point-set-topology

[sleek thicket]

Have you seen the implicit/inverse function theorems and like, vector calculus?

Because I feel like that's a sufficient background (it helps to have more though)

[gritty widget]

yes

but i'll probably just go on and do a few more chapters of rudin and then check out spivak

[sleek thicket]

That's not danworthy lol

it's a good clarification

[gritty widget]

no i mean that was me saying no

ok straight to spivak then

[sleek thicket]

Slim, can you say where the extra analysis background comes up? I just finished a course in smooth manifold stuff and I hadn't done anything like spivak

Ah okay

[honest narwhal]

I generally recommend reading chapter 7 of Rudin on principle fwiw

[sleek thicket]

The rank theorem is way easier to think of for me than the implicit/inverse function theorems at this point

Harder to prove ofc

[honest narwhal]

For me inverse is probably easiest lmao

Like yeah Jacobian is invertible => function is locally invertible

Banach contraction 🙃

But yeah I tend to think of the statement of inverse being easiest to swallow, followed by rank, followed by implicit

[dire warren]

Implicit function theorem is prolly the hardest theorem in baby analysis

[warm mirage]

Does somebody know any books about geometry in the plane and the space, the axiomatic way, not the linear algebra way?

Could be a book of problems in plane and space geometry or a course

Ping me when you answer, thanks

[lyric quartz]

So in Hatcher's proof of Van Kampen's Theorem, he does this thing where he slight perturbs each square, but doesn't justify it why you can do that. I've been told you can use tube lemma to justify it, but I can't see why. Could someone help, thanks!

Like, for example, what if square 6 before being perturbed was exactly $f^{-1}(A_\alpha)$? That would seem to ruin the whole thing of each square being mapped to one $A_\alpha$.

[gentle osprey]

ClearlyHalfAlive743:

[lyric quartz]

I've asked a question about it here also: https://math.stackexchange.com/questions/3701783/question-about-proof-of-van-kampen-theorem-in-hatcher

[supple locust]

https://www.youtube.com/watch?v=0msWLft-lT0&list=PLpRLWqLFLVTCL15U6N3o35g4uhMSBVA2b&index=6 maybe this can help.

[lyric quartz]

@supple locust , I've been watching that actually, he doesn't explain it either.

[lilac nymph]

I'm not sure your example of square 6 applies because they are showing there exists a decomposition into rectangles where every square gets mapped into one open set

not that every decomposition does

and because they are open sets in $f^{-1}(A \alpha)$ you can shift each interval by some $\epsilon$ and they will still be open and contained in $f^{-1}(A \alpha)$

[gentle osprey]

Mrgameraidz:

[gritty widget]

Let $d_1$ og $d_2$ be two metrics on $M$. Show that $D(x,y)=\operatorname{max}\left{d_{1}(x, y), d_{2}(x, y)\right}$ is also a metric on. \
\textbf{Answer} \
Since $d_1(x,y),d_2(x,y)\geq 0$ pr. definition it follows that $D(x,y)=\operatorname{max}\left{d_{1}(x, y), d_{2}(x, y)\right} \geq 0$. \
Symmetriy follows from the proberties of the $\operatorname{max}$ function. \
I have a hard time showing triangular inequality. Any help is welcome 😄

[gentle osprey]

Varimus of Sodom:

[gritty widget]

oh, thanks. that's sounds like it will do it.

[gritty widget]

$$D(x,z)=\operatorname{max}\left{d_{1}(x, z), d_{2}(x, z)\right} $$

$$\leq \operatorname{max}\left{d_{1}(x, y)+d_1(y,z), d_{2}(x, y)+d_2(y,z)\right}$$

I can not figure out how I end up with $= D(x,y)+D(y,z)$? @gritty widget

[gentle osprey]

Varimus of Sodom:

[floral gust]

@gritty widget
Can you show that $\max{ d_1(x,y)+d_1(y,z) ,d_2(x,y)+d_2(y,z) } \leq \max{ d_1(x,y),d_2(x,y) } +\max{ d_1(y,z) ,d_2(y,z) } $

[gentle osprey]

All functions are continuous:

[gritty widget]

hey, i have barely any background in topology, but I am just wondering what would 3-d ball with an empty space in the shape of a torus inside it be homeomorphic to?

[uncut surge]

Hmm, if you think about "deflating" the volume of this object, you'd end up with something which looks like the surface of the torus, but the hole in the middle of the torus still has a 2d-disk inside

oh but that's not a homeomorphism but a homotopy

aw rip

But with that kind of argument, at least you get that this should be homotopic to a 2-sphere with a handle

[gritty widget]

sorry, i am very bad with terminology - is 2-sphere like a "orange with all the flesh inside it"?

[ivory dragon]

er, a 2-sphere is the boundary of a 3d ball

i.e. its what you think of when you hear "sphere"

but only the outside edge

we call it a "2-sphere" since the surface of a ball is 2 dimensional

(as far as manifolds go)

[gritty widget]

ok, cheers)

is any ball in a way equivalent to a point?

@floral gust how do you get the last inequality?

[floral gust]

Arguing by cases should suffice.

[gritty widget]

i do not get this problem to be honest. which case could it be?

are u sure about this? i hope that u do not mislead me, haha 😅

[floral gust]

https://math.stackexchange.com/questions/1671111/elegant-proof-that-maximum-of-sums-is-at-most-sum-of-maximums

@gritty widget

[gritty widget]

how did you found that?

thanks btw !

[lyric quartz]

Still, painfully reading Hatcher's proof of Van Kampen, if anyone could help with the question I asked here, I would be grateful, thanks! https://math.stackexchange.com/questions/3719825/question-about-hatchers-proof-of-van-kampens-theorem-factorization-of-grid-pa

[pine heath]

Excuse me

Can someone pls explain

Why donut equals

One (1)

Cofffs cupsl

[gritty widget]

wut

[lyric quartz]

@pine heath they have the same genus.

[sleek thicket]

it's an open problem

Mathematicians have known this since antiquity but actually explaining why has proven extremely difficult

[lyric quartz]

Of course, I should have known I wasn't smart enough to solve such a problem

[ivory dragon]

the proof, of course, is that someone asked euclid once and he said it's true

[sleek thicket]

proof: folklore

[dim meadow]

Nice

[sleek thicket]

I wonder how hard it would be to formulate "a coffee cup is homeomorphic to a donut"

it should be easy to prove from the classification of compact surfaces

But writing down a description of a coffee cup sounds kind of awful

ig you can also probably prove it pretty explicitly if you can write down what a coffee cup is

[sleek thicket]

oh that's true

hmm, if two manifolds with boundary have homeomorphic interiors and boundaries, are they homeomorphic?

my instinct is not

although that wouldn't apply here anyways since it's just as hard to show the interiors are homeomorphic

[muted swift]

hiya, I can't find a good way to calculate the volume of intersection of two n-balls with just radius and distance

the answer to https://math.stackexchange.com/questions/162250/how-to-compute-the-volume-of-intersection-between-two-hyperspheres has an undefined variable $a$ which is never explained or mentioned, and the paper it cites http://docsdrive.com/pdfs/ansinet/ajms/2011/66-70.pdf doesn't seem to have the right formula

alternatively, if anyone knows an implementation in Python or C, that works too.

[midnight jewel]

But writing down a description of a coffee cup sounds kind of awful
if you’re not worried about smoothness you can straightforwardly write it as something like
∂((solid cylinder ∪ solid torus) \ smaller solid cylinder)
in ℝ³, where the solid torus is placed in such a way that the handle sticks out, and the smaller solid cylinder is placed with one base embedded in the interior of the larger cylinder’s base and otherwise contained entirely on the inside

and then if you really want some parametrization you can read off a piecewise parametrization from its components. it would certainly look rather ugly but it shouldn’t be too bad

to actually show they’re homeomorphic what I’d do is a proof by demonstration: draw a triangulation on the surface of a coffee mug with a sharpie and compute the euler characteristic

[uncut surge]

I think this was even once an exercise in my algebraic topology class; it's really not all that difficult to write down something coffee-cup-like

e.g. finite length cylinder, glue half a circle to the side and take out a finite-length cylinder so that you actually get some empty space in your mug

[sleek thicket]

Hmm that makes sense

[uncut surge]

@muted swift In the math.stackexchange answer that you linked, the parameter a doesn't actually mean anything; the solution to the problem is at the bottom of the answer, where the parameter a has been filled in:

[midnight jewel]

oh and I guess actually the easier way to prove that it’s homeo to a torus is: first show that the cylinder-with-smaller-cylinder-removed is a ball, then use that S¹#T ≅ T

[uncut surge]

Yeah, exactly, you proceed in small steps

It's exactly like the wikipedia gif

[midnight jewel]

(and of course argue that your construction really is just gluing a torus onto a ball)

[muted swift]

@uncut surge i suppose i should have just ignored it, lol. ty

[uncut surge]

that didn't work

[midnight jewel]

yea but we’ve all seen the animation anyway

:P

[uncut surge]

tru

[midnight jewel]

plural of simplex: simplexes or simplices?

[sleek thicket]

simplices

however complex gets pluralized to complexes

I think we should use complices

[midnight jewel]

that’d be hypercorrection

the latin plural of simplex was actually simplicēs, complex comes from complexus

[sleek thicket]

hmm have you considered complices sounds funny

[midnight jewel]

okay fine but only if you start pluralizing phoenix as phoenices

(which is actually etmologically accurate again)

[uncut surge]

my big brain method is just going with what everyone else is using

hence simplices and complexes

[midnight jewel]

which is why I’m asking :P

I’ve seen both simplices and simplexes

and wanted to know which was more popular

[uncut surge]

i might have suppressed it but i don't think i've seen simplexes

[midnight jewel]

I prefer simplices

[uncut surge]

yeah same

[midnight jewel]

I see it right in front of me in a paper from the 70s

[uncut surge]

the 70s were a wild time

i think it is a very good rule in life to never rely on what people did more than 50 years ago

[midnight jewel]

aight lemme stop relying on the concept of wires

[uncut surge]

no i'm being facetious ofc

[midnight jewel]

I know I know

[plucky veldt]

I don't like irregular plurals in English

do you pluralize opus as opera?

[midnight jewel]

I don’t really use opus very often, except in magnum opus, which is kinda always singular by virtue of its meaning

the actually correct plural is opodes btw

[plucky veldt]

wiktionary says "opus (plural opuses or opera)"

[uncut surge]

"I don't like irregular plurals in English" idk fam they're pretty much everywhere

https://www.grammarly.com/blog/irregular-plural-nouns/

it's true that latin and greek words do take a special role, but eh, it's just how language works, people find that "fungi" sounds more reasonable than "funguses"

[midnight jewel]

add to that the ones that are irregular not because they were loaned from another language but simply because they survive from an older form of the language in which they were regular, like ox→oxen or mouse→mice (minus the stupid spelling of that)

[sleek thicket]

hmm, if two manifolds with boundary have homeomorphic interiors and boundaries, are they homeomorphic?
@sleek thicket

I asked this earlier but it kind of got buried

Would also be curious about the smooth case

[midnight jewel]

I’m not convinced it’s true in the continuous case

I don’t see how you can guarantee that you can stitch together the homeos on the interor and the boundary to a continuous map

like you’d have to choose them in some clever fashion

[sleek thicket]

Yeah I don't think it is either, I just can't think of a counterexample

[dim meadow]

I have a counterexample. Take the plane with a small disc cut out of every square [n, n+1]x[m, m+1]

Now fill in the boundary of every disc for one manifold with boundary, and fill in some infinite subset of the boundaries of the discs for the other manifold with boundary

@sleek thicket

Does this make sense?

I think this works

I can draw a picture if you need lol

[sleek thicket]

hmm it sounds like it does

Yeah I think I see

The boundaries will both be a disjoint union of countably many circles

[dim meadow]

Yeah

And the interior will be the same set

[sleek thicket]

Yup, that looks good

and the two sets won't be homeomorphic because no subset of the first space is like, an open annulus

Or something

[dim meadow]

Maybe an even easier example is just taking a disjoint union of countably many closed intervals, and a disjoint union of countably many closed intervals and one open interval

[sleek thicket]

oh yeah that's very simple

and it's easier to show that they're not homeomorphic

[dim meadow]

Yeah

[sleek thicket]

just look at connected components

[dim meadow]

Exactly

I think the annulus idea can be made to work but this is easier

So 🤷‍♂️

[sleek thicket]

Oh I think I have a better idea for the 2d case

Oh nvm I was wrong

[dim meadow]

Lol

Oh actually suppose they are homeomorphic. Then their one pt compactifications are homeomorphic. The one point compactification of the first one is a sphere with a countable number of small holes torn out with the boundary filled in. As such it is a manifold with boundary. The one point compactification of the second one is not a manifold with boundary around the point you add in

Lol I used this sort of argument a lot on my topology qual

[sleek thicket]

I think that makes sense?

[dim meadow]

Although for this to work you need an infinite number of discs with missing boundary in your second manifold

[sleek thicket]

Yeah

[dim meadow]

No I'm wrong

You just need at least 2

[sleek thicket]

okay then I am confused

Why isn't a manifold with boundary?

[dim meadow]

That way any small neighborhood around the point you added in will not be connected

[sleek thicket]

actually, why isn't the one point compactification of the second an open subset of the one point compactification of the first?

[dim meadow]

Because you also need to compactify the discs with missing boundary

[sleek thicket]

ahhh

okay that makes sense

[dim meadow]

So as long as you have 2 of them you aren't going to be connected

[sleek thicket]

I was thinking of everything inside a sphere

Which isn't right

Why wouldn't it be connected? It seems like it would fail to be locally Euclidean

[dim meadow]

Because any small neighborhood of the added point will include an annulus around the discs with missing boundaries

[sleek thicket]

I'm not seeing how that fails to be connected. My current picture is that we take a sphere, remove countably many points, then glue three points together

So lmk if that's wrong

[dim meadow]

Yeah that's not a good picture

One sec

I'll draw something

Okay so this is the picture I had in mind

Suppose you put in a point to compactify it

Then any neighborhood around that point will also be a neighborhood around the discs with missing boundary

[sleek thicket]

Oh you're saying you have two filled disks?

Sorry disks with filled boundary?

[dim meadow]

No that's just part of the picture

Sorry

I have 2 discs without boundary

[sleek thicket]

Okay, so in my mind we fill each of those in with a single point and then glue those three points together

But that's wrong?

[dim meadow]

That's fine

Why do you think that's not locally Euclidean besides for the connected thing

I guess it isn't locally Euclidean actually

You're right

[sleek thicket]

You have six directions to travel

Instead of two

[dim meadow]

Can you phrase that topologically?

[sleek thicket]

Any neighborhood of infinity contains a neighborhood homeomorphic to three open disks glued together at their origins

[dim meadow]

Right and if you take out that point then you are disconnected

But that's bad

[sleek thicket]

ohhh!

I thought you meant the space is disconnected overall

[dim meadow]

that's what I was trying to say lmao

But I was confused

[sleek thicket]

lmao

Okay so I think we understand eachother now

[dim meadow]

Okay great

I like this argument a lot btw

[sleek thicket]

I've never used one point compactification like this

It's pretty cool

[dim meadow]

I used it for 2 questions on my qual lmao

[sleek thicket]

Dang

[dim meadow]

I came up with this argument solving qual problems

And it ended up basically passing my qual for me

[sleek thicket]

I was thinking about taking the manifolds/topology qual just to test myself

Studying for it would probably help cement all the stuff I learned last year

[dim meadow]

Yeah that's a good idea

Here's a couple of problems that this method solves btw

Show that the torus minus a point and the sphere minus three points aren't homeo

[sleek thicket]

Ah interesting. They're homotopy equivalent right?

[dim meadow]

Yeah

[sleek thicket]

right, and if you compactify the first you get the torus, but if you compactify the second you don't get a manifold

[dim meadow]

Yeah exactly

[sleek thicket]

well 2 gives me the suspicion that π1(X) is trivial

lol

[dim meadow]

Yeah

[sleek thicket]

oh right, it's presented by <α|α^2=α^3=1>

By fun stuff about cell complexes

and that's obviously trivial

hmm can I describe this as a disk attached to P^2? I think so

Basically I'm thinking about what happens if I remove a point from X

Can I deformation retract onto P^2?

[dim meadow]

I actually forgot the argument I used for this lmao

[sleek thicket]

Yeah so suppose I attach a disk to a complex Y of dim 2

and then I remove a point from the interior of the disk

Does that deformation retract onto Y?

Or even retract at all actually

[dim meadow]

Oh I know the argument I used actually

I was just like "look at a point on S^1, and look at a small neighborhood around it"

If you go close enough you should be able to get R^2

[sleek thicket]

damn we should name spaces like that

[midnight jewel]

sorry I was trying to read the convo here but I’m confused, what do you mean with “filling in the boundary of the disks” exactly?

[sleek thicket]

Liquid-folds or something

[dim meadow]

Lol

[midnight jewel]

the only way I can interpret it is putting a disk back in btu that would just do nothing

[sleek thicket]

Sascha, compare R^2 minus an open disk vs minus a closed disk

[dim meadow]

Yeah

[midnight jewel]

oh, adding the boundary back

[sleek thicket]

If you remove an open disk that has a "filled in boundary"

[midnight jewel]

yea

[sleek thicket]

right

[midnight jewel]

I thought you wanted to glue something in there

makes sense

[dim meadow]

Anyway the one dimensional example is super easy

[sleek thicket]

The intervals example is better yeah

[midnight jewel]

all of these counterexamples so far are noncompact tho. what about compact ones?

unless you discussed that later I scrolled down eventually

[dim meadow]

Probably compact ones you can't do these sort of shenanigans

[midnight jewel]

if so I’ll read up

[dim meadow]

That's my assumption anyway

[sleek thicket]

definitely the trick we're using doesn't work

[midnight jewel]

obviously any counterexamples for compact ones would have to be more than 2-dimensional

[sleek thicket]

Wait why?

[dim meadow]

Why?

[midnight jewel]

because of the classification theorem

[sleek thicket]

Idk the classification of compact surfaces with boundary

[dim meadow]

Yeah

It's not so straightforward

[sleek thicket]

the interior won't even be compact

you know the boundary is a disjoint union of finitely many disks

[midnight jewel]

compact surface is homeo to genus n surface with k disks removed
or is that only true in the smooth case?

[sleek thicket]

¯\_(ツ)_/¯

Maybe that's true

[dim meadow]

Idk

[midnight jewel]

I know such a statement exists, good friend of mine is using it in her bsc thesis

but idk what assumption she’s using exactly

haven’t read the thesis yet

but we’ve discussed it a bit, she’s been struggling to find a source of the exact formulation of it she wants :P

[dim meadow]

Okay anyway I was finishing my argument with that problem we were talking about shamrock

So you take a small neighborhood of a point on S^1

You compactify, we want to show that the compactification is not S^2

[sleek thicket]

Remove that point and then compactify?

[dim meadow]

No take a nice small open neighborhood and compactify

[sleek thicket]

also Wikipedia agrees that a compact surface with boundary is a compact surface with finitely many open disks removed. Cool!

oh okay

[dim meadow]

Yeah that's cool

So basically the idea is that the stuff on S^1 that's in the neighborhood gets turned into a loop

[sleek thicket]

Ah right, if we do this on S^2 we get S^2

[dim meadow]

But if you remove that loop you should get 2 connected components

But in actuality you get 5

[sleek thicket]

Woah

[dim meadow]

Jordan curve theorem lmao

So yeah

[sleek thicket]

this is the most recent uw prelim. I haven't looked at it before

[midnight jewel]

wait hang on I need to read up on what you wrote above cause said friend of mine asked me a question that might be related to this

[sleek thicket]

I think for 1 it should be like, deformation retract onto a wedge of two circles?

ah sorry

[midnight jewel]

the question was this: if you remove a simple closed curve in a connected 2-manifold, can you end up with more than two components?
I cannot think of a way this could fail but I couldn’t come up with a proof in two minutes either

[sleek thicket]

Liquid's example wasn't a manifold I think

[dim meadow]

Yeah it wasn't

That's what I was demonstrating lol

[sleek thicket]

how does the proof of JCT go?

I would look at that first

if I was trying to prove this

[midnight jewel]

uses homology, idr how it went exactly tho

but the case of a torus is fundamentally different from that of a sphere because non-separating curves do exist

so JCT by itself is already wrong since you can end up with the loop not bounding two components

[sleek thicket]

yeah, that's true

I'm not saying that JCT holds

[midnight jewel]

idk if it could easily be modified, I’d have to read up on it again

[sleek thicket]

I'm saying it's worth looking at the proof

[midnight jewel]

but maybe one can prove it assuming JCT

[dim meadow]

I actually think you can get lots of components

Hmm are we allowing boundary?

[midnight jewel]

the loop doesn’t touch the boundary

but the manifold may have one

[sleek thicket]

oh I think I have an example

[dim meadow]

Oh okay

[sleek thicket]

C\Z

Take little loops around the integers

Connect them with straight lines

Oh sorry

Simple closed curve

[midnight jewel]

that’s not a simple curve yea

[sleek thicket]

Yeah I realized this would work for the plane lol

[dim meadow]

The example I had had the loop on the boundary

[midnight jewel]

yea it’s quite easy to come up with arbitrarily many components that way that was the first thing I thought of :P

[dim meadow]

Yeah

[midnight jewel]

btw if it’s true in the smooth case that’d suffice, but I’m also interested in general

[dim meadow]

Okay I actually believe this lol

[midnight jewel]

I definitely believe it in the smooth case

though I couldn’t come up with anything nice right away, just some wishy-washy argument using a tubular neighborhood to argue that there’s only two components “locally”, i.e. the points neat the curve belong to one or two components. and then you have to argue that any other point on the surface belongs to one of those

which should be true if we assume connectedness of the original manifold

since you can draw a line from any point to the curve

[dim meadow]

I think a homology argument could make sense

[midnight jewel]

in the non-smooth case you’d probably need some more heavy machinery though, yea

(though I guess existence of tubular neighborhoods is kinda heavy machinery?)

[dim meadow]

Eh

[midnight jewel]

I consider everything I haven’t seen the proof of in class to be heavy machinery by default :P

except maybe classification of 1- and 2-manifolds

[sleek thicket]

I didn't understand the proof of the tubular neighborhood lemma and I was like "okay, I'll just blackbox it"

It seemed amenable to blackboxing

[marsh forge]

Classification of 2 manifolds is not so bad actually

If you ignore jct anyway

[sleek thicket]

Then all of my homework problems that week were about slightly modifying the proof of the tubular neighborhood lemma

[midnight jewel]

I mean we covered jct in a class I passed so I am officially allowed to believe it

[sleek thicket]

lol

[marsh forge]

Lmao

I blackbox all the time

Its basically impossible to read papers without blackboxing at least a few results that are either folklore or in another paper

[midnight jewel]

I do wish they’d provide sources for the poor guy who has to write a bachelor’s thesis on the topic tho

[sleek thicket]

Yeah that's something I've been pretty uncomfy about in my REU

[midnight jewel]

cough

[marsh forge]

Esp if the proof is just a computation you could un theory do

In*

[sleek thicket]

there's a lot of knot theory that I don't really have time for

[marsh forge]

Thats the point of the bachelors thesis hahaha

[sleek thicket]

And isn't going to come up directly in what I do

[marsh forge]

You are that person

[sleek thicket]

it's just that nothing I do this summer is actually useful without all the knot theory results

[midnight jewel]

eh, finding sources sounds like a waste of time to me, it’s understanding and making them readable that’s actually valuable

[marsh forge]

Im catching up in my field by just reading all the advanced reu papers

My paper will just be their colimit

[sleek thicket]

U Chicago papers are a blessing

[marsh forge]

Ikr?

Its crazy

[sleek thicket]

Those and Keith Conrad

[marsh forge]

They are great bc they are written by dumb ppl for dumb ppl

[midnight jewel]

what’s special about them?

[sleek thicket]

he has great exposition

[marsh forge]

They are all expository sascha

So they are basically mini textbook chapters

Normally on topics without any good teaching material

[midnight jewel]

ah nice

[marsh forge]

UChicago’s reu honestly deserves funding just for that output

[midnight jewel]

(what does reu mean I think I have to ask at some point)

[marsh forge]

I unironically use them without noticing sometimes

Research experience for UGs

[sleek thicket]

they're summer programs for undergrads to do research

[midnight jewel]

oh that’s nice

[sleek thicket]

(in the US? Also Canada maybe?)

[marsh forge]

I wonder if the uchi reu could like sell out to springer or smth lol

Make a series of the best ones

[strange folio]

Oh Keith Conrad ❤️

[marsh forge]

Zeta do you ever wsnt to write a textbook

[midnight jewel]

my uni doesn’t do anything like that for math though you can hit up any prof and be like hey wanna guide me through some research? whenever you want. but it rarely actually happens cause walking up to a prof and asking that is scary and a lot of them are just too busy

[sleek thicket]

yeah these are usually pretty competitive

And not at your home school

[strange folio]

Yeah, I have a lot of parts of textbooks written. It is only a matter of time haha

[marsh forge]

And paid

[sleek thicket]

Oh that too lol

[midnight jewel]

the best that happens usually is like a reading course where you may be asked to write a summary or sth of what you learned but nothing actually interesting

oh wow

I want

[marsh forge]

A little late to apply this year i think

[sleek thicket]

Yeah people sometimes publish from these, but idk how valuable the research really is

[midnight jewel]

I mean I’m also in ever so slightly the wrong country

[sleek thicket]

A lot of places won't fund people outside the US yeah

some do, I think it's an nsf grant thing

[marsh forge]

Uchicago funds them like small potatoes

Yeah it is

[midnight jewel]

and also not due to be an undergrad for much longer

and I still have exams coming up and a thesis to finish :P

[sleek thicket]

i'm finding it very hard to focus and self direct

[midnight jewel]

my uni considers summer to be a time of learning

[sleek thicket]

research hard, who knew

[dim meadow]

Lol I actually read a lot of Keith Conrad's stuff about PSL(2, Z) today

[midnight jewel]

classes end in june, exams are in august

[marsh forge]

Honestly i find overscheduling helps sham

[midnight jewel]

use your time however you want

[marsh forge]

Like if you give yourself strict hours

Its a lot easier

[sleek thicket]

Yeah I should start doing that

[midnight jewel]

aye I’ve been much more productive last week than like, the entire semester

under the semester I tended to ahve classes starting at like 10

so I just slept in till 9

[sleek thicket]

oh yeah I woke up 3 minutes before my meeting today

[marsh forge]

Ive gotten to the point where im better at math high which has been an experience

[sleek thicket]

because I thought it was Tuesday

[midnight jewel]

now that I don’t have classes I get up by 8 and start working like half an hour later, and that’s been so much more productive

[marsh forge]

I think weed should be an off label adhd treatment

Yeah im shifting to 7 over the next few days

[topaz carbon]

🤔

[midnight jewel]

well let’s not be unreasonable. but some 4 hours of work pre-lunch and some more in the afternoon is plenty for me, I don’t actually have that much to do really

though I do have to somehow learn riemannian geometry

which I both was barely able to follow during the semester and want to ace cause I aced differential geometry I and at that point it’s a thing of pride

and then next semester I won’t be able to do differential geometry III (which doesn’t usually exist at all the prof just felt like doing more) because it overlaps with another large course I wanna take -.-

[marsh forge]

Im out of school

[midnight jewel]

everything overlaps

[marsh forge]

You too right sham

[sleek thicket]

yup

my finals week was last week

So I'm officially out

[marsh forge]

Same

[midnight jewel]

I’ll still be here a while

[marsh forge]

But i had no finals

[sleek thicket]

I had two finals

[midnight jewel]

master’s, then teaching diploma

[sleek thicket]

Did manifolds over the weekend

[midnight jewel]

that’s another 3½ years or so

[sleek thicket]

Cleared my REU schedule to do algebra on Monday

Got 4 hours in and they were both canceled

I was not pleased

[marsh forge]

Oh no hahaha

[sleek thicket]

I mean I didn't have to finish algebra which was nice

But it disrupted my schedule a ton

[marsh forge]

Were the problems cool at least

[sleek thicket]

I didn't do the cool ones yet because they looked hard

[midnight jewel]

well anyway I’mma go to sleep

[marsh forge]

Also sham idk if you will find this interesting but i want to nerd snipe someone else

[sleek thicket]

oh no

[midnight jewel]

gotta actually do that at some point if I wanna stick to my schedule, alas

[marsh forge]

Ive been thinking about the following ‘problem’

[sleek thicket]

Night!

[midnight jewel]

okay I’ll have to tab out now before I see this problem and get sniped

[sleek thicket]

lol

[dim meadow]

Lol

@marsh forge you better not post the problem

I may snipe you

[marsh forge]

Ok let T be a nice subcategory of Top. We have a ton of invariants broadly of the form hT-> A for some ‘algebraic’ category A. Are there functors T->T such that T->T->A can differentiate within a homotopy type?

[sleek thicket]

hmm

I have a dumb example I think

[marsh forge]

Even better, can we make T injective up to iso

Sorry

[sleek thicket]

so my dumb example is to factor T -> Set -> T using the discrete topology

[marsh forge]

T->T

[sleek thicket]

Send a space to like, the number of points where if you remove them you get > 1 connected component

and then let T -> A be the 0th homology group

[marsh forge]

Yeah we want the composite to be factorable

[sleek thicket]

T -> Set -> T -> A is just a number

[marsh forge]

Sorry

Computable*

[sleek thicket]

The number of elements of T -> Set

Hmm fair

[marsh forge]

I guess thats computable

But bijectivity is too simple yeah

So the inspiration

Is that benson farb thinks

That for a some class N if spaces

[sleek thicket]

Is one point compactification functorial? Or is it only functorial on proper maps or something?

[marsh forge]

Taking the configuration space of the N

And then computing pi1

One might be able to yield something

I cant remember what N was lol

But is was a specific group of htpy equivalent spaces that were not homeo

[sleek thicket]

Yeah you need proper maps

To be continous at infinity

I don't know a lot of functors T -> T...

[dim meadow]

How does proper maps make it functorial?

I haven't seen that

[sleek thicket]

you send infinity to infinity

[dim meadow]

Okay

[sleek thicket]

It's continuous at all other points in the codomain

hmm maybe I'm wrong

[dim meadow]

No I think this might make some sense

[sleek thicket]

But I was thinking a nbhd of infinity is like, {infinity} cup Y\K

and this pulls back to {infinity} cup X \f^(-1)(K)

[marsh forge]

Functoriality is a bit strict too

[sleek thicket]

And f^(-1)(K) is compact by properness

[marsh forge]

If its computable on objects and distinguishes well

Functoriality is just icing

[sleek thicket]

well one point compactification can distinguish within homotopy type

Liquid and I were talking about this earlier

The sphere minus three points and the torus minus a point

or the circle and the punctured plane

[marsh forge]

Oh interesting

[sleek thicket]

hmm the universal covering space is going to be homotopy equivalent for homotopy equivalent spaces, right?

Well the homotopy groups will be the same

[marsh forge]

It should be for cw

You should be able to lift and apply WH

[sleek thicket]

Ahh I wasn't sure how to get the map

But ofc you can lift

You can probably maybe prove it explicitly (w/o whitehead)

hmm

oh, compactly supported cohomology?

that's functorial on proper maps too

But isn't preserved by homotopy type

[marsh forge]

Oh interesting

[dim meadow]

Yeah the proper map thing is cool

[sleek thicket]

R^n has compactly supported cohomology equal to Z in dim 1 and n and zero elsewhere

[dim meadow]

I like that a lot

[marsh forge]

We should make a bracket

“What is the best subcategory of Top to work in”

[sleek thicket]

easy, the one generated by the Hawaiian earring and all limit/colimits

[dim meadow]

Lol

[marsh forge]

Shamrock is banned from even entering the bracket

[sleek thicket]

Actually what about the subspaces of Euclidean space and all limits and colimits

you get manifolds for sure

And weirder things

You get nasty spaces

[dim meadow]

Lots of weird shit

[marsh forge]

Do you get CWs too

If you get all manifolds you get the ingrednuents

Then its just pushiuts

[sleek thicket]

what's a space that can't be described in this way? Sierpinski?

I think indiscrete spaces won't work

But it's a big category

[marsh forge]

And therefore bad

Im rooting for CE

CW

[sleek thicket]

I vote CE

[dim meadow]

You can make the Hilbert cube

[sleek thicket]

You have to have computably enumerable open sets

[dim meadow]

Probably you can make every compact metric space?

[marsh forge]

Interesting

The category of Computable Metric Spaces

[sleek thicket]

CW is pretty good for sure

CGHS is also good

Or w/e it is

[dim meadow]

I actually want to study some computable metric space stuff

[marsh forge]

I sent u a link

Did u read it

[dim meadow]

Lol not really

[marsh forge]

pain

[dim meadow]

Lol I have too much to read now

And the person I'm mainly supposed to be reading with ditched me

[marsh forge]

Im telling you the birkhoff stuff

Is begging for a generaization

[dim meadow]

Fine I'll read it

[marsh forge]

Hell yeah

I think its completely morally true that any space with with properties that hold a.e with a decent notion of randomness have that property for all random points

[dim meadow]

How do I call the guy who's ghosting me out for ghosting me

Ghosting me in the mathematical sense

[marsh forge]

Just ask in public

[dim meadow]

I can't

Cause corona

We are doing stuff over zoom

[marsh forge]

Group chat

?

[dim meadow]

We aren't really in any active ones

He's the PhD student of my prospective advisor

And my friend from before that

[marsh forge]

Yike

Maybe ask the advisor if he knows if smth is going on?

[dim meadow]

But we mostly hung out at the campus

Idk

[marsh forge]

Its kinda behind the back

[dim meadow]

Like I don't want to do that

I just want to talk about math

Lol

Not start drama

[marsh forge]

Burnout is real rn

[dim meadow]

With people I actually care about doing math with

Yeah

[marsh forge]

The only reason i can do math today

Is that i didnt at all last week

[dim meadow]

Dude it's so hard for me to work with other people rn

Like I was working with my friend today

[marsh forge]

Yeah i work alone only rn

[dim meadow]

And somehow I expect a lot more detail than I used to I guess

So I kept calling him out on detail stuff

[marsh forge]

Thats kinda extra haha

[sleek thicket]

it's so hard to just, like, communicate. I can't draw pictures!!!

[marsh forge]

I only ever do bit picture in groups

Big

[sleek thicket]

I end up gesturing a lot

[dim meadow]

Idk like if I can't see it in my head I freak out a bit

[marsh forge]

Yeah ipad w stylus is life saving

[sleek thicket]

yeah that's a mood liquid

[dim meadow]

Well the real problem was he kept saying Newton's method

Instead of the Euclidean algorithm

And I was like "is there some discrete Newton's method I don't know about"

So I kept asking him to show me why what he said was true

Eventually I realized he meant the Euclidean algorithm

But we were both pretty annoyed at the time

[sleek thicket]

ugh

[marsh forge]

If i have to communicate details i feel like

Taking a moment alone

And texing it up

Is worth it

[dim meadow]

But there was some other moment when I like drew a shitty picture and was about to talk about the proof

And he was like okay that's the proof

[marsh forge]

Hahahaha thats me

[dim meadow]

And I was like "what do you mean, I haven't said anything yet"

[sleek thicket]

it's hard to communicate with people when you want different levels of detail

[marsh forge]

I feel like theres a bit split between

People who only care about the big idea

And people who only care about details

Like if a claim seems morally true

[sleek thicket]

i have never in my life had a big idea

[marsh forge]

Idc about the proof at all

[sleek thicket]

I've had like, 3 ideas total

[dim meadow]

Lol

[sleek thicket]

and they were all puny

[marsh forge]

Lmao

[dim meadow]

I'm a details person now

I think I used to be an ideas person

[marsh forge]

Im pure ideas ive never even met details

Who is she

[dim meadow]

But that was beat out of me

[sleek thicket]

I'm becoming more of an ideas person

[floral gust]

and then there come people "Well emptyset is a counterexample"......

[sleek thicket]

but like, my starting point was formal verification

[marsh forge]

Dear god

If anyone ever says

[sleek thicket]

so I'm still more detail orientated than 90% of people

[marsh forge]

What about the empty set

In my class

They fail

[dim meadow]

If it's a counterexample it's a counterexample imo

Like you have to account for that sort of thing

[marsh forge]

Idk most of the time the answer is

It works obviously

Or it doesnt work obviously

And you just fix the theorem by specifying

So like

[dim meadow]

Yeah

[sleek thicket]

I think it's awkward when you ignore trivial cases and then you do induction

[marsh forge]

No need to nitpick it

[sleek thicket]

because you run into the trivial case

[marsh forge]

Fix it in ur own head

[sleek thicket]

And it's easy to forget to handle it

Or forget you special cased it

[marsh forge]

Idk is that hard

Thats never taken me more than like

[dim meadow]

I guess I do a lot of finicky things

[marsh forge]

10 seconss

Seconds*

[sleek thicket]

my concern isn't fixing the trivial case

[marsh forge]

You just ask what should be morally true

[sleek thicket]

It's not being aware that it's there

[dim meadow]

Lol

[sleek thicket]

And then having an incorrect proof because you didn't think to check it

[marsh forge]

Oh i dont use induction w/o checking base

I think thats a recipe for disaster

At least check in ur head

[dim meadow]

Suppose that if there are n horses then they are all white

[marsh forge]

Like the base case for H_n(S^n) is the hardest part

[dim meadow]

Clearly for large enough n there are not n horses

[sleek thicket]

Wait but your horse is brown?

[dim meadow]

Therefore all horses are white

[sleek thicket]

Lmao

Contradiction: there are a bounded number of horses in existence

[dim meadow]

Look I don't check my base cases

And if you do you are being anal

[topaz carbon]

🤔

[dim meadow]

Lol

Idk I just get worried

Because I know bad stuff can happen

[marsh forge]

I do not feel fear

I am so certain in my intuition i will cut my head off

[dim meadow]

Like I've spent days on edge cases

[marsh forge]

I ignore edge cases

[dim meadow]

Lol

[marsh forge]

If ur an edge casr

I just add an afjective to my theorem

[dim meadow]

I chase the edge

[marsh forge]

Adjective

[sleek thicket]

here's an example of how worried I am generally about counterexamples and not being fully rigorous

[marsh forge]

I believe all math should be done in pairs

Across this divide

[sleek thicket]

The first time I heard the intermediate value theorem I thought it was false

[marsh forge]

Lmfao

I never doubted it

[topaz carbon]

🤔

[marsh forge]

For a second

As soon as i saw that first picture

[dim meadow]

Imagine if Max was religious lmao

Holy shit

[marsh forge]

Woah

Liquid

You solved it

[dim meadow]

You're a zealot

[topaz carbon]

he doesn't seem that zealot-y

[dim meadow]

Zealous

[topaz carbon]

no i like zealot-y more

sorry

[sleek thicket]

Max has great zeal

[topaz carbon]

max do something zealot-y

i wanna see

[little hemlock]

dumb q but how does expressing L_x as that composition makes the continuity obvious here?

oh yea you can just take projections. (x,g) -> xg still isn't clear to me though. im not sure exactly what you mean by "assumption of top group"

ahghgghh right ofc i forgot already

thanks lol

[cobalt lily]

intuitively, how is a circle a 1 dimensional topological manifold?

my current thought is the loose definition of a homeomorphism could mean flattening the circle into a line, but i dont think this meets the rigorous criteria of the formal definition

would appreciate an @ if anyone has any insights : )

[marsh forge]

draw a circle

mark a small neighborhood

it looks like an interval in the real line

a manifold only has to look locally like R^n

globally it can be wild

(Think about how from where you are standing, earth looks like a flat plane, but from space it is a sphere)

[ivory dragon]

kinda related fact: $\bR / \bZ \cong$ a circle

[gentle osprey]

Namington:

[marsh forge]

(that is kind of misleading because its unclear what the quotient means unless you've seen it before)

[ivory dragon]

eh, fair

[marsh forge]

(it's not quotient Z its the orbit space of a Z-action)

[ivory dragon]

but yeah as max said we only care about local behaviour

if you "zoom into" part of a circle enough

[cobalt lily]

"mark a small neighborhood", along the circumference?

[ivory dragon]

it starts to "resemble" a line

uh

when we say "circle" we mean the boundary

in math

[marsh forge]

A circle is only the circumference

[ivory dragon]

this is given by the definition

[cobalt lily]

well now i just feel like a goof

[ivory dragon]

its the set of points (x, y) such that x^2 + y^2 = 1

[cobalt lily]

crystal clear now

[marsh forge]

hahahaha

happens

[cobalt lily]

TIL what a circle is

[ivory dragon]

we use the word "disc" to refer to the "area inside" a circle

(a "closed disc" includes the boundary/circle, an "open disc" doesnt)

[cobalt lily]

well, now even more things make sense!

[marsh forge]

and sometimes we just say disc and neither audience nor author is sure which one they want

[ivory dragon]

the true topologist's approach

"circle" generalizes into "sphere", and "disc" generalizes into "ball"

so a circle is a 1-sphere (hence S^1) and a disc is a 2-ball

[marsh forge]

broke: circle generalizes to sphere; woke: circle generalizes to torus; bespoke: circle generalizes to wedge of circles

[hexed holly]

how does a circle generalize to a torus

[marsh forge]

$torus \cong S^1\times S^1$

[gentle osprey]

MaxJ:

[marsh forge]

where $S^1$ is the circle

[hexed holly]

right

[gentle osprey]

MaxJ:

[marsh forge]

Sphere = $\Sigma^n S^1$

[gentle osprey]

MaxJ:

[marsh forge]

n-Torus = $\times_n S^1$

[gentle osprey]

MaxJ:

[marsh forge]

Wedge = $\bigvee_n S^1$

[gentle osprey]

MaxJ:

[marsh forge]

so idk i find it a bit arbitrary to say one of these constructions is the true generalization

[hexed holly]

a bouquet is a co-torus

[marsh forge]

lmfao

[ivory dragon]

max thats fair but

i think if i said "higher-dimensional analogue of a circle"

you immediately think "sphere"

[marsh forge]

yeah ofc

[ivory dragon]

even if yes, theres multiple notions of generalizations

[marsh forge]

i was memeing

[ivory dragon]

listen max, im the only one allowed to meme here

[marsh forge]

also its funny to me that the only one of these constructions where the indexing doesnt work properly

is the sphere

i.e. $S^n=\Sigma^{n-1} S^1$

[gentle osprey]

MaxJ:

[marsh forge]

not $\Sigma^n$

[gentle osprey]

MaxJ:

[hexed holly]

sigma^n of S^0

[marsh forge]

so really spheres are generalizations of S^0 and the S^1 thing is a coincidence

[hexed holly]

two points

[marsh forge]

im aware what S^0 is lmfao

[hexed holly]

ok just making sure

[marsh forge]

lmao

[cobalt lily]

the idea of "quotients" came up in the book i am reading, and then you mentioned it, and now you are using the elementary "times" symbol

[hexed holly]

some people think it's three points

[marsh forge]

no one thinks that

[hexed holly]

you can't prove that

[marsh forge]

lol

[cobalt lily]

googling these things was not very helpful even with topology terms, is there a resource you would recommend that covers it sufficiently but also concisely?

[marsh forge]

Hatcher's topology notes, Janich's book, or munkres book

I have listed these in reverse order of popularity

[hexed holly]

never heard of Janich's book

is it good

[marsh forge]

i remember liking it years ago

iir it has more prereqs

such as linear algebra is assumed

but it also doesnt do the boring set theory munkres does

[hexed holly]

are there people who study alg top before linear algebra lmao

[marsh forge]

not really

but point set maybe

point set doesnt really use much LA

the LA comes from the algebraic part of algebraic topology

[plucky veldt]

I think you have to know abstract algebra before algebraic topology

and abstract algebra includes linear algebra

[marsh forge]

you need to know like, a baby version of every basic AA topic

like you need to know groups but you dont really need much group theory

modules are maybe the most extensive thing

but you can blackbox and use vector spaces in a first pass

so like

i would say that if you take a standard college AA course you are most likely overpreppared for intro-level AT

[plucky veldt]

I'm doing topology without tears and in exercise 3.2 #3 it asks me to prove that if every infinite subset of X is dense in X, then T is the finite-closed topology

can't it be the indiscrete topology as well for example?

or am I missing something

[hexed holly]

😢

[marsh forge]

look for errata mzdune but idr want to think about point set in the morning lol

also use something else topology without tears looks kinda bad

[hexed holly]

yes indiscrete topology satisfies this property

[plucky veldt]

why does it look bad?

[hexed holly]

i've tried reading it as well and couldn't get into it

Munkres was much better

[marsh forge]

It looks kinda slow and boring

idk the whole "without tears" thing implies normal pointset is "with tears"

and i dont think thats correct point set is all definition pushing

[plucky veldt]

I like topology without tears, it's accessible and rigorous at the same time

good for me as a beginner

[marsh forge]

fair enough ig

[plucky veldt]

I haven't tried anything else but I like this one

but this is the first case where it seems like there's an error

I was able to prove the T1 one, and the implication that in finite-closed every infinite subset is dense, but the other way around seems wrong

[hexed holly]

you are meant to take (i) and (ii) together, not separately

[plucky veldt]

ah, ok

so Im just stupid

I knew I was missing something, thanks

also, I was wrong that that a T1 space is the finite-closed topology

since it doesnt say in the definition that infinite subsets arent closed as well as singletons

[gritty widget]

Is it possible that, given a continuous function f: R-->R, and a closed set C in R, f(C) is not closed?

[wicked trout]

Yeah - e^x on all of R

@gritty widget

[gritty widget]

I don't see why @wicked trout

I got for example [0, 1]

f([0,1])=[-1,-2.72]

[wicked trout]

a continuous function will always map a compact set to a compact set

so in the case of R, you look for a subset that is closed but not bounded

for which the whole real line is a good contender

oh wait i see what you mean

do you mean EVERY closed set gets mapped to an open set?

(or at least non-closed)

@gritty widget

either way you can just take f(x)=e^x and C=R

because R is always closed in itself, so if you just want a single subset with that property then you’re done

[gritty widget]

(or at least non-closed)
@wicked trout
Yes

[wicked trout]

otherwise you can give R the trivial topology

then every closed subset maps to a non-closed set

i think

[gritty widget]

because R is always closed in itself, so if you just want a single subset with that property then you’re done
@wicked trout
But f(R)=R, doesn't it?

[wicked trout]

no, f(R)=(0,infty)

[gritty widget]

True

My bad

[gritty widget]

Is there any open map f (meaning open sets get mapped into open sets) between topological spaces that is not continuous?

[surreal rain]

There is no one one and continuous function from A onto $(0,1)$ implies A is conpact?? (T/F)

[gentle osprey]

Manas:

[gritty widget]

Is there any open map f (meaning open sets get mapped into open sets) between topological spaces that is not continuous?
<@&286206848099549185>

[gloomy plover]

Depends on which topologies you are considering

(as to construct an example of an open, non-continuous function)

(because generally speaking, open maps are not necessarily continuous)

[little hemlock]

u can make an easy example by mapping an indiscrete space into a discrete space

[gritty widget]

Depends on which topologies you are considering
@gloomy plover
Any

(because generally speaking, open maps are not necessarily continuous)
@gloomy plover
That's what I wanted to convince myself of

Can you state a couple of examples?

[little hemlock]

Let X be a set. Let (X, o) have the indiscrete topology and let (X, t) have the discrete topology and let f be the identity map of the underlying sets. Its open because f(X) = X which is open in (X,t) and f(emptyset) = emptyset is open in (X,t).

But f^-1{x} = {x} is not open in (X, o) for any x in X.

[gritty widget]

Oh I see.. clever example

Thanks!

[little hemlock]

working with the extremes often works out like that :p
np

[weak vigil]

Let X be a topological space and Y $\subset$ X st Y $\neq$ X but Y $\cong$ X (considering in Y the subspace topology). Do such X,Y exist?

[gentle osprey]

Tears:

[little hemlock]

yes. Think about examples in R

[weak vigil]

hm, any clue?

[bitter yoke]

Take X to be R

[weak vigil]

@bitter yoke but there's no But there's no $\mathbb{R}$ subset Y st R $\cong$ Y

[gentle osprey]

Tears:

[little hemlock]

hint: ||exponential function||

[weak vigil]

I'm not seeing where I should go next 😦

[little hemlock]

e^x is a bijection between R and a proper subset of R.

namely, its a bijection from R to (0, infty). We know that e^x and its inverse, log(x) are continuous functions, so it is a homeomorphism

[midnight jewel]

why such complicated example? take X = [0,2], Y = [0,1]. then multiplication by 2 is a homeomorphism

[little hemlock]

oof lol. apparently complicated examples are just the first ones that come to mind.

[midnight jewel]

unless I missed an assumption somewhere

[weak vigil]

That's just what I was thinking about, thnaks 🙂 solved it

[little kettle]

Is there any function f: X --> Y (topological spaces) continuous and surjective such that Y's topology is not the final topology with respect to f?

<@&286206848099549185> <@&681259184582688842>

[little hemlock]

Let X = Y and give X in the domain the discrete topology and give X in the codomain the indiscrete topology. Then let f be the identity

also, pls wait 15 min before pinging helpers

[little kettle]

Okk sorry

What do you mean by codomain?

[little hemlock]

do you call it the target maybe? Its just the space you are mapping in to. i.e. f: (X, t) \to (X, o) where t is discrete topology and o is indiscrete

[little kettle]

Yeap, thanks

Got ya!

[little hemlock]

npnp

[wet shore]

I need some help with erdos' distinct distances problem

can someone explain why KN >= n-1

I've been stuck at this and the paper says that it's clear

please help

<@&286206848099549185>

what do you mean?

I don't think so

It can be arbitrarily small as long as the polygon is convex

yeah, are you participating?

I was just doing some background reading

oooh cool

which problem are you planning on taking

Havent started reading anything tho
@gritty widget Ah cool

yeah, I was also thinking either this or convex geometries

yeah exactly

I mean some topics are totally out of my league

so I already eliminated them

can you explain?

point P1 is fixed so we can have at max n-1 distinct distances, now if say 3 of these n-1 distances were same. K would be n-3 and N would be 3 right?

ah cool, if you get time check this out

nvm, I got it facepalm

[wet shore]

P1 is fixed so K is atmost n-1

and N is also atmost N-1

n-1*

cool 🙂

[gentle osprey]

slimvesus:

[wet shore]

yeah exactly

[surreal rain]

There is no one one and continuous function from A onto $(0,1)$ implies A is conpact?? (T/F)
@surreal rain <@&286206848099549185>

[gentle osprey]

Manas:

[marsh forge]

Can you give some conditions on A

Because this is not true in general

is $A\subset \mathbb{R}$

[gentle osprey]

MaxJ:

[uncut surge]

Maybe we should ask first: Where's the question coming from and what have you tried?

[marsh forge]

Well the first part is more important larto

does not matter what he tried if the statement is false lol

[uncut surge]

depends on what you want to achieve 😄 it also looks a bit like a basic exam question so i don't necessarily just wanna hand out free answers

[surreal rain]

is $A\subset \mathbb{R}$
@marsh forge yes

[gentle osprey]

Manas:

[marsh forge]

Can you post the original question

with all the details

[surreal rain]

We know that there is no continuous onto map from a non compact subset of R to $(0,1)$

[gentle osprey]

Manas:

[dim meadow]

yes

[marsh forge]

Wait are you requiring one-one or onto

[dim meadow]

wait no

the opposite

[surreal rain]

But my question was if f is one one then is it possible?

[marsh forge]

(0,1) is homeo to R so if a compact set lives in R it can be moved into (0,1)

i.e. there is no compact subset of the reals that does not admit such an injection

[dim meadow]

if f is one to one, continuous, onto (0,1) and A is compact then f is a homeo

which is bad

[marsh forge]

Actually your theorem is true because it is vacuous

[dim meadow]

lol

[marsh forge]

Let $f:\mathbb{R}\to (0,1)$ be a homeomorphism. Then if $A\subset\mathbb{R}$ we have a continuous injection $f\mid_A$

oh

[gentle osprey]

MaxJ:

[marsh forge]

there we go

I mean I guess $A\neq \emptyset$

[gentle osprey]

MaxJ:

[surreal rain]

if f is one to one, continuous, onto (0,1) and A is compact then f is a homeo
@dim meadow yes but my Question is : Suppose there is no continuous injection from $ A\subset R$ to (0,1). Then can we say A compact?

[gentle osprey]

Manas:

[marsh forge]

oh no that works too

wrong direction

@surreal rain yes but only in the worst possible sense

no such subset exists

I assumed this was a textbook question or something so I was like trying so hard to see why it was true

until I realized

[plucky veldt]

Prove that $f$ is continuous if and only if for every subset $A$ of $X$, $f(\overline{A}) \subseteq \overline{f(A)}$

[gentle osprey]

mzdunek:

[plucky veldt]

if $f$ is continuous $f(\overline{A})$ is closed since $\overline{A}$ is closed, $f(A) \subseteq f(\overline{A})$ since $A \subseteq \overline{A}$, therefore $\overline{f(A)} \subseteq f(\overline{A})$ since $\overline{f(A)}$ is the smallest closed set containing $f(A)$

[gentle osprey]

mzdunek:

[plucky veldt]

so shouldn't it be the other way around?

[coarse kestrel]

Who told you that continuous functions preserve closed sets?

[plucky veldt]

right, its the other way around

the inverse image of a closed set is closed

[coarse kestrel]

Yes

[plucky veldt]

okay, so

$f^{-1}(\overline{f(A)})$ is closed since $\overline{f(A)}$ is closed, $A \subseteq f^{-1}(\overline{f(A)})$ therefore $\overline{A} \subseteq f^{-1}(\overline{f(A)})$, so $f(\overline{A}) \subseteq \overline{f(A)}$

[gentle osprey]

mzdunek:

[plucky veldt]

thx for helping me spot my stupidity

[coarse kestrel]

Lol it’s not stupidity, it’s more unfamiliarity with the subject

[plucky veldt]

@marsh forge hatcher's notes are only about algebraic topology, from what I've seen

[marsh forge]

No

Hatcher has point set notes

they are separate from his famous textbook

[plucky veldt]

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

interesting