point-set-topology

[plain raven]

I don't do a lot of analysis, so I'll just point out what's clear from flipping through Folland's textbook on real analysis:

• The Baire category theorem holds for LCH spaces.
• If X is an LCH space, then the space of continuous functions on X under the topology of uniform convergence is a Frechet space.
• If X is an LCH space equipped with a Borel measure mu, finite on compact sets, then the space of locally L^1 functions on X with respect to U is also a Frechet space.
• In general, LCH spaces are a reasonable context for developing measures and integration. The chapter in Folland's book on Radon measures is set in the context of an arbitrary LCH space.
• The Riesz representation theorem holds for LCH spaces equipped with a measure

[long hornet]

I don't do a lot of analysis either, but it seems as if Hausdorffness plays a role here? At least in the Frechet space thing

[plain raven]

let's assume the space is compactly generated and Hausdorff.

[long hornet]

How about the existence of a locally finite measure? This seems to me like a property where we "need" a compact neighborhood around a point, not merely that closure can be detected by testing via compact sets

Is a CG space a direct limit of compact spaces?

[plain raven]

it should be.

[long hornet]

I see, so if a property or construction behaves well when taking colimits, and it works for LCH spaces, then it works/holds for CGH spaces

[plain raven]

Yeah, I think so. Like, I think that the space of continuous functions on a CGH space should be a Frechet space

under like, some nice assumption like second countable

[long hornet]

Can you elaborate?

(unrelated: What would the dual notion of CG be?)

[plain raven]

Well, let's say X is a topological space, and we assign C(X) the topology of uniform convergence on compact subsets. For every compact subset F of X, there's an associated seminorm on C(X) given by |f| = sup_F f. These seminorms determine the topology.

A Frechet space has to be generated by a countable family of seminorms so you would want to impose the requirement that like, there is a countable family of compact sets such that their seminorms generate the topology on C(X). I would hazard a guess that this takes the form of an assumption on the number of open subsets, like assuming X is second countable.

[long hornet]

Ohh, I see!

I thought by "Frechet" you meant functions separate points, for some reason

[dusk heron]

Consider a smooth 2-manifold $\Sigma$, and let $\Omega\subseteq\Sigma$ be an open and simply connected subset. Does there exist a smooth coordinate chart which contains the entire closure $\overline{\Omega}$?

[gentle osprey]

gustavn64

[long hornet]

Taking S = M (S for omega and M for sigma), this implies that every simply connected 2-manifold is diffeomorphic to R^2.

[long hornet]

The answer is no. Take M = S^2 and A = S^2 - point, which is simply connected. The closure of A is M, so there is no single chart containing cl(A).

[dusk heron]

What about if instead of requiring $\Omega$ to be simply connected, I require $\overline{\Omega}$ to be contractible?

[gentle osprey]

gustavn64

[long hornet]

I don't know, but it seems like a size issue, not a shape issue

[gentle osprey]

Hausdorff

[long hornet]

It does, both groups are trivial

[gentle osprey]

Hausdorff

[vast estuary]

Why

[dusk heron]

@vast estuary $\mathbb{D}^n$ is homotopy equivalent to $\mathbb{R}^n$

Which is simply connected.

[vast estuary]

Okay, let me try to prove that!

[long hornet]

Removing a point from the n-disk for n >= 3 still gives a simply connected space

[gentle osprey]

gustavn64

[vast estuary]

Are D^n and R^n homeomorphic, by the way

That might be an easier approach, if true

[dusk heron]

@vast estuary Depends on whether you mean the open or closed ball. If you mean the open ball, yes, otherwise no

[vast estuary]

Yep, open always

[dusk heron]

I don't know what kind of facts you already know, but I think the simplest way to see that the open ball is simply connected is to use that it is convex

[long hornet]

D^n - point ~ R^n - point is homotopy equivalent to S^(n - 1), and for n >= 3 this is simply connected. The proof I know isn't trivial.

The key step was proving that a map h : S^1 --> S^(n - 1) is homotopic to a map that is not surjective (this is done using Lebesgue covering number lemma). If we can do that, we are done because then we have a loop in S^(n - 1) - {point} which is R^(n - 1).

[coral pivot]

Standard is probably a quick application of van kampens

[long hornet]

I think it's typically introduced later on

[vast estuary]

Yep agreed

Any two paths will be homotopic via straight line homotopy

but we can't do this if we delete 0

So we'll have to argue some other way for that

[dusk heron]

@vast estuary As @long hornet said, it is enough to show that S^n is simply connected for n >= 2. So take any loop, view it as a continuous function f: [0,1] -> S^n with f(0)=f(1). Then f is uniformly continuous, so we can choose n large enough that on intervals of length 1/n, the curve will not go far enough to go to the antipodal point.

Thus on each interval [(k-1)/n,k/n], where k is an integer with 1<=k<=n, we can view f as a function with codomain S^n - point, which is simply connected, and thus we can deform f to lie in the plane containing f((k-1)/n), f(k/n), and the origin, keeping the endpoints fixed during the deformation.

After doing this for each subinterval, we have deformed f into a loop whose image is contained in the image of finitely many planes, and such a loop cannot be surjective onto S^n.

[long hornet]

Do you think being 2 dimensional matters?

[dusk heron]

I suspect it doesn't matter, but I was mostly curious about the 2-dimensional case.

[long hornet]

Okay, suppose we have our open set S and its closure C. S is also an n-manifold, and the existence of such a chart would imply that S can be embedded in R^n.

What assumptions do we have on S? It's contractible or just simply connected?

[dusk heron]

none other than it being open. we just assume that C is contractible

[empty grove]

@vast estuary Another way to prove the surjectivity Gustav proved: pick a disk on the surface of S^n not containing the base point. Then wherever f enters this disk, you can make it go along the boundary of the disk instead to its exit point to get a new path f'. f ~ f' because each segment of f inside the disk can be straight line homotoped to the corresponding segment in f', and then f' is not surjective because it never enters the disk

[long hornet]

Nicer*

[empty grove]

There is 1 subtle detail which is that f could enter and exit the disk infinitely many times. Proof remains the same, but you need to then prove f ~ f' by transfinite induction/zorn's lemma

There might be an easier way to do this last part

[long hornet]

The boundary of the disk is compact

[empty grove]

I like the much nicer to nicer change after I said this

[empty grove]

How would you use that?

[long hornet]

Not sure. I thought I would consider entire arcs instead of points

[pearl holly]

is this channel still in use or may I ask a stupid question? 👉 👈

[empty grove]

Ask

Mods refuse to give us threads

[pearl holly]

So wtf is H and F_0 here in the bottom?

in the very bottom

[empty grove]

F_0 = H

It's saying if your H is free, then 0 → H →H → 0 is a free resolution

[pearl holly]

oh bruh

[empty grove]

And so Ext(H) will be 0

[pearl holly]

right and H is the R-module?

[empty grove]

Didn't get that

Ah yeah

Since R is a field, H is free

[pearl holly]

but shouldn't H be H_(n-1)(C) because we are looking at Ext(H_(n-1)(C), G)?

[empty grove]

H is a placeholder

So yes it is that

Lol

[pearl holly]

But then how do you know that H_(n-1)(C) actually is the R-module then?

[empty grove]

assuming that it is

If C is a chain with everything a vector space

And maps are all vector spaces maps

Then quotients of subspaces are also vector spaces

And we are assuming that C is like that

[pearl holly]

oh okay I see now

why doesn't Hatcher write this out bruuuh

[empty grove]

[pearl holly]

I'm getting lost by these things very easily lmao

Thank you so much btw!

[empty grove]

[long hornet]

There is a strong counterexample

The Whitehesd manifold M. It's a 3-manifold that is contractible but not homeomorphic to R^3, so there is no single chart covering it.

[pearl holly]

wait if I have a map $i^: C^n(X; G) \to C^n(A; G)$ where $A$ is a subspace of $X$ then what does Hatcher mean when he writes "$i^$ restricts a cochain on $X$ to a cochain on $A$"?

[gentle osprey]

Tokidoki ✓

[pearl holly]

like what am I restricting?

[marsh forge]

The cochain is a function

[pearl holly]

okay and then I'm restricting that function?

ooohh so the i* literally restricts that cochain

lmao

[marsh forge]

@tough imp do you have good examples of separated and non separated morphisms of schemes

I’m trying to write up some notes to force myself to learn this stuff properly

[tough imp]

Most things are separated, but it exists to replace Hausdorff

[marsh forge]

Just FYI I’m also going to ask about like 5 other adjectives

[tough imp]

I think the line with double origin is not separated

Cuz well… the doubled origin

[marsh forge]

Is there some nice stronger condition that is any more intuitive

[tough imp]

But there’s a notion of quasi-separated and fucking everything is quasi-speared

[marsh forge]

Or easy to remember

[tough imp]

Sort of…

So there’s this condition for a map to be separated

Idk if it’s intuitive but it is useful

[marsh forge]

That’s good enough

[tough imp]

If A,B are affine opens such that their image is contained in a single affine open of the target

Then A\cap B is affine

You usually don’t get intersection of affines is affine

So in particular if you’re separated over an affine

Then the second part of the hypothesis is automatic

So you just always have intersections of affines are affine

This is also an equivalence

So if whenever A,B like that exist their intersection is affine, then the map is separated

[marsh forge]

Sorry why is this?

[tough imp]

Cuz the images are contained in the same affine

Namely the target itself

[marsh forge]

Oh I didn’t see that part

Lmfao

So wait why do you need separation

Ah it’s just a part of the hypothesis

[tough imp]

Yeah

I mean I can tell you the proof that separated implies that

Pretty easily

[marsh forge]

Oh no I was just misunderstanding

[tough imp]

Ah okay

[marsh forge]

I focused on separated here instead of “over an affine”

[tough imp]

Ah

[marsh forge]

Now I just have to write up some stuff on finite presentation, flatness, geometric points, and smoothness :)))

And also get some dinner

Probably not in that order

[abstract pagoda]

Why does connected components argument work for showing one space homeo to another?

So like you have topologies A and B

removing a point from both and the argument goes connected componenets of A not equal to connected componenets of B

phrased shorter I think it goes pi_0(A) not iso to pi_0(B) implies A not homeo to B

[marsh forge]

I am not sure what you are asking

[abstract pagoda]

like a way to prove that a space isnt locally homeomorphic to Rn is by removing a point from both the space and Rn

and then saying how spaces differ in connected components

why does that argument work

[marsh forge]

Let f:X->Y be a homemorphism

then f restricted to X minus a point p

is homeomorphic to Y-f(p)

[abstract pagoda]

oh ok

that is simple enough

and i can extend that further

like say X has 4 seperations

[marsh forge]

I mean if you know what this means

then

you should already know how the argument goes

as a homeomorphism induces an isomorphism on all homotopy sets and groups

[abstract pagoda]

it feels so weird doing that tbh

[marsh forge]

?

[abstract pagoda]

I dont know how else to phrase it, but whenever I can just use the fact that a functor induces a certain property. Like using givens in algebra land to prove facts about topology land if that makes sense

I think I know the topological argument,I just need to write it out

[marsh forge]

Okay. But also you should become comfortable with functors if you are learning about them.

It is very easy to show a functor takes isomorphisms to isomorphisms

[abstract pagoda]

I understand it perfectly through functors, but it just feels cheaty in a way

it feels as if its a way to get around understanding topology

[marsh forge]

it is

the entire subject of algebraic topology was designed to avoid thinking as hard about the topology

topology is hard

[abstract pagoda]

feels bad tbh

[marsh forge]

lol

there are a great many cases where it is simple to prove two spaces are not homeomorphic or homotopy equivalent

using algebra

and basically impossible otherwise

[abstract pagoda]

oh no

[marsh forge]

That’s why the subject exists ahahah

I’ve really never seen this take before

[abstract pagoda]

idk the feeling is similar to something else

ah yea

when programming or scripting

over reliance on abstractions from libraries

feels like im doing that with functors

[coral pawn]

Won't I need (at least) two parametrizations for this?

[abstract pagoda]

because its not exactly like all functors take complex object to complex object, but in our case I guess you can have a complex topology to a complex algebraic object, really depends on the object being functored. but i think i dislike the idea of taking a complicated topological object to a simpiler to understand algebraic one and proving facts from there.
surely you can prove it in a topological sense without a functor sometimes? this is all nonsense dont read it btw

[abstract pagoda]

yea

[coral pawn]

I think I got the first one

send x,y to (cos x sin y, sin x sin y, cos y)

What should the second one be?

[abstract pagoda]

what portion of the sphere does that parameterization cover?

[coral pawn]

Everything except the circle?

[abstract pagoda]

what circle?

[coral pawn]

The circle on the xz plane?

[abstract pagoda]

why should i believe you

prove it to me?

[coral pawn]

I've been told I'm a reliable guy?

[abstract pagoda]

i dont believe you

[coral pawn]

Well the second coordinate of what's not in the paramtrication must be 0

This forces the sum of squares of the first and third coordinates to be 1

Also, if the second coordinate is 0 and you are on the sphere, you are on the xz plane

[abstract pagoda]

if the second coordinate is zero then x = npi or y=npi

[coral pawn]

Yes

[abstract pagoda]

what forces you to be on a circle?

[coral pawn]

I'm taking the domain of my first parametrization to be (-pi,pi) x (-pi,pi)

[abstract pagoda]

what forces you to be on a circle?

[coral pawn]

Well

We know this would map to the entire sphere if we did it on [-pi,pi] x [-pi,pi]

[abstract pagoda]

why would it

if it did then wouldnt that imply there is one parameterization?

[coral pawn]

Yeah but a parametrization requires non-singular derivative

Which it would not have at certain points

And a parametrization requires an open set in the domain too I guess

[abstract pagoda]

It doesnt afaik

yea

[abstract pagoda]

Wouldnt this imply there is no possibility of having one parameterization ever?

[coral pawn]

What?

[abstract pagoda]

This is sort of tangential but by that definition it shouldnt be possible to have one parameterization if your domain is bounded

[coral pawn]

I have no idea

This is what he had in the notes

[abstract pagoda]

Yea its besides the point

[coral pawn]

Can I get an answer instead of follow up questions lol?

Not to say this isn't helpful

[abstract pagoda]

uh for choosing a parameterization of the sphere I would use stereograpgic projection

[coral pawn]

Oh god no

[abstract pagoda]

your parametrization isnt clear to me,

but ive seen it before

[coral pawn]

Wait [-pi,pi] x [-pi,pi] --> R^3 given by (x,y) -> (cos x sin y , sin x sin y, cos y) does have the entire sphere in its range right?

Or am I wrong about this

[abstract pagoda]

No?

[coral pawn]

Which point isn't in the range?

[long hornet]

Even if it's not surjective, you can always just enlarge your intervals

[coral pawn]

that won't change it though, you can get all possible values of sin and cos by restricting to just [-pi,pi]

[long hornet]

Yeah, I didn't think lol

[coral pawn]

But I think we hit every point

[long hornet]

We do

[coral pawn]

@abstract pagoda been real quite lately

jk jk

[abstract pagoda]

you need y to be in 0 to 2pi but yea it works

for finding any point on sphere

[coral pawn]

Oh yeah my bad

No wait

Isn't [-pi,pi] the same as [0,2pi] for sin and cos purposes?

You hit every representative

[abstract pagoda]

yeah it is the same but x becomes sorta redundant

you never really need negative values

maybe you can shrink x to half size

but i hope you seeing that this parametrization isnt intuitive

[coral pawn]

I don't understand the point you're trying to make here

[abstract pagoda]

the parametrization you wrote can be derived with calculus, but my point is you need another parametrization because you need dX to be nonzero

[coral pawn]

It will be non-zero if I restrict the domain to (-pi,pi) x (-pi,pi)

[abstract pagoda]

can you show that?

[coral pawn]

Yeah

[abstract pagoda]

and does it still get each point of sphere?

[long hornet]

This map will never be a parametrization though, I mean it won't be a homeomorphism..

[coral pawn]

No

[abstract pagoda]

then doesnt it fail

dont you want to get tangent spaces of sphere

[coral pawn]

It will not contain the xz circle

So we need to pick another parametrization that contains the xz circle

[abstract pagoda]

you can just permute order of elements to get that

[coral pawn]

Omg you're right

[abstract pagoda]

??

[coral pawn]

Why didn't I think of that lol

Thanks

[abstract pagoda]

still should show dX non singular

[long hornet]

Can't we think of this as a composition of phi and a map that permutes the coordinates? Then dX will be nonsingular, being the composite of nonsingular maps.

[abstract pagoda]

sure?

makes sense enough

[lean marten]

You need two patches to get a nice parameterisation right?

So you can just use a square root thingy positive on one unit disk and negative on the other

Obviously you have to translate both to the origin along the way

Actually you'll miss a great circle nvm

[long hornet]

Are S^2 and some rectangle in R^2 bicontinuous?

By bicontinuity I mean the existence of a bijective, continuous function

For example (0, 1] and S^1 are

[lean marten]

Yeah if you take a half open rectangle

[long hornet]

"Rectangle" means a product of two intervals (open, closed or half open)

[lean marten]

Wait no in that case it won't work

Because you want an open rectangle with a single boundary point

[abstract pagoda]

wut

[long hornet]

Oh, neat

I can see why that makes sense

[lean marten]

Like you take (0,1)x(0,1) and then add a single boundary point eg $(0,1)x(0,1)\cup{(0,0)}$

[gentle osprey]

Oatman

[abstract pagoda]

ok

define the map

maybe inverse is obv

[lean marten]

Use the trig one from above

[abstract pagoda]

it has no intution

[long hornet]

It should be the same spherical coordinates one

[lean marten]

Yeah^

[abstract pagoda]

but 0,0 corresponds to 0,0,1

[lean marten]

Then choose a different boundary point

Not sure why that would be an issue either

[long hornet]

S^2 - pt is (0, 1) x (0, 1)

So we can add a point to both sides!

[lean marten]

Yeah can probably use a 1pt compactification or something

[abstract pagoda]

what in the name of quack

[long hornet]

It's weird that it works

[lean marten]

Yeah

[abstract pagoda]

thas no proof tho

[long hornet]

Yeah

But it is reasonable

[abstract pagoda]

its be shoiwing that one point compactified space is homeo to the space described with one boundary point

[lean marten]

No the image of it under the map in my head lol

[abstract pagoda]

im not sure how reasonable

[lean marten]

I'm pretty sure I'm thinking of the trig parameterisation but I might not?

[long hornet]

No, we aren't talking about homeomorphism, but rather bicontinuity

[lean marten]

Oh actually that reminds me of a fun problem

[abstract pagoda]

yes?

whats your point

clearly the problem isnt bijectiveness

[lean marten]

Which manifolds can be expresed as the closure of the image of a cts map $f:(0,1)\times(0,1)\to\mathbb{R}^n$

[gentle osprey]

Oatman

[abstract pagoda]

you argument of adding a point is faulty because of continuity part

[lean marten]

I don't think anyone has really made a super formal argument here

[abstract pagoda]

nope

[lean marten]

It's more of a discussion

But that's fine lol

[abstract pagoda]

adding point is just intuition for a specific map in mind

[lean marten]

Yep

This is a discord channel not a journal

[long hornet]

Hmmmmmm.

[abstract pagoda]

its still a faulty argument

[lean marten]

Ok

[long hornet]

I was thinking as (s, t) approach (1, 1) their images do really approach a pole in the sphere, so we can add a point.

[abstract pagoda]

stil gotta show homeomorphic after you compactifty both spaces

[lean marten]

You can prove continuity with sequences here probably

[long hornet]

We only want a bijective continuous map. Its inverse shouldn't be continuous.

[abstract pagoda]

for what?

[long hornet]

S^2 is obviously not homeomorphic to the space Oatman proposed.

[lean marten]

I believe this problem can be done by considering fundamental plygons?

Oh yeah of course

I'm not constructing a homeomorphism?

[abstract pagoda]

fundemental polygons and more?

[lean marten]

?

[abstract pagoda]

fundemental polygons like the one for the square take the side and add quotient relations

[lean marten]

Yeah

[abstract pagoda]

i think there are other topologies you can get from square than just that

that still work within conditions of question

there definitely are more

[lean marten]

So you can construct all of your nice handlebody type manifolds with a fundamental polygon

[abstract pagoda]

there are more than that me thinks

[long hornet]

For starters, every manifold with M - pt = R^n is one of these manifolds

[lean marten]

And then use CW complexes to at least partly answer the problem I posed

Yes

But what cts function from the square would have this as an image?

Wait actually space filling curves might make this problem kinda fucked

[abstract pagoda]

do they induce topologies?

[lean marten]

Ok add the condition that the map is smooth

Do what induce topologies?

[abstract pagoda]

space filling curves

[lean marten]

I would assume so?

[abstract pagoda]

but like which one matters

why would space filling curves fuck the problem?

[lean marten]

Because you can then get basically any connected subset of Rn you please

So the problem gets ruined

[long hornet]

And has full rank? Also, I'm assuming n >= 2

[lean marten]

Yeah I guess so?

Idk I just made the problem up

[long hornet]

Well, I think that trivializes the problem

[lean marten]

Was thinking about how you can mostly cover the sphere with a rectangle and also a torus

can be done with a square

[long hornet]

When working with fundamental polygons, we don't identify interior points, right?

[abstract pagoda]

no

[lean marten]

No

[abstract pagoda]

think of them as cw complex

[long hornet]

Complex

[abstract pagoda]

but then you add in relations on the 1 cells only

[long hornet]

Okay

[lean marten]

Yeah this problem can be done by assigning a CW complex with a single 2-cell to all the manifolds you think it can be done for

(Which I get the sense is most?)

[abstract pagoda]

cw complex might be too strong me thinks because there are most definitely more

i just cant think of any

the domain is large

and our restrictions arent

[long hornet]

So with fundamental polygons we have F : [0, 1] x [0, 1] ---> R^n which is a closed map?

[lean marten]

I wouldn't worry about constructing F

Once you have a fundamental polygon you're done because the interior is dense in the space

And all manifolds can be embedded in R^n for some n

Otherwise you'll be solving a much harder question lol

[long hornet]

Oh

That's what I was trying to show

[lean marten]

OHHHHH

Ok sorry didnt see the square brackets

[long hornet]

So, so far we know that all 2-manifolds which are quotients of polygons? (all compact ones)?

[lean marten]

Yes

There are more I think

Like the long cylinder

(Not the really long one tho)

[long hornet]

You make f so that it misses a line?

[abstract pagoda]

btw how to construct CP1 with fundemental polygons

[lean marten]

Identify opposite points

[abstract pagoda]

we use square still

[lean marten]

so take a line through the midpoint of your square and then the points on either end get identified

[abstract pagoda]

oh ok

thats super simple

i wanna show this diffeomorphic to S2

but using fundemental polygons probably not the way

[lean marten]

Oh that's not super bad

Uh

Lemme remember lol

[abstract pagoda]

ive been trying via stereographic projection

but inverse maps give headache

[lean marten]

Yeah fuck stereographic projection

Ok so one way to do it is to coordinatise as [z0,z1]

[long hornet]

Both have the same cell structure

[lean marten]

Yeah I'm trying to remember why

[long hornet]

And there is no more than way to glue a 2-cell to a point

[abstract pagoda]

i mean with fundemental polygons

[lean marten]

Btw the cell structure I gave earlier is that of RP^1

RP^2

[abstract pagoda]

S2 is just identifying all sides of square with one point

if you want CP1 with fundemental polygons its probably weird

[lean marten]

Theres a good section on this in topology and groupoids

I believe the chaper is projective and other spaces

[abstract pagoda]

you probably need an octogon or hexagon

well you need octogon

because you are making 2 identifications using 4 variables which corresponds to identifying 8 sides?

z0 ~ tz0’ and z1 ~ tz1’

each z0,z0’,z1,z1’ have two variables each

for real and complex parts

so i think of this as so

for each identification you start off with a square for each side

actually fuck this it wont make sense

[lean marten]

Those identifications need to be done in $\mathbb{C}^2$ so a fundamental polygon might be a bit annoying to construct

[gentle osprey]

Oatman

[abstract pagoda]

yea

im sleeping god damnit

[lean marten]

You already know its a sphere though so just do the normal one of a sphere and then compose with a homeomorphism $h:P^1(\mathbb{C})\to S^2$

[abstract pagoda]

i need to solve this sooner or later

[gentle osprey]

Oatman

[abstract pagoda]

uh

i want a diffeomorphism and you mean compose it with h^-1?

[lean marten]

Oh yeah lol

[abstract pagoda]

yeah thats the thing idk h^-1

i only get stereographic projection way sorta

[lean marten]

If you know its a diffeomorphism you don't need to

Unless they want an explicit function

[abstract pagoda]

i need to prove S1 diffeo CP1

[lean marten]

S1?

[abstract pagoda]

S2 mb

[lean marten]

OOOOOOH

I thought you were trying to do a fundamental polygon for it

Ok

Ummm

I feel like I've seen this proof before one sec

[abstract pagoda]

I could probably use fundemental polygons for it though

[lean marten]

I wouldn't

You should be able to do it directly

[abstract pagoda]

its needless tbh

I think I know chart for CP1 is same as RP1 but complex instrad of real, so phi1[z1,z2] =z2/z1 and phi2[z1,z2]=z1/z2

and inverse is phi1^-1(z)=[1,z] and phi2^-1(z)=[z,1]

and for S2 its probably best to use charts using stereographic projection but the inverse maps are hard to think about

[lean marten]

Right

[coral pawn]

What would be the best charts for this problem?

I was thinking of using the natural map between R^2 and C and then realizing that this is just the function x -> x^3 on C

But I don't think that would work

Doing it by choosing the usual charts would give you all sorts of different cases to deal with

[long hornet]

Is there a recipe for triangulating a surface given by a fundamental polygon?

I want to know how to do that for the dunce cap

[tawdry widget]

Just the normal one… you see the image of (0,1) and (0,-1) are (0,-1) and (0,1) respectively so just use stereographic projection

[coral pawn]

But the preimage of (0,1) isn't just (0,-1)

So you would have to consider the different cases for that won't you?

[tawdry widget]

I don’t get it. Why does it have to be injective?

I thought all you need is two functions R—>U—>V—>R and R—>V—>U—>R where U is S-(0,1),V=S-(0,-1)

[coral pawn]

Well the way you do it is you take a point p, look at its image F(p), pick charts around p and F(p) and then use the coordinate representation of F to compute the pushforward

[tawdry widget]

So?

Just compute the derivative of the above two maps from R to R

Then you get your answer

Since the dimension is 1

[coral pawn]

There are 2 possible charts around p and 2 possible charts around F(p). So we would have to consider 4 coordinate representations of F

[tawdry widget]

Oh I see

4 isn’t very much I guess 😂

[coral pawn]

I guess not

Was kind of hoping to get a slick answer using complex numbers

[tawdry widget]

I don’t think that’s possible since dimension of S^1 is 1, not an even number

[coral pawn]

@tawdry widget Instead of the usual stereographic projection, use the stereographic projection with the "east" and the "west" pole

[tawdry widget]

But you still have 4 maps

[coral pawn]

Yeah I just realized that that does nothing to help

[tawdry widget]

But don’t you think 4 is the best situation already… the only simpler case is when the manifold itself is a chart😂

[coral pawn]

Look at this

And say that again

[tawdry widget]

😂

[coral pawn]

@tawdry widget Final answer

[long hornet]

Funniest answer I've ever seen

[little hemlock]

Just leave computing the compositions as an exercise for the grader, ez

[tough imp]

LOL

Kinda based

[shy moss]

Is this Z^2*Z^2/Z?

[empty grove]

What do you mean by modulo Z there? You'd have to specify that

Because based on how you embed Z, you could have different groups

Eg Z/2Z and Z/3Z are very different even though 2Z and 3Z are embeddings of the same group Z into itself in 2 different ways

[shy moss]

$<a,b,c,d|ab^{-1}>$

[gentle osprey]

Or x1

[empty grove]

Works

[empty grove]

Wait this isn't the answer though

Because you have to take Z² * Z²

So there are relations on a,b,c,d as well

Commutativity relations

[marsh forge]

Whoops wrong channel ❤️

[empty grove]

[shy moss]

$<a,b,c,d|[cd]=[ac]=ab^{-1}=1>$

[empty grove]

Yeah seems correct, though you'd usually write , instead of =

Or also say = 1 at the end

[gentle osprey]

Or x1

[marsh forge]

Both are common

[empty grove]

Yeah but I mean with the equality convention you need to say = 1

[marsh forge]

Oh yes

Was that not there

Before

[empty grove]

Yeah

[marsh forge]

Editing messages is a crime

[orchid forge]

Shouldn't it be something like [ab] = [cd] = ac^(-1)?

Or if you want to use [cd] = [ac] then it should be bc^(-1), not ab^(-1), right?

Actually it's the same thing anyway, nevermind

the fact that the group you wrote down is the same as the group I wrote down is weirdly bothering me, like it looks like you were trying to write the relations for two copies of the torus and then identify two generators, but there is no labeling which yields what you wrote in this way. in your presentation, a,b,c,d cannot all be the image (under inclusion) of the generators of the fundamental groups (what is b?)

But after eliminating the redundant variable we can see that we've got just a central generator and two non-commuting generators either way

[shy moss]

a and b are the generators of fundamental group of the first torus and c and d are the generators for the second torus

[orchid forge]

But you identified a and b

You should have identified a and c, or a and d, or b and c, or b and d

But luckily(?) the group you wrote down is (isomorphic to) the fundamental group anyway

Do you see why a and b doesn't work? You identified two loops in the same torus

[shy moss]

oooh i see

[abstract pagoda]

anyone mind checking my proof for a diffeomorphism between CP1 and S2

Im having trouble coming up with the composition

like I feel as if im supposed to replace z with z=x+yi

and write $\psi_1^{-1}\circ\phi_1(x,y,z)=[1,\frac{x+y}{1-x-yi}]$

[gentle osprey]

bold tifanny

[abstract pagoda]

but that looks fugly

hmm

[abstract pagoda]

rip my life

[marsh forge]

truly crazy how much worse making things smooth is

like the homeomorphism here is at most one line

[orchid forge]

It may help to visualize

Like the atlas for CP1 you've written says that CP1 is two copies of C (planes) wrapped up in a sphere, one omitting the south pole and the other omitting the north pole

That information is encoded in the transition map U1 --> U2

So once you see this the map CP1 --> S2 should be clear, then all the rest is just notation

[ornate lark]

abandon real manifolds and work with complex ones. screw reals

[orchid forge]

There's no particular quirk of real manifolds here, the diffeomorphism is essentially the identity map

[marsh forge]

abandon all manifolds and work with CW complexes

[marsh forge]

and without smooth structure you're done right there lol

[ornate lark]

eek no. i hate simplices

[orchid forge]

You're done with the smooth structure too

It would be very difficult for the identity map to not be a diffeomorphism

The point is that both CP(1) and S^2 are fundamentally just two copies of C = R^2 wrapped up into a ball in exactly the same way

But written with a complex coordinate versus 2 real coordinates

So the diffeomorphism CP(1) --> S^2 is essentially the same as the diffeomorphism C --> R^2

It is nontrivial in name only

[ornate lark]

yeah i guess the turn off is when the formulas come into the picture. pictorially it is pretty clear that you are literally doing the same thing while calling one projective space and the other one a sphere

[orchid forge]

Yeah, that's why I'm saying it will help to visualize so that the formulas are easier to keep track of

[abstract pagoda]

ok

i wrote it finally

[orchid forge]

This shows that F is a well defined map

But not yet that it is a homeomorphism or a diffeomorphism

For that you need to figure out the maps that F induces on the charts

[abstract pagoda]

well fuck

isnt that obv

maybe not the homeomorphism patt

[surreal ocean]

Let $X$ be some topological space. If $B$ is an open subset of $X$ then does the set of all open subsets of $B$ include $B$?

[gentle osprey]

Justinflys

[unreal stratus]

B is a subset of itself

I guess the conclusion doesn't even depend on whether you mean subspace topology or X's topology, lol

[surreal ocean]

oh ok thanks

[gritty widget]

B is always open as a subset of B just by the definition of a topology

regardless of whether it's open in X or not

[unreal stratus]

Yeah, I just wasn't sure if they meant open in the sense of open in X, bit unclear potentially, but either way it doesn't matter

[surreal ocean]

Im new to topology could you explain that in more detail

[gritty widget]

can you tell me the definition of a topology on a set?

[surreal ocean]

a set of open subsets of the set

not 100 percent sure if thats right

[gritty widget]

yeah, that's what you call the elements of a topology. but can you tell me what we require in the definition of a topology?

it's some collection of subsets of X satisfying (various axioms)

what are those?

[surreal ocean]

gimme a sec

union of open sets is open

intersection of any two open sets is open

empty set and the whole set are both open

[gritty widget]

that last one explains what i said

and answers your question

[surreal ocean]

but wouldnt that require me stating "B is a topological space" for that axiom to apply?

[gritty widget]

well you said "open subsets of B" in your question so certainly you have a topology in mind for B

it doesn't really matter which one, though, because no matter what, B will be an open subset of itself

or, wait, do you mean subsets of B which are open in X

[surreal ocean]

It sounds like you are saying that a topology and the set of open subsets of some set are the equivalent statements. Am I interpreting that correctly?

[gritty widget]

yes they're the same thing

[surreal ocean]

in all cases?

[gritty widget]

"open set" literally means "element of the topology"

[marsh forge]

Very very pendantic caveat

[gritty widget]

leaving it to you

[marsh forge]

There are definitions of a topology where the closed sets are the topology

but thats like

silly

for all intents and purposes a topological space is a pair (X,T) where T \subset P(X) is the collection of open sets

Also another technical thing

[unreal stratus]

(potentially pedantic too but when giving the definition of a topology it's pretty important, at least in my eyes, to stress that arbitrary unions of open sets are open and finite intersections of open sets are open)

[marsh forge]

B itself does not have open sets unless you prescribe a topology on B

now, 99% of the time you want the subspace topology

in which case everything tterra said is true

Actually no of course, if you give B any topology B is open in B by definition

[gritty widget]

im gonna make some noodles you guys can take it from here

[marsh forge]

but you do have to give it one

[surreal ocean]

all that is interesting... I'll just state my questions in the order they came to my mind. When I saw you write "T\subset P(X) is the collection of open sets" my immediate thought was "doesnt the power set include all closed subsets of X?" '

i realize that one of the things you guys said about a topology might have explained this but everything got jumbled up in my head so if we could start there that would be helpful

[unreal stratus]

I mean yes, P(X) contains all subsets of X, by definition

The original sentence could be rephrased 'T, which is a subset of P(X), is the collection of open sets of X'

[surreal ocean]

ohhh

oops i was reading that wrong

i see

then my next question is: Starting from the definition of a topological space and the axioms which go along with it, how can one deduce the following... For any set B, the set of all open subsets of B is a topology. If this has already been answered sorry just lmk which message answers this question

nevermind

sorry

got it

[marsh forge]

Wait did my messages send

I was saying that is circular

You need to tell me what sets you want to be open in B

and then I can tell you whether it is a topology

B doesn't come with open sets

[surreal ocean]

does that mean i choose which sets are open in B?

is open-ness not an intrinsic property of a set

[marsh forge]

Not at all

the same set can have many many different topologies

[surreal ocean]

oh wow

[marsh forge]

B starts life as a set

Then you look at the powerset of B to see all its subsets

and if you choose a bunch of those to be open

[feral dragon]

The set of all subsets as open sets is a topology, it's just called the discrete topology.

[marsh forge]

and if those open sets follow the axioms

[marsh forge]

I do not think this is what they meant.

[marsh forge]

sorry, if they follow the axioms then they give you one possible topology on B

[surreal ocean]

i see thats an awesome explanation thanks

[marsh forge]

no problem

[surreal ocean]

i see so when i asked my original question, TTerra assumed i was referring to B as a topology since i used the phrase the set of all open subsets of B. Thus by third axiom B is an open subset of B

[unreal stratus]

Not seeing B as a topology, no, but thinking of 'open subsets of B' as meaning 'elements of B's topology' (as opposed to X's)

When you say 'open', that can be a bit ambiguous in contexts like this, depending on where it is 'open in B' or 'open in X', although usually it's pretty clear what you mean

[hallow swan]

if there were only one possible topology for a set, life would be boring

[marsh forge]

In fact most cool topological spaces have the same underlying set

(Up to bijection)

[surreal ocean]

oh ok thanks i think that clears up everything

and thanks to everyone that helped that was very enlightening and i'm excited to dive further into topology

[cedar pebble]

consider the following two functions:

$\mathrm{Li}2(z)=-\int^z_0\log(1-t)\frac{\mathrm{d}t}{t}=\sum{n\geq 1}\frac{z^n}{n^2}$

[gentle osprey]

nGroupoid

[cedar pebble]

$\Lambda(\theta)=-\int^\theta_0\log|2\sin(t)|\mathrm{d}t=\frac{1}{2}\sum_{n\geq 1}\frac{\sin(2n\theta)}{n^2}$

[gentle osprey]

nGroupoid

[cedar pebble]

We have the following:

$\Im(\mathrm{Li}_2(z))+\arg(1-z)\log|z|\ =\Lambda(\arg(z))+\Lambda(\arg(1-\frac{1}{z}))+\Lambda(\arg(\frac{1}{1-z}))$

[gentle osprey]

nGroupoid

[cedar pebble]

how on earth to see this lol

(this is secretly a question about volumes of hyperbolic 3-manifolds)

[tough imp]

I’m really happy you think anyone on this server can help you

But imma be real with you chief, I don’t share your optimism

[cedar pebble]

lmfao

I ended up figuring it out, and yea I don't think it's something that can be done by hand very easily

: )

[gritty widget]

Chern-Weil theory lets you associate to any any Ad-G invariant polynomial f on \mathfrak{g}, a characterstic class of the principal-G-bundle

is there any reference for what polynomials will give you what characterstic class

for example can i get the euler class like this

i vaugley know that the euler class can be defined in terms of a Pfaffian

[buoyant dew]

is compactness of a space inherited by coarser topologies on the same set? I couldn't come up with a counterexample

ah, it obviously is inherited, sry

[tawdry widget]

what book are you reading? I want to find some materials too

[woven aspen]

hi can I ask something about Geometric Algebra here?

[buoyant dew]

How can one show that $S^1$ is not homeomorphic to any topology weaker than $\mathbb{R}_{std}$? I am stumped.

[gentle osprey]

iruneachteam

[ornate lark]

https://data.math.au.dk/publications/ln/2003/imf-ln-2003-69.pdf

this was what I used. i found it pretty good

[ornate lark]

i dont quite understand this question

[sweet wing]

connected sets remain connected

[buoyant dew]

I'm trying to prove that no topology on R that is coarser than the standard topology is homeomorphic to S^1

[buoyant dew]

I thought about it for a while but couldn't really move on from there

[sweet wing]

just to clarify

If (R,τ_1) is corser than (R,τ_2), then (R,τ_1) -> (R,τ_2) given by identity is continuous

right

[ornate lark]

isnt it the other way around

[buoyant dew]

ah right, that indeed is the case but I can't really see how that would help

[buoyant dew]

i think so, must be a typo

[sweet wing]

uh i cant rmb definitions

LOL

ahh

hm in that case lemme think

[ornate lark]

yeah if tau_2 contains all the open sets of tau_1, then the identity from tau_2 to tau_1 is always continuous

so it has to go from finer to coarser

[buoyant dew]

many thanks to you both, I've been pulling my hair out trying to find how to go on from here

[sweet wing]

the connected open sets in your new topology must be of the form
(-\infty,a)
(a,b)
(a,\infty)

right

[buoyant dew]

yup

[sweet wing]

so

what happens when you remove one point

[buoyant dew]

it becomes homeomorphic to R_std i believe

[sweet wing]

oh hm actly it may not work
(what in trying to do is some form of remove one point -> not connected but circle remove one point is connected)

[buoyant dew]

haha yeah but since it is coarser removing a point need not break the connectedness

[sweet wing]

yea

oof

[buoyant dew]

im positive the proof to this one is beyond my knowledge (and the scope of the course I'm taking)

[woven aspen]

this is still an issue, does anyone know?

[ornate lark]

ask away

[sweet wing]

(just ask, dont ask to ask)

[woven aspen]

so basically i want to know if this way also works for n dimensions

[woven aspen]

didn't want to post in the wrong channel

[ornate lark]

yep. exterior product is the generalization

[sweet wing]

wedge product is perfectly well defined in higher dimensions

[sweet wing]

nah i feel it should be elementary but hard to find kind

[woven aspen]

oh ok I'm quite new to all of this, thanks :D
so that way works nice

[buoyant dew]

a couple of things, we know our new topology R_tau is compact and metrizable, i tried to use limit point compactness but no luck so far

[buoyant dew]

eh don't even bother then

[sweet wing]

oh wait actly?

[buoyant dew]

i mean, R_tau is homeomorphic to S^1

which is the one-point compactification of R_std

[ornate lark]

so you have a topology on \R that makes it homeomorphic to S^1 with what topology

[sweet wing]

oh i thought you meant general spaces oops

mm yea

[sweet wing]

usual topology

=one point compactification of R

[ornate lark]

ah ok

[buoyant dew]

btw I am not 100% sure the statement is correct, that's what I've been told but couldn't find anything about it online

[sweet wing]

ahh

sounds interesting anyways

[ornate lark]

cant you just use the standard topology on \R excluding zero. open sets are the standard open sets that do not contain zero

[sweet wing]

wdym

oh huh i have an idea

[ornate lark]

R_tau is defined as euclidean topology excluding sets that contain zero

[sweet wing]

S1 - {•} = R

[marsh forge]

Not a topology

[ornate lark]

ah ok true. sorry

[sweet wing]

so we need (R,τ)-{0} = R

(select 0 just wlog)

that means we only need to focus on the point 0, what are its nbhd

[sweet wing]

as well as figure if we can have this

[buoyant dew]

btw i figured (-infty,a)U(a,infty) is never open in (R,tau) because otherwise the embedding of R_std into R_tau \ {a} is not connected

[sweet wing]

but it shouldnt be possible

[marsh forge]

Are you asking if it is possible to give R a topology making it homeomorphic to S1?

[sweet wing]

never connected*

[ornate lark]

yes

[marsh forge]

It is.

[sweet wing]

a corser topology*

[marsh forge]

Oh

It isn’t

[sweet wing]

yea

[buoyant dew]

one that is coarser than R_std

[buoyant dew]

Well since we use the point 0 for the one point compactification, its nbhds must be of the form R\C where C is compact in R_tau \ {0}

but idt we have any idea what those might be

[sweet wing]

ah look at R-{0}

it needs to be homeomorphic to R

[ornate lark]

so since we know that the identity from a finer topology to a coarser one is continuous, if there is such a homeomorphism, this is equivalent to saying that there is a sequence of continuous maps $S^1\cong (\mathbb{R},\tau)\rightarrow \mathbb{R}$

[buoyant dew]

yup, but doesn't have the standard subspace topology does it?

[gentle osprey]

Pepa

[ornate lark]

wait the other way around

[sweet wing]

yesh but your things are like (a,b)

not (-\inf,0) cup (0,inf)

[ornate lark]

$ \mathbb{R}\rightarrow S^1\cong (\mathbb{R},\tau)$

[buoyant dew]

do we know every (a,b) not containing zero is open in R_tau \ {0}? Then we're done i believe

[buoyant dew]

does this not assume that R_tau is first countable? lemme check

[sweet wing]

hmm

[gritty widget]

If you put coarser topology on R then its still contractible since continuous paths in R are still continuous

[tawdry widget]

Thanks a lot

[sweet wing]

wait true

[buoyant dew]

oh, but S^1 is not contractible?

ahh many thanks to you all

[sweet wing]

how did i miss that LOL

[buoyant dew]

i was right, this goes way beyond the scope of my course lol

anyways, tysm again you saved me from hours of frustration <3

[teal nova]

Hello guys, i need some help, i have to prove that if $d$ is a metric then $\frac{d}{1+d}$ is equivalent

[gentle osprey]

galois.theory

[teal nova]

equivalent to d

[ornate lark]

this is just shrinking the reals into a unit interval

[marsh forge]

I think there should be a compactness argument

But I just woke up

I don’t think you can make R compact just be removing open sets without screwing up some structure

By*

Like you need to maintain manifoldy ness

Actually that by itself might do it

Anyway gym time

[buoyant dew]

thanks! unfortunately i have no experience with manifolds either lol

[buoyant dew]

have fun!

[sweet wing]

show it is sufficient to consider balls of radius between 0 and 1 and use your usual "2 basis are same" argument

(p.s. you can take this argument further by showing d and min(d,1) are equiv)

[gritty widget]

Because it's most likely wrong

[marsh forge]

No it should work

[sweet wing]

actually idk

[marsh forge]

The oaths will still be continuous as will the path homotopies

Paths*

[sweet wing]

like i kinda prefer if a proof has more of this feeling

oh right

[marsh forge]

You and nine9s will get along lol

[sweet wing]

wait actly

why

actly nvm

right

[marsh forge]

Preimage if open is open

[ornate lark]

i dont know if you can even have manifoldy-ness if you take some arbitrary coarser topology on R

[marsh forge]

There are fewer opens

[marsh forge]

This sounds right but an actual formalization escapes me rn

I mean locally

[sweet wing]

i feel like you can if you have like

[ornate lark]

yeah but R with some arbitrary topology is not even locally euclidean

[sweet wing]

some very cursed long line construction

[marsh forge]

Yeah I doubt it is easy to prove

I feel like there is an elementary proof

I mean

You can identify a bunch of open sets you have to delete

[sweet wing]

yea it feels like a intro point set problem

[marsh forge]

By realizing you have to make R compact

And one of those

Probably causes issues

Like consider the open covers of R with no finite sub over

All of those need to be basically removed

[ornate lark]

ok so if we suppose there is a homeomorphism from $(\mathbb{R},\tau)$ to $S^1$, where $\tau$ is the coarser topology. Then since the identity map from a finer topology to a coarser one is always continuous, we have a sequence of continuous maps $\mathbb{R}\rightarrow (\mathbb{R},\tau)\cong S^1$. Also note that all of these maps are set-theoretically bijective. We end up with a bijective continuous map $\mathbb{R}\rightarrow S^1$

[gentle osprey]

Pepa

[sweet wing]

oh huh that works

yes

[ornate lark]

yep and thats a contradiction

[buoyant dew]

oh hi again

[sweet wing]

hi

[marsh forge]

Hm

Is that map obviously open/closed

[buoyant dew]

tysm for the proof, can you tell me how we know such sequence of bijective functions exists?

[marsh forge]

A bijective continuous map is a homeomorphism iff it is closed or open

But for example

The first map

Isnt open

[ornate lark]

the first one is just identity and the second one we assume exists, so if theres a contradiction at the end, the assumption must be false

[marsh forge]

What is the contradiction

[buoyant dew]

right, so the fact that there is no such sequence of functions contradicts with (R,tau) being homeomorphic to S1?

[sweet wing]

wait crap the proof i was thinking fails

[marsh forge]

Irun I’m not sure this is over

[sweet wing]

damn the details in this qn are

[gritty widget]

We can remove a point from R then S^1 without point is not connected

[marsh forge]

I think that proof is incomplete at best

[sweet wing]

significant

yea

i cant seem to complete the proof either

[marsh forge]

Can you formalize this

How do you know removing a point from R disconnects it

With this new topology

[gritty widget]

The standard topology

Pepa provided a continuous bijection from R to S^1

[sweet wing]

wait this is saying

you have a map from (0,1) -> S^1

that is continous

and a bijection

[gritty widget]

No sorry

[marsh forge]

If you remove a point from R you remove a point from the image and then there’s no contradiction

Oh wait

There’s no contradiction anyway lol

Continuous maps can identify connected components just fine

Or connect disconnected comps

R is not homeomorphic to a closed interval

[coral pivot]

Closed interval is compact

[ornate lark]

oh shit

sorry

[marsh forge]

The fundamental group had a contradiction forever ago

[coral pivot]

R isnt

[marsh forge]

We are trying to avoid using pi1

[sweet wing]

urgh it would be really nice to have [0,1] into S^1

[ornate lark]

why am i missing the simplest things

[marsh forge]

I feel bad I am providing no ideas just shooting them all down

Back to squats come up w something while I’m gone

[sweet wing]

no thats good im also messing up some details here

urgh

so we have basically reduced to asking

is there any continuous functions (0,1) -> S^1 that is bijective point-wise

[ornate lark]

yeah

[sweet wing]

we can extend it to [0,1] -> S^1

[ornate lark]

but then you lose bijectivity

[sweet wing]

that overlaps at two points

but only at two points

wait hm

im hoping to:

your function needs to always go clockwise/anticlockwise along circle

but i cant

wait no

S^1 is usual topology

i can

yes

since it is bijective pointwise

identify S^1 with [0,1]/{0,1}
(i.e. consider the angle wrt horizontal or smt)

[marsh forge]

This is a great exercise if I ever teach alg top

As to why

Alg top is necessary

Lol

[sweet wing]

any continuous map (0,1) to S^1 must be rotating strictly clockwise or anticlockwise to avoid breaking bijectivity on sets

if we extend this to [0,1)

the image of 0 is say x

then extend it to (0,1]

image of 1 must be arbitrarily closed to x but a bit further out clockwise/anticlockwise

but this means the initial function isnt bijective

[ornate lark]

the proof sounds correct but i think it needs to be more formal

essentially you are arguing that R is not its own one point compactification

[sweet wing]

yup

[marsh forge]

There might be some point set lemma no one thinks about that applies here more generally too

Esp if this is in an intro course

Might be worth taking a look at munkres

[gritty widget]

We could try dividing R into equally spaced closed intervals then images of those need to be closed arcs and the function needs to move, say, clock-wise, but then there'll be some point it misses

[buoyant dew]

tysm everyone, i'm positive the prof threw in some random exercise problems from previous years and this slipped in

[sweet wing]

to show

A map f: (0,1) -> S^1 that is be both continuous and bijective on sets doesnt exist

extend this to [0,1) -> S^1 and let the image of 0 be at some point x, we then have S^1 ~ [0,1]/{0,1} where x is sent to {0,1}

hence such a function is equivalent to (after rotating the circle appropriately)
f: (0,1) -> (0,1]
but since it is continuous and bijective then it is strictly increasing then we die

[gritty widget]

Basically I'm saying that this needs to be composition of a monotone map and exp and that's impossible

[sweet wing]

yup essentially

[buoyant dew]

ok, this is kind of unrelated but we indeed can embed $S^1$ into the following topology $\lambda$ on $\mathbb{R}$ with $\lambda\subsetneq\tau_{std}$, correct?\
$\lambda = {\mathbf{U} \in \tau_{std} | \mathbf{U}=-1\cdot\mathbf{U}\wedge (0 \in \mathbf{U}\rightarrow, \exists c ,(c,\infty)\subseteq \mathbf{U} }$

oof

I think this is indeed a topology

It's basically (0,inf) with an extra point

[gentle osprey]

iruneachteam

[ornate lark]

Let $f:\mathbb{R}\rightarrow S^1$ be bijective continuous. Consider $\lim_{t\rightarrow\infty}f(t)$. Since $S^1$ is compact, this point exists in $S^1$. Therefore, since $f$ is bijective, there is a $t_0\in\mathbb{R}$ such that $\lim_{t\rightarrow\infty}f(t)=f(t_0)$. We know that $t_0$ is in $\mathbb{R}+$, because the image $f(\mathbb{R}+)$ intersects an arbitrarily small open neighbourhood of $f(t_0)$. But then the images of an arbitrarily small interval $f([t_0-\epsilon, t_0+\epsilon])$ intersects the image $f(\mathbb{R}_{>t_0+\epsilon})$, contradicting bijectivity.

[buoyant dew]

huh, thanks a lot

this actually seems like the type of answer the prof was expecting

[gentle osprey]

Pepa

[ornate lark]

seems complete now

[ornate lark]

combine with this

[buoyant dew]

🙏 🙏

you're a lifesaver

we indeed can embed $S^1$ into the following topology $\lambda$ on $\mathbb{R}$ with $\lambda\subsetneq\tau_{std}$, correct?\
$\lambda = {\mathbf{U} \in \tau_{std} | \mathbf{U}=-1\cdot\mathbf{U}\wedge (0 \in \mathbf{U}\Rightarrow, \exists c ,(c,\infty)\subseteq \mathbf{U} )}$

oof

[ornate lark]

its just a formalization of ari's idea

$\Rightarrow$ instead of $\rightarrow$

[gentle osprey]

Pepa

iruneachteam

[ornate lark]

seems like a topology but what are you planning to do with it

[buoyant dew]

Then lambda is coarser than the standard topology, and I just wanted to confirm if we can embed S1 into (R,lambda)

[sweet wing]

S^1 anyhow into (R,{{},R}) is continuous btw

[ornate lark]

huh. how so

[sweet wing]

the inverse image of {} is {}

of R is S^1

so

it is by definition continuous

[ornate lark]

ah ok just take any arbitrary surjective map

[buoyant dew]

ah but it's not an embedding right?

[sweet wing]

wait shit

it asks for embedding right

[buoyant dew]

I mean this is not a part of a question, I just was wandering if a weaker form of the previous problem was possible

[sweet wing]

you can embed S^1 into a "funny topology on (0,1] that is corser than (0,1]

[ornate lark]

yeah then all you want is continuity. embedding requires more structure

[sweet wing]

uh actly

[sweet wing]

idts

ill need to

think harder

lol

[buoyant dew]

actually nvm, i think this is an embedding so no problem

thank u again 🙏

[kind cedar]

Sorry if this seems stupid but I don't get why S-{x} is open. An entire space minus a point open by definition?

[gritty widget]

the proof shows that S - {x} is a union of open sets and is therefore open

[kind cedar]

Ah, it makes sense! Thank you

[gritty widget]

also an entire space minus a point doesn't have to be open in general, take {0, 1} with the topology {{}, {0}, {0, 1}}. then {0, 1} - {0} isn't open

[kind cedar]

Ah, I see

The definition of a Hausdorff space is too similar to that of a complete space in my view. Whats is the diference?

[gritty widget]

completeness is a property of metric spaces and all metric spaces are hausdorff. but you could have a non-metrizable hausdorff space (and so it can't possibly be complete)

i don't really see how they're similar

unless we're thinking of different notions of completeness

[kind cedar]

I wasn't remembering that completeness was limited to metric spaces.
How is every metric space Hausdorff?

[empty grove]

Any 2 distinct points have non positive distance

[kind cedar]

non-positive?

[empty grove]

Positive* lol

delta/2

[gritty widget]

yeah yeah

[kind cedar]

I was going to say, delta/2 ahaha. Fine, makes sense

[gritty widget]

although the execution was bad, i hope the idea was there

[kind cedar]

But can I guarantee there is a point between x and y?

[gritty widget]

"point between x and y" is meaningless

[empty grove]

That's not needed

[gritty widget]

k leaving it to moldilocks i gotta go to class

[kind cedar]

Can I show that the intersection of those open sets is non-empty?

[empty grove]

Yeah if you take distance/2 balls

Wait

Why non empty

Hausdorff is intersection is empty

[kind cedar]

Ah, I see where my confusion is coming from

I was understanding a Hausdorff space as "there is always something between 2 arbitrary points", and thus confusing it with a complete space, like "no points missing" in the space

[empty grove]

Ah I see

[kind cedar]

So if there are always disjoint open sets around x and y, this does not mean this open balls are "complete"

[empty grove]

Yeah

There's a T_1 space which is like hausdorff but you remove the disjointness condition

But you don't enforce non empty intersection either

[kind cedar]

But if I have a non-complete metric space, will it still be Hausdorff?

[empty grove]

Yep

Distance/2 thing always works

[kind cedar]

but what if distance/2 is a number not included in the set?

Oh

I see... it must be

[empty grove]

Not sure what you mean

[kind cedar]

If I take a set of, say, integers

[empty grove]

Ah distances are always positive real

[kind cedar]

Could it make an incomplete metric space?

[empty grove]

That's part of the definition of a metric

Like the metric topology is defined by saying that balls of positive real radius form the basis

You could change this to balls of positive integer radius but then it's not a basis for a topology

And generated topology will look uglier

[kind cedar]

Can I make a metric space upon the positive integers, and affirm it's a topological Hausdorff space with a metric, and yet incomplete?

[empty grove]

It will always be complete

Because no Cauchy sequences

If integer valued metric

Distances can't be arbitrarily small without actually being 0

[kind cedar]

So all Cauchy sequences converge?

Or I just can't construct any Cauchy sequence?

[empty grove]

Yes because they are all eventually constant

No non trivial Cauchy sequences

I think you were asking if d/2 works because d/2 may no longer be integer even if you have an integer valued metric?