point-set-topology

[prisma arrow]

Also what are the upsides and downsides of homotopy continuation?

I guess ODEs

[swift fjord]

Wdym? That's the definition of homotopy class. Two maps are in yhe same homotopu class iff they're homotopic, i.e. there exists a homotopy between them

You're not gonna be able to describe this explicitly for 2 general functions

[prisma arrow]

You didnt have to restate the definition, but what do you think are the limitations between knowing if any two given functions are in the same class, and giving an explicit homotopy

Im not sure how to approach the second question but maybe it has to do with really nice descriptions of things like convex spaces

[swift fjord]

I'm sorry, I misread your question as "I DO know they are in the same homotopy class"

So I thought you might be confused about the defn

My bad

[prisma arrow]

dont apologize its fine

[swift fjord]

I think the first question is too broad, and if you want a meaningful answer you'd probably want to narrow your scope a bit

Giving an explicit function in an arbitrary space us nigh impossible

[prisma arrow]

i never thought of this to be honest

and i feel bad because this feels like a natural question

[swift fjord]

It is

I think it's as natural as "when are two spaces homotopic/homeomorphic"

But just like those questions, it's pretty much impossible to give good answers except under some nice special cases

[prisma arrow]

im angry

[swift fjord]

There's just too many spaces, man

And even more functions (well technically not but yaknow)

[prisma arrow]

too many spaces is right

the amount of intractable problems makes me sad

[swift fjord]

Even if we restrict to like, very specific cases, say maps R^n->R^m I still think this question is very hard (maybe unanswered/unanswerable?). I think the best we can do is assume things about either the functions or both the spaces and the functions

And get some partial results

I mean really a lot of theorems of that sort are good because they're applicable to every-day spaces

Pretty much any theorem regarding CW complexes

[swift fjord]

Yea it's a sad reality of math

But if a complete classification was that easy then the definition is probably not very interesting

[empty grove]

R^m is contractible, so any such pair of maps is homotopic

[prisma arrow]

Finding the explicit homotopy part is fine too because linear homotopy I assume

[swift fjord]

Oh oops you're right 🤦‍♂️

[swift fjord]

Yea

I feel like I forgot a lot of topology

I tried thinking of an example that would be simple but not trivial but had a brainfart

[little hemlock]

the difficulty of computing the homotopy groups of spheres is maybe a better example

[prisma arrow]

Im more concerned about

the second question

of knowing when there is an explicit formula for the homotopy

idk what to do if it isnt linear

also i cant think of examples easily

[gritty widget]

@peak crystal I've once said to you that finding Tychonoff spaces which are not T4 is a hard task.
I misremembered - it's finding regular spaces that aren't Tychonoff that's hard.
But I heard there is a process of constructing T3 spaces that aren't Tychonoff from a countable amount of Tychonoff spaces

Hopefully you'll forgive me

[odd flame]

is the fact that this is a homeomorphism trivial?

[hidden crag]

Whats exercise 19.8

[odd flame]

oh i guess i should've checked that before asking here yeah that has it my b

[hidden crag]

[odd flame]

for b) how can it be a bijection if one is finer

this is the questiion

[dry jolt]

the identity map is a bijection on the underlying sets

[odd flame]

ohhhh from X to X

i wass thinking T to T'

but that's like a side effect of the map from X to X right

[dry jolt]

what do you mean?

[odd flame]

i was thinking as the identity mapping as being a map between the topologies

which i guess it kinda is

but the actual map is from X to X

[dry jolt]

Note that the map from the coarser topology to the finer topology won't necessarily be continuous though

[odd flame]

i dont wanna note that

[glass bison]

Kinda stuck on part b here, aren't the conditions for defining this topology the same as part a?

[gritty widget]

What do you mean?

Topology of X' is given you from heavens

[glass bison]

aren't both X and X' spaces defined on metric d?

oh

[gritty widget]

It's not generated by anything

The function just happens to also be continuous for X' in place of X

Hopefully you're not confused anymore

Tbh d doesn't even have to be continuous on X x X, just coordinate-wise

[glass bison]

Ohh I see

Thanks I'll give it another go

[small hemlock]

I'm looking for a counterexample to $\overline{A-B} \subseteq \bar{A} - \bar{B}$. Could I just take A=(-3, 3) and B=(-1,1) in R for example? Then if x=-1 or x=1, $x \in [-3, -1] \cup [1, 3]$ but $x \notin [-3, -1) \cup (1, 3]$?

[gritty widget]

You can do \overline if you want the bar to stretch over the whole thing.

[small hemlock]

I was curious about that, thank you

[gentle osprey]

J.Ross

[gritty widget]

Your example works.

Another nice example is Q in R. You can look for them by letting A be the whole space and B dense, so that the set on the right is empty.

[small hemlock]

Oh that's very interesting, its a nice general result

Thank you for that

[small hemlock]

How could I approach finding a continuous function from R to R with a standard basis element I=(a,b) such that f^-1(I) is not a standard basis element? I understand the idea, but starting a problem like this seems odd

[gaunt linden]

Does "standard basis element" just mean "open interval" here?

[small hemlock]

It does, yes, sorry

[mortal iris]

Not sure, but what book is this? Looks like it has good exercises

[thorny agate]

Ok this is probably going to be an exceedingly dumb question but uhhhh

what exactly is algebraic topology?

I've taken a graduate algebra course and am taking real analysis right now for reference about background (haven't taken a topology course tho)

[tawdry valve]

topological spaces are hard to understand, they got too much subtlety. You often would like to know whether a continuous map exists between spaces, or whether spaces are homeomorphic, but it’s hard to tell. If two spaces are the same, then you maybe have hope of just writing down a homeomorphism. Otherwise, it can be really tough to show no map exists.

algebraic topology deals with extracting data from spaces and continuous maps. It’s kinda like how a data set is complicated, so we learn about it from simpler things like mean, standard deviation, etc. We do something analogous where we start with a topological space and get something more tractable, like a number or maybe different groups that are associated to spaces (e.g., homotopy and homology groups).

Then, we’re interested in what continuous maps do to the associated groups. The cool thing is “functoriality”, an idea of category theory invented for alg top. Just as we extract groups from spaces, we extract group homomorphisms from continuous maps, and this process respects composition.

Let me just give one vague example. We might be interested in whether the 2-sphere and the 3-sphere are homeomorphic. Your intuition probably tells you no, but justifying that intuition turns out to be a little tricky. This is where homology helps you. It turns out that the 2nd homology group of S^2 is Z (the integers), but the 2nd homology group of S^3 is 0. If these spaces were homeomorphic, the homeomorphism would turn into a group isomorphism but clearly Z and 0 are not isomorphic. Thus, there does not exist a homeomorphism

this is at least how algebraic topology starts out being useful in like an intro class. I’m sure there are more mature perspectives to it too

[gritty widget]

Or just calculate their dimension

[unreal stratus]

Ew

Well how do you mean calculate dimension here

[gritty widget]

Well, it's very easy tbh

A sphere S^n is union of two closed subspaces homeomorphic to [0, 1]^n

From subset and sum theorems its dimension equals that of [0, 1]^n

Which is n as standard computation shows

To show dim [0, 1]^n = n you still need Brouwer fixed point theorem and a bit of work, but it's doable

So yeah, no need for all the AT machinery to show this

AT is still crazy important of course

[gritty widget]

I just learned about ultraconnected and hyperconnected spaces

that just screams bad separation properties

[coarse night]

what's ultraconnected

[gritty widget]

any finite intersection of non-empty closed sets is non-empty

[small hemlock]

What's connected in general?

In a topological space

[little hemlock]

when ur space is not the union of disjoint open sets

[small hemlock]

So is R with the standard topology connected since you cannot express is as the union of disjoint open sets?

[little hemlock]

ye

[dense wren]

before i drop in with the problem i'm facing for the next couple weeks, is the topic of set/multiset diversity functions ok for this channel or should i look somewhere else
i can't really tell since i'm a compsci major taking a math elective purely by necessity >.<

[small hemlock]

Nice, that's pretty interesting

[little hemlock]

there are a few equivalent definitions of connectedness. one cool one is that a space is connected iff X and the empty set are the only clopen sets in the space

[unreal stratus]

and if X is not empty 😼

[little hemlock]

connectedness can also be a bit pathological though (see topologist's sine curve for example). There is a stronger condition called "path connectedness" which often captures the intuitive notion of connectedness better.

[small hemlock]

Oo that equivalent definition in unique, I like it. Does that imply that the trivial topology is always connected and the discrete topology is never connected?

And path connectedness, okay. I've heard about these terms, we just haven't made it that far in the class yet, but they're nice definitions

[unreal stratus]

Correct outside trivialities

(singletons with discrete top are connected)

[gritty widget]

I never heard of something like that in my life

[dense wren]

trust me, me neither

[gritty widget]

just ask, if it's wrong channel for this then we'll correct you

[small hemlock]

Ahhh of course. Topology keeps getting weirder, there are so many very general and basic definitions with so much embedded in them, it's nice

[dense wren]

but basically, i'm doing a project to develop a metric (in its formal definition) used to compare 'how far', as in 'how much do you have to change one set to turn it into the other set', sets/multisets are from each other and looking for any useful resources

more specifically it's finite multisets of finite sets of anything that you can check equality for (technically no ordering, or i guess no specific ordering), but let's assume natural numbers for simplicity (and implementation) sake

if it's fine to post here about then i can continue on what i tried so far

[gritty widget]

connected is not strange at all. You should see ultraconnectedness or hyperconnectedness

And I wouldn't say connectedness is pathological or unintuitive

nor that path-connectedness reflects our intuition somehow

arc-connectedness is a little better
fortunately for a lot of spaces arc-connected and path-connected are the same

that said, topologist sine curve is actually intuitive

it's a good example to get rid yourself of some bad intuition

what connectedness does tell you, and what it doesn't

[small hemlock]

I'll definitely look into these definitions, there seem to be a lot. You said that ultraconnectedness means that any finite intersection of nonempty closed sets is nonempty. Does that mean R with the standard topology is not ultraconnected? Just take two disjoint closed intervals

[gritty widget]

Nah. I haven't known what ultraconnected/hyperconnected means since today

those properties look very pathological and probably only something like algebraic geometry uses them

[small hemlock]

Ahhh okay fair enough

[gritty widget]

I mean - pathological in a sense that they're not what we usually think about when we think about nice spaces, like the metrizable ones

I'm sure they're nice properties to have

[small hemlock]

I'm definitely not there yet, but maybe one day. I covered metric/normed vector spaces in analysis, but not topology yet. In topology we finished separation axioms and closure and are finishing up continuous functions this week.

A nice problem I haven't been able to solve yet is "Show that there exists a subset A of R and disjoint subsets C and D that are closed in A such that the closures of C and D in R are not disjoint."

[gritty widget]

what kinda separation axioms did you talk about already

[gritty widget]

Try A = R without 0

[small hemlock]

T1 (singleton sets are closed, and an equivalent definition), Hausdorff, Regular, and Normal is all we covered

[gritty widget]

T0?

[small hemlock]

I know it exists but am not sure what it is

[gritty widget]

completely normal, perfectly normal, Tychonoff?

[small hemlock]

Nope, unfortunately

Also I'll look into the A=R thing real quick

I imagine there's a nice reference of all the separation axioms I could hang on my wall, or type up

[gritty widget]

wikipedia

[small hemlock]

Oh there are so many

[gritty widget]

it's alright, no need to remember them all
Normal, Hausdorff, Tychonoff are probably the most important

[small hemlock]

That's relieving, there are a lot of definitions as is

[gritty widget]

Normal because of Tietze and Urysohn theorems, Hausdorff for obvious reasons, and Tychonoff because they're the topologies which can be generated by a family of pseduo-metrics/which can be given a uniformity/which can be Cech-Stone compactified/which embedd into the generalized "Hilbert cube" [0, 1]^k for some cardinal k

and regular too ig, because of the Urysohn metrization theorem

all (Hausdorff) topological groups or vector spaces are always Tychonoff for example (but not necessarily normal)

I don't think there'll ever be a moment when you're happy that your space is regular. It's just less conditions to check for Urysohn metrization theorem

[dense wren]

just as a reminder - i guess it's fine if there was no correction? i'll prob get back to it later (since i'm doing something else right now)

[gritty widget]

cardinality of symmetric difference maybe

[small hemlock]

I'll understand all of that one day I'm sure

[dense wren]

hm, i initially thought of symmetric difference, but in a different way now that you mention it, in the form closer to levenshtein distance where i allow substitution of elements (so basically ceiling of cardinality of symmetric difference divided by two, if i'm thinking correctly)
the bigger problem that i'm having here is using this to 'measure distance' between multisets of sets, and proving it is indeed a metric, but I may be missing something here

i'll try just using the cardinality itself instead of some offshoot of it for now and get back later if i get stuck, thanks

[gritty widget]

It's definitely a metric

[dense wren]

that's my intuition, or at least it was close enough I suppose, but putting it into a formal proof is going to give me some trouble i feel

[gritty widget]

It boils down to $A\Delta C\subseteq A\Delta B\cup B\Delta C$, $A\Delta B = \emptyset$ iff $A = B$, and $A\Delta B = B\Delta A$

[gentle osprey]

Blitz

[gritty widget]

first property makes triangle inequality work, second makes d(x, y) = 0 iff x = y work, and third is symmetry

I'm talking about finite sets of course since I don't want to dive into multisets

From topological standpoint this metric is very boring - the topology is discrete

if we replaced cardinality with a different notion of measure, we might get something interesting

[dense wren]

well it's a binary "the element is there or not"

though hm, maybe i used the wrong words here
if it's a set to set comparison, I can handle that, I meant it more on the line of comparing a "multiset of sets" to another "multiset of sets"
the initial idea i had is to use that offshoot of cardinality of symmetric difference, order up the sets in multisets arbitrarily but both in the same way, if one of the multisets has lower cardinality pad it with empty sets (and adjust the final distance), use that ordering to compute distances between sets pairwise then sum them up

not sure if understandably so or not, but proving triangle inequality here proved troublesome for me

maybe i just need more background to understand what's going on here

either way thanks for suggesting some ideas

[gritty widget]

I live to serve

[jagged sage]

If I have X and Y with cofinite topology does X x Y need to have cofinite topology? I think not but can’t think of counterexample

[gritty widget]

Take X, Y infinite then for any finite F, X x F is closed but not finite, nor the whole space

in fact, X can be infinite and Y have at least two points

so if X, Y have cofinite topology then X x Y has cofinite topology only in trivial cases

[jagged sage]

Thanks way simpler than I thought

[gritty widget]

I've seen the definition of "dimensionality" for vector spaces, but is there also a notion of "dimensionality" for metric spaces or topologies or something? If we have a continuous function f on [0, 1] where f(0) = f(1), I know that R^3 is the only R^n where we can divide such paths into equivalence classes of knots, so is there a sense in which a topology that has distinct kinds of "knots" be considered 3D?

[ornate berry]

Yes, there are notions of topological dimension, but these tend to be somewhat specific to the kind of topological space you're looking at. I'm familiar with Krull dimension for the Zariski topology, but this is not suited to manifolds – I think covering dimension is one that is.

I think the 3d knot thing isn't the best idea as if you just choose a topological space with nontrivial fundamental group, you get a nontrivial "knot"

[gritty widget]

Are these things you're talking about part of algebraic topology? I've only seen the basic point-set stuff covered in undergrad analysis.

[ornate berry]

Oh, I think when one studies a topology course you'll see the fundamental group.

It is indeed part of algebraic topology.

[gritty widget]

Thank you!

[ornate berry]

No worries

[gritty widget]

Covering dim works good for any separable metric space

There are also cohomological dimensions

For normal spaces this is a pretty good notion of dimension as well, but we can do a better (equivalent) one

Which works relatively well for any Tychonoff space

[gritty widget]

Usually only at the end of some courses

[gritty widget]

Most notably small/large inductive dimensions and covering dimension.
Those are the first 3 classic dimensions for topological spaces

For subspaces of Euclidean spaces they are the same, and can be described axiomatically

Result of Hayashi

They have to satisfy Menger's axioms and a decomposability axiom, about decomposing an n-dimensional space into n+1 zero-dimensional subspaces

Moreover, they are the largest dimension on Euclidean subspaces from the ones satisfying Menger's axioms

I feel like nobody reads what I write

[small hemlock]

I typically can't understand what you write yet

[gritty widget]

I don't know any of those terms. Thank you anyway for explaining.

[coarse night]

Do we know what’s H_0(S^n) is

Using just the axioms not any particular theory of homology

[unreal stratus]

Uh I mean doesn't the zeroth group very much depend on your theory

Well

[coarse night]

Thats what i am asking

[lunar yoke]

Yeah from the axioms you can show that H_0(S^n) = H_0(pt) for n > 0 and of course H_0(S^0) = H_0(pt) + H_0(pt)

im assuming by axioms you mean at least homotopy invariance, excision, and the LES

you then get finite additivity for free

[coarse night]

Yes with dimension axiom and disjoint union

,av @lunar yoke

[gentle osprey]

Phil P#6615's Avatar

Click here to view the image.

[hidden crag]

Legendary bookcover

[gritty widget]

#bots

[coarse night]

How do you get S^n?

Ok nvm IK how to do that

Well I know reduced homology of Sn

[lunar yoke]

well then you always have isomorphism $H_n(X) \cong \tilde{H}_n(X) \oplus H_n(pt)$

[gentle osprey]

Phil

[lunar yoke]

also $\tilde{H}_n(X) \cong H_n(X,x)$

[gentle osprey]

Phil

[lunar yoke]

for some point x in X

[coarse night]

I thought that’s for H0 only

[lunar yoke]

for the higher ones it tells you that reduced homology coincides with unreduced

since by dimension axiom H_n(pt) = 0 for n > 0

also makes sense when you look at the LES

[coarse night]

Yes then how we get this exactly

[lunar yoke]

depends on how you define reduced homology

[lunar yoke]

oh its as in your image actually

you have a split SES $0 \to \tilde{H}_n(X) \to H_n(X) \to H_n(pt) \to 0$

[gentle osprey]

Phil

[coarse night]

That should for n=0only ?

[lunar yoke]

well idk what your definition of reduced homology is. In my lecture we defined it as the kernel of H_n(X) -> H_n(pt)

this gives you the SES for free

and it splits because as long as X is nonempty X -> pt is a split epi

meaning pt -> X -> pt is the identity

[coarse night]

In bredon it says we only define it for n=0 so i am confused

[lunar yoke]

well you can define it like this and get a homology theory that coincides with the one you started with in all higher degrees

we used it here and there, but i guess its more of a notational convenience

[coarse night]

Okay

[lunar yoke]

but yeah you can basically compute the homology of S^n by induction on n

you can cover S^n by two contractible open subsets U and V, (S^n without northpole, S^n without southpole).

Then H_k(S^n, pt) = H_k(S^n,U) = H_k(S^n - pt, U - pt). Note that S^n -pt is homotopy equivalent to a point, and U - pt is homotopy equivalent to S^{n-1}

so the LES is $H_k(pt) = H_k(S^n-pt) \to H_k(S^n-pt, U-pt) \to H_{k-1}(U-pt) -> H_{k-1}(S^n-pt) = H_{k-1}(pt)$

[gentle osprey]

Phil

[lunar yoke]

say k-1 > 0

then using $H_k(S^n) \cong H_k(S^n,pt) \cong H_k(S^n-pt, U-pt)$ we have a SES $0 \to H_k(S^n) \to H_{k-1}(S^{n-1}) \to 0$ and hence an isomorphism $H_k(S^n) \cong H_{k-1}(S^{n-1})$

[gentle osprey]

Phil

[lunar yoke]

and like this, once you know that H_0(S^0) = Z + Z and H_k(S^0) = 0 for k > 0, you get by induction that for n > 0 we have H_k(S^n) is Z for k=n or k=0 and H_k(S^n) = 0 otherwise

as one would expect with the intuition that H_k counts k-dimensional holes

[wicked yew]

It is connected right? I used the f: X -> {0,1} where f is continuous to show this but i'm not sure if my argument proves the conclusion

[gritty widget]

Yes, it's arc connected

Since p(t) = tf+(1-t)g gives an arc between different f, g

[wicked yew]

How does this argument look so far

i want to show there is maybe an intersection somehow and so one of the preimages is empty

[gritty widget]

What's fg(x)

[wicked yew]

oh i forgot to write that " let f:C[0,1] ---> {0,1} be continuous"

and I kept switching between X and C[0,1]

[gritty widget]

My point stands

[wicked yew]

f applied to the function g in C[0,1] ?

[gritty widget]

Then how is that in X

[wicked yew]

because it is continuous?

or am i bugging

[gritty widget]

What is

[wicked yew]

fg(x)

[gritty widget]

?

What's fg(x)

[wicked yew]

f(g(x))

[gritty widget]

Huh?

[wicked yew]

wdym

[gritty widget]

That doesn't make any sense

[wicked yew]

im sending a function to a value 0 or 1

using f

then composition of continuous functions

[gritty widget]

Which functions?

f and g?

[wicked yew]

yeah

[gritty widget]

How can you compose them

f ○ g doesn't really make sense

[wicked yew]

g maps from [0,1] to the reals

then f cts on reals

[gritty widget]

f is not a function on real numbers

[wicked yew]

oh yeah fuck my bad

🤦‍♂️

you were about to say something?

[gritty widget]

I've thought maybe we can fix your argument but I figured out that's too much hassle

It's a lot easier to use that C[0, 1] is convex

[wicked yew]

would I use that to show that X and the empty set are the only open and closed subsets of X

we can form a sequence in S that tends to an element in S_c

in*

[jagged sage]

To prove that for a connected topological space X, any non empty proper subset has non zero boundary, how do I prove backward direction

That non zero boundary implies connected

Other way was fine with contradiction but idk here

[rancid umbra]

convex implies path connected implies connected

[rancid umbra]

if there is a subset C of X with empty boundary, then C is both open and closed since bd(C) = cl(C) \ int(C)

[wicked yew]

is there anyway to prove it strictly using definitions of connected?

^ proving my question

i think path conected is the next chapter but i want to see if i can prove it without using the path connected argument

[rancid umbra]

my first instinct would be to show that if i have a non-empty clopen set then it has to be all of C([0,1]).

[wicked yew]

yeah that's what im going for now thanks

[rancid umbra]

i’ll try and work out some details

[jagged sage]

Yeah, I understand this part but how does that lead to connected conclusion

[rancid umbra]

what def of connected r u using

[jagged sage]

X not connected if there exist non empty U,V disjoint open sets where U union V is X

[rancid umbra]

okay. so i have a non-empty closed and open subset C of X

and recall that X = C U (X/C)

[jagged sage]

I see now

Clopen sets make up separation thanks

[rancid umbra]

non-empty, proper clopen sets* but yea

[jagged sage]

I just thought about premise and how does X not connected if boundary is 0 mean X is connected if boundary is not 0? Isn’t it not necessarily true

[rancid umbra]

uhh

rephrase this just a bit

X is connected if and only if the only subsets of X with empty boundary are X and 0

[jagged sage]

That part makes sense and then ur explanation makes sense. But directly, would empty boundary implying non connected mean that non empty boundary implies connected?

[rancid umbra]

thats the converse, not the contrapositive

the converse is not true. take X = [0,1] U [2,3]. [0,1] has boundary {0,1} but X is not connected

the contrapositive says that connected implies that every non-empty, proper subset has non-empty boundary

[odd flame]

im being asked to prove that a collection of finite sets with the FIP has nonempty intersection - doesnt this follow directly from the definition?

[rancid umbra]

yes

[odd flame]

okay then

[rancid umbra]

[odd flame]

nothing changes if the collection is infinite right?

i feel like im losing my mind why is this on my homework

[gritty widget]

It's not true then

[odd flame]

oh but that's just still by the actual statement of FIP

[gritty widget]

It isn't

[gritty widget]

It doesn't follow in a straightforward way as you're suggesting

You have to fix a finite set A and for every element x of A find a set B_x which doesn't contain x. Then look at intersection of A and all of B_x

[odd flame]

what im doing for the proof is considering a collection X = {X1, ... , Xi, ... , Xn}, taking subcollections {X1, ... , Xi} {Xi, ... , Xn} and trying to do something w that

[gritty widget]

It doesn't have to be finite

[odd flame]

ahhh that's why i was asking before

[rancid umbra]

if the collection is finite before hand then its trivial

[odd flame]

o

that explains why i thought it was obvious then

[odd flame]

by FIP any subcollection containing A and any B_x must have a non-empty intersection

hmmm but actually if $x \in A$ and $x \not \in B_x$ the FIP tells us there must be some $a \in B_x \cap A$ but that's not enough to show that the intersection of the whole thing is nonempty

[gentle osprey]

stμ₂dying

[wicked yew]

@gritty widget is there a way to prove my problem without path-conectedness

I'm trying hard to formulate an argument but it isn't coming through

[unreal stratus]

The point is to use path connectedness

Lol

So just use that

[wicked yew]

is there no other way?

[unreal stratus]

Well I mean there probably will be but idk why you'd want it

Proving path connectedness is more conceptual and stronger anyway

and everyone I know proved path connectedness (it's the expected thing)

[wicked yew]

ah fair enough

[unreal stratus]

a rough like rule of thumb imo is like

it's often easier to show stuff is path-connected and to show something is not connected than the other way round

Because in either case you can explicitly give a path or a disconnection

(Ofc the two notions are not necessarily equivalent)

[wicked yew]

oh thats fair

[unreal stratus]

but often if you want to show smth is connected you should see if you can prove the stronger statement that it's path connected

For example uh so like

[wicked yew]

so they are just easier approaches

[unreal stratus]

why is the sphere connected?

There are various approaches but to me the most obvious is just that you can take a great circle through any two points

[wicked yew]

ahh right

thanks

[unreal stratus]

np

[jagged sage]

Sorry to reply so much later, but then the statement that if X is connected if and only if every non empty proper subset of X has non empty boundary isn’t true?

Doesn’t that example contradict the direction where empty boundary of proper subset implies connected X

[rancid umbra]

yes because i made an oopsie

the boundary of [0,1] in X is empty since [0,1] is clopen in X.

[jagged sage]

I got u. To prove it then do we need to consider contrapostive. Rn I suppose boundary of A is non empty and assume that A is clopen. This motivated by fact that only clopen set can form separation (?). However this contradict since then boundary empty so A not clopen I.e. no seperation and X is connected

[sturdy notch]

This is a kind of random question which popped up in my head because I’m learning about singular homology and group cohomology right now but.

If a group G acts on a topological space X, then this induces an action of G on H_n(X) and turns H_n(X) into a G module.
Is it interesting to study the group cohomology of H_n(X)? Does this tell us something about X, does this have a name or is it just kind of irrelevant/uninteresting?

[cedar pebble]

yes this can be quite interesting and is related to equivariant cohomology (which is a joint generalization of singular cohomology and group cohomology)

e.g. if G is a finite group acting on X then one has H^i(X/G,A)=H^i(X,A)^G (G-invariant subspace), but the latter is just H^0(G,H^i(X,A)). The higher group cohomologies H^j(G,H^i(X,A)) will measure the extent to which this taking G-invariants is not exact, so it measures something about how cohomology of G-quotients behaves.

you do always have a map like H^i(X/G,A)->H^i_G(X,A) where H^i_G is G-equivariant cohomology, this is an isomorphism when G acts freely

[sturdy notch]

Oh wow I didn’t expect there to be a positive answer to my question!

I haven’t learned singular cohomology yet but ig this is something I have to look forward to now, it sounds awfully interesting

[cedar pebble]

singular cohomology is quite similar to singular homology (and the universal coefficient theorem gives you a precise relation between the two)

everything here should have a homology analogue, it's just that group cohomology is sorta more natural than group homology (since taking G-invariants is more natural than G-coinvariants)

[sturdy notch]

I see, Ill definitively have to read up on that, though I probably have a few missing prerequesites to understand anything in depth

[odd flame]

question about FIP - it states that any finite subcollection of some collection A has non-empty intersection, that subcollection doesn't need to fully cover the whole collection A does it?

[rancid umbra]

what do you mean by cover the whole collection A?

[gritty widget]

Yes

[odd flame]

oh i guess i phrased that wrong

[gritty widget]

The intersection doesn't need to be non-empty

Even in a compact space

[odd flame]

wait nvm wdym

actually i can answer my own question from before easily, the subcollection of A needs to cover the same elements as A otherwise you can definitely find an empty intersection

i think that's what you meant

[unreal stratus]

Wdym cover here

Covers are not directly relevant to the statement and there are no restrictions on what the union of your closed sets is

[gritty widget]

suppose that there is some continuous bijection f: A -> B and there is some continuous bijection g: B -> A. does it imply that A and B are homeomorphic?

[gritty widget]

uh... some online friend googled for me https://mathoverflow.net/questions/30661/

[rancid umbra]

https://www.mathcounterexamples.net/two-continuous-bijections-no-homeomorphism/

also this

[gritty widget]

Appeared here many times

[gritty widget]

well may I ask if moreover A and B are metric spaces does it imply? @@

I should better not to do too much google... @@

[rancid umbra]

i believe the pictoral answer on the mathoverflow post you sent is a counter example

[gritty widget]

The example you gave in the link is a subspace of R

[coarse night]

We know that any compact CW is finite and so $H_n(X)=0$ for $n\geq N$ for some N. It is true for general compact topological space that higher homotopy eventually =0?

[barren rune]

i was having some trouble figuring out where to start this problem that I've been assigned
i feel like I don't relaly have a good idea of where to even start looking in terms of thms that might be helpful rip

[lunar yoke]

what does closed cell mean here?

[barren rune]

closed cell means homemorphic to closure of unit ball B^n

[lunar yoke]

ok nice

so you can assume wlog that D = D' = closure{B^n}

and then you can even manually extend the given function on S^{n-1} to closure(B^n)

although I concede personally I like the homotopy group argument better as well

[barren rune]

why can we assume D = D' if they're not of the same dimension? would the homeomoprhism not die

[tawdry valve]

I have no idea how to prove this but apparently the answer is no. If you do a 2 dimensional analogue of the Hawaiian earring, you get arbitrarily high dimensional nontrivial homology.
https://en.everybodywiki.com/Barratt–Milnor_sphere

https://www.ams.org/journals/proc/1962-013-02/S0002-9939-1962-0137110-9/S0002-9939-1962-0137110-9.pdf

[lunar yoke]

oh yeah sorry ofc they dont have to be the same dimension

but the important thing is that you can assume they are the unit balls in some R^n

and then you can use the coordinates there to manually extend your function

[coarse night]

I see tks

[wicked yew]

is the set X ={ (x,y) : x in Q and y in R\Q} open is R_2

[lunar yoke]

no

at least not in the usual topology

any ball around (0,sqrt(2)), no matter how small, will not be contained in X

[wicked yew]

ah yes

thanks

[barren rune]

I still don't think I'm quite following >//<
are we saying closed cells are unit balls in R^n, R^m respectively ==> continuous map between boundary of cells is continuous map between S^n-1, S^m-1 circles ==> (something about how the interior gets filled in by construction of F, this is what I'm missing and also feels like the meat of the proof rip) ==> the function can be extended continuously (also continuously is kinda ?? but it will probably show up in the middle step somewhere)

[lunar yoke]

well you said closed cell means homeomorphic to closed unit ball

so if i manage to solve the problem for closed unit balls, then i solved the original problem by pre and postcomposing with the homeomorphisms to the cells

I guess you need that they send boundary to boundary

but thats given when you cell closed cell i reckon

So you are given a continuous function S^n -> S^m and you want to extend it to a function D^{n+1} -> D^{m+1} that sends interior to interior

call the given function f, then you can extend it by defining F(tx) = tf(x) for t in [0,1] and x in S^n.

[odd flame]

in the forwards direction (second paragraph) why do we require that y \neq f(x)

ohhh wait y is just "y-coord" of the pt outside the graph

[gritty widget]

If y were equal to f(x), then (x, y) would be in the graph.

[odd flame]

yeah i was once again just being a goober

[unreal stratus]

Another way to think about this is in terms of the diagonal being closed btw

[barren rune]

another question -- I was trying to prove that a CW complex is locally compact iff it's locally finite
for compact implies finite I tried doing my proof like
X locally compact => each point has precompact => each point of e has precompact means closure of e is compact => e is covered by finitely many precompact nbhds. Because of this define U as the compact nbhd of e in X, U is contained in a finite subcomplex so it intersects a finite number of open cells and is thus locally finite
but (1) i'm not sure if that's actually correct and (2) I have no idea how I might start the reverse implication

[wicked yew]

im not sure why i need a suitable point a unless im wrong

[gritty widget]

The projections are open

[wicked yew]

isn't R_2{a} path connected regardless of the point a

^ that uploaded so late

[wicked yew]

what does this mean

[unreal stratus]

Yes

Which is quite key to this

[wicked yew]

but then why does it say for suitable a

[unreal stratus]

That's for you to work out

Although actually it really doesn't matter, lol

[wicked yew]

but wouldn't it work for all a

[unreal stratus]

Yes

[wicked yew]

since R will just be split into two open sets

[unreal stratus]

Take a = 0 let's say

Well

[wicked yew]

I mean R\b

will be disconnected

once we take a from R_2

[unreal stratus]

Okay so I'd be slightly careful I guess

So like you need to be slightly more careful because what's to say that the image of R^2 under f isn't (0,1] and you're taking away 1

you need to rule out all things like that

[wicked yew]

hmm

[unreal stratus]

does that make sense

like you seem to assume f is surjective in the above ^

[wicked yew]

yeah i get what you mean I'm just thinking of how to rule it out

[odd flame]

what's an example of a closed set with no finite subcovering

[gritty widget]

Do you mean to ask for a closed set for which it's not true that every open covering has a finite subcovering?

You can always find a finite open covering of a subspace. Take the entire space.

[odd flame]

yes my b

working on d,e here

[gritty widget]

Then any closed but not compact subset of R^n will do.

Closed set which is not bounded.

For part e, try to think of a closed unbounded set (in R^2 for concreteness) which gets arbitrarily close to some neighborhood of it. This will ensure any epsilon-neighborhood of the closed set intersects the outside of the neighborhood.

For part d, you just use compactness and the fact that metric open balls generate the topology.

[odd flame]

dense moment but im not sure i see this, can you elaborate please

[ornate berry]

Try R.

[unreal stratus]

as an elabortion like your question is "does closed => compact" and so the suggestion is just to try R or R^n where it's particularly nicely because compact is equivalent to closed and bounded

[odd flame]

would a clopen set in R suffice? (0,1] isn't compact

[ornate berry]

It's not clopen

(0, 1] is neither closed nor open in R with the usual topology

As I said, try R.

[ornate berry]

A comment on this: clopen sets should generally be thought of as 'walled-off' parts of a space. In R, the only clopen sets are {} and R. If you take the subspace [0,1] u [2,3] of R, the clopen sets of that include [0,1] and [2,3].

[gritty widget]

dense

[wicked yew]

coming back to his question. can we rule it out due to the nature of the interval (0,1] or is it something else?

[tame iron]

(0,1] is neither open nor closed xd

its complement is not open nor closed either

[odd flame]

im glad someone got it lol

[unreal stratus]

Well the point is you have to pick a such that you don't take off an endpoint or something like that

[wicked yew]

I'm confused

so why is it not valid

isn't that the point I said before about there being two open, non-empty disjoint sets in the image after removing a point

[gritty widget]

@wicked yew take any three points in R^2 and remove the middle one

(middle one after mapping to R using f)

this way you'll make sure the sets you are dividing the image into are non-empty

[unreal stratus]

nah

you said that you could take away any point

and I said you don't know that you'll disconnect the image with that (with the example being that theoretically the image could be (0,1] for all we know, with you just taking away 1)

But we can always pick a point a which will lead the image of the restriction to be disconnected

[wicked yew]

ohhhh

right

im slow my bad

[unreal stratus]

nah dw

[wicked yew]

if I added that together with what I said earlier would that be formal enough to show that such an f doesn't exist

[unreal stratus]

Well depending on how you do it sure I mean

[unreal stratus]

Ig you gotta justify this

But you know roughly what the image of any such f would look like anyway

[wicked yew]

yeah

so if I said a is the preimage of the "midpoint" of an interval within R

that f maps to

[unreal stratus]

Yup, that'll definitely work (of course you can also assume the interval isn't a singleton for obvious reasons)

[wicked yew]

yep

[gritty widget]

Every continuous map of a closed disk to itself has a fixed point, for if not, you could find a retraction of the disk onto its boundary.

It might help to ask a more specific question. What about the theorem do you need explained?

Do you want an explanation of the statement?

So you want the definitions that appear in the Brouwer fixed point theorem explained. Any particular ones?

"Every continuous function from a closed disk to itself has a fixed point."

Which?

I don't want to converse in DMs.

Post it here.

If you need to talk in DMs for whatever reason, someone else can help.

It is in words.

You need to be more specific.

Do you want the meanings of continuous function, unit cube, and so on explained?

Could you please just point out which specific part you need explained? It's still unclear.

Which words.

Is the phrase "Brouwer fixed-point theorem" the confusing part?

it's a theorem, not a definition

The "definition" of the Brouwer fixed-point theorem is the statement you see in the page you linked. (Or, with the cube replaced by a closed ball, or a compact convex set, or whatever).

Theorems are not said to have definitions.

i give up

good luck

the word has a definition but the Brouwer fixed point theorem isn't a definition

I'd like to think that with as many years of mathematics training as I have, I'm at least alright with "basic mathematical definitions" and the like. The same is true of Blitz. You're just not doing a good job of communicating what you want.

Blitz is right.

They have a good reputation in this channel, so you should listen to them.

It's literally right there on the page!

You linked it!

I do.

And I offered multiple times to explain the theorem.

My spiritual guidance is that you should criticize whatever you see, including what I write. In short, you can listen to me but don't take it for granted

On the theorem?

[gritty widget]

You seem pretty consistently good, so I'd listen to you at least.

I haven't seen you type anything blatantly wrong.

No. Why would you do that?

No, I'm not. Feel free to, though.

I was perfectly willing to explain the theorem, but now that you're threatening me with moderator action and such, I'm suddenly not really interested.

[odd flame]

the fuck happened

[ornate berry]

A strange person

Did you manage to find a closed non-compact set btw stu?

[odd flame]

(x, 1/x) in R^2

not R like you said but that's what i submitted for my hw

[ornate berry]

What's x?

Is this like, the set {(x, 1/x) | x in R\{0}}?

Bc that's homeomorphic to two copies of R

[odd flame]

[gritty widget]

The guy probably got banned, which wiped their messages.

[arctic relic]

Kind of a stupid question but when we try to show the product of two compact spaces is compact, why cant we just say that the product of the finite subcollection of A and B is a finite subcollection of the product of the open coverings of A and B? Why the tube stuff?

[gritty widget]

You're proving that A x B is compact, provided A and B are?

[arctic relic]

Yep

I cant find why the “naivest” solution isnt acceptable

[gritty widget]

So you start with an open covering of A x B, right?

Because that's how you prove compactness.

This gives an open covering of A and an open covering of B, just by projecting.

[arctic relic]

I cant start with an open covering of each individual space and product them?

[gritty widget]

You can extract finite subcovers of each. However, you may not recover elements of your open cover of A x B by simply taking the products here.

The open sets which you have assumed to cover A x B may not necessarily be products of open sets covering A and open sets covering B.

Not all open sets in the product topology are products.

[arctic relic]

Thank you; last statement is what I wanted to hear

[gritty widget]

So you need a slightly more delicate argument, which is what the tube lemma stuff does.

[arctic relic]

Yes I see it now thanks again

[gritty widget]

Nope. When you prove that a set is compact, you start with an open covering of that set. As this example shows, you might not be able to build arbitrary open covers out of specific ones.

In the utmost general setting, you can only start with the set itself.

[halcyon dock]

hi so i don't really know anything about topology or what a metric is. can anyone point me in the right direction for how to even understand this problem

[bitter smelt]

probably just read whatever book your class is using

if you're not in a class, any point set topology book or real analysis book

@halcyon dock

[graceful abyss]

Essentially the metric defines the open balls (points at distance < r), an set is open if there is an open ball around each point of the set that is entirely in the set

And a set is closed if complementary is open

Finally, continuity is defined as usual (limits: for all e, there is d a.th. if d(x,a)<d then d(f(x),l)<e, which can be reformulated in terms of neighbourhoods)

[potent sky]

right the preimage of f will take open sets to open sets

[wicked yew]

How is my argument?

[unreal stratus]

"preimage of f^-1(r)" is a typo I'm sure and you could streamline this a bit more but yes the argument is correct in essence

Also this is incorrect, sorry

If it were true, you could take U = X and you'd already be done

Do you see what it should be instead?

(@wicked yew in case you don't see)

[wicked yew]

proper open subset?

@unreal stratus

[unreal stratus]

No, the quantification and stuff isn't quite right - try giving it another go

[wicked yew]

ok

also just another quick question. When we refer to a connected component of X containing x could this also just be X?

[unreal stratus]

It is indeed just X

[wicked yew]

the definition is a bit weird in the notes

so what's the point

[gentle osprey]

potato

[unreal stratus]

wdym?

[wicked yew]

oh so X may not be connected

makes sense

[unreal stratus]

Exactly

[wicked yew]

Thanks I was losing my mind

[unreal stratus]

It matches with the more intuitive notion really.

I think for example you are asked to prove that R^2 minus the diagonal (i.e. points of the form (x,x)) has two connected components (the obvious ones)

[unreal stratus]

np

[wicked yew]

just above and below the line?

[unreal stratus]

Indeed

[wicked yew]

ah

[unreal stratus]

though there are very different cases

[wicked yew]

but in this case R_2 \diagonal is still connected

[unreal stratus]

Uh it's not connected

[wicked yew]

actually no

[unreal stratus]

otherwise there'd be just one connected component anyway

[wicked yew]

i was thinking that any R_2\A where A is countable is connected

but ig thats not the case

actually wait i'm bugging

(x,x) isn't countable

[gritty widget]

R^2?

yes, that's correct

in fact < size of continuum is enough

to see this we can prove it's arc connected

consider a straight segment between two points x, y of R^2\A

it might not be a desired path, but we can modify this

[wicked yew]

concatenating lines?

[gritty widget]

consider the line passing through (x+y)/2 and being perpendicular to this segment

call the line L

now for every z in L, we can consider a path from x to y made from first traveling from x to z, and then from z to y

since A is of size < continuum, and all of those paths are intersect only at points x, y, continuum many of them need to be disjoint from A

in particular, one of such paths works

[wicked yew]

damn that's smart

how do you think of such constructions

It never comes to my head

[gritty widget]

I saw it before

[wicked yew]

Ah

Thanks for this

[gritty widget]

the same proof works for R^n with n > 1

[wicked yew]

do I have the right definition of f?

that for all x in U f(x) = k

[unreal stratus]

i mean that isn't a definition and needs more detail

But yes, for each x, there is some k in R and some open U containing x such that f(U) = {k} (equivalently for all y in U, f(y) = k)

[wicked yew]

and why cant U be X

i'm missing something

[unreal stratus]

The word some in what I wrote

this is why it is called locally, rather than globally, constant

[wicked yew]

Ohhh

[wicked yew]

from this can I then take largest open subset U_k for each k

and their union must be X

and then my argument should follow?

[unreal stratus]

Yeah

Another way to see it is that f^-1({r}) is clopen for each r, I guess

[wicked yew]

yeah

[unreal stratus]

but it's essentially the same lol

[wicked yew]

yeah they are equivalent

thanks for helping me spot that

[unreal stratus]

np

[wicked yew]

I need to watch for these things

[wicked yew]

if a finite set of A_i are connected in X then their intersection should be connected right?

I want to use the argument of f:Z -> {0,1} being continuous

[unreal stratus]

How would your argument go?

I'd recommend you do some pictures on R^2 or something to get some intuition

[gritty widget]

No, intersection of connected sets need not be connected

it fails for two sets already

in R you won't find counterexamples, but in R^2 you will

because in R connected is equivalent to convex

in R^2 we can find two simply connected domains, whose intersection is not connected

[ornate berry]

Example:

[gritty widget]

They're hugging.

[gritty widget]

can we do uncountably many components?

[ornate berry]

I think so!

If we can find some uncountable set of disconnected points in R, we can

[gritty widget]

no because R has countable cellularity

at least, we can't do so for two open sets

[ornate berry]

Yeah this is true, open sets would be impossible

My thought was just having points. Not isolated points of course...

Ah, connected or path connected?

The latter is probably easier

[chilly mica]

yea, without requiring open sets, you can like, take A = {(x, 0) : x irrational} union {(x, y) : y > 0}, and B = {(x, y) : y <= 0}

with open sets, should be possible in really crazy topological spaces, is my guess

[ornate berry]

This was the image I had in mind, but I'm not sure that the irrationals are disconnected – although they are obviously not path-connected

[chilly mica]

they are, because between any two irrationals is a rational, and R minus that rational has two connected components

and different components dont merge together if you take subsets of them

[ornate berry]

Very nice argument

ty

[chilly mica]

np

[gritty widget]

lol. They are totally disconnected

In fact they are zero-dimensional

(which means that they have a basis of clopen sets - see intervals with rational endpoints)

the name zero-dimensional comes from the fact that it's equivalent to ind X = 0 where ind is the small inductive dimension

for Ind X = 0 we say that X is strongly zero-dimensional, and for Lebesgue covering dimension it's the same as strongly zero-dimensional

so there are those two concepts of zero-dimensionality

(strongly zero-dimensional because for any good enough space, say T1, ind <= Ind)

[coarse night]

Can anyone help me see why reduced zeroth homology of S0 is Z? Just using the axioms. It’s defined to be the kernel of H0(X) ->H0(pt)

[cedar pebble]

H_0(S^0)=Z^2, the map H_0(S^0)->H_0(*) is just the map Z^2->Z sending (a,b) to a+b. So the kernel is Z.

[little hemlock]

the map \phi is 1 \mapsto (2,-2) since the boundary circle of a Mobius band wraps twice around the core circle

tbh, im not exactly sure what question to ask, except to say I am a little uncomfortable with this reasoning hm

[coarse night]

i am not sure why the map sends {a,b} to a+b

[wicked yew]

would it be easier to prove this by contradiction or a direct proof

idk how to formulate that there is a convergent subsequence in the union

[viral atlas]

For metric spaces, sequential compactness is equivalent to compactness (in terms of open covers and finite subcovers)

Can you use that instead?

[viral atlas]

Okay, this can be addressed too. Take any sequence in the union: there are only finitely many sets in the union, so at least one of them must end up with infinitely many terms of the sequence, i.e., you get a subsequence of the original sequence in one of the A_i.

[wicked yew]

Oh yeah that’s smart

Thanks

[gentle osprey]

Nobody

[coarse night]

So then those two definitions become equivalent, I see. Thank you

[wicked yew]

do i take the "unit vectors" in a closed ball to show that the general closed ball is also sequentially compact?

[unreal stratus]

That is part of the idea sure

[gritty widget]

Idk what's the point of saying sequentially compact here

Also, this just follows from B being continuous image of S x [0, 1]

V is complete because V = R^n tbh

[wicked yew]

wait how is B the continuous image of that product

[gritty widget]

just multiply an element of S with an element of [0, 1]

[wicked yew]

but how is that in B

[0,R] u mean?

[gritty widget]

my bad I thought they were asking about R = 1

still, it's the same thing, the two are homeomorphic

from the unit closed ball, multiplying by R gives a homeomorphism

[wicked yew]

yeah i was trying to find that exact map but didn't know how to verify that it was continuous

[gritty widget]

any normed space is a topological vector space, meaning that

1. It's Hausdorff
2. the multiplication K x V to V given by (k, v) to kv is continuous
3. The addition V x V to V given by (v, w) to v+w is continuous

the condition 1) isn't always assumed but there's no reason not to assume it tbh

[wicked yew]

damn

[gritty widget]

anyway, we are using the property 2) when restricted to [0, 1] x S

[wicked yew]

I think topology is just not for me

[gritty widget]

what is for you

[wicked yew]

is this a definition?

[gritty widget]

it's a definition of topological vector space

it's basically what we expect a vector space with topology to satisfy

[wicked yew]

good question lool

ahh

[gritty widget]

I mean yes, it might not be for you. Idk, I didn't find anything as cool as topology in math yet

Functional analysis being a close second

[wicked yew]

fair fair

i don't think i'll pick the topology course next term

it would be a horror show

[gritty widget]

I'm sorry

[wicked yew]

😔

[wicked yew]

how did you find complex analysis

[gritty widget]

most people love it, I didn't

[wicked yew]

ah

[gilded crane]

I can’t figure out why F is continuous

any hints?

[gritty widget]

F is just multiplication on R

restricted to some subspace

multiplication is continuous

[wicked yew]

for the second part which says to show V is complete, would a direct proof or a proof by contradiction be easier?

[midnight echo]

Hi, Trying to do this question but i'm not sure how to approach it. I think the goal is to show that kernel of rho is the identity, but not sure how to get there

[unreal stratus]

I don't think you ever really want to show completeness by contradiction (not sure how it'd work really)

But if you want a hint: why is B complete? and how can you use that to show V is complete?

[rancid umbra]

hint: show that cauchy sequences in V are bounded

[wicked yew]

right now what I have for why B is complete

• suppose a cauchy sequence in B does not converge

• then limit must be outside of the ball

-however B is a closed ball and so all limit points must be within the ball

and so every cauchy sequence must converge in B as a result due to it being bounded

[rancid umbra]

for the second point, since B is sequentially compact then there is some subsequence of your cauchy sequence which converges to some limit L in B

just show that the entire cauchy sequence converges to L

[wicked yew]

but then what do I do for V? union of all balls?

[rancid umbra]

this

[wicked yew]

hmm ok

will probably take me a while

[wicked yew]

but does it not follow from B being closed

or?

[little hemlock]

all you can say when B is closed is that a convergent sequence in V whose points lie in B converge in B

[rancid umbra]

no

[wicked yew]

oh i think i finally got it

thanks

every cauchy sequence is bounded in V

so we can find a B in which the sequence resides

and so it converges since B is complete

[little hemlock]

ye

[wicked yew]

how do you guys formulate arguments when you are answering questions? Like how do you piece things together

[rancid umbra]

uhhhh

[gritty widget]

That's vague and not topology

[rancid umbra]

i’ll refer you to nike

“Just do it” ✅

[wicked yew]

is a metric space sequentially compact iff it is closed and bounded?

[gritty widget]

no

[midnight echo]

Is the whole plane R^2 star shaped by definition since there are straight paths from any point to 0?

[gritty widget]

every metric space is closed in itself

and every metric can be made bounded

are you asking if pseudocompactness implies sequential compactness

for metric spaces

[wicked yew]

yeah ig

[gritty widget]

yes, it's equivalent to compactness

[wicked yew]

how do you go about showing that

do you use open covers

[gritty widget]

if X isn't compact then there exists an infinite sequence x_n with no convergent subsequence
Now take f(x_n) = n or something

extend using Tietze

[wicked yew]

tietze?

[gritty widget]

If A is a closed subset of a metric space X, and f is continuous from A to R, then there exist a continuous extension of f to whole of X

[midnight echo]

Is this true?

[rugged swan]

Yes

[midnight echo]

Thanks

[unreal stratus]

Indeed, note convex => star shaped more generally too

[wicked yew]

i've shown that X is complete but I'm struggling to show that it is totally bounded

then i can conclude with X is sequentially compact iff it is complete and totally bounded

@gritty widget

can i get a clue

[gritty widget]

yes?

clue is that this characterization of compactness is mostly useless anyway

not in any proofs you'll need anyway

Now, it might be sometimes useful to throw random ideas

but it might be equally useful to stop and think

What's your first intuition to prove this? Mine is to assume X isn't sequentially compact and try to construct a continuous function which would be unbounded

and this approach indeed works

[wicked yew]

yeah i did this prove it was closed

to ^

oh so you used it for the whole claim

i tried that before but i didn't get to the end. I'll try again

In the contradiction, did you use the fact that there exists a sequence in X such that there is no convergent subsequence in X?

[gritty widget]

yes

This implies the sequence can be chosen to consist of different terms and the set itself is closed

[wicked yew]

huh

what do you mean by different terms

and are you using the assumption the set X is closed

[gritty widget]

There could be repeats in x_1, x_2, ...

but we can assume there is none

[gritty widget]

no, there's no such assumption

That X is a subset of R^n is basically irrelevant

but maybe it'll help you to construct this function in some simpler manner than what I am telling you here

[wicked yew]

but how do you come to the conclusion that the set of elements is closed

[gritty widget]

That x_1, x_2, ... is closed?

[wicked yew]

since there is nothing that contradicts the definition of it being closed?

[wicked yew]

yes

[gritty widget]

I mean, I've just realized that since we're in R^n we have a very straightforward way of proving this

even though it holds very generally

after all, compactness is equivalent to being closed and bounded

so it's enough to check that we get a contradiction if X is not closed or not bounded

[wicked yew]

yeah I did it that way but I want to still know how you went about it

[gritty widget]

If it's not bounded then we can just take the norm of R^n

well, I used an extension result by Tietze

this says that if you have a continuous function on a closed set then you can extend it to the whole space

the sequence (x_n) is closed so we can do that

[wicked yew]

is it closed because there is no convergent sequence?

[gritty widget]

the only assumption in Tietze theorem is that your space is normal, and all metric spaces are normal

yes

[wicked yew]

damn thats smart

I haven't covered Tietze so I probably will not use it

[gritty widget]

we've basically proven that if a metric space isn't compact then it contains a closed copy of natural numbers

where x_n corresponds to n

[wicked yew]

umm

when you say not compact do you mean not bounded and not closed?

in general

[gritty widget]

no

bounded and closed can only be used for R^n with standard metric

[wicked yew]

so how did we come to this conclusion

[coarse night]

If we only assume path connectedness then given two covering maps $p_1: X_1 \to X$ and $p_2: X_2 \to X$ and $F:X_1 \to X_2$ s.t. $p_2 F = p_1$, it is true that F is also a covering

[coarse night]

It’s true if we assume X is locally path connected,

If not can I get a ce

[gritty widget]

well maybe you didn't but it boils down to checking that x_n is additionally discrete as a topological space

no need to over-analyze this though

[rancid umbra]

@wicked yew consider the norm function on X to show that X is bounded. if X is not closed, think about using 1 / |x| or something along those lines to reach a contradiction, showing that X is closed

in this case, closed and bounded iff compact iff sequentially compact

[gritty widget]

yeah, 1/|x-a| where a is the point in cl(X)\X

[wicked yew]

Yeah that’s what I used

I was just interested to see how your other proof went

[gritty widget]

the sequence with no convergent subsequence can be treated as a sequence convering to some cl(X)\X where X is embedded in some larger space

at least you can interpret it that way

Tietze theorem shouldn't be that bad to prove once you know Urysohn lemma

and for metric spaces Urysohn lemma has an easier proof involving the distance from a set function

[unreal stratus]

(tietze is very much unnecessary for any question i've seen in this course btw for op lol like but still cool)

[gritty widget]

I never had a course like that and it's what naturally comes to my mind when I see this question

[slow halo]

Let M be a non-invertible complex matrix of order n. I was wondering if $(GL(n,\mathbb C)\cup{M})\setminus\det^{-1}(\mathbb R^{\ast})$ is connected... 🤔

[unreal stratus]

Ofc - just saying it for context lol

It is a shame tietze etc was never mentioned

[gentle osprey]

Gillian Seed

[gritty widget]

isn't that just M

[rancid umbra]

lol

no

[unreal stratus]

Matrices there can have complex non zero determinant

[gritty widget]

sorry, complex matrices

[rancid umbra]

i want to say yes. GL(n,C) is path connected

[slow halo]

What about GL(n,C)\det^{-1}(R*) ? 🙂

[lusty trench]

IIRC, if you take a connected manifold of real dimension n, and remove a submanifold of real codimension >= 2, the resulting space remains connected, right?

I am trying to solve exercise 14-C from Milnor and Stasheff's “Characteristic Classes”. The exercise asks to show that, given a complex vector bundle $E \to M$, the $(q+1)$-th Chern class $c_{q+1}(E) \in H^{2q+2}(M, \mathbb Z)$ is the primary obstruction to the existence of $n-q$ linearly independent sections $M \to E$.
I already know that this primary obstruction actually lives in $H^{2q+2}(M, \pi_{2q+1}(V_{n-q}(F)))$, where $F$ is the typical fiber of $E \to M$.
I also already know that $\pi_{2q+1}(V_{n-q}(F))$ is isomorphic to $\mathbb Z$, but in principle this bundle of coefficients could be “twisted”. Why is it not twisted?

[gentle osprey]

Automorphism

[lusty trench]

It probably has something to do with the fact that $V_{n-q}(F)$, being a complex manifold, has a preferred orientation, right?

[gentle osprey]

Automorphism

[coral pivot]

Unrelated but are you using the Tex version of the book?

[lusty trench]

I have a physical copy.

[coral pivot]

I see

[lusty trench]

But I stumbled upon the TeXified version less than an hour ago. (And, in fact, that is what landed me here.)

[slow halo]

oh

[coral pivot]

Ah nice

[lusty trench]

Was it the same person or group that TeXified Milnor's Morse Theory book?

[coral pivot]

Different group (my group) but we had help from the Morse theory group.

[lusty trench]

Oh, nice. Good job!

[wicked yew]

Q n [0,1] is compact right?

since there is always a finite union of open sets in an open cover that give the set

[gritty widget]

no

[unreal stratus]

I mean one way to see this is uh

is Q closed in R? You should see it is very far from being so; indeed, what is the closure of Q?

[wicked yew]

The closure of Q is R?

Or something between Q and R

[lusty trench]

R is the Cauchy completion of Q, so by construction Q is dense in R.

More concretely, if you have any irrational number, then the sequence of truncated decimal expansions gives you a sequence in Q that converges to it.

[unreal stratus]

If you want a concrete example of a subcover with no finite subcover you can consider $(U_\epsilon){\epsilon > 0}$ where $U\epsilon = [0,1]_{\mathbb Q} \setminus [a-\epsilon, a + \epsilon]$ and $a \in [0,1]$ is your favourite irrational

[gentle osprey]

potato

[jagged sage]

Why can’t a finite T1 space have limit points

[lusty trench]

A finite T1 space is automatically discrete, right?

I mean, every point is closed. Hence the union of all points but one, which is a finite union, is closed. Hence every point is also open.

[rancid umbra]

because point sets are closed

sniped

[jagged sage]

Ok thanks both of u guys

[unreal stratus]

Indeed it may be helpful to note (somewhat tautologically) that the cofinite topology is the smallest topology you can give a space to make it T1 more generally

[next crystal]

would knowing linear algebra help with understanding point set topology better or are the two more or less unrelated?

[lunar yoke]

basically unrelated

pset topology has basically no prerequisites besides maybe knowing how to work with sets

[lusty trench]

Linear algebra is more likely to help you with algebraic topology, if you decide to study that at some point.

But point-set topology, probably not so much.

[next crystal]

thanks so much

i'm trying to take pset topology next semester without having taken the linear algebra prereq but wanted to make sure it wasnt a terrible idea

[lusty trench]

Usually the prerequisite for a first topology course is having taken a real analysis course. Because, in real analysis, you learn the topology of R^n, and that gives you a starting intuition before working with more general topological spaces.

[unreal stratus]

yeah it won't rly be necessary initially but linear algebra is definitely something i'd recommend anyway XD

[lusty trench]

That being said, when I took topology, the professor used function spaces as examples quite a lot, e.g., “space of continuous functions” or “space of differentiable functions”. And it would have hurt a lot not to know linear algebra. In one take-home test, a question was “Let V be the vector space of continuous functions [a,b] -> R. Show that the map that sends f \in V to its integral over [a,b] is continuous.”

[next crystal]

wouldnt a (first) real analysis course just cover the topology of R?

in like baby rudin for example

but either way the prereqs listed are calc 1 and 2 and linear algebra

[lusty trench]

Mmm, then two courses on real analysis. Just the topology of R doesn't give you enough intuition to bootstrap an understanding of topology, IMO.

[little hemlock]

Do the projections of the torus onto each of its factors induce an isomorphism $H_1(S^1 \times S^1) \to H_1(S^1) \oplus H_1(S^1)$?

[gentle osprey]

kxrider

[little hemlock]

Trying to get a handle on generators for H1(S1 x S1)

[gilded crane]

Doesn’t that last line imply that tau_T is just the discrete topology? I can draw an open disc around any subset of T

[gritty widget]

No

What generates the topology on T are open arcs

[gilded crane]

i dont think i follow

all subsets of T are arcs

or unions of arcs

so wouldn't it still be the case that tau_T is the discrete topology?

[gritty widget]

Proof?

If you prove it to me then I'll consider it

[wicked yew]

Idk why but this response is just so funny 😂

[unreal stratus]

Mathematicians when someone confesses their love for them

[gilded crane]

ok yeah i can think of subsets which aren't arcs, sorry

[wicked yew]

🤣

[stoic eagle]

Hey, can someone help me with this bit in Hatcher? im probably misunderstanding how the boundary map works

[dim radish]

is the set of continuous bijections from R^2 to R^2 a connected set in the space of functions from R^2 to R^2

i would think it to be disconnected as you should be able to split it into sets which preserve the orientetation of the unit circle and others which reverse it

[gritty widget]

with what topology

[dim radish]

Product topology

[coarse night]

Look at the simplicial strucure he has given before

[stoic eagle]

I think i figured it out already but thanks

[coarse night]

Oh lol

[coarse night]

Bump?

[wicked nova]

.

[gritty widget]

What happens when we attach D^n to D^n by their boundary? Do we necessarily get S^n?

I think it's true for n = 1 and n = 2 but I'm really not sure for n = 3

I guess you can write a homeomorphism explicitly

[coarse night]

Yes

That is one of the CW construction of Sn

[jagged sage]

I have proven that for i=1,…,n that Ai connected subspace where each Ai intersect A{i+1} is connected that union of union i=1 to n Ai is connected. But not sure if this result can be extended to infinite sequence (Ai)i in N of connected sub spaces

I don’t see why it couldn’t but not sure

[unreal stratus]

It does work for the N case and works under more general things provided that you can uhhh

Well basically there is no need to do it by induction. One way to do it is to use the characterisation that a space X is connected iff every function into the two point space (call it 2) is constant

[jagged sage]

I proved first part by induction already. Doesn’t that just affirm this second part

[unreal stratus]

How would it?

Proving a statement for all finite n doesn't somehow prove the infinite case if that's what you mean

[coarse night]

say A = U ∪V is a disconnect and Ai are open then Ai ⊆ U or V, but since they intersect, all of them must lie in one contradicting other one is nonempty

[unreal stratus]

Yeah I was gonna say if you have a function into 2 then the restriction to each A_i is constant and the constants agree on overlaps by then restricting to intersections (which are also connected)

[coarse night]

You do

[unreal stratus]

sorry yes

[jagged sage]

Ok I think I see thanks

[odd flame]

goober moment but im having a hard timie understanding the definition of the product topology

[unreal stratus]

which definition are you using?

[odd flame]

munkres

[unreal stratus]

like in terms of basis elements

[odd flame]

subbasis i think, one sec

[unreal stratus]

the intuition / a nice way to define it is that it is the smallest topology such that the obvious projection maps X x Y -> X and X x Y -> Y are continuous

[unreal stratus]

Yup

Okay sure so what I said about works

Smallest topology you can put on the product set such that the projection maps are all continuous

that means in particular that all the sets in those collections must be open in the product.

So then we just consider the topology those sets generate to get the smallest one

[odd flame]

i wanna translate to english first though

[unreal stratus]

What I just said was english XD

[odd flame]

the collection is projections of open sets into the product right

[unreal stratus]

No, for each open $U_\beta \subset X_\beta$ we put its preimage under the map $\prod_{\alpha \in J} X_\alpha \xrightarrow{\pi_\beta} X_\beta$ in $S_\beta$

[odd flame]

[gentle osprey]

potato

potato

[unreal stratus]

*subset lol

[odd flame]

so X1 x ... x U_beta x ... is an an element of the collection right

[unreal stratus]

Well that is ambiguous and not really how I would format it but I get what you mean, sure xd

[odd flame]

like the full X_i in all indices except beta, where it's just an open set of U_b of X_b

[unreal stratus]

Yup

Exactly

[gentle osprey]

potato

potato

[odd flame]

and that's just the box topology!

[odd flame]

so are all elements in this subbasis of the same form?

[unreal stratus]

indeed

[unreal stratus]

yes up to permuting indices since they are preimages of opens under the projections

so yeah that is one way to describe it more explicitly

(just it's a) notationally simpler and b) more conceptual to write it as preimages under projections)

[gritty widget]

well, we can say at least that what you're looking at is the subspace of all homeomorphisms of R^2

[dim radish]

i think the orientation idea should work, but im not sure how i would show that a homeomorphism from R^2 to R^2 has a well defined orientation

[odd flame]

that's this right?

[gritty widget]

not quite

[odd flame]

"a map into a product space is cont. iff it is cont. for every component of the product"

i see it's not the same as what potato said just making sure that that's correct now

[unreal stratus]

Yeah it's not the same but is very related

[jagged sage]

How does a partition of a topological space yield some new geometric space

I’m given example of X=[0,1] x [0,1] and Y partition consisting of two points sets {(x,0),(x,1)} for any 0 leq X leq 1 and singleton sets (x,y) for any 0 leq x leq 1 and 0<y<1

That sounds reasonable but don’t think I understand what the partition and subsequent quotient topology are actually doing

Idk how though. I understand the first part of partition takes two opposite edges of the unit square and that if you can glue these together to obtain a cylinder but not sure how this is concluded from just this info

[little hemlock]

You can take another way to represent a cylinder: i.e. [0,1] x S^1 and construct a homeomorphism

[jagged sage]

What is the significance of the part of the partition containing singletons

Homeomorphism sounds reasonable but I think I don’t understand the relevance of partition and quotient topology