergent

and a 3 point set is clearly compact

can someone help me see why if $U_n:=\bigcap_{k=1}^n A_n$ then $U_m \subset U_n$ for all $m \geq n$

[gentle osprey]

MyMathYourMath

[pseudo coral]

-cause youre intersecting more sets

so does that make the intersection smaller ?

[gritty widget]

Doesn't have to be

Maybe I'm using wrong terminollogy, I mean accumulation points

[gritty widget]

A sequence on a 3-point set will have a constant subsequence

[gritty widget]

$U_m = U_n \cap \bigcap_{k=n+1}^m A_k$

[gentle osprey]

Blitz

[wicked yew]

for dii) I tried to use the hint and here is my thought process( probably going in the wrong direction). I said that every sequence x_i(k) is bounded as given in the definition and so by Bolzano-Weirestrass in R, we can find a convergence subsequence for each i. Then at maximum we have summation 1/n^2 and we know this is convergent?

I've worded this very weirdly and idk if that's even right

I think my problem is that if I take a subsequence for each particular co-ordinate, how would I don't really have an idea if all the subsequences coincide and give a limit

[gritty widget]

This is about Cantor diagonal argument

Take a subsequence such that first coordinate converges. Then a second. And so on. Take the diagonal

The resulting subsequence is convergent

[wicked yew]

so a subsequence of a subsequence of a subsequence ..... ?

[gritty widget]

Yes

[wicked yew]

but since the sequence is infinite is this valid

[gritty widget]

We have a sequence of subsequences

And this won't converge to a subsequence in any sense of the word

But we can place them one under another and take the diagonal

The result will be a subsequence for which each of the coordinates converges to some point

This is your convergent subsequence

[wicked yew]

ah that's clever

thanks

[sinful hazel]

I think this is the right place... so I have a collection of facets that make up a continuous non-self-intersecting surface, each defined by their vertices and a normal vector. I want to pick a point in space and determine if I am inside of or outside of the body, but I have no idea how to do that, besides maybe checking all the centroids of the faces and checking if the closest facet is facing away or not like obtuse angle criterion type deal, any help?

[pseudo coral]

could i ask for proof verification here

on sequence of nonempty sets having nonempty intersection given their closed and compact

lemme make it sound better

here

took me a while but i managed to get it!

i think

if im correct that is

and for proving that union of connected sets with nonempty intersection does this suffice: Let X=U_nX_n with\cap_nX_n \neq \emptyset. Then there exists an x such that x \in X_n for all n

then suppose X. is disconnected then X = A u B with A,B clopen disjoint and non empty

then x \in A or in B

WLOG suppose x \in A

but B non empty so there exists some b \in B so b \in X_m for some m \in \Z^+

but then x \in X_m as well

but X_m is connected and can lie only in A or only in B

contradictign connectedness of X_n's for all n

its been so long im gonna step away for a sec so @ me if answering to either question plz

[pseudo coral]

any1??

[pseudo coral]

????

this channel dead or whaaa lol

[untold lily]

it's an advanced channel...

waiting hours for a question is extremely normal

[untold lily]

I'm also a beginner to topology so take my words with a grain of salt, but I don't see why the U_n are supposed to be open

your connectedness proof seems correct to me, although you might want to expand on why X_m can lie only in A or only in B (esp if this is for an assignment)

though again, a grain of salt because I don't feel like I have a strong enough grasp on topology to be helping u out, just doing so because you seem to be in a hurry?

[pseudo coral]

ive shown that separately

[pseudo coral]

cause their the complement of closed sets

[untold lily]

U_n = Y_1 - Y_n = Y_1 cap Y_n^C

Y_1 is closed, complement of Y_n is open

so isn't this just the intersection of a closed and open set?

suppose y1 was [-1,1] and y2 was [-1/2,1/2]

unless I am doing something terribly wrong y1-y2 does not seem to be open

[pseudo coral]

its open in the subspace topology i believe @untold lily

you can write [-1,-1/2) U (1/2,1]

[untold lily]

right, it is open in Y_1 but that is a very important distinction

and you need to mention that

[pseudo coral]

as union of the space and standard basis element

true

[untold lily]

you unambiguously write U_n is in tau

tau is topology on X

[pseudo coral]

true good point to point out

its open in the topology Y_1 inherits as a subspace

[untold lily]

yes

there are also problems with your last statements, although the main idea seems to be correct

you actually shouldn't have bothered with a contradiction tbh, a direct proof is much more clear here I feel

[green zinc]

Can anybody tell me how to learn topology?

Is it by books or video or course??

[plain raven]

You can look at the books "Topology" by Munkres or "Topology without tears". I am sure there are lecture courses for topology on youtube

[silk ember]

Question about first and second countability axiom. I am given a theorem in the book stating that A subspace of a first-countable space is first-countable and same for second-countable. Is it not enough to proof only second-countable as it implies first-countable?

[tawdry valve]

It is not enough. If you only proved only for second-countable, then you can only say that a subspace of a second countable space is first countable. You also want the theorem that says that a subspace of a first countable space is first countable.

[silk ember]

But doesn't second countable not imply first countable?

Or am I going wrong somewhere with my logic?

[plain raven]

Second-countability is a stronger notion than first-countability.

[wispy veldt]

Proving subspace of first countable is first countable starts from a weaker assumption than second countablity.

For example
If second countable implies second countable subspace , does that mean first countable implies second countable subspace?
Point is just because second countablity implies first countability doesn't mean results that are implied from second countablity will be implied by first countability.

Because the former is a stronger notion as mentioned.

[silk ember]

Oh yeah right. I see my mistake. Thanks!

[gritty widget]

First try to learn metric space topology so you can have intuition.
See "Topology of metric spaces" by Kumaresan

[gritty widget]

This is direct consequence of compactness of Y_1 and Y_i being closed

Recall that a space is compact iff every family of closed sets with finite intersection property has non-empty intersection

Here Y_i is such family in Y_1

Hence the intersection is non-empty

[untold lily]

but it can happen in a topological space that...

isn't this obviously wrong since the whole space will always be a nhood of both x and y?

I looked it up in stackexchange and the most reasonable interpretation seems to be excluding the whole space

even then it makes no sense though... weird

[viral atlas]

Yeah it does sound slightly careless for the reason you mentioned

[gritty widget]

They mean it's not in some neighbourhood

Wrong quantifier used is all

What the author had in mind is how cl({x}) can be properly contained in cl({y})

Such as in the Sierpiński two point space

[unreal stratus]

Sierpinski space 🤝 counterexamples

[fair idol]

Not sure where to ask this but this is more so a question about point set topology I think. Can someone give me a reason why we would like open sets of a topological space to be closed under arbitrary unions?

I understand that making sense of limits, continuity, and compactness are reasons for using a topological space. Is there somewhere that arbitrary unions of open sets being open are used for these? I would like to get an intuition for this property.

[gritty widget]

for me open sets are generalizations of unions of open balls

so naturally, unions of unions of open balls are unions of open balls

¯_(ツ)_/¯

[untold lily]

yeah, if you want to understand why we defined a topological space to have X property it's best to look at why it is the case for metric topologies

[plain raven]

One way to come at this perspective is to take the notion of "neighborhood" as fundamental, so that topology assigns each point x in a set X a family of subsets of X, each containing x, called "neighborhoods", the intuition being that if A is a neighborhood of x, you can move x around a little bit and you will still be in A.

The set of neighborhoods is required to satisfy certain axioms, for example the intersection of two neighborhoods of x is a neighborhood of x, any set containing a neighborhood of x is a neighborhood of x, the whole set is a neighborhood of x, and so on.

Then an "open set" is any set which is a neighborhood of each of its points. Thus, for every point in an open set, even those close to the edge, you have a little wiggle room to maneuver and you will still be in the open set.

From this perspective it is obvious why the union of open sets is again open, if you take this as the definition of an open set it is easy to prove

[fair idol]

Thank you so much for this perspective. I think further that the proof that A is open iff A=intA uses the closure of arbitrary unions of open sets is open

[prisma arrow]

i forget what second countable means visually

it is it just an amazing boundness condition

[fair idol]

Visually you imagine there is a collection of countable subsets that can generate all the subsets in your topological space

[gritty widget]

visually you can imagine (the definition)

[fair idol]

I guess I would draw a shape and turn it into patches and close my eyes and say "countable..." A few times in my head

[gritty widget]

i'm teasing cause i'd probably say the exact same thing lol

[plain raven]

The collection is countable, not the subsets

A countable collection of subsets, not a collection of countable subsets

[gritty widget]

you're countable

because i can count on you

[plain raven]

❤️

[prisma arrow]

dont hate they tried lol

o u said jk

im just going to say visually it means too many balls

because ig there isnt much visual intuition for understanding countability besides the idea of having gaps

[gritty widget]

as a gay man, you can never have too many

[prisma arrow]

are you gay?

if so my conjecture reigns true

[gritty widget]

yeah i have the honorable role

[prisma arrow]

#discussion message

[gritty widget]

i have that channel blocked

screencap

[gritty widget]

lol

yes i am gay

second countability to me is kinda like a way of restricting how ""big"" the space can be and it's a useful technical condition that gives you stuff like partitions of unity

i don't really think of it visually but maybe that's useful

[gritty widget]

(the latter obviously for nice enough spaces only)

(and i care about such nice enough spaces)

[untold lily]

I also think of it as how "big" a space is, in a certain sense

it is similar to cofinality of cardinals

there are probably some very important set theoretic concepts I am missing here that would expand upon that line of thinking

[gritty widget]

i would be interested in these set theoretic concepts

[untold lily]

what I am saying is, while the actual size (i.e. cardinality) of the topology might be massive, we can play around with the topology purely in countable land and be okay

I've only recently started studying topology tho so this might be a pedestrian take

[gritty widget]

yes, there is a concept of so called cardinal functions in topology

one of such functions is the weight of a topological space

[untold lily]

yeah I've seen those skimming engelking

[gritty widget]

there is a huge amount of inequalities between them, and this is part of set theoretic topology

if you're interested you can check out Handbook of set theoretic topology

[prisma arrow]

maybe there is intuition in terms of metric spaces

[untold lily]

well isn't that just the rationals

I think rationals are great intuition

[gritty widget]

second countable metric spaces are precisely those which admit a metrizable compactification

so in this sense they are small

[untold lily]

while set of reals are uncountable, they can be approximated to an arbitrary precision by a countable subset

so in some sense, the reals are shackled to countability

[gritty widget]

separability is a lot weaker property than that of second countability

it's really the latter that we usually care about, but the two notions are often equivalent, which gives us an easy condition to check if a metric space is second countable

[pseudo coral]

question about this proof

first off, canw e find the disjoint open sets U,V because all metric spaces are normal and A,B ebing open means theyre both closed ? since theyre complements??

and secondly how to theyc onclude that X_i \cap A and X_I \cap B are nonempty?

[pseudo coral]

that and metric space gives you second countable, right?

[dry jolt]

We're assuming that A and B are closed. But yeah, normality of metric spaces imply the existence of disjoint open sets U and V.

X_i \cap A is nonempty because A is nonempty, and if x is in A then x is in the intersection of all X_j, hence also in X_i.

[pseudo coral]

and siimlarly for B right

[dry jolt]

yes

[pseudo coral]

whats the point of creating the U,V and F_i

oh to show you cant have a separation of an individual X_i

[dry jolt]

right

[gritty widget]

Hey, I am reading "A Combinatorial Introduction to Topology". I've got a question about a vector field along a path and counting the "winding number" of points:
In the figure, the point R on γ_1 passes through the vertical "clockwise". Can anybody explain how it passes "clockwise"? As a counter example, R, T and U on the path γ_2 pass "counter clockwise".

[rain ether]

I would assume that you go along the path counter clockwise (conventionally positive direction) and look at how the parallel transported vectors pass through the vertical.

[gritty widget]

I think I get what you mean. For example, right from R (in γ_1) the arrow points north-west, and left from R it points north-east, so it moved clockwise?

[rain ether]

Yeah

[west spindle]

trying to do this problem

looking at these corollary and theorem from Spanier as suggested by @plain raven and trying to think how to modify them for my purposes

all i can think of is that maybe i should work with F2-homology groups

but i cant think of any way to proceed from this

like

maybe i could express my hypersurface as the sum-modulo-2 of some hyperspheres...?

but then i don't know how to work with that

like, i guess my "goal" is to show \tilde{H}_0(S^n - B; F_2) is nontrivial

but i dont know how to do that

if someone responds please ping me

[coarse night]

@west spindle I know you can prolly do this with Alexandar duality, but can you explain what's a "hypersurface in S^n" actually mean?

[west spindle]

what's alexandar duality?

i mean a manifold of dimension n-1 embedded in S^n

a manifold-without-boundary, as i would understand it

[coarse night]

AD is under "suitable" condition, $\tilde{H}i(S^n \setminus A) = \tilde{H}{n-i-1}(A)$

[west spindle]

i doubt that this was covered in class

we are studying by hatcher's book

oh, wait.

[coarse night]

hatcher has it

[west spindle]

hm

well

[coarse night]

hmm uses pincare duality

[west spindle]

"locally homeo to R^(n-1)" certainly qualifies for "locally contractible"...?

[coarse night]

yes

[west spindle]

but then how do i apply this here

[coarse night]

I believe this is the classical statement

and the one you shared is using PD

[west spindle]

hmm

kinda confused now

wait so like

$\tilde{H}_0(S^n - A) \overset?\neq 0$ \textit{is} the right idea, yes?

[gentle osprey]

Ann

[coarse night]

applying that we just get $\tilde{H}_0(S^n \setminus K) = \tilde H^{n-0-1}(K)$

[coarse night]

yes thing is I am getting that to be 0

[west spindle]

but isn't the reduced homology group trivial iff the space is connected?

[coarse night]

path connected, yes

actually, have you done Poincare duality in your class?

what we get is that for $n>1$ $\tilde{H}_0(S^n-K) = \tilde H ^{n-1}(K) = H^{n-1}(K) = H_0(K) = R$

[coarse night]

assuming K is connected, otherwise work on path components

but maybe someone else can give an argument without using PD

[west spindle]

i dont know. i think we did?

i missed a lot of class because of unforeseen circumstances

ive basically had to self study from the book alone

and a classmates notes

handwritten notes so i can't ctrl+f them

[coarse night]

[west spindle]

what is R btw

[coarse night]

coefficient ring

[west spindle]

right

[west spindle]

i think if K is disconnected then i can just take one connected component of K and show that S^n minus that is disconnected

and that will do it

[coarse night]

yes

[west spindle]

so the first two equalities are by poincaré duality, and the middle one is... what exactly

what's reduced cohomology? is it just cohomology with the adjustment at 0? or like

the thing with $C^{-1} = \bZ$

[gentle osprey]

Ann

[west spindle]

the modification of the chain complex for reduced homology

with the extra thing at degree -1

[coarse night]

first is AD, 2nd definition of reduced (co)homologies, for n>1 they are same and 3rd is PD

[coarse night]

ye just the dual of the augmented (singular) chain complex

[west spindle]

you mean the chain mplex?

@coarse night this look ok?

[coarse night]

No I meant the augmented chain complex

[coarse night]

yes ig

but this only works for n>1

[west spindle]

well i don't need n=1 anyway

it's trivial

[coarse night]

otherwise \tilde H^n-1 neq H^n-1

[west spindle]

S^1 is a circle and a hypersurface therein would just be a bunch of points

[coarse night]

is it?

[west spindle]

well ok no

[coarse night]

what if just one point

[west spindle]

can you have a smooth 0 dim manifold

[coarse night]

[west spindle]

i think its no big deal

can i ask you about another one

diligentclerk and i tried to go through it but not much luck

[coarse night]

If I can help then sure

[west spindle]

one mo

[coarse night]

The other way is clear, if Y is a m fold covering the char is m multiple of char of X

for the other way

[west spindle]

well yeah

i

wait

no, it's not clear to me

euler char is the alternating sum of cell counts per dimension, and also the alternating sum of ranks of homology groups

[coarse night]

ok I can only show for CW complexes

[coarse night]

yes that's the idea

[prisma arrow]

oh lol

[coarse night]

can't we argue in the similar manner? since it's compact polyhedra, the char is just alternating sum of i-faces

[prisma arrow]

you can decompose the map into exact sequence and use additivity of euler characteristic here?

like m points injecting into Y?

or no i phrased this poorly and it isnt applicable cuz kernels

[coarse night]

each point having exactly m preimages and being piecewise linear should(?) imply that 1-faces go to 1-faces?

[west spindle]

yeah but how do we know faces get sent to faces

1-faces go to 1-faces...?

but can't we have one edge in Y mapping into half of an edge in X

[coarse night]

what exactly they mean by piecewise linear map/

never worked with these so I am unaware of the def

[prisma arrow]

just like sharp turns

uh a picture is good

but abs x is an example

[coarse night]

ah i see, I think we need to use simplicial approximation to subdivide Y such that faces to faces

i-faces go to i-faces, the "exactly m preimages" should prevent degeneracies

[prisma arrow]

thats what i was assuming

but idk how to prove that

[coarse night]

one thing you can do is assuming m=1 we can show it's homeomorphism. for m>1, I want to claim there are disjoint copies of X (assuming X is just one compact convex poly)

@west spindle help me out on this one

[west spindle]

[coarse night]

I am assuming X is just one compact convex poly

[west spindle]

yeah but we can't do that can we

[coarse night]

so Y is again union of convex compact polys right?

[west spindle]

what if X is some kind of thing shaped vaguely like the union of two balls

[west spindle]

maybe? im not sure

well

oh

yeah

it is a union of convex compact polys yes

[coarse night]

by the def it's convex compact right?

so say x ∈X has 2 preimages lying in single path component

say y, y' then by linearity, f(ty+(1-t)y')=x means it contradicts exactly m preimage

so like no 2 preimage of x is in the same pc?

is this correct?

that is because connected components of Y are convex, by def

assuming that's correct we get m disjoint subspaces of Y each homeomorphic to X so char formula follows

@west spindle

[west spindle]

sorry, got distracted

wait but X itself need not be convex

also how do you know y and y' are in the same "linearity component"

[coarse night]

just working on one component

[west spindle]

so. wait

you aim to show that the preimages of any x ∈ X lie in distinct path components, yes?

[coarse night]

Y is union of compact convex subspace right?

[coarse night]

yes that's what I was trying

[west spindle]

so then, let me try to follow along

we have y, y' ∈ Y in the same path component such that f(y) = f(y') = x

you wish to say that there exists a convex polyhedron in the decomp of Y that contains both y and y', whence the entire segment [y, y'] collapses into x, contradicting the problem statement

did i understand you correctly?

[coarse night]

yes

and that's true because any pc component of Y must be a convex poly

[west spindle]

is that correct?

i'm not sure about that

what if Y is shaped something like this

[west spindle]

why's that the case?

we could have Y be something like 7 cubes glued together in a Pi shape, this would be path connected but not convex...

[coarse night]

oh ok that's not true, the def says just union, not "disjoint union" fuck

have to find another argument

somewhat like this? take a path from y to y' and vary t

[west spindle]

i think we can adapt this to saying that any two preimages of x must lie in different faces or something

[coarse night]

we get, for t and 1-t same image, but t=½ gives only one preimage, but continuity should give 2

ok have to formalize this

(nvm)

[gentle osprey]

MyMathYourMath

[pseudo coral]

does it follow from stating that since the X_n are compact subsets of Hausdorff space then theyre all closed thus their intersection is closed and a subset of X_1 which is compact and closed subsets of compact need be compact?

[gritty widget]

$\bigcap_n X_n$ is a closed subspace of $X_1$, hence compact

[gentle osprey]

Blitz

[pseudo coral]

yeah thats what i meant by theyre a subset of X_1 lol

[gritty widget]

yes, that X_i are closed follows from Hausdorff

that's the only place where we use that assumption

[gentle osprey]

MyMathYourMath

MyMathYourMath

MyMathYourMath

MyMathYourMath

MyMathYourMath

MyMathYourMath

[gritty widget]

intersection here, open in the intersection

[pseudo coral]

Oops

Meant intersection

[pseudo coral]

Meant $A \cup B = \bigcap_n X_n$

[gentle osprey]

MyMathYourMath

[gritty widget]

yeah looks ok

[pseudo coral]

sweet thabks

[pseudo coral]

i have a question of proving a compact Hausdorff space is regular, and thus normal as well. (Its similar to the proof of showing compact subspace of Hausdorff is closed)

[gentle osprey]

MyMathYourMath

MyMathYourMath

MyMathYourMath

MyMathYourMath

[pseudo coral]

And as finite intersections of open are open, we have that

[gentle osprey]

MyMathYourMath

MyMathYourMath

MyMathYourMath

[pseudo coral]

and the same trick for two closed sets shows its normal ?

[pseudo coral]

is open as its a finite intersection of open

the trick for normality given regulaity is to take A,B as closed sets

then for each y \in B there exists open sets U_y,V_y such that A \subset U_y, y \in V_y

then take the intersection (finite) of the U_y then A is still in the intersection and take the finite union of the V_y which covers B

igtg so @ me with a yes or no plz

[limpid hedge]

To show that the set [0,1]^[0,1] is not sequentially compact for sup metric, is it ok to justify that it is not by picking f_n(x)=x^n? This sequence converge pointwise to discontinuous function and thus can't converge uniformly (i.e for sup metric)

[gritty widget]

Yeah

[limpid hedge]

thank you

[west spindle]

having some issues with this problem still

depicted is my partial solution

i'm not quite sure where i'm going with this, or how to continue, or whether this is going to be a productive idea

https://cdn.discordapp.com/attachments/1058317440385294366/1058318245955899402/image.png

if needed, the definitions of "polyhedron" and "piecewise-linear map"...

[mental island]

Hi! Any good book or internet course about topology for beginners?

[gritty widget]

Topology of metric spaces by Kumaresan

[mental island]

Thanks! Can you add any other?

[coarse night]

topology without tears

[viral atlas]

There is an extensive set of lecture notes from University of Toronto

Let me see if I can find them

See "Resources" here: http://www.math.toronto.edu/ivan/mat327/?resources

[warm hedge]

is euler characteristic of a cylinder with two boundaries which dont go to infinity equal to zero?

[coarse night]

boundaries are open? in that case it's just homotopic to S¹ and you already know what's the EC is

[warm hedge]

What if the boundaries aren't open ?

[coarse night]

homotopic to S²

I realise that the "open" boundary is very vague

what I mean is like it's it like a soda can, top and bottom closes or just a cylinder like you get by rolling a paper

like you can put your hand through it,,, lacking terminology

[warm hedge]

Oh yeah I am bad with English terminology. I mean my cylinder has open top and bottom, as you said you can put your hand through it

[coarse night]

lol

[gritty widget]

is it easy to prove the uniqueness of the dimension of top mfd

for example there is no homeomorphism between Rn and Rm or sth like that

[coarse night]

without using homology theory? yes

you can show for lR and lRⁿ (n>1) using point set argument tho

[gritty widget]

lR

damn u have good tricks

lN

€

[coarse night]

IR

that's better

[unreal stratus]

Lol nice

[hollow wigeon]

Hello, I am a mathematician. Nice to meet you all

[hidden crag]

Hello

[rain ether]

Hi

[gritty widget]

Hello

[gritty widget]

this is not a discussion channel

[warm hedge]

could we say something about euler characteristic for compact surfaces with boundaries ? like for example it cant be over 0 or it cant be odd number (i am not saying these are true just giving an example 😛 )

[coarse night]

any compact odd dimensional manifold has EC 0

for with boundary, idk

[warm hedge]

oh ok

yeah i am trying to see what difference boundaries are making in the euler characteristic

[cedar pebble]

boundary components shouldn't make a difference as long as you're not putting punctures along the boundary components

[warm hedge]

i mean like if we take a torus and put two boundaries then from 2 we are going to zero for euler characteristic, is this working in general ? like from something if we cut a hole so add a boundary then the euler characteristic will drop by 1 ?

[cedar pebble]

every time you add a puncture or a boundary component the Euler characteristic drops by 1

if you have a genus g surface with b boundary components and n punctures not on those boundary components the Euler characteristic is 2-2g-b-n

[warm hedge]

oh ok thank you

[limpid hedge]

Yo, I have one question. Let (X,t), (Y,t') 2 top spaces with cofinite topologies. I need to show that if X and Y are finite then product topology on XxY agrees with cofinite top on XxY. But as X,Y are finite, in both cases it's just a discrete topology on XxY, no?

[coarse night]

Yes

Also you have typo

[limpid hedge]

thank you!

[coarse night]

@west spindle about that polyhedra thing, I suppose I have an answer but it uses simplicial approximation. So the polyheadras are simplicial complexes. The statement of SAT is for any map f: Y → X of simplicial complex, after subdividing Y, we can get a map g: Yⁿ → X that is homotopic to f and is simplicial, that is sends vertices to vertices and is affine when restricted to each simplex.
Note that subdividing doesn't change euler char as it's chain homotopic to id hence a quasi-iso.
So what we get is g( <v0, ..., vn>) = <g(v0), .., g(vn)> but all g(vi) may not be distinct. But say g(v0)=g(v1) the it will contradict the m-images. So g indeed sends i-face to i-face. Now m-images condition gives there are exactly m many i face of Yⁿ for an i face of X. So char Yⁿ= m charX, but char Yⁿ= charY.

[west spindle]

right

[coarse night]

sorry for not using LaTeX. Also for reference, check out Bredon section 21, 22 of ch4

[west spindle]

i can read your notation just fine

[gritty widget]

Hey! This is a figure from the book "A combinatorial introduction to topology". Can anybody explain why the vertices (0-simplexes) on the boundary d(C) (second column) appear at those locations (so first and second rows)? I would assume only the "endpoints", but there are drawn more than the endpoints.

[coarse night]

the diagrams are not really that readable tbh

bold lines are hard to see for the first figure

[gritty widget]

if there is an open set that contains all the isolated points of a space is that set dense.

[coarse night]

Not necessarily

like take (0,1) of lR, lR has no isolated point

[gritty widget]

A dense set needs to contain all the isolated points though

[feral copper]

Depends on the mod 2 number of lines that start at those points

Like, the rule is:

--+--   --->    ∅
  |

--+--   --->    .
  |

--+     --->    ∅
  |

--+     --->    .```

In the first and second cases, you can decompose the 1-chain into the sum of two chains. In the first case, they don't have endpoints at the "crossing", and in the second, exactly one will have an endpoint at the crossing. In the third and fourth cases, you directly have one 1-chain whose boundary you can compute locally

[warm hedge]

a question about the formula of the area of a triangle in hyperbolic geometry. I have two of them, one says the area is equal to pi minus the three angles of the triangle and the other one says its pi minues the three angles multiplied by r^2. Both of them are correct ?

[feral copper]

Idk, but it seems like it's more of a #diff-geo-diff-top question though 🙂
Bit rusty here, but I don't remember there being an r^2 term. What I remember is that the sum of angles is \pi-Area

[warm hedge]

sorry if i am in wrong channel

it gives me a trianlge with its area and its angles but the sum of angles is pi-area doesnt work

[warm hedge]

ah i think r is the curvation so for hyperbolic plane its -1 but in general it isnt always -1

an other question i have is that if i know the Euler characteristic of a topology how can i identify which surface it is ?

[feral copper]

A surface is uniquely determined by:

• its orientability
• its number of boundary components
• its Euler characteristic once you glue disks to each boundary component
  So if you know that the surface is orientable and has no boundary, then its Euler characteristic uniquely determines its genus by the relation \chi=2-2g

[warm hedge]

ohh i see

thank you

[feral copper]

Note that this is only true for surfaces

For 3-manifolds or more, or even for topological spaces in general, this is far from true

[sudden spire]

Closed surfaces?

Or does this allow for punctures also

[feral copper]

In manifold topology, closed means compact without boundary

So no punctures, no nothing

[warm hedge]

if we say compact for a surface, can it have boundaries ?

[feral copper]

Yes

[sudden spire]

Yeah so this is a classification for closed surfaces

No punctures no boundaries

[feral copper]

But then the Euler characteristic is not enough to distinguish between surfaces (e.g. the annulus and the torus both have zero Euler charecteristic)

Re-read my message, it works for all surfaces (compact I should say though!)

There are three things you need, not just the \chi

[sudden spire]

Oh I see oops. So compact surfaces possibly with boundary

[feral copper]

Yes, and possibly non-orientable too 😉

Oh and connected too x')

[warm hedge]

also if i know the orientability of my surface and that its compact and also uniform geometry and i have a specific triangle with specific angles has a specific area, can i find its euler characteristic with a formula ?

[sudden spire]

Yeah I originally read boundary components as just components lol

[feral copper]

Eeeeh, maybe, I'm no geometer
But I'd still say no. The flat torus (embedded in C² with constant zero curvature) is not the disk, which also happens to have zero curvature

Curvature is purely local

And the triangle only relates angles and area to curvature

[warm hedge]

i think if its uniform geometry the curvation is the same everywhere

[feral copper]

Also, a triangle is local too (it's basically a disk, i.e. an open set of a chart)

Idk what that means, but again this is more of a #diff-geo-diff-top question and people will surely be more knowledgeable than here (or at least than me xD)

[warm hedge]

ok i will ask there too 😛

thanks for your help

[feral copper]

What comes to my mind is genus at least two surfaces
They are all Riemann-covered by the hyperbolic plane, and I think the curvature is the same (the geometry of these surfaces are very similar), so maybe that answers no to your question?

[warm hedge]

i see 🤔

i will see if i can find something more myself and if not i will come back in the #diff-geo-diff-top

[feral copper]

Sure

[gritty widget]

Sorry! Is this better:

[gritty widget]

Do you mean that if an intersection has 2 or 4 edges, it won't result in a vertex, if it has 1 or 3 edges it will result in a vertex (0-simplex / boundary)?

[rain ether]

How is the boundary defined in the book?

[tame path]

hey, i have an idea of a proof that R and R^2 arent homeomoprh

I thought of a proof a couple minutes ago, but I have a better one now
D = {(x,y) : x in R and y = 0} is homeomorphic to R since f : D -> R, f (x,y) = x is continuous, bijective (in D) and f-1(x) = (x,0) is continuous
for D and R^2 to be homeomorph, there must be a continuous bijective mapping g : D -> R^2, so we can write g as g (x,y) = (a(x,y),b(x,y)) where a : R^2 -> R and b : R^->R
for any point (x,y) in D, g(x,y) = (a(x,0),b(x,0))
for g to be continuous and bijective, both a and b must also be bijective
and continuous
Say that there's a point (u,v) that can be mapped using g, so (u,v) = g(x,y)
so (u,v) = (a(x,0),b(x,0))
Can the point (u,v+1) be mapped then?
(u,v+1) = g(x',y') = (a(x',0),b(x',0))
since a is bijective, x = x'
so v +1 = b(x',0) = b(x,0) = v
since that implies 1 = 0, this means that there are points in R^2 that cannot be mapped using g
so g is not bijective
since g describes all possible continuous mappings, theres no g that is bijective
so R^2 and D arent homeomorphic and, therefore, R^2 and R arent homeomorph

Intuitively, the proof is based on the idea that g has only one degree of freedom (since y is fixed) so it can only describe curves and not surfaces

[coarse night]

invariance of domain go brr

[gritty widget]

"for g to be continuous and bijective, both a and b must also be bijective"

that's your error

[tame path]

can you give me a counter example?

[gritty widget]

(x, y) to (x, y) is continuous and bijective, but the projections (x, y) to x and (x, y) to y aren't bijective

[tame path]

what if I say that a : D -> R and b : D -> R

instead of R^2

[gritty widget]

how you can prove that R and R^n for n > 2 aren't homeomorphic is by removing a point from R

[tame path]

yeah, i read that because R is no longer connected

I just havent studied connectedness yet, so i thought i could try using that I knew

[coarse night]

There's another way, you can try to show no continuous map from R² to R is injective

but connectedness is the easiest one

[tame path]

I tried using to show that it coulnt be injective at first

then I thought of showing it coulndt be surjective

[coarse night]

like just restrict your map f:S¹ → IR, then show by IVT there must exist one t s.t. f(t)=f(-t)

||use characterization of S¹ and look at function g(t)=f(t)-f(-t)||

[coarse night]

There are surjective maps from R to R² tho, they are called space filling curves

Here's another problem if you want to try

show that there's no continuous bijection from IR to IR². ( Not saying homeomorphism btw)

[tame path]

I mean, cant i just say that R and R^2 arent homeomorphic?

oh wait

yeah thats not enough

[coarse night]

That's the point right

[tame path]

wait hang on

that isnt enough right?

[coarse night]

There can exist continuous bijection without being homeomorphism

(X, discrete) → (X, indiscrete)

identity map

[tame path]

I really dont know

this is all very new to me, i study physics

[coarse night]

maybe keep this question in your mind and comeback to it later.

oh

then NVM

[tame path]

ikr

like, i had to learn the concepts that you guys learn in analysis in the same subject

ill probably stick to the easiest ways of solving problems

thanks though

[pseudo coral]

does this suffice in showing path-connected implies connected: suppose the space is disconnected then X = A u B where A,B are open and disjoint and non empty. let a \in A and b \in B with path such that f(0)=a and f(b)=1 but the image of the unit interval is connected so it must lie entirely in A or entierly in B, we cannot have f(0) \in A and f(1) \in B as f([0,1]) \subset A or \subset B. I know this may be a silly question to ask but just wanted to make sure I have the logic right

and if youre a connected subspace of A u B then you lie entirely within A or entirely within B (I call this the connected lemma)

[gritty widget]

I'm curious, why study topology?

[pseudo coral]

so by the connected lemma the image of unit interval under the path need lie within A entierly or within B

[gritty widget]

ur proof seems fine

[pseudo coral]

thanks ^

[gritty widget]

you can also just do it directly by taking the preimage of the separation

[pseudo coral]

which must be disjoint and open and closed

[gritty widget]

how I'd prove it is that image of a connected space is connected, and union of connected sets with non-empty intersection is connected
so consider images of paths from some fixed point of your space to any other point, which is the whole space, and by what I said, is connected

[pseudo coral]

but there is no separation of unit intrval

[tame path]

we use it in qft and deformations as far as I know

[pseudo coral]

but my proof works fine, right?

I see that method

so you union over all paths from a given fixed point to other points and their union is the whole space

and they share the fixed point in common

[gritty widget]

TTerra already read it so I don't have to

[pseudo coral]

lol fair enough

[gritty widget]

how do you know i read it?

[tame path]

hello

U is open

can anyone help?

[balmy field]

what is A?

[tame path]

both U and A are subsets of X, where (X,t) is a topological space

but U is open

[gritty widget]

Take some point x which is both in U and closure of A

that the point x is in closure of A, means that any neighbourhood of x intersects A

therefore, U intersects A

[tame path]

thank you

[pseudo coral]

let me fixate one thing hang on

[gritty widget]

"for each x and n there exists j" is what you meant

[pseudo coral]

yes

here

[pseudo coral]

thats exactly what I was fixating on lol

[gritty widget]

yeah good

no wait

first line

"Let Y be a subset"

delete that

[pseudo coral]

oh ur right were CONSTRUCTING such a Y

[gritty widget]

after I claim that, the equalities there are wrong

[pseudo coral]

wdym

isnt the closure of the x_nj's all of X

[gritty widget]

first set is Y

this is how you want to define Y

[pseudo coral]

i should denote that set as Y first huh

[gritty widget]

then you write = cl(Y)

?

you see what I mean

[pseudo coral]

yeah

i didnt define a Y

then just introduced its closure

is my Y the set of all the x_nj's

and their closure is all of X

[gritty widget]

yes

[pseudo coral]

there we go! thanks!!

is this correct

[gritty widget]

yea sure

[gentle osprey]

MyMathYourMath

MyMathYourMath

MyMathYourMath

[pseudo coral]

and X \setminus B \subset U

and conversely, suppose $E,F \subset X$ are disjoint closed sets then $U:=X \setminus F$ is open and contains $E$ thus there exists $A \in \tau$ such that $E \subset A \subset \overline{A} \subset X \setminus F$ then $A,X \setminus \overline{A}$ are disjoint. open sets containing $E,F$ respectively.

[gentle osprey]

MyMathYourMath

[sand dust]

hello 👋
am first year in college nd am totally new to topology, can you recommend me some good books/teachers or any sources to get started in ?

[pseudo coral]

topology by munkres

and intro to topology by gamelin and greene

[sand dust]

i'll check that now, thank you!

[unreal stratus]

Yeah just examine how paths lift

[pseudo coral]

silly question but

[gritty widget]

the answer is yes

[gentle osprey]

MyMathYourMath

[pseudo coral]

by DeMorgans Laws

right?

yes i get both set containments

[feral copper]

So it is defined exactly as I said, there's not even a proof to be made...

[feral copper]

Use this then

[gentle osprey]

MyMathYourMath

MyMathYourMath

[wicked yew]

sorry this probably seems very trivial but I just want to make sure I'm on the right lines.

i and ii) are not complete since the sequence of 1/n is cauchy in both of these metric spaces however does not converge

is that right?

[pseudo coral]

i wanna say the first one IS Cauchy the latter is NOT, but i could be mistaken have you tried proving it ?

[golden gust]

do you mean the other way around?

as in, I agree that the first one is complete but the sequence 1/n is not cauchy

[wicked yew]

taking epsilon = 1/N doesn't work?

then for n,m greater than N |1/n-1/m| < epsilon

[golden gust]

for the first one, d(1/n,1/m) = |n-m|

[wicked yew]

ohh its 1/

i'm dumb

[pseudo coral]

could someone help me on this one

[pseudo coral]

is THIS true

is the idea the following: the preimage of singletons in Y are compact in a Hausdorff space which are closed then can I separate these disjoint closed sets with disjoint open sets and map them back into Y?

my only question is do distinct y values in Y map backwards to disjoint preimages in X

[coarse night]

yes

easy set theoretic argument

[pseudo coral]

ok and their compact subspaces of a Hausdorff space so their disjoint and closed

[rancid umbra]

if r^2 = r is r a retract?

i don’t think it is but i’m having some trouble producing a counter example

[coarse night]

it is, assuming r is continuous.

retracts are just like projections on linear spaces, characterized by P^2 = P

[unreal stratus]

Lol t these past papers r giving me flashbacks

[wicked yew]

haha

I keep messing up these simple questions and definitions

[pseudo coral]

Same here 😒

[unreal stratus]

Also like a nice trick with these is uh

[rancid umbra]

but it’s a retract on the image of r, correct?

[coarse night]

yes

[unreal stratus]

Well okay so for the first, this is isometrically isomorphic to [1,infty) with normal metric

Which we know is complete

For the second, this is isometrically isomorphic to (0,infty) with the standard metric for the same reason, I think

[golden gust]

I think given the hint for that question "you may assume R is complete" they want you to explicitly say it's a closed subset of R

[unreal stratus]

Well sure

I wasn't writing out complete proofs there

complete lol

[golden gust]

nice

[unreal stratus]

How is ur project btw

[golden gust]

not bad I've been doing some work on it recently

wbu

[unreal stratus]

Noicee

[golden gust]

topological k theory right? I've had encounters with it bc one of the main constructions of my thing comes from the index theorem

[wicked yew]

oh yes i forgot about that

[placid cave]

Hello! I have a problem with proving that $S^n$ is a covering space of $RP^n$. Given $y=q(x) \in RP^n$ we have that a set $y\in U$ is open iff $x\in q^{-1}[U]$ is open in $S^n$. Define $A$ as the upper hemisphere and $B$ as the lower hemisphere of $S^n$, then $x\in A\cup B$ . Where should I go from here?

[gentle osprey]

よろしく

[placid cave]

The map q is the canonical projection onto the quotient, that is q(x)=[x]

[unreal stratus]

yeah that's part of it

i'm basically doing some old papers in alg top / htpy theory so i will work on either like j homomorphism or hopf invariant one now probs + basic chromatic homotopy theory

how does the index theorem crop up in what you're doing? (@golden gust)

[golden gust]

I have no idea what those things are but they sound cool

[golden gust]

the index theorem gives the "virtual dimension" of some moduli space of curves cut out by a fredholm map

[unreal stratus]

I have no idea what those things are but they sound cool

[golden gust]

haha fair enough

[unreal stratus]

i mean i get some of the pieces but idk what fredholm maps are

or are these just fredholm operators ig

[golden gust]

the linearisation of a fredholm map is a fredholm operator

[unreal stratus]

but yeah so how does this relate to what you're doing?

ah

what is a fredholm map then

hm i might look at the index theorem more properly lol

[golden gust]

so um I'm considering a space of pseudo-holomorphic curves, and there's a clever way to interpret this space as the zero set of some section of a banach bundle. this section turns out to be a fredholm map and so (in a way I haven't figured out yet, but can probably just blackbox) the dimension of the zero section is given by the index of the linearisation

[gritty widget]

is there such a things as a connected set in topological space?

[unreal stratus]

Because a major theorem in K-theory (idk how much you know) is Bott periodicity and there was a proof of this using the index theorem

[unreal stratus]

Well connected space I guess

But yes

I can explain the definitions and stuff if you want + examples?

[golden gust]

yeah I know the rough statement, but nothing more than that

[unreal stratus]

yeah i mean the important thing being that reduced real (resp. complex) K-theory is 8 (resp. 2) periodic

[golden gust]

why 8

[unreal stratus]

hm okie inch resting

[unreal stratus]

Well okay good question I can't really answer lol

So like the 8 and 2 come from the spaces O and U themselves

But then ofc it's like well why the 2 and 8 for them lol

Maybe this is clearer from the proof though idk

[gentle osprey]

potato

[golden gust]

Omega^8 is the loop space of the loop space of the ... of the loop space? (8 times)??

[unreal stratus]

Yes

[golden gust]

what the hell

[unreal stratus]

Well there is also \Omega^{\infty}

[golden gust]

😵

[unreal stratus]

But idk I mean repeated suspensions/loop spaces are common

Just you kinda give up visualising them in the same way i guess,

[golden gust]

I've never seen them I guess

[golden gust]

yeah I suppose you have to

[unreal stratus]

Yeahhh oof

Idk about say suspensions too cause like lol

You can visusalise the suspension of like a 2D object okay i suppose as like a couple of ocnes

but like beyond that and with repeated suspensions lol

[golden gust]

"taking loop spaces shifts the homotopy groups" is there a way to see this or is it deep?

[unreal stratus]

Well so like

[hidden crag]

is this bott

ah wait nvm

[unreal stratus]

One way to see this (at least ignoring the group operation) is that if $[A,B]$ denotes the set of htpy classes of pointed maps $A \to B$, then we have a bijection $[\Sigma X, Y] \simeq [X,\Omega Y]$ and in particular taking $X = S^0$ this gives us that $\pi_n(Y) = [S^n, Y] \simeq [\Sigma^n S^0,Y] \simeq [S^0, \Omega^n Y] = \pi_0(\Omega^n Y)$ at least on the level of sets.

[gentle osprey]

potato

potato

[unreal stratus]

Dw if you don't know what a fibration is, like it is basically a generalisation of a fibre bundle

[golden gust]

ah yeah nice

[gentle osprey]

potato

[golden gust]

oh that is cute

[unreal stratus]

yee :)

I think also uhhhhh

So higher homotopy groups of points vanish

[gentle osprey]

potato

[unreal stratus]

So like all higher homotopy groups of S1 or the torus etc vanish lol

[unreal stratus]

No, Bott is a person, not an image

[golden gust]

sick

I wish I had fun facts like this

[unreal stratus]

Aw lol

[next crystal]

this is from munkres (in the ss "range" is what we usually call "codomain" so a set containing the image)
i'm gonna use codomain so its less confusing
Why do we have "g○f is defined only when the codomain of f equals the domain of g"?
Isn't having image(f) ⊆ dom(g) or codomain(f) ⊆ dom(g) enough for g○f to be defined?
so it seems like it should be (codomain(f) = domain(g) implies g○f defined, instead of the converse)

I know in the book this is a definition but if what i'm saying makes sense isnt this definition way more limiting than it has to be?

[gritty widget]

although you're reading this from a topology book, this isn't a topology question

[plain raven]

yeah it's fine

you're right but just like, accept his definition lol

Essentially you can think of him as working in a formalism where there's a law that associates to each pair of spaces (A, B) the set of continuous functions Cts(A,B) and to each triple of spaces (A, B, C) a composition law Cts(A, B) x Cts(B, C) -> Cts(A, C)

or it associates to each pair of sets (A, B) the set of functions Fun(A, B) and to each triple of spaces (A, B, C) the composition law Fun(A, B) x Fun(B, C) -> Fun(A, C)

You're not wrong but like. if it makes you feel better you can think of him as working in a more restricted formalism than ZFC set theory, everything's interpretable in ZFC but for him every function has a clearly specified domain and codomain as part of its data

[next crystal]

okay thanks that makes sense

[placid cave]

Ive got a question concerning the definition of a covering map, we know that R is the covering space of S^1 via the exponential function, by definition the exponential function restricted to a connected component is a homeomorphism, i.e., restricted to an open Interval is a homeomorphism, but that would imply that the fundamental group of S^1 is trivial, isnt that a contradiction?

[plain raven]

the middle part is not true

"the exponential function restricted to a connected component is a homeomorphism"

"restricted to an open interval is a homeomorphism"

this is not true

When we say that the exponential function is locally a homeomorphism onto its image what we mean is that for each point x there exists an open interval (x - a, x + a) such that e^{i\theta} is a homeomorphism onto its image when restricted to that interval

not that this is true for arbitrary a

Your claim is only true for sufficiently small open intervals, but the function is not injective on larger open intervals

as $e^{i\theta} = e^{i(2\pi+\theta)}$ for any $\theta$

[gentle osprey]

diligentClerk

[plain raven]

so $e^{i\theta}$ is not injective on any connected interval $(a,b)$ of length greater than $2\pi$

[gentle osprey]

diligentClerk

[unreal stratus]

(also i'm not sure how this is "by definition")

Probs also worth saying like covering maps are always local homeomorphisms

and every sufficiently nice space has a universal cover (i.e. a simply connected covering space) so the case of S^1 isn't unique here

[placid cave]

But part of the definition is that, for example here, r restricted to an Interval of that kind is a homeomorphism right?

[unreal stratus]

onto its image sure

[placid cave]

Ohhhhh yes, i'm SO blind xD
Thanks!

[unreal stratus]

np

[old oak]

If it is homeomorphic to the cantor set, by moore kline it needs to be totally disconnected right? but since any 2 open sets are non disjoint (at least one of their projections need to be \Sigma), the space itself is not disconnected.

so how can this be true?

oh its not an iff statement

[fair idol]

This is a basic topology question.

If f:X->R is a continuous function from a compact topological space X and the real numbers R
Then f is bounded and attains it's max and min values.

I'm pretty sure the idea is just to use that since X is compact, so is f(X) and as a compact set in R, Heine borel says f(X) is closed and bounded as a subset in the standard topology of R.

But for attaining it's min and max I'm not entirely sure. My guess is that min and Max's are limit points of f(X) and as it is a closed set we are done.

Is this the right idea?

[gritty widget]

compact subsets of R have smallest and largest elements

bounded gives you an inf and a sup, and closed tells you they're in there

[gentle osprey]

MyMathYourMath

MyMathYourMath

MyMathYourMath

MyMathYourMath

[pseudo coral]

are nets just generalized sequences where your indexing set is some directed set lol im new to learning about nets

[plain raven]

yes

[pseudo coral]

the continuous image of compact is compact and compact in R as you mentioned is closed and bounded.

[pseudo coral]

thx, is my proof correct for showing convergent nets get mapped to convergent nets implies f is continuous

[plain raven]

probably but i'm not awake yet lol

[pseudo coral]

lol

[plain raven]

up to boolean algebra of subsets

i think your proof would be clearer if you spent less time on inverting subsets and stuff

is there a way you can rewrite it that does not involve doing this or like

maybe you can prove that the definition of net convergence can be inverted in a way that's closer to what you're trying to prove

[gritty widget]

say there is a point q for some point p such that every open neighbour of p contains q what is the releation between them?

is threre a name?

[unreal stratus]

Not what you asked for exactly, but if this relationship is symmetric in p and q you can call them topologically indistinguishable