point-set-topology

[unreal stratus]

Well like I would do it all kinda recursively basically ye

Like okay you have a small ball about γ(0), use that to correct the path on like [0,ε] to a straight line, then keep going

[pseudo coral]

oh i see

so I AM picking an epislon so that the difference of any two points in the subdivision are less than that radius

[unreal stratus]

ye ig lol

[icy zenith]

Maybe a stupid question but can we write $$[0,1] = \bigcup_{i \in I} [a_i,b_i] $$ where I can be infinite ?

[gentle osprey]

Julien J

[gritty widget]

take all a_i to be 0 and all b_i to be 1

[icy zenith]

sorry forget to mention that a_i and b_i cannot be 0 and 1 respectively

[gritty widget]

well no then the union won't contain 0 or 1. at best you'll get the open interval (0, 1) (by taking, for example, a_n = 1/(2n) and b_n = 1-a_n)

[untold lily]

don't think you can, unless I'm missing another pathological counterexample

if the point 1 belongs to the union on the right, it has to belong to one of the sets being unioned over

no way around that

[icy zenith]

yeah seems very logical

I am trying answer this by an argument using simpley-connectedness of [0,1] but I am not really getting anywhere... Can someone give some help ?

[gritty widget]

is the torus simply connected?

this is overkill btw

you can give a much more elementary argument. my favorite trick is to think about what you have to remove from one space to disconnect it but not the other

[icy zenith]

I am trying to show that it is not but I am not succeeding

[gritty widget]

so do the easier argument

[icy zenith]

Mmmh I can remove {1} from [0,1] but not from S^1 ?

[gritty widget]

those aren't the spaces you're working with, though

you're working with [0, 1]^2 and (S^1)^2

[icy zenith]

can we get the same argument with the product or something similar ?

[gritty widget]

i suggest you argue with pictures

[hidden crag]

maybe drawing a picture helps

[gritty widget]

lol

[hidden crag]

oh come on

[gritty widget]

great minds think alike

[hidden crag]

i already deleted two of my messages and reacted with instead because you said it while i was still typing

and now another snipe

[stark fog]

think of an open set of [0, 1]² as a subspace of the product topology of R² and see if it can be mapped by a continuous function to S1 x S1 (if the resulting set will be open as well)

[hidden crag]

[stark fog]

dunno if it helps

[gritty widget]

i'm on mobile too timo

just typing too fast i guess

[gritty widget]

there obviously exists a continuous map from [0, 1]^2 onto (S^1)^2, so i don't see where you're going with this

[0, 1] has one onto S^1 so take squares

the question is asking whether the spaces are homeomorphic; whether any such map can have a continuous inverse

[icy zenith]

A possible open subset of [0,1]^2 can be [0,1) x [0,1) right ?

[stark fog]

sorry, but I was thinking that maybe we need f(U) open whenever U is open in X for continuous f to be a homeo

[hidden crag]

I suggest you think more about what tterra said

that's definitely the easiest way to do it

[gritty widget]

i'll elaborate. [0, 1]^2 is a square and (S^1)^2 is a torus. what can you remove from the square to disconnect it? what if you remove that from a torus?

[hidden crag]

i saw that tterra

[stark fog]

same

[gritty widget]

[icy zenith]

I can remove the point (1,1) for example for the square and a circle for the torus

[hearty jacinth]

Ig it's sufficient to show [0,1] is not homeomorphic to S^1?

[hidden crag]

that's basically the same thing

[gritty widget]

that's not going to disconnect either of them

so it says nothing about the spaces

[icy zenith]

ah yes you are right

I can remove 2 points then

[gritty widget]

nope

[0, 1]^2 is a filled in square, it's not just the boundary

what do you need to remove to disconnect this thing?

(pretend it's completely filled in)

[icy zenith]

oh I am tired I forgot this is filled 😅

I remove the diagonal for example nah ?

[gritty widget]

that works

so you can disconnect [0, 1]^2 by removing a line. can you disconnect the torus by removing something homeomorphic to a line?

[icy zenith]

no because we will need 2 circles to disconnect a torus ?

one circle does not disconnect our torus

[gritty widget]

you could disconnect the torus with just one circle

just have it cut off a little part of the surface

[icy zenith]

how ?

[stark fog]

yeah, that's what he meant with cut

[gritty widget]

rather than winding around the whole thing

but the solution i have in mind is indeed that removing a line (something homeomorphic to a line in R^2) will not disconnect the torus

[gritty widget]

you can think of it as the same solution to this, but everything has the dimensions up by 1

sorta

maybe.

[icy zenith]

yeah I understand thank you

[gritty widget]

if you want to use simple-connectedness, just note that the torus has some loops that can't be deformed to a point on the torus

in symbols: pi_1(square) = 0 and pi_1(torus) = Z^2

[icy zenith]

yeah okay I get it

To be sure (0,1) x (0,1) is open in [0,1]^2

[gritty widget]

true but i don't see how it's relevant

[icy zenith]

no sorry just an independant question nothing with the previous discussion

[gritty widget]

ah

yeah it's true. it's the intersection of an open set in R^2 with [0, 1]^2

[halcyon dock]

i remember that $\sup A+\sup B=\sup(A+B)$, but it won't necessarily be true that
$$\sup_{x\in\bR}|f(x)-g(x)|+\sup_{x\in\bR}|g(x)-h(x)|=\sup_{x\in\bR}(|f(x)-g(x)|+|g(x)-h(x)|)$$
right? can we at least say that it's $\geq$?

[gentle osprey]

nilpotent nix

[stark fog]

I think that's the triangle inequality on a space of bounded functions with uniform norm

[halcyon dock]

yeah i'm trying to prove d is a metric on bounded real functions but i'm not sure about my proof

[stark fog]

Yeah, the other two properties are trivial to check

[untold lily]

yes you can say it is greater than, the sup of the stuff you are taking over is a subset of the original one

[stark fog]

maybe this do the trick, but yeah it is good to make sure whether the equalities or inequalities hold for all functions

[untold lily]

I mean greater than in here is correct

[halcyon dock]

okay great thank you both! @untold lily @stark fog

[warm hedge]

maybe stupid question but whats the difference between D2 disk and any B disk ?

[plain raven]

It depends on context.

[warm hedge]

I will try to figure it out from the rest then

[keen cliff]

probably D2 centered at origin whereas B can be centered anywhere

[warm hedge]

oh ok, yeah it could make sense if it means it like that

thanks

[stark fog]

or maybe D² for the unitary disk centered at the origin, then B any disk of any radius at any point

[bitter estuary]

Is the union of D_1(0) with D_1.1(2) open? It's from complex analysis but is certainly a topology question, which I'm not super versed in. D_1(0) is a disk of radius 1 centered at 0, if notation is not clear.

[gritty widget]

any union of open sets is open

so if these mean open disks, you're good

[bitter estuary]

Ah yeah of course

Thank you

[nocturne basalt]

in Hatcher's description of RP^n as a cell complex, he first identifies it with S^n/(v ~ -v) and then says that this is equivalent to the quotient space of a hemisphere D^n with antipodal points of its boundary identified. how does he get from the first to the second? isnt a hemisphere of D^n equal to saying the last coordinate is greater than or equal to 0 (when embedded in the obvious way) so then antipodal points of its boundary would only lie on the "flat part" that is D^{n-1}? and if so how does this equivalent to the quotient space from earlier?

[coarse night]

you are discarding the lower open hemisphere

since it's being quotiented out anyway

and identifying the remaining part of the upper hemisphere, which is the boundary D^n-1

Another way to think about it is you are quotienting in 2 steps, first when the last coordinate is <0, this step will leave you with a closed hemisphere. Then you quotient when the last coordinate =0. That's exactly the boundary

[unreal stratus]

Another description is that the map D^n -> S^n/(x ~- x) is a surjective closed map (clearly surjective and closed as domain compact, codomain hausdorff, say) and it identifies points iff they are antipodal on the boundary of D^n

Also more generally we are taking a closed saturated subset ig so it should work

[gritty widget]

How can you write the lein bottle as an S^1-bundle over S^1

[empty grove]

Take the mobius strip as a line bundle over S¹ and 1 point compactify all the fibers

[gritty widget]

that gives RP^2

[empty grove]

oh

Right

Well you know the polygon you use to glue the Klein bottle

Hold up I forgot which way the gluings go

Like one pair is the same way and other the wrong way right?

[gritty widget]

yes

[empty grove]

Hmm then I don't see why this don't work

But you take that polygon

And you treat one of its edges as the S^1

[gritty widget]

the 1-point compactification is the same as the Thom space (for compact base spaces)

and the Thom space of the mobius band is definitely RP^2

[empty grove]

And projection onto that edge as the bundle map

[empty grove]

Thom space also identifies all the points at infinity

[gritty widget]

yes

but for compact base spaces

you don't need to first do it fiberwise

you can just 1-point compactify the total space

but yes in general you should compactify fiberwise first and then identify the points at infty

[empty grove]

Yeah but I'm 1 point compactifying fibrwise

Does that give the same thing?

Really doesn't seem like RP² to me lol

[gritty widget]

are you 1-point compactifying fiberwise

[empty grove]

I don't think RP² is even a circle bundle over S¹?

Not sure

[gritty widget]

but then not identifying the points at infinity

no RP^2 is not

RP^2 is the Thom space

[empty grove]

Ye

[gritty widget]

this will still give RP^2

[halcyon dock]

so a function is continuous iff the preimages of an open set in the codomain is open, right? then how does one negate that definition? like would that mean there is an open set in the codomain that has a preimage that is not open?

[ornate berry]

Yes

[stark fog]

yes, for example a function from the real numbers with standard topology to the real numbers with the lower limit topology

[pseudo coral]

so im still trying to show R^n \ {0} for n > 2 is simply connected. For the path connected, if I pick two points that are not multiples of each other then i can use the straight line homotopy right? And for two points that are scalar multiples of eachother, can I pick a third point which lies say above the origin, then straight line homotop from the starting point to this third point then to the terminal point? so in R^3 say (x1,y1,z1) and (x2,y2,z2) are two points that are not scalars of eachother then the path t(x2,y2,z2)+(1-t)(x1,y1,z1) works , if however theyre scalar multiples then 0 is between them then can i straight line homotop (x1,y1,z1) to say (x1+\epsilon,y1+\epsilon, z1+\epsilon) then from this point to (x2,y2,z2)?

running 2t from point 1 to the epsilon point and 2t-1 from the epsilon point to the 2 point?

or should I map (x1,y1,z1) to (\eps,\eps,\eps) then to (x2,y2,z2) as (\eps,\eps,\eps) is just hovering above the origin

[pearl holly]

just do the linear homotopy thingy

[pseudo coral]

but what if the two points are scalar multiples of one another, then 0 lies on that line segment between them..

so do i straight line it from (x1,y1,z1) to (x1+\eps,y1+\eps,z1+\eps) then to (x2,y2,z2)?

[pearl holly]

idk man tbh I'm too tired to read the notation, idk why I responded in the first place I'm sorry

[pseudo coral]

lol its cool

[pearl holly]

tbh I don't see what the problem is. For path connected, just do a line. For trivial fundamental group, just do a linear homotopy

[pseudo coral]

sorry i meant R^n \ {0} where n >2

[pearl holly]

oh bruh

[pseudo coral]

lol sorry

does my idea work up top ^^

[pearl holly]

I mean I guess that works for path connected? Like you pick a third point, like you said

for the simply connected part I would look at what R^n\0 is homotopy equivalent to

[unreal stratus]

For the path connected it suffices to do n =2 and then it is an easy proof by pictures

😎

[pearl holly]

yeah I mean fr, just draw a picture

imagine doing explicit homotopies

[unreal stratus]

Or even cooler: C^x is path connected as it is the image of exp: C -> C^x

Lol

[pseudo coral]

i wanted to show simply connected just by brut force if possible by having the loop trace along the boundary of some epsilon ball about the origin as the path nears the origin, i know that R^n \ {0} is homotopy equiv to S^n-1 x (0,\infty) and the fund group of prods is the prod of fund groups

[pearl holly]

oh okay yeah I'm sorry

[unreal stratus]

Lol

If ur using homotopy then just use 0,infty contractible too

[pseudo coral]

so my idea for path-conn works ? have the first point get mapped to some point plus epsilon of original point?

and then to the terminal point

[unreal stratus]

I don't think doing simply connected by hand will be too tractable

[pseudo coral]

lol

[unreal stratus]

Mostly because it is hard enough for spheres and those r htpy equivalent to these

[pseudo coral]

does my idea work for path-connectedness using a third point and labeling it (x1+\eps,y1+\eps,z1+\eps)

[hollow geyser]

Question, can two closed (nonempty) sets form a partition of a connected topological space? This isn't hw, just curiosity related to something else.

[gritty widget]

what's the definition of a connected topological space?

[hollow geyser]

Ah I had forgotten

Thanks. I feel silly now

[gritty widget]

it happens to everyone

[gritty widget]

If there exists a homotopy of paths between f1 and f2, how does one show there exists a homotopy of paths between their inverses?

[versed geode]

Can you express the inverses in terms of reparameterizing the original function?

[gritty widget]

I see, precomposing the homotopy with the parametrization yields the required homotopy.

[versed geode]

Uhh what do you mean by that

[gritty widget]

So if H is the homotopy between f_1 and f_2 and since f_i^(-1)(t)=f(1-t), then H composed with p gives a homotopy between the inverses, where p is the reparametrization p(t)=1-t

[versed geode]

Yeah

[odd flame]

can i get some pointers on this, im not sure how to see this

tbh im not even sure i can really see a projective plane

[versed geode]

Do you know the fundamental polygons?

Try the m,n=1

Also, have you learned the classification of surfaces?

[odd flame]

this is chapter 1 of massey's book

which covers that

so im still grappling with it

if i can bother you, can we talk through what a projective plane is first

the text has this but i dont see how this corresponds to the standard form abab

[hollow wigeon]

@odd flame

If you want, I can help you.

I am a mathmatician.

[odd flame]

i mean that would be nice sure

[odd flame]

im assuming this is different than the platonic solids?

[hollow wigeon]

So you didn't understand the above text?

[odd flame]

are you referring to the projective plane one?

[hollow wigeon]

sure.

first Do you understand the homogenous

concept?

[odd flame]

do you mean homeomorphic?

if not then no

[hollow wigeon]

No

[odd flame]

then wdym

[hollow wigeon]

Do you have the class and classification concept?

[odd flame]

all i have is classification of surfaces

S^2, T # ... # T, and P # ... # P

and the notions of triangulation and the euler characteristic

[hollow wigeon]

okay.

I will explain.

if two points are on the same line through the origin, we will say that they are in the same class

so

in mathematical expression.

got it?

[odd flame]

[hollow wigeon]

so when we think about the RP1, we can think the circumference with the radius 1.

Do you know what the RP1 mean?

@odd flame Are you there?

[odd flame]

yeah im here

trying to read the book at the same time is all

i havent seen that notation but i probably know what it is

[hollow wigeon]

I will explain.

it is easy

1-dimentional Real projectory space

[odd flame]

i can guess what that is yeah

[hollow wigeon]

when we consider RP1, we give the 'homogenous relation" in R2 (2-dim real linear sphere)

If two points are on the same line through the origin, we say that they are in the "homogenous relation"

got it?

[odd flame]

[hollow wigeon]

Good.

And we consider all the points in the "homogenous relation" as one point.

got it?

[odd flame]

so it suffices to consider one hemisphere right

[hollow wigeon]

yeah, but not exactly.

First let's consider about the Rp1

so the semi-circumference

okay?

with the radius 1.

[odd flame]

okay

[hollow wigeon]

thanks.

[odd flame]

yeah i see that

[hollow wigeon]

Do you understand my image?

we have to identify the (1,0) and (-1,0).

Because they are in the same class.

[odd flame]

same with every point along the equator too right

or well, every class i guess

[hollow wigeon]

yeah.

so in the case of Rp2, as you know we think the semi-hemisphere.

but we have to consider about the circumference on the xoy plane.

[odd flame]

[hollow wigeon]

you got it?

It looks like the wallet with zip.

[odd flame]

i follow so far

[hollow wigeon]

So Do you understand the whole meaning?

[odd flame]

oh is that it>

i must be overcomplicating it then

[hollow wigeon]

Sorry but what do you mean?

[odd flame]

i'll be right back i wanna write some stuff down

[hollow wigeon]

pls take your time.

see you.

[odd flame]

what do you think about this description

also @hollow wigeon im feeling better about the projective plane

could we go back to the question from before

[hollow wigeon]

So Are u going to understand above description?

let me see...

[odd flame]

i do

i think i was just overcomplicating it, walking through it slowly helpde

[hollow wigeon]

so you want me to make it more simple

right?

[odd flame]

this was the original question

i know that a torus summed to a projective plane is P # P # P (let P be a projective plane)

but im not sure about n,m many if that makes sense

[hollow wigeon]

It is somewhat complex.

Are you a student?

[odd flame]

yeah

undergrad

[hollow wigeon]

which university?

Do you know what the result of sum of 2 tori is?

[odd flame]

lurking timo

i wont doxx myself for now, math dept is small and very googleable

[odd flame]

isn't it just a double torus

or one w two holes

i dont know the formal name for it but i know what it looks like

[hollow wigeon]

pls say.

I think this problem is too difficult for you.

now.

[odd flame]

nah

i can do it

[hollow wigeon]

Okay.

Sorry.

[odd flame]

what does it have to do with what uni i attend

[hollow wigeon]

Then pls say first.

what is the result of sum of tow toris?

what is the result of sum of two projective planes?

[odd flame]

i know the two tori give a "double tori"

i have the planar model for it in my notes

and i know the sum of n projective planes is represented by a 2n-gon

[hollow wigeon]

Do you know RP2#RP2=K?

K mean klein bottle.

[odd flame]

yeah

i have it in my notes as well

[hollow wigeon]

Good.

In topology, Imaginary is very important because it is the abstract math branch.

I will right back.

[odd flame]

@hollow wigeon could it be enough to use that T # P = P # P # P

if you have m tori and n projective planes

[hollow wigeon]

Maybe,

It is also a challenge for me.

So Do you master the maths?

[odd flame]

god i wish

but ok if m = n

you get 3n projective planes summed to each other i think

connected sums should be commutative

if we have more tori than projective planes, m > n

you still just get some connected sum of projective planes

[hollow wigeon]

I think you misunderstand the problem.

[odd flame]

possibly

[hollow wigeon]

we have mtori T#T#T#T#T.....

and P#P#P#....

And we are going to glue two objects now.

right?

[odd flame]

it's asking what the connected sum of those two surfaces is, no?

isn't that just the connected sum

oh i guess not necessarily

oh i think my picture is wrong too actually

T#P should be a 6-gon

[odd flame]

is the act of "gluing" not necessarily the same as a connected sum

[hollow wigeon]

Maybe, but I think we have to think about the order.

Do you think T#T#P#P is equal to T#P#T#P?

[odd flame]

yeh

im 99% sure it's commutative

[hollow wigeon]

I don't thinks so now. I think you have to check about that.

[odd flame]

https://mathoverflow.net/questions/87705/connected-sum-of-surfaces

[hollow wigeon]

As a mathematician, we have to consider all the possible conditions, even the 1% possiblility.

[odd flame]

it's commutative

[hollow wigeon]

Okay. thanks.
My major is the Algebra and geometry.

not the topology.

But I am interested in the topology.

[coarse night]

[odd flame]

ryu

sun salutations

[coarse night]

[odd flame]

this is the problem if you wanna take a look

i think it's just different numbers of copies of P summed to each other tho

https://tenor.com/view/fall-kirby-drop-falling-deep-gif-17081698

ryuuuuuuuuuuuuuu

[empty grove]

Why though

[odd flame]

holy shit moldi

[empty grove]

sub seppp

[odd flame]

wanna look at my shitass question pls

[odd flame]

.

[empty grove]

Nah looks shitass

[odd flame]

i think it's just P # ... # P a certain number of times depending on if you have more tori or more projective planes

but that feels too easy

[empty grove]

idk

[odd flame]

https://tenor.com/view/crying-men-four-crying-men-crying-men-anime-crying-meme-crying-men-meme-gif-26190576

embed failure moment

and now it embeded?

ok discord

alg topology is proving hard moldi

[empty grove]

I feel like it is the Klein bottle because if you start with the mobious band and glue the "opposite edges" together you get the Klein bottle, right? And that's what fibrewise 1-point compactification is doing

[empty grove]

It is so easy

[odd flame]

are you saying that because youre cracked

[odd flame]

$\amogus$

[gentle osprey]

sebbb

[empty grove]

@gritty widget https://en.wikipedia.org/wiki/Klein_bottle#3D_pinched_torus_/_4D_Möbius_tube this is what I am seeing lol

[coarse night]

Why not working for me

[odd flame]

lol

[coarse night]

Is that a custom command?

[odd flame]

i can send definition if you want

yeah

[coarse night]

Sure lol

[odd flame]

tex devs did not, in fact, write an amogus command into their language

here or dm?

[coarse night]

Anywhere is fine

Maybe #latex-testing

[odd flame]

i just wanna check this

for this problem

[coarse night]

[odd flame]

i know T # P = P # P # P

so i should just be able to use that as much as i want right?

no worries if you dont know

[coarse night]

I forgot is RP² orientable?

[odd flame]

it is not

[coarse night]

I had done a similar ex, so the only orientable surfaces are surface with n-genus, and non orientables are RP²#RP²#... so if you attach one RP², you'll get a non orientable one so it's otta be one of the form RP²#...#RP²

something similar to that

I cannot prove the classification theorem but what you can try is draw the n-gon and try to cut paste some sides, see if you get RPⁿ s

[odd flame]

my general idea was to start from T # P = P # P # P

[odd flame]

we did it in class so i dont have to do that

[coarse night]

that's how I did that so you'll eventually get it as well

[odd flame]

but then here

if you sum another P it's obviously just more projective planes

i use P it's easier lol

but adding another T you can just rearrange bc it's commutative

imma just go with that

,ti

[gentle osprey]

The current time for stμ₂dying is 03:27 AM (EST) on Tue, 31/01/2023.

[coarse night]

ya probably do some cut-paste on the identification diagram

[odd flame]

"proceed by induction"?

also what

what is the difference here

oop nvm

[odd flame]

isnt 8.9 the torus for even and odd n????

[fiery token]

could someone verify if my proof is correct?

other two definitions used:

note that we have not yet defined open and closed sets

[icy zenith]

Is $\mathbb{Z}$ with the p-adic metric compact ?

[gentle osprey]

Julien J

[fiery token]

haven't defined compactness yet either

(which also means I don't know what that is)

having had a quick look at the wikipedia article and assuming 'p-adic metric' means this:

I'd say no, $\mathbb{Z}$ is not compact with respect to the p-adic metric

[gentle osprey]

thou_art_an_egg

[fiery token]

...hold on

this is a separate question, isn't it?

my bad

[weary herald]

in a metric space, convergent sequences are sufficient to describe continuity: if every sequence x_n converging to x is such that f(x_n) converges to f(x), then f is continuous at x.

Is there a (relatively simple) counterexample for this in a non-metric space ?

[gritty widget]

this holds true in, more generally, any first countable space, so you want to look for counterexamples in spaces which aren't that

[weary herald]

do you know of an example?

[gritty widget]

In X = [0, w_1] any sequence convergent to w_1 must be eventually constant. So if you take f from X to itself with f = id on [0, w_1) and f(w_1) = 0 then for any sequence x_n in X convergent to x, f(x_n) converges to f(x). But f is not continuous

[weary herald]

what is the topology on X?

[gritty widget]

Order topology

[weary herald]

right, haven't seen that

w_1 is the first transfinite ordinal?

[gritty widget]

w_1 is the first uncountable ordinal

[weary herald]

my bad

does a simpler example exist?

something not involving ordinals

I'm guessing not, in which case our professor will not ask us this type of question on our exam

[gritty widget]

nope, but i knew where to look so i hoped that would help

[untold lily]

not sure you can hope for much simpler examples since you are looking for topologies that aren't first countable

[next crystal]

in the last paragraph, why does U x V \subset (X x X) \ diagonal imply that U and V are disjoint?

[gritty widget]

Is a lifted closed loop still a closed loop?

[plain raven]

are you talking about like, in the fundamental group

and covering spaces

if so, the answer is no

[gritty widget]

yes

[plain raven]

yeah so consider the covering space map R -> S^1

[gritty widget]

Ahh of course

It'll just be an open curve

the lifted path, that is

[plain raven]

[gritty widget]

thank you

[paper wedge]

suppose its not

what happens

as in suppose U and V are not disjoint

so there exists an element both in U and V

what would happen inside U x V

[next crystal]

ah that makes sense now
if they shared a common element x, then (x,x) would be in U x V, but then U x V isnt a subset of (X x X) \ diagonal

[paper wedge]

cool\

[weary herald]

why would every sequence in A converge to a point in A?

[unreal stratus]

Wait no mistake ignore

[stark fog]

literally a question in my topology exam today

[next crystal]

I need to prove that if for each pair of points of X, each has a neighborhood not containing the other, then X is $T_1$.\
Suppose the hypothesis and that $x \in X$. Then by hypothesis, for all $y \in X \setminus {x}$, there is an open set $U_y$ such that $y \in U_y \subseteq X \setminus {x}$. So $\bigcup_{y \in X \setminus {x}} U_y \subseteq X \setminus {x}$. I want to also show that $X \setminus {x}$ is a subset of the union, so then we have that $X \setminus {x}$ equals the union of open sets, completing the proof, but i'm not sure how to do this

[gentle osprey]

michαel

[stark fog]

I think this goes by construction. Notice that you're taking a union of open sets containing a point of $X\setminus{x}$ for each point in this set. So this set is contained in the union.

[gentle osprey]

SubGui

[stark fog]

its like you taking a covering by open sets of it

[next crystal]

that makes sense

[stark fog]

(more interestingly but I might be wrong here is that if this open cover had a finite subcover, then X\ {x} would be compact, but {x} is closed because it is its own cover in any topology, so either X is finite or the topology on X is the discrete topology)

[stark fog]

like I am saying this but I didn't even had lectures on compactness

[next crystal]

we're covering compactness in a few weeks

if i remember i'll come back to this then lol

[pseudo coral]

Does anyone know a short proof for this:

[gritty widget]

yes

[pseudo coral]

If C \subset complex is compact and contracible then it’s complement is connected

[coarse night]

One point compactify C then from Alexandar duality
$\tilde{H}_0(S^2 \setminus K) \cong \tilde{H}^{2-0-1}(K)=\tilde{H}^1(K)=0$ since K is contractible

[coarse night]

As short as it gets

ok wait K may not be "locally contractible"

[tidal lynx]

I'm trying to prove that in the set (0, 1), the sequence p_i = 1/i for i > 0 is Cauchy

and to do this, I need to show that for any eps > 0, there exists an N > 0 such that for all i, j > N, d(p_i, p_j) < eps

I chose N = ceil(2/eps), but I'm having struggle with proving that this works without referring to 0, which isn't in the set

If I was able to refer to it, I could just say:
for all i, j > ceil(2/eps), we have that d(p_i, p_j) < d(p_i, 0) + d(0, p_j) < eps/2 + eps/2 = eps

but we can't, so this seems to be an issue

---

so if we were working in R and not (0, 1), the above would be a valid proof. thus, can't I say that because it is cauchy in R, it must be cauchy in (0, 1) as well?

[untold lily]

you can refer to 0 lol

specifically, you are using the induced standard metric from R

so whatever you can deduce from R about distances will hold in the induced metric for (0,1)

[tidal lynx]

but doesn't the induced metric "forget" everything not in (0, 1) ? specifically, the metric is just a partial function of the original metric

[tidal lynx]

or are you saying something like this

1. working in R, we know that p_i = 1/i is Cauchy
2. thus it must also be Cauchy in (0, 1) (removing p_1 of course)

[untold lily]

the induced metric forgets, you don't

call p the metric on (0,1) induced by the standard metric d on R

you know that d(p_i,p_j) < epsilon

by definition, d(p_i,p_j) = p(p_i,p_j)

there is really nothing mystical here, you are working with 2 different functions

unless someone banned you from using d

[tidal lynx]

wait lol

u probably didn't wanna choose p as the other metric

wdym by d_i and d_j

[untold lily]

oh mb

fixed

didn't mean to quote that either

[tidal lynx]

makes sense, thanks

[untold lily]

you can say this but it all comes down to the reasoning above, a sequence being cauchy in a subset depends on the distances between points on that subset

and if you are using the induced metric, these by definition agree

[tidal lynx]

I see

are there any nontrivial complete metric spaces where it's easy to prove completeness without using high powered theorems

ok not even "high powered theorems", I just mean something not as hard as the proof for R

[silk tapir]

What does the identity element of C_n(X) (the free abelian group with basis the singular n-simplices on X) look like?

Will it just be the zero map?

[coarse night]

no

the identity of C_n(X) does not come from a map from Δⁿ → X, it's a "formal" element

[tidal lynx]

am I crazy or does this not just work?

all the proofs I saw on MSE were like three times as long

[unreal stratus]

To me this seems fine lol

[hearty jacinth]

Yeah seems fine

[stark fog]

yeah, like the inclusion map takes an element of the open set and send it to the same element in the top. space. So if you take the pre-images of these open sets, you will always find an open set containing the point in the subspace top. means the intersection with the open set on X with Y

I think so

[coarse night]

Am I missing something here. So every CW complex is paracompact, and there's an obvious CW structure of $\vee_n S^1_n$. But finding an open refinement of the open cover ${ S^1_n \setminus , N() }$ doesn't seem plausible.

[unreal stratus]

(what does N mean here - a small nhd of the wedge point or smth?)

But yes I think you could refine it just by taking some smaller bit of each S^1

So like uhhh I can draw a picture lol to show what I mean

idk if this is clear enough lol

But like you just take out a bigger section of the S^1 \ *

This is clearly an open refinement and there's a neighbourhood of the wedge point which only intersects one open set, i.e. N(*) in ur notation

This should go through without any problems to the general (i.e. n > 2) case

@coarse night

[coarse night]

What’s the nbd of *

also isn't any nbd of * is of the form U1 v U2 v ... where Ui is a nbd of * in S¹_i

so O don't see why that intersects only finitely many of those

@unreal stratus

[unreal stratus]

The red ting is the nhd of 0

I assume that's what you're given

[coarse night]

yes

[unreal stratus]

Yeah sure

So that's why I made the other sets smaller

[coarse night]

but that's wedge of countable nbds of * in S¹

[unreal stratus]

I haven't changed the neighbourhood of *

I just changed the other sets

I can write it more formally if you'd like

[coarse night]

that still intersects nbd of * no?

if you make them too small then it'll stop becoming a cover

[unreal stratus]

Yes, the point is that there is some fuzziness as I showed in my picture ^

sorry if that was unclear lol

[coarse night]

so won't be a refinement

[unreal stratus]

I'll make it better lol

Okay so I mean really all we need to do is do the n=1 case in a sense like

[coarse night]

ok just draw for one circle maybe

[unreal stratus]

If we look at the intersection with each circle, then uh

let's say for ease that N(*) is some open arc of angle φ, say

Then we can replace S1 \ * with S1 \ {closed arc of angle φ/2 about the point}

[coarse night]

oh lmao I've been working with a wrong definition of locally finiteness

[unreal stratus]

Oh lol

Yeah the point is that the point ahs some neighbourhood

not some neighbourhood in the cover

[coarse night]

Been thinking thay N(*) has to intersect finitely many

[unreal stratus]

Ah okay

Well that means this was a good question then!

[coarse night]

I feel dumb now, thanks

[unreal stratus]

No worries, literally just means you were a bit unsure w a def that's calm

And gave me good practice cause I have never rly considered examples lol

[coarse night]

well

[old violet]

When Munkres say we only need to show it for the interval $[0, 1]$, why is that true? How does the general case follow?

[gentle osprey]

Cookieman

[old violet]

Is it because the proof presented works for the interval [a, b] instead of [0, 1]?

[marble socket]

because you have nice homeomorphism between [a, b] and [0,1]

[old violet]

huh

[marble socket]

find f for [0, 1] and then compose this with the map g : [0, 1] --> [a, b] given by g(t) = a+(b-a)t

so in the end you get X --> [0, 1] --> [a, b]

you can then check that this composite sends A and B respectively to {a} and {b}

not uwu?

[old violet]

right sorry, I'm just thinking about how to adapt this into my proof.. This makes sense. Thank you!

[old violet]

Also just a sanity check, the proof presented for the interval [0, 1] does work for all $a, b \in \mathbb{R}$, right?

[gentle osprey]

Cookieman

[old violet]

I don't see how the proof utilized anything special about 0 and 1, except from they make the proof looks cleaner

[versed geode]

Yeah

[versed geode]

This is all

[marble socket]

ig there is a subtlety

endpoints of [0,1] are rational, but for [a, b] this need not be the case

i don't remember exactly how the proof went, but ofc this could be fixed

[old violet]

Ah interesting..

[marble socket]

but yea afair you only needed a countable dense subset which included the end points, so doesn't matter that much, ig having the endpoints isn't really a requirement, but the "same" proof won't work

[split orchid]

Im starting a topology class and we are currently talking about the fixed point property.

We proved that the fixed point property was a topological property and that seemed pretty straightforward, the book then follows this up by asking if a torus has this property, and Im struggling to intuitively understand what this means.

A continuous function f: from the surface of the torus -> the surface of the torus must have a point where x = f(x) right? Im not sure where to even begin for proving this. Are we supposed to unglue the object into a 2d plane and try to answer the question there?

[versed geode]

What examples of surfaces with/without the fixed point property have you covered?

[split orchid]

In class we talked about [0,1] vs (0,1)

Where [0,1] had the property but (0,1) did not because of f(x) = x^2

[versed geode]

Hmm, okay, do you recall the definition of a torus

[split orchid]

I dont think the book covers it, but we did it in class it was something like a 2d plane with X/a gluing it into a cylinder, and then X/b gluing the cylinder into a torus. But idk if thats the actual definition or just a thing we did on the board for intuition or somethsing

[versed geode]

Is there anything you know homeomorphic to the torus?

[gritty widget]

they're asking if every continuous function has a fixed point

but rotate a torus 180 degrees and you see this map has no fixed points

[split orchid]

coffee mug? 😅

[split orchid]

Im not sure what you mean by this

[versed geode]

what blitz said is a good hint

Fixed property needs to hold for any continuous function

But if I imagine the donut from a birds eye view and imagine the continuous map that rotates this 180 degrees clockwise

[paper wedge]

S^1 x S^1

[split orchid]

So i thought that the maps in question had to be movements across the surface. like a path or something. So im not sure how to translate that to rotating the object

[split orchid]

Oh I see

I think i actually get what ur saying

The map is basically reflecting the x to its adjacent point on the other side of the torus right?

[versed geode]

That's exactly in the 180 degree case

But note even "rotating" by 1 degree is enough

[split orchid]

right

that makes sense

Okay is that valid as a proof? Like can topologist just say "imagine the rotating torus. QED" lol

[versed geode]

Idk depends on your prof tbh

For a more rigorous version

I was trying to hint that the torus is homeomorphic to S^1 x S^1

Now since fixed point property is preserved under homeomorphisms the questiosn becomes

Does S^1 x S^1 have the fixed point property?

[hearty jacinth]

You can write a formula for the function

[split orchid]

for S1 x S1 im imagining infinite circles place around in a circle and eventually forming a solid torus, and inline with the rotating torus, you could assign f(x) to be the same point x in the next circle. but idk how to rigorously work with this stuff

[split orchid]

I find it just as hard to answer this question, as the torus question lol

[versed geode]

Okay simpler case

Does S^1 have fixed point property

[split orchid]

okay wait

I over complicated this

S1 does not have the fix point property because you can define a function f to the points on the opposite side of the circle and finding one mapping is enough (thats the part i was forgetting)

[versed geode]

Yeah

Having fixed point property is quite "strong" in the sense it must hold for every continuous function

[split orchid]

Right I have no idea what I was trying to solve for in my head, but finding one mapping is not that hard i dont think in this case. It just requires encoding the rotation into a function.

I think for the circle f(x,y) = (-x,-y) should be reflection along the center.
and similiarly for the torus i think f(x,y,z) = (-x,-y,-z) should work.

I think ™️

[onyx raft]

I am trying to show that the equivalence will collapse the disjoint union down to \Delta^n

[plain raven]

what do you have so far

the basic idea is that for any degenerate simplex in the disjoint union, there's a map from that simplex to some lower dimensional simplex and you're going to glue along that map

so the degenerate simplex is going to disappear in the sense that it just gets identified with its image in the lower dimensional simplex

and so at the end of the day, only the nondegenerate simplices really appear in the complex

[zinc siren]

...is there another definition of the torus besides as S^1 X S^1 endowed with product topology?

[zinc siren]

why even use Cartesian coordinates? more natural to use the two circle angle cords.

\forall a != 0,
f(\theta, \phi) = (\theta+a, \phi) where we take angles mod 2\pi can't have a fixed point because that would imply \theta = \theta + a but a != 0.

[gritty widget]

identifying points of the square [0, 1]^2

[zinc siren]

ahhh

[gritty widget]

for example

you can also define it using polynomial equation... or something

[zinc siren]

eww

[gritty widget]

I'm sure algebraic geometry can't be this bad

[fathom flare]

group actions?

[gritty widget]

(x^2+y^2+z^2+1/2)^2-4(x^2+y^2)=0 for example

[tidal lynx]

this is a proof for why cl(A) x cl(B) ⊂ cl(A x B), where A, B are subsets of a topological space X and the product topology is considered.

why is it that because A is closed, U and A are not disjoint?

[gritty widget]

"by the definition of closure points, U intersects A at some a'."

i don't know where you're reading what you wrote about A being closed

[tidal lynx]

oh yeah that's wrong

I still don't understand though

[gritty widget]

what is the definition of the closure of A

[tidal lynx]

these are the defns of a closed set I know:

• contains all its limit points
• complement is open

and the defn of closure I am using is a set along with its limit points

[gritty widget]

what is a limit point

[tidal lynx]

a point p is a limit point of a set A if all neighborhoods of p intersect A infinitely number of times

or just once I think

[gritty widget]

so a is in the closure of A and U is a neighborhood of a

so a is either in A or is a limit point of A

do you see how the definition of a limit point gives you a point in the intersection of U and A?

[tidal lynx]

I see that now, but what if a is not a limit point

[gritty widget]

think about what you are trying to prove

if a is not a limit point, then it's in A, and it's in U, so...

(so we can take a' to just be a)

[tidal lynx]

---

lemme try to reproduce the proof from the beginning

given a point (a, b) \in cl(A) x cl(B), in order to prove that (a, b) \in cl(A x B), I need to show that every neighborhood of (a, b) intersects A x B at least time (other than (a, b)). so I need to find a point (a', b') in this intersection

let W be a neighborhood of (a, b). then there exists open U, V such that (a, b) \in U x V ⊂ W (standard basis).

assuming a and b are limit points:
because cl(A) is closed, we know that the neighborhood U of a must intersect cl(A) at some other point a'. similarly we can define b'. then (a', b') \in ((U \cap cl(A)) x (V \cap cl(B))) ⊂ U x V, so this point indeed works

if either are not limit points, then

wait I understand the defn of a closed set, but how can I check that a point p is a member of a closed set?

because

given a point (a, b) \in cl(A) x cl(B), in order to prove that (a, b) \in cl(A x B), I need to show that every neighborhood of (a, b) intersects A x B at least time (other than (a, b)). so I need to find a point (a', b') in this intersection
isn't really true right? I basically just said that in order for a point to be in a closed set, it has to be a limit point of that set, which is wrong

[tidal lynx]

---

I realized the problem is I don't understand the concept of a closure well

Let X be a set, and let X' be the set of limit points of X
I want to show that cl(X) = X \cup X' is a closed set, i.e., it contains all of its limit points. To do this, I can show that all limit points of cl(X) are also limit points of X, so by the defn of cl(X) this will suffice.

Suppose p is a limit point of cl(X). This means that every neighborhood U of p contains a point x \in cl(X) other than itself. I will show that p is also a limit point of X, and to do this, I will show that in every neighborhood of p, there is a point other than itself contained in X.

Let U be a neighborhood of p. If the other point x \in cl(X) contained in U is in X, then we found our point.
Otherwise, if x \in X', then by the definition of X', x is a limit point of X. Moreover, x \in U means there is another neighborhood V of x such that V ⊂ U. But since x is a limit point of X, this neighborhood must contain a point in X. Thus we have found our point.

---

Lol the explanation of this everywhere is 2 lines but I only understand it if I write out everything

ok now I can tackle the original problem

---

given a point (a, b) \in cl(A) x cl(B), in order to prove that (a, b) \in cl(A x B), I need to show that either (a, b) \in A x B or every neighborhood of (a, b) intersects A x B at least once (other than (a, b)). so I need to find a point (a', b') in this intersection

now on with the proof. let (a, b) \in cl(A) x cl(B). if (a, b) \in A x B we are done. otherwise, for every neighborhood of (a, b), we need to find a point (a', b') also in this neighborhood that is in A x B

let W be a neighborhood of (a, b). then there exists open U, V such that (a, b) \in U x V ⊂ W (standard basis).

assuming a and b are limit points (because (a, b) is not in A x B, but still have to check when one is a limit point and the other isn't*****):
because cl(A) is closed, we know that the neighborhood U of a must intersect cl(A) at some other point a'. similarly we can define b'. then (a', b') \in ((U \cap cl(A)) x (V \cap cl(B))) ⊂ U x V, so this point indeed works

*****if one is a limit point and the other isn't, WLOG a is a limit point and b isn't. then (a', b) is our other point.

finally!

[tidal lynx]

now I finally understand why this works 👍

[tidal lynx]

does every infinite sequence (of distinct points) in the zarski topology converge to every point

yeah seems so

[coarse night]

@odd flame were you able to resolve your problem?

[odd flame]

i'll find out tomorrow when i get my hw back

[coarse night]

lol anyway I can refer you to some material where the algorithm is given

I was actually reading it so thought might refer you as well

[odd flame]

[coarse night]

John M Lee Introduction to topological manifolds, look at chapters on classification of compact surfaces

[odd flame]

do you have pdf or link?

[coarse night]

wait

https://link.springer.com/book/10.1007/978-1-4419-7940-7

[odd flame]

thank <3

[next crystal]

Suppose X and Y are topological spaces and f: X -> Y. "for every closed subset V of Y, the preimage f^{-1}(V) is closed in X" implies that f is continuous
Pf:
Suppose U \subset Y is open. Then Y\U is closed in Y. So f^{-1}(Y\U) is closed in X by hypothesis. Therefore X\f^{-1}(U) = f^{-1}(Y\U) is closed in X. Therefore f^{-1}(U) is open in X

How do we get that X\f^{-1}(U) = f^{-1}(Y\U)?

[coarse night]

it's a set theoretical fact that $f^{-1}(A^c)=(f^{-1}(A))^c$

[coarse night]

try picking an element and show they are same, basic exercise

[gritty widget]

hi, is this proof right?

[haughty cedar]

Im trying to get a grasp of a compact set. I have a definition that if for every x_n in S there is a convergent subsequence in S, then S is compact.
Now, would the set such as [2, 3] \union [4, 5] be compact?
I can't find a proof that it is not.

[marble socket]

yep that is compact

what examples of compact sets can you think of so far?

[haughty cedar]

since i know from borel theorem, that compact is bounded and closed, I can think of all the sets that are bounded and closed.

[marble socket]

so is the above set closed and bounded in R?

[haughty cedar]

Yes

But, the reverse implication is incorrect AFAIK

bounded and closed => compact

[marble socket]

ahh, wait i read it wrong

heine borel theorem says that this is true for subspaces of R^n

but you'll see that soon ig

[haughty cedar]

cool

that will fit me for tomorrow's finals :p

[marble socket]

[haughty cedar]

thanks!

[marble socket]

you could prove that the above set is (sequentially) compact directly by definition

[haughty cedar]

That's a good idea as well.

[gritty widget]

Too much text

alright bet

f^-1(empty) = empty, f^-1(Y) = X so f is continuous. Is what I'd write. Add some flavour to make it look like a proof. Then qed

yeah i suppose thats all you need to write, i think for me though i wrote out like everything i was thinking sense when i first looked at it i didnt realise f^-1(Y) is X

[haughty cedar]

Suppose we have a function, f which is differentiable on (a, b). How do we know that inverse of this function is differentiable?

[gritty widget]

this seems like a #calculus yes?

[gritty widget]

https://en.wikipedia.org/wiki/Inverse_function_theorem

next time put it in a different channel, if it's related to differentiability then this is probably not the place to post it

[coarse night]

believe that's not the question they asked.

[coarse night]

it's not always true, take (-1, 1) → (-1, 1) by x ↦x³

[gritty widget]

i read it as what blitz interpreted it. it seems like theyre wondering what conditions need to be met for the inverse of a function to be differentiable over an interval.

[old violet]

Can we actually expect the Locally Compact Hausdorff Version of Urysohn's Lemma to do this?

In particular, I am wondering if the statement after "Moreover" is possible

It seems like the original proof for the Urysohn's Lemma for LCH spaces wouldn't go through. In particular, if we require continuity, then we won't have compact support and if we require compact support, then continuity fails.

This is the proof I am following: https://www.math.ksu.edu/~nagy/real-an/1-05-top-loc-comp.pdf

[old violet]

Also I hope this is not duplicated question from yesterday. I believe composing by the original homeomorphism suggested by @marble socket, unfortunately, won't help in this case. In particular, everything outside of the support $f$ after composition would be raised by $a$, making $f$ not compactly supported.

[gentle osprey]

Cookieman

[gritty widget]

it's a typo (or someone didn't think this through for all I know)

they really want f-a to have bounded support

[old violet]

I see what you mean, but then can they get f to be still a on F?

[gritty widget]

all they are doing is sending [0, 1] to [a, b]

so whenever f would output 0 in the first part of the theorem, they send it to a

[old violet]

Oh I see

[gritty widget]

yep

a+t(b-a) is the homeomorphism

[old violet]

Lol okay, so are we asking too much if we also ask for compact support?

[gritty widget]

so you basically do a+f(x)(b-a) where f is the compactly supported function

[old violet]

Is there a counter example?

[gritty widget]

yep, it really should be "constant outside of a compact set"

just take anything like a = 1

and X not compact

[gritty widget]

this is what the author really meant

[marble socket]

me was tagged

i think it's resolved now uwu

[gritty widget]

maybe idk. I didn't get an answer

[old violet]

Mmm yeah I agree. Sorry I had to head to a class earlier and wasn’t able to reply. Thank you!

[gritty widget]

It's fine! I was just saying you might still be confused

[pearl holly]

anyone knows what this notation means (I am referring to the smash with a subscript)? It's never defined and I can't find sources that define what this is

at first I thought it was some orbit space but idk now

[lunar yoke]

Yes schwede uses them for the orbit space of the diagonal action

So here you can think of it as inducing up the action from the subgroup 1xS_n to S_m+n

[median sand]

Quick sanity check: D\subset X discrete => D\cap A\subset A discrete for all A\subset X:
If B\subset D\cap A, then B\subset D and it's open in D, so B=D\cap O for open set of X. Then B=B\cap A=(D\cap A)\cap (A\cap O) is an open subset of D\cap A => D\cap A is discrete in A.

[gritty widget]

D cap A is discrete as a topological space

being discrete means just that

(though sometimes they impose that it needs to be closed, but that's uncommon)

[median sand]

Well that's exactly what I wanted to check my logic on.

You're being tautological.

[gritty widget]

D cap A is a subspace of D

[median sand]

Oh like that you mean.

[gritty widget]

subspace of discrete space is discrete

[median sand]

OK sure

If you have two subspaces A,B of X and a subspace C of A\cap B, then the topology of C doesn't depend on whether you consider it as a subspace of A or B, right?

[gritty widget]

since subspace topology is transitive

in the sense that if $X\subseteq Y\subseteq Z$, $X$ has subspace topology from $Y$, $Y$ has subspace topology from $Z$, then $X$ has subspace topology from $Z$

[gentle osprey]

Blitz

[median sand]

Yeah, I know.

[gritty widget]

we have that C has, in both cases, subspace topology from X

[median sand]

Thanks.

[gritty widget]

this transitivity property doesn't always hold for some properties of a topological spaces so I thought I'd write it as a fact

[median sand]

If I want to show V=R^n x C^m (as an R-vector space) is Euclidean (in terms of its topology), is it enough to say
"|(x,z)|=max(|x_i|,|z_j|) defines a norm on V [it does, right?] and since all norms on a findim space are equivalent, V is Euclidean"?

[gritty widget]

C is homeomorphic to R^2

from properties of product of topological spaces, V is homeomorphic to R^(n+2m)

[gritty widget]

well if you're going to introduce the concept of finite-dimensional space as a topological space

then you ought to show it's well-defined beforehand

[median sand]

I know, but I want to show it differently.

Does what I say work?

[median sand]

The Euclidean topology is the only topology on R^n that makes it a topological vector space and all findim R-VS have the same topology induced by a basis (change of basis is a homeomorphism).

[gritty widget]

like, you're trying to prove something but I don't even know what you're proving

[median sand]

Yeah, maybe it's a little ill-posed.

[gritty widget]

if V is n-dimensional R-vector space, with any topology giving it topological vector space structure, then V is homoemorphic to R^n

[median sand]

NVM, I'll just treat V as R^{n+2m} (which it essentially is).

[gritty widget]

this follows from V being linearly isomorphic to R^n so that it induces a topological vector space structure on R^n

[median sand]

Yeah.

[gritty widget]

which then is homeomorphic with standard topology on R^n

[gritty widget]

such that the linear isomorphism becomes a homeomorphism

so

topology of finite-dimensional space is uniquely defined by the dimension

in the sense that, any two n-dimensional topological vector spaces are homeomorphic

[median sand]

Thanks.

[pastel coral]

Hello, I just want some reality check. I conjecture that it's true that the circle group $\mathbb{T}$ is isomorphic to quotient space $\mathbb{R}^2/\sim$ where $(a,b)\sim (c,d)$ if and only if $ab-cd\in \mathbb{Z}$. I think this is true by considering the quotient mapping $\rho: \mathbb{R}^2 \to \mathbb{T}$ via $\rho(a,b)=exp(2\pi i ab).$ Are there any obvious reasons why this may fail to be true?

[gentle osprey]

blackiris

[unreal stratus]

Well i guess you've just gotta check whether this is indeed a quotient map, but assuming that is true this holds fine

But yeah I imagine this is an open map anyway

[gritty widget]

if I have X is the set of two disjoint nonempty closed subspaces, X is connected? because something is not connected if it's the union of two nonempty open disjoint subspaces. like rn I'm visually like two closed balls in R2 and it seems like by the definition it's connected? but it feels weird to call them connected

this is definitely disconnected

it's the union of two non-empty disjoint open subspaces!

do note that in the definition of connectedness it doesn't matter if you use open or closed subspaces

if X is a disconnected topological space, X = U \cup V with U, V open and disjoint, then U is X - V so it's closed as well

same for V

or maybe i should phrase it like: if X is a disjoint union of U and V, then U and V are both open if and only if they're both closed

ohhhh

yeah I was thinking about the topology on R2

[stark fog]

there is also that result that X is connected iff its only clopen subsets are the empty set and itself

[forest barn]

Is If A is in a topological space then we call A open (in that space)?

[stark fog]

you mean as a subspace?

or as an element of the topology?

[forest barn]

element

[stark fog]

the elements of the topology are called the open sets of the space

you can also define a topology where the elements are closed

but for the subspace case, it may not be true, for example [0, 1] as a subspace of the real line, it is open in the subspace topology but closed in the topological space

[lunar yoke]

yes, as part of the axioms of a topology, A must be an open subset of the space A

[stark fog]

note: as a subset of the space A

because in the subspace topology, $A = A\cap X$, where $A\subset X$ is a subspace of the top. space $X$.

[gentle osprey]

SubGui

[gritty widget]

that's not called a topology

I don't think it has any name

[stark fog]

maybe I messed it up, I was thinking of a example where you have a discrete topology on a finite set satisfying T1 axiom

very specific btw

[gritty widget]

[stark fog]

[gaunt linden]

There are some contexts where it's most convenient to define a topology by describing what the closed sets are, rather than what the open sets are. (One example is the Zariski topology in AG). It sounded like that was what you were talking about.

[gritty widget]

I wouldn't say it's part of axioms of topology. We define topology to satisfy certain axioms and simply call set open if it belongs to the topology

[stark fog]

most likely, but I possibly messed it up

thanks nevertheless

[lunar yoke]

its is literally an axiom

also all these nitty gritty special case examples are super useless for someone that is just learning the definition of a topological space

[gritty widget]

what is

[lunar yoke]

like you cannot seriously imagine that this is in any way helpful

[gritty widget]

its a definition

a set is open if it belongs to the topology

[lunar yoke]

jfc

we need a split of pset and algtop after all

[gritty widget]

the original question was asking about subsets of topological space

[gaunt linden]

Do you have a definition in mind that does not make it an axiom that the entire space must be an open set?

[stark fog]

[gritty widget]

the question is open up to interpretation

what stronq was asking that is

I don't think they were asking about whole space being open at all

[gaunt linden]

There's nothing up to interpretation here. If you want to claim there is such a defintion in anything resembling common use that doesn't make this an axiom, the onus is on you to provide it.

[gritty widget]

Is If A is in a topological space then we call A open (in that space)?

is that not open up to interpretation?

[shadow charm]

I agree with blitz here question wasn’t clear

[gaunt linden]

No, the answer is unablguously yes beacuse the axiom requires that.

[shadow charm]

Nowhere was it specified that A is the topological space in that question

Or what A even is relative to that space

[gaunt linden]

Oh shoot.

I somehow didn't see the word "in". Sorry.

[stark fog]

yeah, that's why I asked if it was as the subspace of the space, but it didn't mention if A was a subset to begin with

[shadow charm]

If A is a subset of X, A is open in the subspace top on A

But not necessarily open in X

If that was the question

[gritty widget]

I think the one answering the question should have just confirmed what the question really is

because we're all guessing at this point what it could mean

[gritty widget]

It's okay, at least you weren't rude about it

[void gazelle]

Hi, guys, why this is a homotopy? How to see that immediately $F: I\times I\to Y, F(t,s) = h_t\cdot\varphi_tf\cdot \overline{h_t}(s)$ is continuous?

[gentle osprey]

cc0101

[steel glen]

are you asking how you’d prove it or more intuitively?

[void gazelle]

i can see why intuitively this is true

[steel glen]

path products (when endpoints are shared) are continuous by the pasting lemma, and these paths all share endpoints for arbitrary values of t

[void gazelle]

Yes, this is true. My main concern comes from that if F: X\times Y\to Z is map, and F: X\times {y}\to Z is continuous, F: {x}\times Y\to Z is continuous, for all x,y, then we cannot conclude that F: X\times Y\to Z is continuous

so for every t, F(t,s) is continuous does not make F into a homotopy

[steel glen]

well you can use the product topology to show this is continuous. i believe, in particular, this amounts to using/proving the tube lemma, but that could be wrong

[steel glen]

oh are you saying F(t,s) continuous is not enough?

the rest of the conditions should follow really easily

[void gazelle]

yeah, i worried about that, F: {t}\times I \to Y is continuous is not enough

[steel glen]

h_t is continuous as it is a composition of continuous maps. same reasoning for phi_t of f

i don’t know if there’s anything else left to show? you have continuous maps from continuous compositions, and you’re taking their product which is also continuous (as they share endpoints)

[void gazelle]

Sorry, i am a little confused, do you mean if we have a map F on X times Y. We want to show it is continuous, we just need to show F is continuous on X by fixing a point in Y, and F is continuous on Y by fixing a point in X?

[steel glen]

no

i’m saying that F is a product of paths, where each path is continuous as it is a composition of continuous functions

so F is continuous itself

[void gazelle]

Oh I see

Sorry about that

Thank you

[steel glen]

no no problem

[void gazelle]

Hi, sorry to bother, but I looked up the definition of homotopy from wiki. I am not sure about the last sentence contradicts to what you said

[steel glen]

we’re not claiming h_t is continuous for each individual t, we’re saying h_t is continuous over the whole space X x I since it is a composition of continuous maps.

they’re saying it is not sufficient for h_t to be continuous at each X x { t }

[echo oyster]

Can someone help please to understand the question 7(b)

[gritty widget]

what's not clear about it?

[echo oyster]

Do i need to assume some functions continuous from R to Rl , then show it is continuous from R to R ?

[gritty widget]

The question doesn't ask you to show anything

It just asks you to ponder on what those functions could be

[echo oyster]

Oh

Constant functions are continuous from R to Rl

Does it help to conclude something about f : R→R

[stark fog]

recall that the open sets in $\bR_{\ell}$ are of the form $[a,,b)$

[gentle osprey]

SubGui

[swift fjord]

slight nuance: those are the basic open sets

open sets in R_\ell are going to be unions of these

[stark fog]

mb, yeah the union of these

[elder loom]

I never use this topology ever

like

where does it come up?

[stark fog]

notice you're taking a limit as x approaches a from the right, so x is in a set [a, b) for some b > a

now you apply the definition of continuity

[untold lily]

topology is all about counterexamples, the sorgenfrey line is home to many of those

so it is important to a topologist

[unreal stratus]

eh intro pointset top is about counterexamples

[untold lily]

advanced pointset top is not?

I am talking about general topology though

[prisma arrow]

counterexamples are important but there is more to the field than this

the types of questions i see are usually about classification if anything

[untold lily]

okay I apologize for exaggerating, topology is not literally all about counterexamples

[stark fog]

counterexamples are a dense set in general topology

[river granite]

intro pointset is sort of about classifying spaces and finding out precisely which hypotheses you need for interesting theorems (e.g. Tietze) to hold, so counterexamples are needed to draw the line so to say

[gritty widget]

just like in any field of math, huh (comment was edited so what I said might not make sense)

counter-examples gives us crucial insight no matter where we are headed

part of general topology is about classifying spaces

so its important to know what classes of spaces overlap... and which is contained in which etc.

that's the only reason why there's more counter-examples in intro to general topology than other introductions to a math field

that's my take anyway

[untold lily]

[stark fog]

maybe there is a God of topology and it doesn't want us to find it

[ruby crown]

A refinement D of a cover A does not necessarily have to be a cover right? If you think about it empty set can always be an refinement, as the definition only requires that for all elements in the refinement there exists an element of A such that the element in the refinement is a subset of that element of A. Since, for all statement of an empty set is always true the empty set is always a refinement, which is of course not a cover for any set except for the empty set

[gritty widget]

it does have to be a cover

[ruby crown]

Is that part of the definition?

[gritty widget]

if it doesn't follow from your definition, then take it as an additional assumption

it should be in the definition of refinement you're using, though

[stark fog]

Yeah, I think the definition of refinement of a cover also says "still covers"

[ruby crown]

I see. I am also looking for a way to interpret paracompactness, namely an intuition for what it means for there to exist a locally finite open refinement for every open cover. The notion of open refinement is pretty easy to understand as it is just about having the possibility of a cover with less/smaller sets than the one you started with. Locally finite may be interpreted as you do not want a neighbourhood of a point in the space to collide with infinite many elements in your cover. However, I still do not have a clear idea of why we would need this notion of paracompactness based on its definition. Does anyone have an intuitive way thinking about this concept?

[prisma arrow]

you know how hausdorff is a way to talk about seperating two points?

[gritty widget]

maybe it'll be insightful to think in what way is paracompactness used
we usually have a local property, and using paracompactness we can "glue it" obtaining a property which is global

[ruby crown]

Yeah, that you have two disjoint neighbourhoods for any two points

[unreal stratus]

(also if it didn't need to be a cover, it'd be a bit of a pointless definition)

Was about to say smth but Blitz beat me too it :)

Well, perhaps to add to it, importantly a space X is paracompact hasudorff iff all open covers admit subordinate partitions of unity, which allows one to often do what Blitz says in the last message ^

[prisma arrow]

you can interpret it as a bound on the open sets of the cover that makes things more localized

i can try and think of an example to illustrate this

[gritty widget]

as for definition of paracompactness itself, I don't see much intuition for it, really, a locally finite open cover obtained from using paracompactness, will probably be pretty wild in general

[unreal stratus]

Yeah to me it just feels like "here are some arguments we want to make - what is the weakest hypothesis possible that allows us to do them"

Which is fair but yeah feels inch resting

[gritty widget]

yeah, you want it rephrased asap

if to gain an intuition for it, is what it feels like

[unreal stratus]

Yeah

[ruby crown]

A concrete example would be helpful for me to grasp it.

[gritty widget]

I mean, compactness even, to see why its defined that way you really want to go and look into the history

[ruby crown]

I'm trying to see what it meant with local property, what would be an example of that?

[gritty widget]

I suspect to gain intuition for the concept of paracompactness, other than it just being similar to compactness, you need to look at how it was obtained in the first place

[gritty widget]

manifolds are a good example

well, by locally we really mean using partitions of unity in a clever way to derive global properties from local ones

for example, using property such as local metrizability

local properties usually come in two flavours

things like local compactness, and things like how usually the term is used

here we mean that there is a neighbourhood which is metrizable

[ruby crown]

I see, I'll try to understand why that turns that into metrizability with paracompactness

[gritty widget]

Hausdorff paracompact locally metrizable spaces are metrizable

because Hausdorff paracompact makes the local metrizability into a global property

i.e. metrizability

[stark fog]

coincidentally I asked a friend if they studied paracompactness in his courses and he couldn't tell me what was it useful for

thanks about these

[gritty widget]

its okay

[shadow charm]

How do you extend local properties for just general hausdorff paracompact spaces? I know for manifolds you can use partitions of unity but how does it go in general?

[gritty widget]

.

[tidal lynx]

graders said that this wasn't enough?

but isn't showing "U x V must be open for open U \in X, V \in Y, when the two projection maps are continuous" equivalent to the the product topology being the coarsest?

[plain raven]

I think your logic is sound here but you did not make that explicit.

"Let U \subset X be open. Since alpha is continuous, alpha^{-1}(U) = U x Y is also open" In what topology?

In the product topology or in another topology?

You probably would have gotten full credit if you added a sentence at the beginning where you said "Let tau be an arbitrary topology on X x Y such that the projection maps alpha, beta are continuous. Let U be open in tau" Otherwise it seems like you might just be talking about the product topology, which is the only topology mentioned explicitly in the question

[tidal lynx]

In X's topology. X is given as a topological space so isn't it fairly implicit?

how could a subset of X be open in the product topology anyways

that's a categorical error

(I posted this like a week ago) this was the previous part to that question, and I got full credit for it

[tidal lynx]

I guess it's because I included the "That is..." part similar to what you said, but I thought I could just leave it out of the next part rip

[plain raven]

Sorry i meant U x Y

and X x V

not U and V

the topology of X and Y is unambiguous.

What's ambiguous is the topology of X x Y.

I edited my original post to clarify.

[eager herald]

are these the T_0 T_1 kind of things?

i searched it up and the separation axioms seem kind of different from these

it later proves the equivalence of (i) and (ii)

and defines any top space satisfying these to be separable

is there any relationship between separable spaces and the separation axioms?

i'd imagine there is, but not sure how

[coarse night]

These are odd terminologies. A space is called separable if it has countable dense subset and called 2nd countable if it has a countable open basis. In general 2nd countability implies separability and converse is true for metric spaces.

[unreal stratus]

Bruh