point-set-topology

[gritty widget]

the sets A, B don't need to be bounded

[candid hedge]

i am dumb

[gritty widget]

what I'm saying is that you'd probably want them to be unbounded

[candid hedge]

sooo n and n+1/n?

in R

does that work?

[gritty widget]

You mean, A = {n : n natural} and B = {n+1/n : n natural, n > 1} ?

[candid hedge]

yes

[gritty widget]

it does work

[candid hedge]

what did you have in mind?

[gritty widget]

just two curves in R^2 approaching each other at infinity

I thought it's better to visualize there

[candid hedge]

x^2 and x^2 +1?

[gritty widget]

well this probably works but a better example might be something like 1/x and -1/x

where they approach each other at the line y = 0 as x goes to infinity

[candid hedge]

oh yeaaa

thanks a lot blitz i understand things better

[shadow charm]

in hatcher it's asked to show that homotopic maps induce the same homomorphisms on reduced homology groups. Now this follows by homotopy invariance for n>0, but the problem with n=0 is that no induced maps f#:Z->Z were defined in the book and i dont see any obvious way to define this which isnt just the identity in which case the exercise is more than trivial

[candid hedge]

What are the conditions for a sequence to be a closed set?

[gritty widget]

It definitely cannot have limit points which aren't already members of the sequence

I think this is sufficient for spaces which are, say, Hausdorff

[candid hedge]

so a convergent sequence with its limit is a closed set

in a normed vector space

[gritty widget]

yes

but if we were in a general topological space then we need to watch out for constant sequences which can converge to multiple different limits

[candid hedge]

to show this we have to show that any infinite sequence taken from the convergent sequence will converge eventually to the same limit?

is there an easier approach?

[candid hedge]

ouff

[gritty widget]

Take a sequence of elements $y_k\in {x_n : n\geq 1}\cup {x}$ where $x_n$ is your sequence and $x$ is its limit. If $y_k$ has finite amount of terms then it has a constant subsequence, so we are done. If it doesn't then we can take a subsequence of $y_k$ that will also be a subsequence of $x_n$. Then it converges to $x$

[candid hedge]

we can show that A=A_ right? by taking an element of the closure we show easily that its an elemnt of A

[gentle osprey]

Blitz

[candid hedge]

why we can take a subsequence of yk that is also a subsequence of xn. Shouldnt we show this somehow??

i mean i know its obvious

but its the proof that they want

[gritty widget]

do it by recursion

[candid hedge]

yeah. i am gonna have to construct it

[gritty widget]

first take some y_1 = x_(n_1)

[candid hedge]

can i do this?

[gritty widget]

then there exists y_(m_2) with y_(m_2) not being equal to x_k for k <= n_1 and not being equal to x

and m_2 > 1

this is because if we have y_n then there always exists m > n with y_m =/= y_n

and so on

so y_(m_2) will be equal to x_(n_2) for some n_2 > n_1

now the sequence constructed y_(m_k) will be a subsequence of x_(n_k) and so it will converge to x

and being a subsequence of y_n, the limit of y_n must be x

[candid hedge]

okaaaaay yeaaa i seeeee

this is amazing

[gritty widget]

It's not hard if you learn how to intuitively view those things
Then the only hard part becomes justifying them

[candid hedge]

why if it has a finite amount of terms then it has a constant subsequence?

ik its true but i dont see how

[gritty widget]

Say y_k can only take values from a finite set A. Then {k : y_k = a} for a in A partitions N into a finite amount of sets. One of them has to be infinite

then using this infinite set, we have a constant subsequence of y_k

[candid hedge]

that is genius i am glad i asked

[rich bison]

I am doing an course on topology by UB (on youtube). Just wanted to know are these topics enough. Here's the list of topics :-
(lemme know if this is the wrong place to discuss this)

1. SOME SET THEORY
2. METRIC SPACES
3. OPEN SETS
4. BASIS, SUBBASIS, SUBSPACE
5. CLOSED SETS
6. CONTINUOUS FUNCTIONS
7. CONNECTEDNESS
8. PATH CONNECTEDNESS
9. SEPARATION AXIOMS
10. URYSOHN LEMMA
11. TIETZE EXTENSION THEOREM
12. URYSOHN METRIZATION THEOREM
13. METRIZATION OF MANIFOLDS
14. COMPACT SPACES
15. HEINE-BOREL THEOREM
16. COMPACT METRIC SPACES
17. TYCHONOFF THEOREM
18. COMPACTIFICATION
19. QUOTIENT SPACES

[gaunt linden]

"Enough" is an extremely fuzzy concept. Enough for what?

[gritty widget]

yes, this should be enough for most purposes

[rich bison]

Enough for competitive exams and as an introduction to AT

Few things like fundamental groups and CW complexes are missing but I think that would more like intro to AT

[gritty widget]

yes, that belongs to an AT course

people sometimes give a brief introduction to fundamental group in the end of a topology course, though

if you're going with Hatcher then he has something like... topology chapter 0 I think it's called

and you can fill up all the holes in learning there

and then you should be good to go with Hatcher

[ornate berry]

That chapter 0 is really really intense (of AT)

I must warn you

People's opinions really strongly differ on this, so take mine with adequate salt, but I don't think Hatcher's book on AT is a good book.

[gritty widget]

I didn't mean chapter 0 of his book on algebraic topology

[rich bison]

I did take an AT course in my last semester using Hatcher and I vaguely remember it but then covid came and it was all a mess

[gritty widget]

https://pi.math.cornell.edu/~hatcher/Top/TopNotes.pdf

you can look at his notes to fill up the holes in learning

[ornate berry]

Oh whoops haha mb

Sorry

[gritty widget]

no, it's my bad, I thought he made a book which is named like that

but he only made notes

[rich bison]

Ah, cool. I might also look a bit to Lee's book to further fill the gaps. Have heard much about it

[gritty widget]

for manifolds?

I have to warn you that the versions you can find online have lots of errors so it's better to get a physical book for that one

[ornate berry]

Lee's book is really good but not for an intro to topology.

Maybe as a 2nd topology book :)

[rich bison]

Yeah, the table of contents makes it look quite like it's written for graduate topology with some AT

[gritty widget]

he does cover a lot of the needed topology briefly in the appendix

the most difficult concept there is probably paracompactness

[rich bison]

Is the same true for versions downloaded from libgen?

[gritty widget]

yes

I saw Jack Lee speaking about it once on math.se but I don't know where it was

[rich bison]

It's 40 usd here online in my country which seems reasonable for a hardcover but people have complained about the print quality

[gritty widget]

(It's his name on stack exchange)

[gritty widget]

I had some nice books about smooth manifolds recommended to me before, I'll try to check for some

@rich bison how about Hirsch - Differential Topology

This was recommended to me by a retired professor that was working in the field

There's also Milnor - Topology from differentiable viewpoint, iirc it touches upon some connections between AT and differential topology as well

[rich bison]

I feel that's not what I wanna do but not sure. My route is: Topology -> AT -> Homotopy Theory. I am not exactly sure how well diff topology sits in this. I might be completely wrong though

[gritty widget]

I'm just giving you alternatives to check out instead of Lee's Smooth Manifolds

There's also Guillemin Pollack which does the things in R^n

I don't know either, algebraic topology/differential manifolds are not my prime interests

sounds good to just go through Hatcher and think about it later

or other book about AT, if you prefer

[rich bison]

My main interests in the book are cause he does quite a lot of topology and AT judging by the table of contents. But I guess I'll try Hatcher first, it's kinda the standard recommendation

[gritty widget]

Not everyone likes his books, but yes. It's the primary recommendation

[rich bison]

And ofc, there's also good old Rotman and Tom Dieck

[gritty widget]

or Spanier

[rich bison]

I think it's this one - https://math.stackexchange.com/questions/1628180/trouble-with-exercise-from-lees-introduction-to-smooth-manifolds

[gritty widget]

yes

[quick delta]

How does this argument work? It just feels a bit hand-wavey and doesn't really convince me that it's not a cell complex

infinite cell complex is defined by removing the word "finite" from the definition

[bitter smelt]

You have a few options. You can prove as a Lemma that a compact subspace of a cw complex is contained in a finite subcomplex. Alternatively you can prove as a lemma that CW complies are locally contractible.

I agree that the argument you have there is not very rigorous. But it was just an example, so you're probably expected to work out the details yourself

If you struggle with either of these lemmas, they're somewhere in hatcher I think. Maybe relegated to an appendix, he does that for a lot of CW stuffs

You should try it on your own first though, it's relatively easy

The lemmas that is

[quick delta]

Is the first lemma almost obvious via contains an infinite subcomplex => non-compact? If so then I think I can prove it

But I think the problem is that I understand the arguement that it can't be a cell complex, but don't see what part of the definition it doesn't fit

[bitter smelt]

There is a little more nuance but it should be easy enough

[bitter smelt]

What do you mean? I'd you prove as a Lemma that CW complexes satisfy condition X and the hawaiin earring does not satisfy condition X then just look through your proof of the Lemma to see what part of the definition hawaiin earring defies

[candid hedge]

how do we prove that 1/x is closed?

[unreal stratus]

Well the graph of a continuous function into a Hausdorff space is always closed

[gritty widget]

You can intersect it with arbitrarily large closed balls. Then they are homeomorphic to bounded closed intervals which are compact. Since it's closed under intersection by arbitrarily large closed ball, it must be closed

[gritty widget]

I guess if you interpret it as a function from R to the one point compactification of R

We're in R^2

[unreal stratus]

Yeah lol mb

[candid hedge]

so this is false?

[unreal stratus]

This is true

Just would need work to apply here

[candid hedge]

why 1/x isnt a continuous function into a hausdorff space?

[gritty widget]

It is but the theorem says that its graph is closed when interpreted as a subspace of (R\{0}) x R

[unreal stratus]

Ye

[gritty widget]

Which is not necessarily closed in R^2

(but here it is of course)

[candid hedge]

is there another proof

[unreal stratus]

I suppose one other method is to consider {(x, y) | xy =1} which is closed and its intersection with closed right half plane is the graph we want

[candid hedge]

why xy=1 is closed?

[unreal stratus]

Multiplication is continuous

[candid hedge]

ah ok

And?

[unreal stratus]

this is the preimage of 1 under the multiplication map

[candid hedge]

right

okay

why is 1 closed again

lol

[unreal stratus]

R is Hausdorff

(Well all you need is T1 ofc but yes)

[gritty widget]

Let (x_n, 1/x_n) converge to (x, y). Then x_n and 1/x_n is bounded, so in particular x =/= 0 and so 1/x_n converges to 1/x = y

[candid hedge]

so?

[gritty widget]

4 different proofs of the same thing
Nice

[candid hedge]

how is this proof?

that its closed

[unreal stratus]

A subset of a space is closed iff it contains all its limit points

[candid hedge]

AH any convergent sequence converges with the image?

[candid hedge]

Riggght

okay

amazing

if A is a compact of E, for any x in E there exist a y in A such that (normed vector space): d(x,y)=d(x,A)

if i can prove that there exists a y such that d(x,y)=<d(x,A) then its done

i want to say that d(x,yn)-1/n=<d(x,A) then if yn converges to y i am done

but this is obviously wrong

[gritty widget]

functions from a compact set to R attain their maximum

And minimum

[unreal stratus]

Also quantifiers aren't in the right order here

[candid hedge]

its <=?

[unreal stratus]

That isn't a quantifier

[candid hedge]

ah ik what you mean

[candid hedge]

i fixed it

[unreal stratus]

There we go

[candid hedge]

right :/ any other proof?

i am so dumb

[unreal stratus]

Sounds like what you were doing is almost there

[gritty widget]

Let f:A to R be defined as f(y) = d(x, y). There is a sequence a_n in A such that f(a_n) converges to inf(f). From compactness of f(A) we can assume that f(a_n) converges to f(a). Then inf(f) = f(a).

[candid hedge]

So wait this is correct?

But i didnt use compactness i only used the property of closed subsets that for any y there exists a seq that converges to y

thats why i said this is obviously wrong

[candid hedge]

i dont see how the compactness of f(A) was helpful here?

[gritty widget]

It allows us to replace f(a_n) with a subsequence of it which converges to some element f(a)

[candid hedge]

Ohhhhh we want to show that f(a) is reached right?

[gritty widget]

inf(f) is

[candid hedge]

okay

[candid hedge]

if for all x,y llx-yll>r for some r>0 then its a discrete topology?

[candid hedge]

if a function is continuous, then f (closed set)=closed set?

[woven sundial]

they way you've worded this is slightly ambiguous but yes
if there exists r such that d(x,y) > r for all x,y then the topology is discrete

[candid hedge]

because for any element x in a closed set there exists an xn such as xn converges to x. then f(xn) converges to f(x), so all sequences of the image converges within the set of images and the proof is done

[woven sundial]

this is not true!

err that's not a good counterexample let me think

[candid hedge]

take your time

[candid hedge]

And telling me why this proof is wrong would be even better

[woven sundial]

oh it does work
consider the image of Z under sin(x)

[candid hedge]

i dont see how this contradicts

[candid hedge]

and i cant tell why my proof isnt enough to prove this

[woven sundial]

the first statement ignores isolated points.

[candid hedge]

isnt is enough to consider constant sequences?

for isolated points

[woven sundial]

uh I don't get how this is meant to work as a proof

[gritty widget]

you need to start with a convergent sequence in the image and show its limit is in the image; it's not clear that all of them should come from convergent sequences in the domain like that

[candid hedge]

a set is closed if it contains all its limits

[gritty widget]

an easy example of a continuous function taking a closed set to a non closed set is arctan on R: the domain is closed, the image is not

[candid hedge]

Yes

still

Ok so arctan takes a bounded closed interval to an unbounded open one

[gritty widget]

i should be more precise

$\arctan(\bR) = (-\pi/2, \pi/2)$ is not closed, despite $\bR$ being closed and $\arctan$ being continuous

[gentle osprey]

ТТерра

[candid hedge]

how does compactness fix this

f(compact)=compact

lets take a convergent sequence y1...yn in the image set

there exists an xn=f-1(yn)

[woven sundial]

i wouldn't use sequential compactness

[gritty widget]

but they are, so let them

[candid hedge]

its the only one i learned

[woven sundial]

ok fair

[gritty widget]

now you have a convergent subsequence

[candid hedge]

right

xk_n converges to x then f(x k-n) converges to f(x)

and we're done right?

[gritty widget]

back here, there was no guarantee the sequence in the original domain you take converges (let alone has any convergent subsequence); this is how compactness fixes things

[candid hedge]

yeaaaah

amazing thanks a lot :)

now when is f(open)=open

xd

is there a condition on the space?

if f is only continuous

[gritty widget]

[candid hedge]

i dont think so nvm

[gritty widget]

nope. Those maps are called closed

[gritty widget]

nope. Those maps are called open

[candid hedge]

I understand

[gritty widget]

conditions for when a map is open? For linear operators you just need them to be surjective. In some other cases you can have open maps too

[gritty widget]

it might be a good exercise to think about this a bit (for example, could you conclude anything using what you know about closed maps? just as a starting point)

[woven sundial]

this is a less clear example than tterra's but what i meant was that the image is dense in [0,1] but is countable so cannot be the entire interval

[candid hedge]

so its not closed?

[woven sundial]

ye

[candid hedge]

why?

[gritty widget]

I think it's easier to construct some conditions for when a map is closed than when a map is open. Just because of things like compactness

[woven sundial]

bcos it is dense in [0,1] its closure is [0,1]

[candid hedge]

oh yea x(

god when will i have the reflex

[candid hedge]

ill try

[gritty widget]

why don't you construct some bitches

[gritty widget]

why don't you stop being toxic and get a life

[candid hedge]

wth

[woven sundial]

another characterisation of compactness is that X is compact if every open cover of X has a finite subcover

(an open cover is simply a collection of open sets such that each point of X is in at least one set from the collection)

[candid hedge]

yeah ive seen that on youtube. Ive only learned topology on normed vector spaces untill now in uni

so it wasnt deemed necessary

[unreal stratus]

Also worth emphasising that what twiceshy just said is the typical definition and indeed is different from sequential compactness for general topological spaces

[woven sundial]

yes, thanks potato.

[gritty widget]

something something only equivalent in metric spaces

[unreal stratus]

Yeah I believe compact implies sequentially compact for first countable spaces (hence including metric spaces) whilst the reverse direction is more subtle

[candid hedge]

right

[gritty widget]

all the spaces any reasonable person cares about are metrizable anyways

[unreal stratus]

Algebraic geometers are unreasonable

[woven sundial]

zariski topology go nyoom

[gritty widget]

zariski "topology"

topological spaces were never supposed to be non-hausdorff

that's not true at all. In functional analysis you constantly consider non-metrizable topologies

just consider weak topologies of a Banach space

[candid hedge]

how to show that an open ball inteersects all open balls around a point on the border of it's closure

ik it's definition of border but i need a rigorous proof

[vocal wharf]

well, start by giving your relevant definitions

[candid hedge]

i did that

B(x,r')inter B(a,r) for all r'>0 is what i want to show

where x is at the border

and B(a,r) is my open ball

[vocal wharf]

where is your definition of border (i assume boundary?) and closure?

[candid hedge]

llx-all=r ig

[vocal wharf]

i mean the point on the boundary itself is in the closure, so ...

[candid hedge]

no i want to show that the closure is equal to the closed ball

B'(a,r)

[vocal wharf]

whats your definution of closure then?

[candid hedge]

any of em. The smallest closed set containing the open ball for example

or the set containing all of it's limit points

[gritty widget]

You can translate the ball so that it has 0 as its origin and use multiplication by scalars

[candid hedge]

the ball around x?

[gritty widget]

To show that the closure of an open ball is a closed ball

[candid hedge]

can you elaborate?

[gritty widget]

B(x, r) is open ball with center x and radius r right

[candid hedge]

right

[gritty widget]

for any point y from the closed ball, lets denote it by D(x, r), for 0 < s < 1 we can write x+s(y-x) which will be in the open ball

but as s -> 1, x+s(y-x) -> y

if we were to shift it by x then this would look a little simpler

because x = 0 then

the purpose of s is to make the the point y a little closer to the center so that it's contained in the open ball

[candid hedge]

yes

[gritty widget]

and that proves it because the closed ball is closed

[candid hedge]

how would we translate?

[gritty widget]

the function f(y) = y-x is a homeomorphism for every point x

so we have things like, f(cl(B(x, r))) = cl(f(B(x, r))

you know, it commutes with the closure for example

if you were to show that cl(B(0, r)) = D(0, r) then shitfing it forward by x, we get D(x, r) = x+D(0, r) = x+cl(B(0, r)) = cl(x+B(0, r)) = cl(B(x, r))

so we can just prove it in case x = 0

similarly we could have assumed r = 1

[gritty widget]

it's not necessary, this formula just looks a little worse

[vocal wharf]

im still confused about your definitions lmao

[vocal wharf]

like this doesnt work

i guess you are in R^n with standard topology?

[candid hedge]

nice

[candid hedge]

Not R^n

a normed vector space

but i understand the confusion

i meant just in the case of an open ball

[vocal wharf]

hm does this definition work then

[candid hedge]

this is great thanks a lot

[candid hedge]

i mean yeah. the border is the set of x that verify llx-all=r

[vocal wharf]

doesnt it break for trivial norm

[candid hedge]

which one is the trivial

[vocal wharf]

then the boundary of every ball should be empty

where am i wrong?

[gritty widget]

what's the trivial norm

[vocal wharf]

0 for 0, 1 otherwise i guess

[gritty widget]

discrete metric?

[vocal wharf]

ye, such that you get discrete metric

[gritty widget]

Pretty sure that space wouldn't be normable

[vocal wharf]

oh ok

what breaks?

[gritty widget]

I don't think it's even a topological vector space actually.

[vocal wharf]

oh yeah, thats true

i mean {0} still works i guess?

[gritty widget]

yes

[vocal wharf]

its just such a weird definition of boundary

but what you said is the simplest way then

[gritty widget]

But if it has any non-zero vector x, then (1/n)x for example should converge to 0 by continuity, but it also has to be eventually constant since we're going for the discrete metric

but (1/n)x is never non-zero

[vocal wharf]

i still dont trust that defintion of boundary though 😛

[gritty widget]

If C is a convex neighbourhood of zero, then if p is the Minkowski functional of C, we should have that the points of boundary correspond to the points at which p = 1

[vocal wharf]

(i was originally thinking this is supposed to work in metric spaces, its my mistake to not read history some more)

[gritty widget]

which is an analogue of the situation with the open ball centered at zero, because then the Minkowski functional should be the norm times some constant

[gritty widget]

No, it's totally okay to be confused, I just guessed it because @candid hedge was working with normed spaces the past days

it's good to know that this property fails horribly when we're not in a normed space

[candid hedge]

yes. Ive only worked with normed vector spaces so far

But i watch a lot of youtube

xd

show that the interior of a convexe is convexe

we know that in an interior all points are centers of some open ball completely contained within the set

so tx+(1-t)y is contained in some open ball in the set is what ive got to show

[candid hedge]

oh and theres also the question to show that the interior of a closed ball B'(x,r) is the open ball B(x,r)

I am almost done through the sheet of questions. Sorry for asking too much but this is really important for me to understand and i have a shitty uni

[unreal stratus]

Dw, topology can take some time to get used to

ig that's true of any area of math(s) xd

[candid hedge]

yeah for sure

[candid hedge]

can you help me show that the interior of a convex set is a convex

having some trouble

[candid hedge]

just a general guideline on how to do this

[unreal stratus]

Is this for any normed vector space?

[candid hedge]

yes

[candid hedge]

its a normed vector space

i should get in the habit of specifying that xd

[gritty widget]

take x, y in (the interior of) your convex set C
take an open ball B centered at 0 such that x+B and y+B are contained in C
Conclude

It should that for any t in [0, 1], tx+(1-t)y+B is also contained in C

[candid hedge]

i am still unsure how to do this

im thinking

have no idea what to do

[unreal stratus]

Oh interesting, the idea I had was fairly different but possibly wrong then lol

[candid hedge]

and i am not sure what the translation achieves

[candid hedge]

what did you have in mind

[gritty widget]

you should think about it on your own a bit

[unreal stratus]

I was going to say you could probably take the line between the two points x,y (which lies in C) and consider the sup of t in [0,1] such that tx + (1-t)y lies in the interior

Or some perturbation of that idea

[candid hedge]

oh thats really good

but idk how it would work here

[candid hedge]

Ig since the original set is convexe then tx+(1-t)y lies in C but i cant get further than this

i need a hint i cant spend more time on this quest

[gritty widget]

you won't learn if we give you all the answers

[candid hedge]

yes thats true lol

[candid hedge]

but ive been stuck at this since 3pm

[shadow charm]

You’ll want to show that for interior points the line between them lies in the interior too

[candid hedge]

yeah i realized this

[shadow charm]

Have you tried showing it’s everywhere locally convex

If I’m not mistaken this is equivalent to convexity

[shadow charm]

And this is easier to show I’d assume

[candid hedge]

i have not tried it

i think i got it

a bit

[gritty widget]

I never heard of locally convex refering to subsets of a topological vector space

[shadow charm]

I thought we were talking about subsets of a normed vector space

[gritty widget]

probably not because if you draw two disjoint balls then they'd be "locally convex"

I only ever heard locally convex refered to locally convex topological vector spaces

[candid hedge]

tell me if this statement is true: if there exists an r such that B(x,r) and B(y,r) lie in the set then B(tx+(1-t)y,r) lies within the set

[gritty widget]

and never as a property of subsets

[shadow charm]

Locally convex as in there is a neighborhood that is convex

[shadow charm]

Hmm that’s fair though maybe connectedness is enough?

[candid hedge]

on my drawing i see it works but i cant tell if its always true

[gritty widget]

This is way easier than this and I gave a hint above. It basically solves the question

[shadow charm]

Ah i didn’t see your hint

[candid hedge]

AHHH FUCk

[candid hedge]

this is this no? xd

[shadow charm]

Right yeah lol

Nvm me that makes it much easier

[gritty widget]

sure

[shadow charm]

Yes

But you should show it of course

[candid hedge]

ofc i take any point in the ball and show that its in the set

its a nice property of convex sets that i didnt understand until now

so let z be tx+(1-t)y, then a a point in the open ball B(z,r). lla-zll<r then...

i have no idea

help

OHHH i got it

Its incredible

i am soooo fucking dumb

thanks a lot

[candid hedge]

if the distance between a point and a set is 0 then its a part of the set's closure?

[gritty widget]

try proving it

[bitter smelt]

a related a good exercise is to prove:

It is possible for two closed sets $A, B$ to have $d(A,B) = 0$ but $A \cap B = \emptyset$

[gentle osprey]

Migillope

[bitter smelt]

and if we require one to be compact it is no longer possible

(in a metric space)

[unreal stratus]

It's quite hard to have a distance function if you're not in a metric space I guess

But jokes aside, yet another related thing is that you can characterise the closure of A as the set of points distance 0 from A - that's a good thing to prove and obviously a strengthening of the original thing

[dry jolt]

Forgot to bump this but I asked someone irl and they said this sounded correct :)

[swift fjord]

Oh yes free action on finite set => finite group. Even faithful action is enough I think since it determines an injective map into the symmetric group on the cells

[unreal stratus]

That's cool

[dry jolt]

I agree

oh yeah, yesterday I learned about a special case of Lefschetz fixed point theorem and used it to prove that Z/2 is the only nontrivial group that acts freely on even dimensional spheres :)

[unreal stratus]

That's exactly what came to mind when I saw your post aha

[dry jolt]

I'm working through Brown's book on cohomology of groups, have you read it?

[unreal stratus]

Nah, it's just this is done in Hatcher

[dry jolt]

ah I see, never made it past like the first section on simplicial homology

[unreal stratus]

how are you finding that book though? group cohomology isn't really something I've looked at

Yeah fair enough

[dry jolt]

I'll probably go back to hatcher once I actually take an alg top class lol

[swift fjord]

John is using this in his homalg course

[unreal stratus]

Yeah fair enough

I need to review cohomology tbh

[dry jolt]

the book is fun, I haven't gotten to any of the core material yet because there's an introductory section that just talks a little bit about free resolutions of Z over group rings

[unreal stratus]

Kinda skimmed through it to get to chapter 4 (on intro homotopy theory)

Ah ok

[dry jolt]

I talked to clerk about this a little while ago, but there are really cool interactions where you use topology to construct the bar resolution for Z over a group ring

Basically you construct a simplicial complex where n-simplices are (n+1)-tuples of elements in the group. There's a natural free action of G on the simplices, and the space is contractible, so if you look at the induced map on free abelian groups generated by simplices, you get a free resolution of Z

The context for my question was that if you have a free action of G on an odd dimensional sphere, then the Lefschetz thing says that G acts trivially on the non-trivial homology groups, so you get an exact sequence 0 -> Z -> C_{2k-1} -> ... -> C_1 -> C_0 -> Z -> 0, and if you just chain these together you get a periodic resolution of Z over your group ring

neat stuff, I'll probably talk to you once I get to the actual subject material

[small hemlock]

Are Munkres topology and Gallians contemporary alg good prerequisites for Hatchers alg topology? Because I have those first two, and want to go into alg top. I'm not sure, other than groups and maybe some knowledge of other general structures, what abstract alg I need

Also, for those of you who answer in this channel a lot, I'm taking my first introductory topology course this semester (jr. in ugrad), and I'll probably be asking a lot of questions on intuition and the like. I apologize in advance lol

[onyx sonnet]

You’ll have a good foundation for alg top if you go through Brown’s book

He connects lots of things back to homo-cohomo of top spaces

[onyx sonnet]

Imo On the algebra side of things, if you’ve gotten up to free groups I think that’s enough to start out with

Wait does Gallian even go that far?
Edit:it does

[river granite]

first few chapters of Munkres should be fine, Hatcher himself also has some free notes with the point-set topology required for his AT book: https://pi.math.cornell.edu/~hatcher/Top/TopNotes.pdf

don't worry about asking questions btw that's the whole point of this server

[small hemlock]

Up to free groups and the first few chapters of munkres, I can do that. And thanks for those notes, I'll most likely reference those a lot.

The professor this semester is focusing on the more point-set side of things, and we aren't touching on geometry much, or working towards classifications of surfaces. We're mostly just learning what's required to really prove stuff

There's the contents for the semester, so it's definitely more general topology

Okay Hatchers notes touch on quotient spaces, good

[icy schooner]

is there some relation between π_n and H_n

been watching some videos on homology of simplicial complexes
pi_0 and H_0 both measure connectivity
pi_1 and H_1 holes

[unreal stratus]

You may want to look at the Hurewicz theorem(s)

[surreal lantern]

for a path connected space the abelianization of the fundamental group is iso to H_1

[unreal stratus]

The main results are that the first homology group is the abelianisation of the fundamental group (for path connected spaces) and that if n>1 and X is n-1 connected (path connected and first n-1 homotopy groups vanish) then the nth homology group equals the nth homotopy group

[icy schooner]

I see

ty

[unreal stratus]

This is quite a specialised hypothesis, though it does let you compute the first n homotopy groups of the n sphere for example

[icy schooner]

honestly some of these results are impressive they feel like black magic

[shut moat]

Maybe a silly question, but in homology theory, do you lose that much (at least when dealing with manifolds and other nice things) by requiring the p-simplices to be embeddings rather than just continuous maps?

the intuition of kindof lassoing holes in your topological space with the simplicies would be slightly nicer this way I feel

[plain raven]

Take the projection map R^2 -> R^1 onto the first coordinate. The first space has lots of embedded 2-simplices; the second space doesn't have any embedded two-simplices. So smooth maps between manifolds don't carry embedded simplices to embedded simplices. In order to get a functorial theory you would have to consider only embeddings between manifolds or something like that.

You would run into a similar issue with homotopy, I think. Only subspace embeddings X x I -> Y would lend themselves to establishing a nice relationship between homotopic simplices. Thus the homotopy invariance of homology is thrown off. It is difficult to see in this theory how you would prove that an arbitrary contractible space has trivial homology groups.

[shut moat]

Thanks, this makes sense! Do you by any chance know if on a fixed manifold (so that functoriality isn't a concern), a hypothetical embedded homology theory would capture more or less topological data than the singular homology?

Essentially what I'm going for is that If the singular homology is in that sense the same as the embedded homology, then I can morally think about singular homology in terms of the embeddings, and I can just consider the singular simplices as a trick that makes it naturally functorial

[small hemlock]

Is Q ⊆ R with the subspace topology not discrete because if you have some p in Q, then p=A ∩ Q for some A open in R. So that means that p is in A, and so (p - r, p + r) is in A. But since Q is dense, there is some q in Q such that p < q < p+r, and so q is also in A ∩ Q? Which means A ∩ Q cannot be a singleton set?

[gritty widget]

yes

[small hemlock]

Okay nice

[unreal stratus]

I guess you should say {p} = A \cap Q but sure

[gritty widget]

the informalities weren't that bad so I payed no mind. But yeah

[unreal stratus]

Yee

[small hemlock]

So could I show that ${\frac{1}{n}}{n \in \mathbb{N}}$ is discrete in R by taking some ${\frac{1}{n_1}}$ and showing that ${\frac{1}{n1}}=A \cap {\frac{1}{n}}{n \in \mathbb{N}}$ where $A=(\frac{1}{n+1},\frac{1}{n-1})$ is open in R? I wanna make sure I have the idea correct for these kinds of proofs. I imagine I could always just show whether or not a singleton set is open in a subspace topology

[gentle osprey]

J.Ross

[gritty widget]

yes. And the last sentence doesn't make sense to me, this is precisely what you're already doing

[small hemlock]

Oh I suppose I worded it weird. I meant in general, if I'm trying to show that a certain subspace is discrete in a topological space X, I would just have to show that an arbitrary singleton is open in the subspace?

[gritty widget]

yes, it's equivalent

[unreal stratus]

Here a cute alternative is to use the fact N is discrete in R and that x -> 1/x is obviously a homeo between that and this

[small hemlock]

Oooo, I like that

[vague brook]

Is there some theorem about quotients of quotients

Like mobious band is some quotient of the square

And klein bottle is the same quotient with some more relations added

[long grail]

If A is open in X and B is open in Y, is A x B open in X x Y? proof: let A be open in X and B open in Y. Then if a is in A and b is in b there is an neighborhood N_a that is contained in A and a neighborhood N_b that is contained in B. Then N_a x N_b is a subset of A x B, for any u in N_a is in A and any v in N_b is in B. Let (u,v) in N_a x N_b. Then (u,v) in A x B, so A x B is an open set in X x Y. Is that all??

[swift fjord]

How do.you define the topology on XxY

Brcause depending on your definition, this is true by definition

[long grail]

a topology on X x Y is the a set of open sets W subsets of X x Y such that UW is in the topology and finite intersections of W is in the topology, X x Y is in the topology and empty set is in the toplogy

really I'm trying to prove A closed in X and B closed in Y means A x B closed in X x Y but I just need to know what an open set of X x Y is like

this is doing my head in these proofs feel so weird to me

[vague brook]

lol you get used to it

[tiny obsidian]

That's a topology on X x Y

There is a product topology which is usually the one assumed

[swift fjord]

So by this definition AxB is definitionally an open set

Try and characterise what a set of the form (AxB)^c looks like

To see why AxB is closed when A,B are

[long grail]

ok will do

thanks math cloud

[quick latch]

It seems to me that these go in opposite directions? A neighborhood near x implies a neighborhood near f(x) on the left, but it's the other way around on the right. Which one is correct?

[river granite]

on the left you're taking an epsilon-nbhd of f(x_0) to find a delta-nbhd of x_0.

note that f(U) contained in V (right hand side) just means that if x is contained in U => f(x) is contained in V

[quick latch]

I think my confusion stems from this example:
$\rho = \theta\mapsto e^{i\theta}:[0,2\pi)\to S^1\subset\mathbb{C}$ where $S^1$ is the unit circle on the complex plane

[gentle osprey]

chic-chicago

[quick latch]

here rho is not continuous?

[tender halo]

looks continious to me

[quick latch]

what if I fix a neighborhood around f(0)? there is no corresponding neighborhood around 0 because you can't just jump to 2pi

[tender halo]

then you'll have some nbhd of 0, some [0; r)

[river granite]

a neighborhood needn't be connected, in this case you'll get something like [0, r)U(2pi-r, 2pi)

if you take the preimage of e.g. an open ball around f(0)=1+0i

[river granite]

and this is open in the induced topology over [0, 2pi)

[tender halo]

also consider that this function is also continious as a function R -> C

uhh

i mean that if you take the function r -> e^ir then its continious as a function from R to C

[quick latch]

hm, ok. rho inv is supposed to be discontinuous

[river granite]

yes

[quick latch]

could you explain why? my intuition was that the neighborhood has to be connected in some way

but clearly it's wrong

[river granite]

it's a good example of a function that is continuous, bijective but not a homeomorphism for that reason

and yes a straightforward way to check that rho^{-1} isn't continuous is by a compactness argument

if you've seen that

[surreal lantern]

i deleted it and went back to check the question again

[river granite]

otherwise you can also play with preimages of open sets around 0 and directly check that rho^{-1} isn't continuous by definition

[surreal lantern]

i misread a while ago on a similar example

didn't wanna do the same mistake again

[river granite]

lol it happens

dw

[quick latch]

so intuitively, is the fact that [0,2pi) is bounded contain some information that the unit circle can't represent?

as in, we can't represent that cutoff on a circle

[tender halo]

im also interested how does a compactness argument look like

[surreal lantern]

its a bijection

[unreal stratus]

If X is compact and Y homeomorphic to X, then Y is compact, but [0,2pi) isn't compact whilst S^1 is

[surreal lantern]

hence the image of the inverse is all of [0,2pi)

but compact sets (like S^1) are mapped to compact sets under continous functions

[0,2pi) is not compact

[river granite]

no boundedness doesn't really have to do with it -- after all, both [0,2pi) and S^1 are bounded, and boundedness isn't a topological property

[tender halo]

oh, duh

thanks

[river granite]

one reason is because continuous maps f: X -> Y map compact sets to compact sets

[surreal lantern]

to add to this, (0,1) is homeo to R

[river granite]

and another is that the preimage of any open set of the form [0,r) around 0 isn't open in S^1

[surreal lantern]

lmao potato (deleted )

[unreal stratus]

Another way is to note there are many continuous maps S^1 -> S^1 (like a rotation) which are clearly not continuous after composing with rho^-1

[quick latch]

this neighborhood isn't an open set of R (I think), so how could it be for [0,2pi)?

[tender halo]

it's not, but its an open set of [0; 2pi)

cuz it an intersection of an open set of R and [0; 2pi)

for example, (-1; r)

[quick latch]

how though. it has two disconnected parts. Say we stick with whole numbers.
The way I understand {0, 2} wouldn't be an open set because no intersection of {0, 1, 2} with an open set of whole numbers results in it

[tender halo]

both [0; r) and (2pi - r; 2pi) are open in the subspace

{0, 2} wouldnt be an open set in which space

[quick latch]

if you intersect [0,2pi) with (-1. r) don't you get [0,r)?

[tender halo]

yeah

[quick latch]

and what about the rest (2pi-r,2pi)

[tender halo]

and then (2pi - r; 2pi) is also open in the subspace

since its also an intersection of an open set of R (namely, itself) and [0; 2pi)

they are both open means their union is open

[quick latch]

i think i need to go back to square 1

[tender halo]

if you want an explicit example of the set, it would be (-1;r) \cup (2pi - r; 2pi)

another way to look at it is that since topology on R is generated by an order on R, topology on its subspaces is also induced by the induced order

and [0; r) \cup (2pi - r; 2pi) is open in the order topology

[quick latch]

I haven't learned about compactness or order topology yet

[river granite]

order topology on R is just the usual topology

[quick latch]

yeah i didnt realize the union of open sets is an open set in R. I get why rho is continuous then. As to why rho inv is discontinuous...

[surreal lantern]

is there something unclear?

[tender halo]

since you have a bijection on your hands, its enough to find an open set whose image is not open to prove discontinuity of the inverse

[quick latch]

yeah I know. I'm just unable to imagine it xD

so like, [0,1] is not open right

[tender halo]

in [0; 2pi)? no

[quick latch]

so any interval I pick in R or C has to be open for it to be an open set

[tender halo]

uhh

what?

[unreal stratus]

You need to specify the topological space it's open as a subspace of

An "interval in C" doesn't really make sense

[surreal lantern]

we're assuming the standard topology here i suppose in C and R

[quick latch]

yes

[unreal stratus]

Yeah, just the preceding discussion was focused on [0, 2pi)

[surreal lantern]

what's the definition of openness you are working with chic

[tender halo]

yes, interval in R has to be an open interval to be an open set

[quick latch]

Definition. A set U of real numbers is said to be open if for all x ∈ U there exists δ(x) > 0 such that (x − δ(x), x + δ(x)) ⊂ U.

[unreal stratus]

Indeed

That's fine for subsets of R yes

[surreal lantern]

oh this wasn't directed towards your preceeding messages sorry

i agreed with your points

[unreal stratus]

Ye dww

[gritty widget]

So it’s the standard topology/Euclidean topology/topology used in analysis

[surreal lantern]

now think about why [0,1] is not open

[quick latch]

so to show rho inv is discontinuous, I need to find an open set in [0,2pi) that does not have a corresponding open set in S^1

[surreal lantern]

where could the parts lay where that definition fails

[quick latch]

no, I get that, I wrote closed by accident

[surreal lantern]

oh ok

[surreal lantern]

have you not covered compactness yet

[quick latch]

no

[unreal stratus]

covered

[surreal lantern]

intended

[tender halo]

well, "does not have" is a strange way to phrase it, more like whose image is not open

[unreal stratus]

Nice

[quick latch]

could you give me an example? I feel like im truly stuck

[tender halo]

you have a continuous bijection, only thing that stopping it from being a homeomorphism is that rho^-1 is not continuous. rho^-1 is not continuous, by definition, means that there is an open set A in [0; 2pi) such that its preimage in S^1 under rho^-1 is not open

since rho is a bijection, it just means that the image of A under rho is not open

[quick latch]

yes, right. I can't think of an example A. I think it comes down to the definition of openness on S^1

[tender halo]

well, what do you think would mess with the continuity in this case

[quick latch]

I would assume at 0

[tender halo]

yes, 0 is a bad point

namely, you come close to rho(0) from 2pi side

if you want a hint, ||A includes 0||

wait uhh

i meant to say the opposite :)

[quick latch]

ok now im back to square 1

if 0 is included in an open set of S^1, the pre image isnt an open set in [0,2pi) right

so rho is not continuous

[tender halo]

it is continuous

the pre-image is an open set

we are arguing the opposite

that some open sets of [0; 2pi) dont have open images

[quick latch]

yes right, but what about 0 then?

we have an open set in S^1 that contains 0

but no open set in [0, 2pi) contains 0

[tender halo]

that never happens

for example, [0, 2pi) is an open set (in [0, 2pi)) that contains 0

[quick latch]

but [0? doesn't that contradict the defintion I sent

[tender halo]

whats [0? means

[quick latch]

well there is no point less than 0 within the interval
Definition. A set U of real numbers is said to be open if for all x ∈ U there exists δ(x) > 0 such that (x − δ(x), x + δ(x)) ⊂ U.

[tender halo]

yes, and because of that [0, r) is not an open set in R

but it is an open set in [0, 2pi)

[quick latch]

aha right, back to the subspace idea

look, I think the ideas are written out. I'll try again later. thanks again

[tender halo]

do you want to hear the set A?

[quick latch]

sure

[tender halo]

its any [0, r)

since it would only contain points from rho(0) from one side

[quick latch]

why can't [0, e^(irpi)) be an open set for S^1

[tender halo]

uhh

you mean rho([0, r))?

[quick latch]

yeah sorry

[tender halo]

cuz rho(0) does not have an open nbhd in it

it contains points only from one side

[quick latch]

so does this come down to intersection between induced topology of C and S^1 vs the induced topology of R and [0,2pi)?

yeah ok, thanks.

[tender halo]

you worded it wrong, but yeah, those are not homeomorphic spaces

[quick latch]

don't waste your time w/ me. I'll go back to definitions. thanks again for all the help!!

[broken mica]

Hi I am unexperienced in topology. Is there a theorem stating that the boundary of a bounded connected subset of the plane is a connected curve (or a set of connected curves if this subset has holes). Please @ me if you have any knowledge on this. Also tell me if i am not precise in what I am asking. Thanks in advance!

[unreal stratus]

What do you mean exactly by curve here? Just to check

Because under some interpretations this has simple counterexamples

[broken mica]

I think the term is jordan curve? like the boundary of a disk

[unreal stratus]

Then no, e.g. just take a line segment

[broken mica]

ok what if this subset is homeomorphic to some n-holed annulus?

[unreal stratus]

Well boundary of an n-holded annulus is then a set of connected curves so the image under the homeomorpuism will remain as such I guess?

It is a bounded connected subset

And its interior is empty; it is its own boundary

[broken mica]

thanks I think this is all I need

[odd flame]

does an arbitrary topological space (say it has an infinite # of elements) have an infinite number of topologies that could be defined on it

[gritty widget]

if X is any set and S any subset, {empty set, S, X} is a topology on X

[opaque cloud]

trivial topology?

[odd flame]

what i mean is like

the set of all possible topologies on a topological space

if X is infinite, then that ^ should be infinite right

oh wait

[gritty widget]

use this

[odd flame]

yeah i realized you said yes without saying yes

[gritty widget]

just an observation on this construction

no actually nevermind

i changed my mind

[odd flame]

[opaque cloud]

would the set of all possible topologies be a set of all subsets of the powerset that satisfy the axioms

[gritty widget]

i was going to say something about the problem of finding the number of topologies on a set, but there were a couple of details i wasn't sure of, so i didn't want some pedant to come by and correct me later

[odd flame]

i asked cuz im thinking about a) here

dont spoil tho

[opaque cloud]

but you could use that pedant to do the work for you, like ur not sure of the details, that pedant will just provide you the details

[gritty widget]

you overestimate my willingness to deal with people

[odd flame]

moment of appreciation for tterra for helping me with so much algebra and topology lol

a counterexample is enough to prove that the union of a family of topologies isn't necessarily a topology right

[gritty widget]

one counterexample is good enough

[odd flame]

cuz it's not asking to show that it's never the case

[gritty widget]

union of topologies could or could not be a topology, but to show it's not true in general, one counterexample works

if you wanted to explore more, you could try to figure out when the union of topologies is a topology

so like, go through trying to prove that the union is a topology, and see exactly what fails

[odd flame]

,ti

[gentle osprey]

The current time for nitezba is 01:06 AM (EDT) on Mon, 12/09/2022.

[odd flame]

perhaps tomorrow lol

[gritty widget]

sleep is advisable

[coarse night]

they have already given an example

[gritty widget]

health > mathematics

[coarse night]

🗿

[odd flame]

this isnt topology but i never know who's advice to listen on this

on the one hand deadlines, gpa, expectations, i should stay up and finish my homework

on the other hand, i need to function as a human being, health, etc.

idk ive never been able to say "fuck it" to my grades, except for when it's really bad

[gritty widget]

if you don't take care of yourself, you're probably going to see worse results in the long run

balance is necessary

[odd flame]

this is true

yeah i bit off a bit more than i can chew this semester but i can't really drop anything so here we are

my hope is that once im caught up understanding-wise things will smooth out but we'll see - i'll conclude that intermissiono there dont wanna get off topic much

[odd flame]

for b) shouldn't it just be the power set and the trivial topology?

oh well

i guess the smallest containing all collections T_a should be the intersection of all topologies, by part a)

[odd flame]

,ti tterra

[gentle osprey]

The current time for TTerra is 01:19 AM (EDT) on Mon, 12/09/2022, the same as nitezba!

[gritty widget]

all my classes start later in the day

plus i have none on mondays

[odd flame]

that's so nice damn

[gritty widget]

staying up to finish a bit of work and then hang out with someone when they get back

basic operator theory review for my QM class

[odd flame]

QM looks cool

out of my league for now tho

[gritty widget]

it's just functional analysis with physics words

[coarse night]

just called an entire branch physics obsolete

[odd flame]

[tender halo]

experimints

[long grail]

Will someone review my proof?

It seemed too easy

[quiet pilot]

It is not true in general that $(A \times B)^c = A^c \times B^c$

[gentle osprey]

AoiKunie

[long grail]

No?? Woah ok

That's a bummer

[quiet pilot]

It's not even true in $\mathbb{R}^2$ try drawing a picture of a rectangle

[gentle osprey]

AoiKunie

[quiet pilot]

Essentially, the complement of a rectangle is not a rectangle

[gritty widget]

$(x, y)\in A\times B$ means that $x\in A$ and $y\in B$. Now take negation

[gentle osprey]

Blitz

[long grail]

All right

I'll give it a try thanks!

I understand

[quiet pilot]

It is however true that e.g $(A \times Y)^c = A^c \times Y$

[gentle osprey]

AoiKunie

[quiet pilot]

Which is potentially useful for your proof

[long grail]

Yes it is

Thanks

That changes everything

I was thinking it was A^c × {} derp

[woven sundial]

no it isn't

it's A^c x Y \cup A x Y^c

[quiet pilot]

Y is the whole space

[woven sundial]

oh 😵‍💫 that's my bad apologies

[swift fjord]

||If it wasn't the whole space then you forgot to add A^c x Y^c||

[candid hedge]

nilpotent matrices are a closed set of Mn(R)?

and R^n projectors form a closed set of L(R^n)?

and how to reason when asked such questions

[coral pivot]

I would use limits

Like show the limit of a sequence of nilpotent matrices, if they exist, is nilpotent

[candid hedge]

how to even construct a seq of nilpotent matrices

and how would i deduce this

my logic was simply: consider a nilpotent matrice of order 3. Now consider the function g that takes f and gives f^3, the nilpotent matrices are simply g^-1(0)

so they're closed

[coral pivot]

Hmm that works too as you only need to take a finite union

[candid hedge]

but that's for finite orders of nilpotence. Idk what to do with infinite

[coral pivot]

If A is nilpotent A^n is zero

[candid hedge]

wdym

you mean infinite union?

[coral pivot]

No look at this

[candid hedge]

but id love to see how this would work

to gain a new way of thinking

i suck at using sequences

[candid hedge]

like for example here. A projector is simply a function that satifies pn(pn)=pn

so can i take a sequence (pn) and simply say it converges towards p and then p(p)=p

so i am done?

[coral pivot]

Yes

[candid hedge]

HOW tho

i dont understand how that proves anything

it feels like i cheated

[coral pivot]

Closed means the set contains all its limit points right

[candid hedge]

why does the limit has the same property as the seq?

no i know the def of closed i just dont understand how we have the right to say p(p)=p

given only that pn(pn)=pn

[coral pivot]

So if you have a sequence that converges, and what it converges to is in that set, this shows closed under limit point

Well multiplication is continous

[candid hedge]

here its composition

[coral pivot]

So like, limit splits over it

[candid hedge]

no?

[coral pivot]

Sure same thing

[candid hedge]

composition is continuous?

idk what that means xd

[coral pivot]

Like, you have a function sending f,g to fg and this is continous, I’m pretty sure atleast lol

[candid hedge]

okay

[candid hedge]

you were saying something here

[coral pivot]

You can use your idea there

Take the pre image of p under the raising to the power n map

Of 0

But also if you wish, take a sequence of nilpotent, their n power is 0 so obviously their limits n power would be 0

[candid hedge]

WHY tho

LOL its not obvious at all

[coral pivot]

Matrix multiplication is continous

[candid hedge]

WHO told you this

why is it continuous

[coral pivot]

It’s the same reason the limit of product of sequences is the product of the limits in like, R

It’s just polynomials entry wise right

And ofcourse polys are continous

[candid hedge]

i dont know what that means

[coral pivot]

Well I mean how would you write matrix multiplication as a function?

[candid hedge]

AHH

yeah

its a polynome

[coral pivot]

Yeah

[coral pivot]

I am not sure this is right actually

[candid hedge]

I am asking in discussion

[coral pivot]

Oh lmao it is, L is like bounded operators on Rn right

Or did you mean something else

[candid hedge]

L means linear functions of R^n

[coral pivot]

Right and all linear functions are continous right

On finite spaces that is

I’ll let you fill in the rest about why composition is continuous

[candid hedge]

I cant

my brain is fried

cant think of shit

[coral pivot]

Yes you can you haven’t even tried lol

[candid hedge]

everything entering is getting out xd

[coral pivot]

(If u wish u can also identify these with like matrices and conclude there)

[candid hedge]

yeah

if matrix multiplication si continuous then so is function compostiion

no?

[coral pivot]

Mhm

You would have to do a bit of work showing it’s the same topology ig but yeah

[candid hedge]

No no it would show that composition on linear functions are continuous

[coral pivot]

Lol

[woven sundial]

im an idiot

[small hemlock]

Okay, I have a question, because my intuition seems to be lacking severely here. We've defined a basis on a set S as a collection $\mathcal{B}$ of subsets such that 1.) For each $x \in S$ there is some $B \in \mathcal{B}$ such that $x \in B$ and 2.) If $x \in B_1 \cap B_2 $, where $B_1, B_2 \in \mathcal{B}$, then there is some $B_3 \in \mathcal{B}$ such that $x \in B_3 \subset B_1 \cap B_2$, and I'm told that we define a topology generated by $\mathcal{B}$ to be the collection of all subsets U of S such that for all $x \in U$, there exists some $B \in \mathcal{B}$ such that $x \in B \subset U$. Now, I have to prove that this collection of sets is a topology, but I'm for some reason really lacking on how to show this

[gentle osprey]

J.Ross

[small hemlock]

The empty set is open in S vacuously, and S is open in S because if x is in S, then there exists some B in B* such that x is in B by definition

But for finite intersections and arbitrary unions, I think I'm just not thinking clearly enough

[woven sundial]

if you take an arbitrary union of B, will it be of the form U as given?

[small hemlock]

It must, I suppose, since any element of the union is in Some B in the basis, and thus satisfies that openness requirement?

[woven sundial]

yep

[small hemlock]

And I guess for finite intersections, that's just by the definition of a basis, that if there is an element in the intersection of two basis elements, there is a new basis element contained in the intersection such that it contains the element, and thus satisfies that openness requirement

Wow, sometimes I really just have to sound things out

[broken mica]

Let U be the union of open balls such that each open ball intersects U without the ball. Is there a theorem proving that the boundary of U is a set of jordan curves?

please @ me

[hard wind]

Both me and the professor are a bit confused on when to used reduced homology with relative homology. In the definition of relative homology, they give this long exact sequence of homology groups. However, when applying it to find the homology groups of a sphere, it randomly swaps to reduced homology. Huh?

How do we know we can use reduced homology groups here?

[hard wind]

Interesting, Hatcher did not introduce homology of a triple

I'll take a look, thank you

[gentle osprey]

Joe 1

[bitter smelt]

What would you want f’ to be so that the second space is obviously contractible?

Then prove f and f’ are grocery

Yep, I obviously meant grocery, thanks Apple. htpc*

[candid hedge]

show that if E a normed vector space has a finite dimension and A is a closed set of E that for all x in E there exists a point y in A such , d(x,y) =d(x,A)

[shadow charm]

Question i asked a bit back already, but if I’m asked to verify that $f\cong g$ implies $f_\ast\cong g_\ast$ for induced homomorphisms of reduced homology groups, what is here the definition of the induced homomorphism $f_#:\mathbb{Z}\to\mathbb{Z}$ (and similarly for $g_#$)? I see no obvious way to define this other than taking the identity, in which case the question is a trivial consequence of the same result for standard homology groups

[gentle osprey]

𝓛ittle ℕarwhal ✓

[silk tapir]

Given a based homotopy between $\gamma * \eta^{-1}$ and the constant loop at $x_0$, how can one construct a homotopy between the based loops $\gamma, \eta$?

[gentle osprey]

adi

[candid hedge]

is the transpose function continuous on Mn(R)?

[swift fjord]

What do you think?

I believe you could argue for/against this yourself

There's several arguments that would work here

[gritty widget]

It just switches few coordinates around. So of course

[unreal stratus]

lol these are pretty much exactly in our problem sheets too

just they are two parts of the same question

[unreal stratus]

Well note the attaching map is just a map S^1 -> S^1 in this case, right. So you just need to show that this is homotopic to the identity map S^1 -> S^1

So really you just need to visualise what the attaching map is

I'm viewing the dunce cap as being what you get from attaching a 2-cell to the circle

which I think is a bit clearer

Though if you do have a map S1 -> [0,1], then it's nullhomotopic anyway and so you're pretty much done

Hm I think it was mostly intuition and the fact it works lol

Wdym

I'm just imagining like you take a circle and then glue a 2-cell in such a way that you have the right identifications on the boundary

I don't fully get your CW complex structure

Oh okay hm

I wasn't thinking of it geometrically like that at all aha

I was just thinking of taking a circle and using an attaching map which gives the right identifications (without drawing any pictures rly)

Yeah, that's how I meant, though I don't think that picture is enough to describe the attaching map

Well the attaching map I had in mind is one that loops round the circle a few times in different directions, to give the necessary identifications

To get you started, I mean that as you go through the first third of the circle in the codomain, its image goes round the entire circle (that identifies two of the points)

Yup, exactly! Which is just the identity map S^1 -> S^1

And so really all you've got is a disk, up to homotopy equivalence

[queen shale]

it's weird how we can make separable spaces but we cannot cut manifolds

"haha snippity snip"

but when i try to take a scissor to a manifold

"DEATH BE ON YOU"

[vague brook]

realized you can kinda visualize fundamental group of circle by looping headphone cable around a number of times and then plugging it in. And you can't change the number of loops on the cable without unplugging

[shadow charm]

There’s not much that needs to be said : D^2 x [0,1] def retracts onto D^2 x {0} u S^1 x [0,1] and onto D^2 x {1} u S^1 x [0,1], si you get an induced deformation retract by fixing X and just working within the attached cylinder (it works because you’re fixing the borders of the cylinders which are the only parts that get attached to X)

[swift fjord]

The question of the level of detail is one you should probably ask your prof/TA

Not really something we can answer as we don't know their rubric

If you're just self studying you can be as formal or nonformal as you like as long as you understand what you're doing

[shadow charm]

Yeah this is the correct answer actually

[swift fjord]

Usually the explicit construction of the homotopy is superfluous

Imo

Unless it's very nonobvious

[wise ruin]

In this problem, what exactly is meant by the two topologies being the same - their underlying sets aren't even the same?
Is it meant that the natural bijection between the two is not an identification?

[gritty widget]

We can naturally consider tau_1 as a topology on V

If you have the disjoint union of V_1 and V_2 consider function which injects V_1 and V_2 in that disjoint union to V

then after quotiening, this becomes a bijection

they basically mean to say that this bijection is not a homeomorphism

[unreal stratus]

Lol do we just have the same problem sheets

[unreal stratus]

Yeah true

[unreal stratus]

Okay so @uneven hawk I've been writing up my own solution - the key idea is to consider the maps on fundamental groups that the inclusion/retraction would induce and think about the generators of fundmaental groups of the the boundary of the strip vs of the strip

Your intuition about looping once vs twice is the key idea actually, I believe - think about how that corresponds to the maps between fundamental groups!

(and thanks for reminding me to do this lol)

So letting $X$ be the Möbius strip, the inclusion $ \partial X \hookrightarrow X$ induces a homomorphism $ \mathbb Z \xrightarrow{\cong} \pi_1(\partial X) \to \pi_1(X) \xrightarrow{\cong} \mathbb Z$

[gentle osprey]

potato

potato

[unreal stratus]

And then of course as Z is cyclic, it suffices to consider a generator and consider its image

[wise ruin]

This is from my prof's lecture notes. Is this not false? For if it were true then every set is open as X \times Y is a cover of that set.

[gritty widget]

They mean that it's union of such sets

should be "is a union of"

[wise ruin]

gotcha

[unreal stratus]

Ah, well one way to do it is to show that X retracts onto the 'equator', which is homotopy equivalent to S^1

Yes

Yeah, it's like a particularly nice form of homotopy equivalence

I hope what i wrote above is clear enough?

Well an injective homomorphism, sure

But any group map Z -> Z besides the 0 map is injective, so that doesn't tell you much - that's why you should think about what the map is more concretely (e.g. in terms of generators)

Not quite actually. Once you have a description of the map Z -> Z induced by the inclusion, it's obvious no such r_* can exist

Indeed

and when do they admit a right inverse?

Can e.g. multiplication by 2 on Z be undone?

Yeah

Yup

So if we can show that the map Z -> Z induced by inclusion doesn't send a generator to a generator then we're done

(Tbh this is a cool question as it's more delicate than "different fundamental groups hence not homotopy equivalent" lol)

Yes (under the obvious identifications between Z and the relevant fundamental groups)

No, what you just said is the key idea

Formalising your statement finishes up the problem aha

Yeah exactly

So the smallest non-trivial loop is actually a generator of π1( boundary X), which isn't too hard to see if we view the boundary as a copy of S^1

Now you need to show that its image in π1(X) "loops round twice" by which we mean it's double a generator

So what I did was: pick a basepoint of boundary X and use the same one for X. Fix a loop L with [L] generating π1(boundary X) and a loop M with [M] generating π1(X).

Now the image of [L] in π1(X) is again represented by L, and this is homotopic to some composition of M with itself (since [M] is a generator!)

If you pick L and M nicely enough, it's clear that L is homotopic to M * M (relative to endpoints), or the inverse thereof depending on orientations. This then shows the corresponding map Z -> Z is multiplication by 2 (or -2)

Sorry, this is a bit hard to give hints for without just drawing a picture... which would spoil the problem entirely

Yeah exactly

Well I drew pictures to show what I was using as L and M

and to show a homotopy

Don't think so

The solution is actually pretty short once you write it up which is nice

np!

I mean I only wrote it out like an hour ago or so so it's fresh lmao

[shadow charm]

Anyone have a hint on how to find a basis for $H_1(\mathbb{R},\mathbb{Q})$? I’ve already shown it’s free abelian using a long exact sequence, the fact that $H_0(\mathbb{Q})\cong\bigoplus_{i=1}^\infty\mathbb{Z}$, and that subgroups of free abelian groups are abelian. Finding a basis I’m not so sure about

[gentle osprey]

𝓛ittle ℕarwhal ✓

[shadow charm]

I’m thinking it might be easier to visualize by looking at $\tilde{H_1}(\mathbb{R}\cup C\mathbb{Q})$ but idk

[gentle osprey]

𝓛ittle ℕarwhal ✓

[shadow charm]

Is all linear paths between pairs of points in Q too naive?

Now that I think about it

[dim radish]

i remember seeing this somewhere

[shadow charm]

Yeah that should work: take $p,q \in Q$ and $\sigma_{p,q}: I \to \mathbb{R}$ the path (1-simplex) $\sigma(t) = (1-t)p + tq$. Then $\sigma_{p,q}$ is homologous to $\sigma_{p’,q’}$ iff $p=p’, q=q’$

[gentle osprey]

𝓛ittle ℕarwhal ✓

[dim radish]

i think you were supposed to look at the long exact sequence and find it from that

[shadow charm]

Then these form a basis of representatives for $H_1(\mathbb{R},\mathbb{Q})$?

[gentle osprey]

𝓛ittle ℕarwhal ✓

[shadow charm]

To find the basis?

[dim radish]

yeah

[shadow charm]

The problem is I don’t know how the injection $H_1(\mathbb{R}, \mathbb{Q})\longrightarrow \bigoplus_{i=1}^\infty\mathbb{Z}$ really looks so I can’t just take preimages of the generators of the latter right?

[gentle osprey]

𝓛ittle ℕarwhal ✓

[dim radish]

you know that H_1(R, Q) is isomorphic to the kernel of the map from \bigoplus Z -> H_0(R) = Z

[swift fjord]

oh god I thought this was group homology for a second

[shadow charm]

Fair but that’s also a map for which it is unclear how the quotienting process is done if you get what I mean

I’ll think about it though thanks

[shadow charm]

Group homology looks kinda lit

[swift fjord]

it is pretty lit

[shadow charm]

Ah no you’re right since this is induced by inclusion it’s easier to work with

[dim radish]

i think it maps individual generators of the Z in the direct sum to 1

which is pretty nice

[shadow charm]

Yeah each generator gets mapped to the homology class of a point in R

So yeah I just have to look at the pairwise differences of the generators

That’s just a generating family though you should get a basis out of it by fixing some basepoint or something

[dim radish]

i can believe in that

[shadow charm]

Thanks for the help 🙏

[dim radish]

lol i remember not getting this exercise in my algtop course

[unreal stratus]

It's funny when there's like one major textbook for a topic and so the same problems crop up

[hidden crag]

Which textbook is it? (The one the exercises you and Joe are doing are from)

[unreal stratus]

Oh I mean the one narwhal was doing was Hatcher as you may know, but Joe and I were just doing uni psets which happened to share problems

[hidden crag]

Ohh okay I see, they look interesting

[unreal stratus]

Sure, tho now I'm scared I'll have made a mistake xd

[unreal stratus]

Ok i wrote it out more properly and it is longer than I expected lol

[unreal stratus]

@uneven hawk

[swift fjord]

Potato at least use camscanner

[unreal stratus]

I hope it's readable lmao, hard to know what the best way to draw paths is

[unreal stratus]

Ok mb

bruh says camscanner is 45 a year

[swift fjord]

I find this a bit hard to read

?

Camscanner has a free version

[unreal stratus]

Oh okay

Okay yes I mean I need to write it out more nicely ugh

[swift fjord]

Also if you sign up with ur student email you should get free premium

It's mostly the angle and the light it's a bit hard to read

[unreal stratus]

Just not sure how to give lots of detail without it being a couple of pages lol

Ah okay yes

[unreal stratus]

cool :)

Yeah I think it's weird to write stuff like this properly because I want to give lots of details (e.g. more details on the paths and all) but it'd be a pain right

Also wonder if it can be compressed into a page or less...

[long grail]

In a Hausdorff space, every finite point set is closed. So its complement is open. Is this enough to show an order topology is Hausdorff?

[gritty widget]

No, that only shows it's T1

Being Hausdorff is T2 which is a stronger axiom

Or are you asking, if order topology is T1, is it T2

[long grail]

I see. I got it twisted

The question is: if it's an order topology, Show it's Hausdorff

And I have the theorem every finite point set in Hausdorff is closed

[gritty widget]

You just need to take two points a < b and consider two cases

[long grail]

Ok

[gritty widget]

1. there is c with a < c < b
2. there is no such c

[long grail]

Oh

So..

Let me think about it. Thanks

[gritty widget]

👍

[long grail]

So I need to find disjoint neighborhoods around a and b. So in the first case is it sufficient to say c is in the closure of an open set containing a, and c is in the closure of an open set containing b? Basically I'm using {c} as a boundary to define disjoint open sets

[vague kite]

i dont know if this is the right channel but does anyone know if this is a knot?

sorry for bad drawing

[vague kite]

is it feasible for me to do that if idk shit about topology or knot theory

[bitter smelt]

no

but barrier for entry for knot theory isnt so high

its a popular reu topic

[vague kite]

thank you!!

im preeetty sure its a prime knot according to this https://mathworld.wolfram.com/SatelliteKnot.html

[wanton terrace]

Hey Im learning about the Zariski topology in the context of affine sets, and I'm a bit confused on its definition and how it relates to an example. The example is taking an affine space to be the set of all 2x2 matrices. The claim is that the set of invertible matrices is open, proven by showing its complement, the set of non-invertible matrices is closed. What I dont understand is how one proves this using the definition of the zariski topology.

Closed sets are subvarieties according to the zariski definition, so is it enough to say that the set of non-invertible matrices is closed, because it is a subset of matrices with determinant greater than or equal to 0?

[sturdy notch]

The set of non-invertible matrices is the set of matrices with determinant 0

[gritty widget]

The order topology is generated by open rays. If such c exists then we have two rays {x < c} and {c < x}, they are disjoint, a belongs to one, b to the other.
If such c doesn't exist then we can simply consider {x < b} and {a < x} instead.

[wanton terrace]

@sturdy notch Yes, I know. Hence it being a subset of the set of matrices of determinant greater than or equal to 0. Perhaps I should have been more clear in my phrasing

[sturdy notch]

Right, but why are you looking at the set of matrices with positive determinant, a priori your field does not have an order?
The determinant is a polynomial map from k^nxn to k for some fixed field k, so that might help

[gritty widget]

Yes, it's the right channel, knot theory is very much topology.

[wanton terrace]

I'm just trying to figure out the definition of a Zariski topology. Did you read my post? Im not sure if it being a subset is enough to say that it is closed under the zariski topology or not