point-set-topology

[fading vale]

Maps(X smash Y, Z) cong Maps(X, Maps(Y,Z))

Have you seen the tensor product for like vector spaces

[pearl holly]

oof no, is that similar to tensor product for modules?

ye

[fading vale]

Yes same thing

[pearl holly]

okay then I have

[fading vale]

So you know linear maps out of tensor product correspond to bilineat maps out of the cartesian product

Replace linear with continuous base point preserving and bilinear with base point preserving in each component

[pearl holly]

okay and the smash product will have this property?

ye okay you just said it lmao

ye okay I see that's nice

[fading vale]

Yes

The relationship to Hom/Maps is most important

And equivalent iirc

[pearl holly]

I see I see

[fading vale]

The really important example is like

You know how X smash S1 is suspension

[pearl holly]

ye

[fading vale]

And maps(S1, Y) is the space of loops in Y b definition

Like pi_1 but not up to htpy

Makes sense?

[pearl holly]

ye like the weird omega right?

[fading vale]

Yep

So Maps(X smash S1, Y) cong Maps(X, Maps(S1, Y))

That is

Maps(Suspension X, Y) cong Maps(X, Omega Y)

Take X = S^n and youll see why this is a big deal

[pearl holly]

wait I don't see it

[fading vale]

Whats suspension of S^n

[pearl holly]

I legit don't know

I forgot the definition

[fading vale]

S^n+1

[pearl holly]

oooh okay that's kind of cool

[fading vale]

So quotient up to homotopy and you see that pi_(n+1) (X) cong pi_n(Omega X)

[pearl holly]

yeee right

that makes sense geometrically too

[fading vale]

There are lots of other applications like commutativity of higher htpy groups

[pearl holly]

oh that's like eckman hilton right?

I remember moldi talking about thsi

[fading vale]

Yes

[pearl holly]

wait so you can prove it with this fact?

[fading vale]

I can talk more abt it in like 20 minutes once im home but yeah

[pearl holly]

yeah okay, I think I will be asleep 20 minutes from now on tho lmao but thanks a lot!

cool stuff

[fading vale]

Sweden moment

[tough imp]

@pearl holly if u wanna be really cool and swag

You should learn that thing which happens when you take AT and replace the T with a G

[novel acorn]

Corrupting the youth

[gritty widget]

What’s the big picture of stable infinity categories

Is it just, the ideal setting for doing stable homotopy theory

From what I’ve learned infinity categories seem to be good because they

Allow us to neatly package higher homotopical information

[nova fjord]

Can I ask a question about Morse theory here?

[gritty widget]

Yes

[nova fjord]

So the standard result in Morse theory is that when you pass a critical point of the Morse function, the homotopy type changes (i.e. you attach a cell whose dimension equals the index of that critical point)

But there's also the idea that you want to find a Morse function with the fewest possible critical points and want to cancel certain critical points in pairs. Is the idea that these critical points contain redundant information and that their cancellation doesn't affect the homotopy type of the smooth manifold? This is probably pretty obvious, but it wasn't to me and I was wondering if there's an intuitive argument for it

[gritty widget]

I think it’s not obvious that this does work out

If you draw some examples it’s “clear” it should workout

But I think this actually takes quite a bit of work to make sense of

Like for example if you do a homological treatment of Morse theory

Take the normal sphere

And take this deformed sphere

The Morse complex of these things will be different

The latter will have some degree 1 terms

But when you take the homology it cancels out

I only really know that, the Morse complex does not depend on the Morse function

• the homology of the Morse complex does not depend

The Morse complex certainly does

So this can be one way to show what you are asking

But perhaps not a]satisfying

I think maybe what you are looking for you might find in references on Cerf theory and “birth death singularities”

Please @ me if you reply I probably won’t see this until tomorrow

[tight agate]

it's a better framework (compared to triangulated cats) for things that look like the stable homotopy category or the derived category of something

[gritty widget]

Okay this confuses me because it seemed we needed to talk about triangulated categories

To talk about stable infinity

[tight agate]

[gritty widget]

Now maybe I just mixed up the motivation

[tight agate]

yeah you dont need to mention triangulated cats at all

the homotopy cat of a stable infinity cat is triangulated

but being stable is just a property of an infinity cat

there are several advantages from what I could gather

1. better cones

2. being stable is a property, and not structure that you put on the cat

3. theyre better when you have families of categories

for example, you can take limits and colimits of infinity categories in the infinity category of infinity categories

so it's a good setting for descent problems

if youre only interested in derived cats, I think you can get away with just working with dg-enhanced derived cats

I dont know this stuff very well, you might (will) get better answers in the AT server

[tight agate]

You might have seen how cones in triangulated categories cause problems when you're trying to relate the derived category of the heart of a t-structure and the original triangulated cat. There's no obvious functor going either way. Similar issues turn up when youre trying to glue t-structures in an exact triple of triangulated categories.

also when you do have a functor from D(heart) --> C, it's only an equivalence when Ext is generated in degree one

or something like that

[tight agate]

this is helpful when you're putting extra structures on you cat, such as a tensor product etc. The extra structure will usually be automatically compatible with the triangulated structure on the homotopy cat. You don't need to check compatibility because everything is canonical.

[gritty widget]

There is an AT server?

Can you hook me up with an invite

[tight agate]

this sort of thing often turns up when you're studying things like the category of bundles over each open set of some space.

or something like U \mapsto D^b(Coh(U))

notions like "sheaves of categories" make sense in the infinity cat setting

idk if this is an advantage, but you dont have to deal with all the weird zig-zags that you see while dealing with model cats

[tight agate]

DM'd

there's a good answer on this by Denis Nardin on MO

most of the things I said above are from that answer

lime_soup does AT so they probably need it/will need it soon

[gritty widget]

I have to give a presentation on stable infinity categories

So in a very real sense

I

I need it

[tight agate]

oh yeah lime soup you asked for an easier reference

gaitsgory-rozenblyum

I forgot about it yesterday

[gritty widget]

Yeah I think nobody mention that

[gritty widget]

I’m really enjoying this

[tight agate]

idk if it's easier, but it goes a lot faster

by skipping all the details

[gritty widget]

I find

I really have to have n]an idea of the big picture to learn something

[tight agate]

did you work with classical spectra before

or triangulated/derived cats

[gritty widget]

I’ve seen spectra

Not until too much detail

And no on the derived/triangulated cats

[tight agate]

reading about them might help

and their deficiencies

[tough imp]

[They don’t get enough vitamin C :(]

[tight agate]

vitamin cone

[tough imp]

Lmao

[gritty widget]

Okay

Thanks very much

I’m going to check this stuff out

[tight agate]

I think some of the problems in gelfand and manin ch4 are very good at pointing these out

all of the ones they point out comes from the non-functoriality of the cone

but yeah Cat being a bad cat is also an issue

model cats are hard to work with in families

that issue can be fixed using model cats

and a lot of people are perfectly happy working with them

the definition I read requires functorial factorizations

Quillen's original definition does not iirc

any quillen model cat?

ignoring size

interesting

idk

oh boy derivators

[quasi forum]

I was about to ask why the rationals aren't locally compact, but I figured it out. So thanks for the help. :p

[quasi forum]

It's interesting how local compactness does not carry over subspaces.

[plain raven]

wild

[cursive spade]

problem: every net has a subnet that's a ultranet .
could u give some examples of nets that might give some idea of solving this

[quasi forum]

So on the topic, is local compactness not an invariant of homeomorphisms?

Yeah. Now that i think about it, that makes a lot of sense

[cosmic beacon]

Can anyone tell me what the setup of this question may be hinting at me to do?

[quasi forum]

I could use a bit of help with this. I have a bit of a difficult time understanding limit point compactness to be able to work with it effectively.

[trail tiger]

Does anyone know why "super" as in supersymmetry, superalgebra, and so on equips an object with a graded tensor structure? What's with the nomenclature.

[tough imp]

I think it’s derived some sort of physics thing

Super symmetry or something in physics and it was Z/2Z graded and then it just stuck

¯_(ツ)_/¯

I dunno tho I’m just a chmonkey

[quasi forum]

I think I understand the hint part. For the {x_n} set, a limit point does not exist. If it did, then it would not be contained in the cover of X, which is the issue

Wait wait, that isn't quite it. If we were to look at a point x not in {x_n}, then there exists a k s.t. x is in U_k, which is disjoint from x_n when n>=k.
For the remaining elements, by the T_1 axiom, we can create a neighborhood around each x_n that does not meet x.

So {x_n} has no limit point

[rancid umbra]

im trying to work through this one with you lol. i havent ever seen this result

[quasi forum]

I think I got the 2nd direction down (I'll share in a sec). I would like help on the first direction (if you can).

Here is the 2nd direction. Ignore the picture, I thought it'd be useful xp

Mine?

Oh, yeah i dont mind

There

Thank you

Okay, back to the issue at hand. I understand the main idea. If we let A be a subset of X with no limit points, we show that A must be finite.

I know how to do it for compactness, but not with a countable cover...

Here is the proof when X is compact. Maybe this will help someone

[rancid umbra]

i think closed subsets of countably compact spaces are countably compact

that would fix the proof ps

[quasi forum]

Almost. The cover we were taking in this proof is not necessarily countable

[rancid umbra]

yea, but it would be now

[quasi forum]

But yes, you are right. A should be countably compact

[quasi forum]

Why?

[rancid umbra]

oh wait

yea A may be uncountable

[quasi forum]

The cover we did in this proof was take neighborhoods of A. Their union is a cover, but it is not necessarily countable

Yea, so that isnt quite it.

[rancid umbra]

so we just gotta find a countably compact open cover of A

[quasi forum]

Yep, that is somehow built by the elements of A

[rancid umbra]

okay

lets see how this turns out

Suppose $X$ is not limit point compact. Then, without loss of generality, there is a countably infinite subset $V$ of $X$ with no limit point in $X$. Since the set of limit points of $V$ is empty, then $V$ is closed, so $X\setminus V$ is open. For each $v\in V$, there is a neighborhood $U_v\ni v$ such that $(U_v\setminus{v})\cap V=\emptyset$. Then ${U_v\subseteq X:v\in V}\cup{X\setminus V}$ is a countable open cover of $X$.

[gentle osprey]

c squared

[rancid umbra]

No finite sub-cover of this covering can exist, because each U_v only contains one point from V, which would mean that V is finite

contradiction

so X is not countably compact

[quasi forum]

Ah, you did the contrapositive again here too

[rancid umbra]

ye

nowhere did i use T1 assumption either i dont think

[quasi forum]

And you just chose a set that is countably infinite to do something similar to what the compact->limit compactness proof did

[quasi forum]

Yea, I think T_1 is only necessary for the other direction.

[rancid umbra]

yea pretty similar to that one

im still looking over the proof you gave for the other direction cus i couldnt get it, so i guess we both gave each other one direction of the proof lol

[quasi forum]

For sure! Thanks for the help c^2

[rancid umbra]

yw! was fun

[quasi forum]

I got another one I need to bite into if you are interested

[rancid umbra]

yea fs

[rancid umbra]

these are from munkres right?

[quasi forum]

Yes it is

[rancid umbra]

oh thats cool lol

[quasi forum]

A good book, but dang is it challenging

[rancid umbra]

yea i have it. one of my favs

what do you have so far for it?

[quasi forum]

So before we get to this new one, why can you assume a countably infinite set has no limit point.

What if only uncountable sets had no limit point?

[rancid umbra]

because every infinite set contains a countably infinite subset (assuming some stuff about dedikind infinite stuff and AoC)

[quasi forum]

AoC?

[rancid umbra]

Axiom of Choice

yea

[quasi forum]

Okay, so I am starting to process the hint a bit more.
So I think the x_n's are supposed to be a compact set.
Then by EVT, we know it attains its bounds, so we just need to 1) ensure that the infinimum of d(x_n,x_m) is non-zero and the x_(n+1)=f(x_n) doesnt produce repeats

[rancid umbra]

surjectivity is the hardest part here

[quasi forum]

Okay, I understand where the hint is leading (and why this question is in this section).
Since X is compact, it is sequentially compact, so it has a convergent subsequence. But all convergent sequences are Cauchy, so there is an N, so that d(x_n_,x_m)<e when n,m>=N.
There is the contradiction

[rancid umbra]

i think the other parts are pretty easy: injectivity (immediate), continuity (almost immediate), and continuity of inverse if it exists (pretty much almost immediate)

[rancid umbra]

ah

i was trying to look at the sum of d(x_n,x_{n+1}) for some reason

but d(x_n,x{n+1}) = d(x_1,x_2) for all natural numbers n

so it didnt work

[quasi forum]

Lol, let me save you time :p.

On the other hand, I need to show that e is an upper bound of d(x_n,x_m)

[quasi forum]

Ah, I see why.

[quasi forum]

Which is what I am having a hard time with

[rancid umbra]

well, d(x_n,x_m) = (n-m+1)d(x_1,x_2) > d(x_1,x_2) > e

since a = x_1 is not in f(X) and x_2 = f(a) is

[quasi forum]

Where did the n-m+1 come in?

[rancid umbra]

ah shit

[quasi forum]

Isnt d(x_n,x_m) just equal to d(x_1,x_2)?

[rancid umbra]

no its only for consecutive n,m

[rancid umbra]

but this still feels like crankery

and also doesnt help with the upper bound

lol my b

[quasi forum]

So, can we show d(a,x_{m-n+1})>d(a,f(a))?

[rancid umbra]

yea

[quasi forum]

Actually, no need.

We know d(a,x_{n-n+1}) is at least epsilon away because x_{m-n+1} is in f(X)

[rancid umbra]

yea. also, thats true for any index bigger than 1, doesnt need to be x_{m-n+1}

[quasi forum]

Yes, but if we let m>n, we can simply d(x_n,x_m) to that

[rancid umbra]

okay, got it

[quasi forum]

And that's our ticket

And of course, since we have surjectivity and we are mapping X onto X, injectivity is immediate

[rancid umbra]

okay. so just to put it all together. we assumed that f wasnt surjective.
we got a sequence of points x_n that had a convergent sub-sequence of x_n.
but we just showed that the distance between any two points of the sequence was at least epsilon

so the sub-sequence of x_n cant be convergent, since its not cauchy any more, which is a contradiciton

[quasi forum]

Yep, bingo!

[rancid umbra]

cool. for continuity of the inverse,~~ it suffices to show that f is a closed map~~

actually. u can do it differently too, using the isometry condition
use the isometry assumptions and Lipschitz continuity. its pretty nice how it works out

[quasi forum]

1 sec. Still writing the proof down

Okay, I am ready

[quasi forum]

Oh yeah, continuity is pretty automatic :p

Heck yeah! Solid finish

[rancid umbra]

nice!

[quasi forum]

So I am pretty sure I understand 4.
Question, does uniform mean the box topology?

Oh no, it is much worse than that

I think that's a good sign to stop for the day xp

Thanks for the help c squared. I feel like I have a much better understanding of limit compactness now.

[rancid umbra]

just one quick check, i think there exists such an epsilon because f is a closed map, in particular, f(X) is closed

[rancid umbra]

oh and yea yw. me too

[rancid umbra]

i think uniform topology stuff is on page 124

[quasi forum]

Yea, I saw. I may have to deal with that beast tomorrow. It seems rough

[rancid umbra]

yea i have to look at the definitions again for R^omega and local compactness. it do seem rough

[quasi forum]

In regards to 3), I understand the issue is that an open neighborhood of x may not be open in f(X). But I am having a hard time coming up with a counter example

Question. Can we find a continuous map from R to Q?

[rancid umbra]

suppose you had one

[quasi forum]

Damn, not a good start

[rancid umbra]

what is one thing that you know is topologically different from R than from Q

[quasi forum]

Oh good lord... Q isnt connected

Okay, so that wont cut it

Why are counter examples such a pain to find sometimes

[rancid umbra]

lol this is kinda my favorite part sometimes

finding exotic c.e.'s are cool

[quasi forum]

I always suck at finding counter examples. Its the bane of my mathematical existence.

The bane of my uniqueness is a whole nother issue that we wont get into

[rancid umbra]

hmm. what are some not locally compact spaces

Q

any subset of Q

[quasi forum]

The irrationals (using the same logic)

Yep, I take it back

[rancid umbra]

lol

the set {1/n : n in N} is not locally compact

i think near zero you get issues?

lemme check

[quasi forum]

No, I am pretty sure it is

[rancid umbra]

yea ur right it is

dang

[quasi forum]

You had me excited for a moment.

[rancid umbra]

me too haha

[quasi forum]

For starters, compactness implies local compactness, so we need two non-compact spaces to get anywhere

Wait.....

God we are silly

[rancid umbra]

how so?

[quasi forum]

Q is countable, so perhaps we can find a continuous map from N to Q

[rancid umbra]

any function from N to Q is continuous

[quasi forum]

Well then we're done

[quasi forum]

Care to share why?

It is not obvious to me

[rancid umbra]

what do you know about discrete spaces and their open sets

[quasi forum]

All sets are open

[rancid umbra]

yea

[quasi forum]

....

I see

Well, then we found it

[rancid umbra]

oh shit u right lol

[quasi forum]

N is locally compact, but Q is not

[rancid umbra]

i was about to bust out such a nasty ce

[quasi forum]

Haha. And if f is an open map, it does work, since U in C means f(U) in f(C) where f(U) is open

[rancid umbra]

okay, awesome

wait. so quick run through if f is continuous and open

let y be a point in f(X), so y = f(x) for some x in X

X is locally compact, so there is a compact set C containing an open neighborhood U of x

f(U) subseteq f(C) and f(U) is open and f(C) is compact

nice

[quasi forum]

Yep, that's all there is to it. :D

[rancid umbra]

lmao all sequences are continuous

that just doesnt feel right to me for some reason ig lol

[true garden]

Hello. Why is there no function $f\colon \mathbb{R} \to \mathbb{R}$ that is continuous precisely on a countable dense subset of $\mathbb{R}$?

[gentle osprey]

MathPhysics

[rancid umbra]

https://arxiv.org/pdf/1809.06453.pdf
@true garden

if you let f be uniformly continuous, then the proof is a bit easier

[finite heath]

Hello folks

Anyone here know how to prove that f: R^2 - {0} --> S^1 x (0, infinity) with f(rcostheta, rsintheta) = (costheta, signtheta, r) has an inverse f^-1 that is also continuous? [big boy problem]

[rancid umbra]

is R^2{0} suppose to be R^2 without the origin?

[finite heath]

oh yeah

meant to have the \ in there

yeah

without the origin

im trying to prove this

:[]

[wise sigil]

have you had any ideas for the inverse?

[finite heath]

well, that's where im confused

couldn't inverse literally just be

f^-1(costheta, signtheta, r) = (rcostheta, rsintheta)

[wise sigil]

thats right

[rancid umbra]

yuh

[wise sigil]

theres a few ways to prove continuity

[rancid umbra]

easiest is prob composition of continuous functions

[finite heath]

well

i was told that since this function maps product spaces

that I can prove that all the components, say, costheta, sintheta, and r, that if they are all continuous then the entire function is

something to do with analysis

[rancid umbra]

yuh

[finite heath]

but i never took that

but i my prof would get it

[finite heath]

is that this?

[rancid umbra]

no

[finite heath]

so that's like

f: X-->Y is Homeo and g:Y-->Z is Homeo, then p:X-->Z is Homeo?

[rancid umbra]

if f : X --> Y and g : Y --> Z continuous, then g o f : X --> Z continuous

[rancid umbra]

this is true if p = g o f, but its stronger than what we need

[rancid umbra]

i forgot about component functions. if you can show that the component functions
f_1(r cos(t), r sin(t)) = (cos(t), sin(t))
f_2(r cos(t), r sin(t)) = r
are all continuous then ur good

[finite heath]

so as in

f(rcos(t), rsin(t)) = (cos(t), sin(t), r) was my original function

and so

f_1(r cos(t), r sin(t)) = cos(t)
f_2(r cos(t), r sin(t)) = sin(t)
f_3(r cos(t), r sin(t)) = r

then I show that function is homeomorphic?

or simply that f is continuous

[rancid umbra]

you show that the two functions above are continuous

since those are the component functions of f

there should only be two. lemme fix it rq

the first component function is from R^2 - {0} to S^1 and the second component function is from R^2 - {0} to (0, infty)

[empty grove]

You can break the S¹ component further into 2 components

Checking continuity into S¹ and into R² is the same when S¹ has subspace top

[finite heath]

f_1(r cos(t), r sin(t)) = cos(t)
f_2(r cos(t), r sin(t)) = sin(t)
f_3(r cos(t), r sin(t)) = r

^ proving these continuous proves R^2 - {0} to S^1 and the second component function is from R^2 - {0} to (0, infty) to be continuous?

[empty grove]

Yes
So (cos t, sin t, r) really is more like
((cos t, sin t), r)
Where (cos t, sin t) is a point of S¹

Because then it's a pair, a point in S¹ and a point in (0, infty)

So you have to check continuity of 2 coordinate functions

But the continuity of the first coordinate function breaks apart as checking continuity of 2 coordinate functions again

By same reasoning

Because you can think of S¹ as sitting inside R x R

[finite heath]

ngl im hella confused but gonna shleep and wake up and look at this

but okay and so

i essentially prove that f: R^2 - {0} --> S^1 x (0, infinity) with f(rcostheta, rsintheta) = (costheta, signtheta, r) is continous

how do i prove inverse is cont.

[empty grove]

Yeah I'm really struggling with explaining this. We are also implicitly using the fact that the punctured plane is homeomorphic to a quotient of (0,infty) x [0, 2pi] but that's too much work and there should be a simpler way to do this

[finite heath]

no ur good im just super novice

[empty grove]

[finite heath]

this is hard!

are u gonna be online in 3-4 hours?

:[]

[empty grove]

Will be in the middle of my exam

[finite heath]

v

i knew it

[empty grove]

F

[finite heath]

i guess i use plan B

plan B: stay up all night and go to sleep when moldilocks offline

[empty grove]

Lol just sleep, there should be some analysis brains 🤡 around here that can figure out nice proof

[finite heath]

🤡

[empty grove]

I just don't want to think about it because it looks so annoying

[finite heath]

yeah wait me neither

[true garden]

Thanks.

[gentle osprey]

Nobody

[wise sigil]

Just a small nitpick but your f is a function from RxR but we need a g from S^1 x R. i guess from f =g circ q, where q:RxR -> S^1 x R is the quotient map, it follows that g is continuous as well

cant think of an easier way without using a quotient map

[swift fjord]

Does anyone know where I can find a proof of homotopy of simplicial maps induces chain homotopy on the induced homomorphisms? My prof didn't wanna get into the details but I wanna find a rigorous treatment for my notes.

[pearl holly]

Lmao I’ve been thinking about learning sum AG but I have to do some more AT first

[quasi forum]

Hi all, I could use some help on #4

Here is the def for the uniform topology

[tough imp]

Yeah but have you considered homotopying the T into a G and learning that instead?

Like rn

At this very moment

[pearl holly]

Lmao

I mean not right now

I want to learn some more stuff before I move on

[tough imp]

Yeah but

Imagine if

You replaced T with G and replaced Hatcher with Hartshorne

https://tenor.com/view/think-about-it-digibyte-dgb-meme-high-iq-gif-19108212

[pearl holly]

Yeah okay sure I will think about it but AG might even be too hardcore for me

[tough imp]

Haven’t you been like mucking around with shitty CW complexes for a month now

[pearl holly]

Like I have no idea what it is like

[tough imp]

It’s local commutative algebra

It’s actually so sick

[pearl holly]

Ye but AG might be completely different

[tough imp]

You expand commutstibe rings

In a way they couldn’t before

The geometry let’s you glue affine schemes to make new schemes

But an affine scheme is just a commutative ring

So you get to glue commutative rings along geometric lines which the ring itself doesn’t originally seem to have

It’s so cracked

And bro like a sheaf dude 🤯

[pearl holly]

Yoo okay that sounds cool

[tough imp]

They’re like just bundles brooooo

A module dude

It’s like a fiber bundle brooooo

[pearl holly]

Oh no not bundles

[fading vale]

Chmonkey shits up innocent channel demanding swede study his memes

Sad!

[tough imp]

My memes are supreme tho

And toki loves me

They backed me up on the Kanga thing

Before anyone else did

Anyway

Point is I love commutative rings

And AG let’s you like do even more epicer stuff

Like bro the tensor product is so cool

And a fibered product is like a mega tensor product

It’s like brooooo

[pearl holly]

Yee okay I will think about it

[tough imp]

It’s like a tensor product of these omega rings (a scheme)

And then it’s like bro this shit makes no sense wtf is going on

Then it makes sense and you’re like brooooo

It’s like bro all rings are rings of functions bro

It’s like yo dude if you’re not zero at a point you’re not zero on a nbd bro that’s why the support is closed bro (oh wait this is kinda backwards)

[quasi forum]

Okay, so the way I need to do this is to take a ball around x and I need to show it's closure is not compact.
But I am having trouble finding a cover that fails

[tough imp]

Wtf is the uniform topology

[quasi forum]

Here is the definition

[tough imp]

Point set

[quasi forum]

Yes, point set topology

[tough imp]

Maybe you want to look at some fucked up ooint like

(1/n) in the n-th spot

And something goes wrong

I dunno

I think the ball itself will matter less than the point it’s centered on

[quasi forum]

Why would that be

[tough imp]

Idk

Any sufficiently small ball should do

I mean an epsilon ball and an epsilon/2 ball are basically the same thing so

This is just my intuition

My chmonkey intuition

[quasi forum]

I think you are right. But I think the issue is going to be near the endpoints.

[tough imp]

That’s why I was thinking

1/n in the n-th spot

It’s kinda near the edge

But in a cool mysterious way

[quasi forum]

Lol, I got you at a very strange time

[quasi forum]

Is a set where in one coordinate, there is an x,y s.t d(x,y)>=e and for all other coordinates, the distance is strictly smaller.

Is that something that could be open here?

Actually, I think this will work. Let U be an e nbhd of 0 and consider it's closure. Then for the coordinate y_i s.t. d(0,y_i)=e, create a ball of radius epsilon around (0,...,y_i,...). These balls+ B(0,e/2) cover the nbhd of U.

But the moment you remove 1, then you are missing (0,...,y_i...) or 0, which is in the closure.

[swift fjord]

Bump

[orchid forge]

it's probably worth writing down yourself to make sure you understand the definition of a simplicial homotopy

[swift fjord]

He didn't really explain simplicial homotopy, he just said we're taking a homotopy between the 2 induced maps on the underlying spaces and triangulating the product space with some triangulation preserving the top and bottom

[orchid forge]

when you say simplicial maps

do you mean just in the sense of simplicial homology, a map from a standard simplex into X? or the generalization in terms of simplicial sets

[swift fjord]

Maps between simplicial complexes that preserve simplices

And induce.continuous maps on the underlying spaces

[orchid forge]

oh ok

[heady grove]

Anyone know how to do a?

[swift fjord]

Ty will do

[finite heath]

any tips for how to begin this?

I feel like I am missing something

[gritty widget]

how to begin this
look at your favorite definition of continuity and try to show that it holds here

[finite heath]

alright TT

i want to show that the preimage of x + y is open in R^2

or wait

[gritty widget]

(that said it might be easiest to do this with epsilon delta)

[finite heath]

i would have to show that the preimage of every open in R is open in R^2

[gritty widget]

so if you wanted to use the definition of continuity involving preimages, it'd be enough to check that the preimage of any open interval (a, b) is open

@finite heath

that may simplify the process

[finite heath]

oh word okay

[remote marsh]

Is Int(A∪B) ⊆ C(A)∪Int(B) true?

Int is interior and C is closure

[quasi forum]

For starters, is it true that int(AuB) is a subset of Int(A)UInt(B)?

If so, the answer to your question comes automatically (the closure contains the interior)

[viral atlas]

Had the same idea but was too scared of a potential counterexample

[remote marsh]

The answer is no, that’s not true

Take [0,1] and [1,2]

[quasi forum]

Ah, it is the other way around

[gusty wagon]

are the box topology and product topologies on R infinity the same?

[quasi forum]

No. They are only the same on R^n

[remote marsh]

No but quit distracting from my interesting question

[gusty wagon]

sorry lol

[quasi forum]

Okay then, here is a question. Is the only reason Int(A)Uint(B) a subset of int(AuB) because of the intersecting limit points?

[remote marsh]

1. Int(A∪B) ⊆ C(A)∪Int(B)?
   If so, can it be further weakened to
2. Int(A∪B) ⊆ A∪Int(B)?

[quasi forum]

For instance, $(0,1)^{\omega}$ is open in the box topology of $R^{\omega}$ but not in the product topology.

[gentle osprey]

dackid

[gusty wagon]

im talking about r infinity not r omega

[viral atlas]

Uhh they're the same?

[quasi forum]

They are the same. Countably infinite

[gusty wagon]

r infinity is a subset of r omega. all eventually 0 sequences

[quasi forum]

Oh that

My bad

[viral atlas]

[quasi forum]

Euler, if this is the main issue for these two not being equal, then I think you can arrive at your answer.

[gusty wagon]

i think that inclusion is because int(A) is an open set contained in A and thus contained in AuB and thus contained in int(AuB), and same for B

[gusty wagon]

why are they the same?

[quasi forum]

I'm more interested in why it is not an equality. Not necessarily a counter example, but a reason as to why.

[gusty wagon]

i think its because int(AuB) can contain points that are on the boundary of both A and B, but int(A) u int(B) doesn't

[quasi forum]

The counter example euler brought up is because of the limit points they share

[gusty wagon]

keyword can

[quasi forum]

Yes, I think so too. Then if that is the case, the Cl(A)UInt(B) definitely contains int(AuB) because the limit points are included in the former.

[gusty wagon]

hmm

[quasi forum]

Actually yeah, that is the reason why it is not an equality. So this holds

[gusty wagon]

https://math.stackexchange.com/questions/1068915/interior-of-union-of-two-sets try reading that

it might answer your question

is the function from r infinity to r infinity defined by f(x) = 2x continuous in either topology?

[obtuse meteor]

operads are hard

[flint cove]

Shamelessly stealing these as exercises for myself. Nice questions.

[remote marsh]

Int(A∪B) ⊆ C(A)∪Int(B)

Proof:
If Int(A∪B) is empty we are done
Therefore let x belong to Int(A∪B).
If x belongs to C(A) we are done.
So let x belongs to B but not C(A)
Since x is in Int(A∪B)nC(A)^c
Which is an open set inside B. This x is in the interior of B

They’re really nice questions. Inspired by trying to invent modal topological logic

I need the first statement to be true as it’s the modal K axiom in disguise

[gusty wagon]

why is int(AuB)nC(A)^c contained in B?

[remote marsh]

Interior of AUB only contains points from A or B

And C(A)^c has no points in A

[gusty wagon]

kk

actually, better question, is r infinity contractible?

[remote marsh]

1. Int(A∪B) ⊆ C(A)∪Int(B) is true.
   Conjecture:
2. Int(A∪B) ⊆ A∪Int(B) is false

[quasi forum]

It was taking me a minute, but no. An open set in the product topology would look like $U_1\times \dots \times U_k\times \R \times \R \dots$.
Of course, order does not matter, but this suffices. We can bound the open sets so that none of them contain 0.
In contrast, the box topology has open sets that bound the coordinates after the $k^th$ one, where the product topology does not.

[gentle osprey]

dackid

[flint cove]

2. No, it holds. Using int(A\cup B) \subseteq int(A)\cup int(B) and then int(A) \subseteq A, this follows.

[remote marsh]

2. Is false. Take (0,1) and [1,2].
   LHS is (0,2) RHS is (0,1)U(1,2)

[flint cove]

Nevermind I switched directions

(┛◉Д◉)┛彡┻━┻

[quasi forum]

So the box topology is a finer in R infinity than the product topology

[remote marsh]

It’s so delicate

[gritty widget]

nobody can do point set today

[quasi forum]

Hey! I did just fine 😆

[flint cove]

point, set, lose (nah this doesn't work)

[gusty wagon]

A = [0,1), B=[1,2] is a counterexample to 2.

[remote marsh]

Yeah that works too

[quasi forum]

Yep. Once again, it comes down to the limit points

[flint cove]

There must be a good axiomatic way to see it
Kuratowski wouldn't be proud of us at the moment

[gusty wagon]

this doesn't make any sense to me lol

idk what bound means

[remote marsh]

There isn’t one I bet, else modal logic wouldn’t need it as an additional axiom too

[quasi forum]

Bound may not be the right word, but every coordinate after the k^th one is free to go anywhere in the product topology, but not the box topology (your open sets are not necessarily R)

[remote marsh]

Well let’s try

I guess there has to be

[gentle osprey]

Nobody

[gusty wagon]

yeah but why is that map continuous?

whats a colimit

[remote marsh]

I’m gonna prove it from the axioms

[pearl holly]

can someone plz help I'm so stupid

like has someone read about the construction of steenrod operations on space level?

from Hathcer or something

there's so much notation everywhere that I don't want to type out

PLZ

[gusty wagon]

:c

[gusty wagon]

pls

[flint cove]

Tried it for a bit then gave up on it. glhf if you're still on it

[finite heath]

Ur a big boss

Hoss

[pearl holly]

so Hatcher is trying to prove that $Sq^i(\alpha) = \alpha \smile \alpha$ when $\alpha \in H^i(X; \mathbb{Z}_2)$ and I don't understand the argument. I know that if you restrict $\lambda(\alpha)$ to $X \wedge X$ then you get $\alpha \otimes \alpha$ which by an isomorphism can be represented as $\alpha \times \alpha$, the cross product. So I'm guessing that $\nabla^(\lambda(\alpha)) = \Delta^(\lambda(\alpha))$, $\Delta$ being the diagonal map. The problem I see is that $\Delta$ and $\nabla$ have completely different domains and codomains. Also you can't just compose $(\alpha \cross \alpha) \nabla$ the image of nabla is different from the domain of the cross product so I assume that you restrict the image of nabla somehow but I don't see it

[gentle osprey]

Tokidoki ✓

[pearl holly]

most of the notation is unnecessary to understand what's going on here I think

so like $\Lambda X$ has $X \wedge X$ embedded in it so I can restrict $\lambda(\alpha): \Lambda X \to K(\mathbb{Z}_2, 2n)$ to $X \wedge X$

[gentle osprey]

Tokidoki ✓

[pearl holly]

so $L^\infty = S^\infty/\mathbb{Z}_2$ is the infinite real projective plane. $\Gamma X = (S^\infty \times (X \wedge X))/\mathbb{Z}_2$. $\Lambda X$ is the quotient of $\Gamma X$ with $S^\infty/\mathbb{Z}_2$

[gentle osprey]

Tokidoki ✓

[pearl holly]

and I'm stuck at the very end of the picture I sent btw

and the fact that $\lambda (\alpha)$ restricts to $\alpha \otimes \alpha$ is just something that Hatcher showed before

[gentle osprey]

Tokidoki ✓

[pearl holly]

so my confusion lies in the fact that like sure, I know that $\lambda(\alpha)$ restricts to $\alpha \times \alpha: X \times X \to K(\mathbb{Z}_2, 2n)$ and I know that I have to use this fact to show the property, but the composition $(\alpha \times \alpha) \nabla$ doesn't make sense and even if I changed the codomain of $\nabla$ to be like $X \wedge X$ which is close enough I guess then I would just get a weird map $\mathbb{R}P^\infty \to K(\mathbb{Z}_2, 2n)$ and I don't see how this is the cup product

[gentle osprey]

Tokidoki ✓

[pearl holly]

plz help

[orchid forge]

Woah actually yes

[pearl holly]

omg

but where did you read it from? Hatcher?

do you maybe recommend some other sources?

[orchid forge]

No, basically just papers and personal correspondence

[pearl holly]

ah okay I see

[orchid forge]

I'll see if I can answer your question but the setup is a bit new to me

I'm only used to the axiomatic characterization and a spectrum level construction for a different homology theory

[pearl holly]

Oh yeah okay I see. Sorry that I didn’t respond lmao I took a quick shower

[finite heath]

Why is the union of lines of form

y = -x + b

in R^2

where b in some interval (c, d)

open in R^2 :D?

i cant figure it out cause i know individual line in R2 is closed

[sonic hill]

Cheeky answer: after rotation by 45 degrees which is a homeomorphism, the set becomes a product of two open sets

[finite heath]

are u responding to me hehe

[sonic hill]

Yes

[finite heath]

i dont understand =[

[sonic hill]

=[

[pearl holly]

=[

[orchid forge]

Ok, so it is true that $\nabla^(a \otimes a) = a \smile a$, when $a \otimes a$ is identified with an element of $H(\Lambda X)$ via the cross-product isomorphism. You don't want to say $\nabla^ \lambda = \Delta^* \lambda$ in general, since like you said the domains don't even make sense

[gentle osprey]

Kogasa

[orchid forge]

It may be necessary to just stare at the definition of $a \otimes a$ as an element of $H(\Lambda X)$

[gentle osprey]

Kogasa

[orchid forge]

I'll see if I can write it out in a bit when I'm free

[pearl holly]

yeah okay I see, I will think through this tomorrow

but the fact that you took your time and read all of that is amazing lmao, thank you so much!

[pearl holly]

oh the name of the channel changed lmao

Point-set topology (topological spaces), algebraic topology (homotopy/homology/etc.), geometric topology anime where

[finite heath]

Can someone help me in voice chat with Provide a formula for a homeomorphism between R and the interval (—oo, a). I feel so silly.

[finite heath]

Topology is frusterating folks.

(respectfully)

[rancid umbra]

play around with the exponential function

like scaling, shifting, and possibly reflecting it to fit the constraints

[lean marten]

I would suggest doing (-oo,0) firsg

first

then you can compose with a translation or something to get the rest

[rancid umbra]

yea thats pretty much the same idea

[lean marten]

Yeah basically

[rancid umbra]

there is a way to do this without using exponential functions as well

if stuck, you can try looking at the function (-infty,0) --> R given by || x - 1/x|| to get started

[finite heath]

word okay

[rancid umbra]

finding the right map from (-infty, a) to R is just a matter of finding a nice map from (-infty, a) to (-infty, 0)

then you can algebraically solve for the inverse

[orchid forge]

Rip, I'm gonna have to pick a topology channel to sit in because i cba checking all of them

[pearl holly]

Pick this channel

[remote marsh]

@flint cove

A < AuB = (A-B)uB
C(A)<C((A-B)uB)=C(A-B)uC(B)
C(A)-C(B) < C(A-B)
C(A)nC(B)^ < C(AnB^)
C(AnB^)^ < [C(A)nC(B)^]^
i((AnB^)^) < C(A)^ u C(B)
i(A^uB) < i(A^)uC(B)
Let B=X and A^=Y
i(XuY)<C(X)ui(Y)

^ means compliment

[tidal cedar]

based channel

[dreamy smelt]

Yo, I've been thinking a little about calculus on 0-manifolds

the largest distinction I can conceive of is connected vs. disconnected, and so maybe we'd take R with the discrete topology in the disconnected case

regardless

is there any sort of integration here

[flint cove]

You gotta put backslashes before the *

[remote marsh]

I'm too lazy

[flint cove]

Thanks for pinging me tho and gn

[orchid forge]

0-forms are smooth functions, and integration over an oriented 0-manifold (discrete set with orientations) is just summation (with a sign given by orientation).

You can't exactly have a connected 0-manifold with more than 1 point, since by assumption it is discrete, so {x} and {X \ x} would be a separation

[dreamy smelt]

Understood

thank you

[obtuse meteor]

Omg it happened

We’re alg top 🥺

[orchid forge]

yeah um... cup product

uhh... braided monoidal category

very cool and algebraic

[dreamy smelt]

So in this case, then

[gentle osprey]

insising

[dreamy smelt]

||Sorry for the ping @orchid forge||

[orchid forge]

you would need f(p) to be zero at all but countably many points, and the sum to converge

[dreamy smelt]

Understood c:

[orchid forge]

as far as i know there's no meaningful difference between integrating over a 0-manifold and just regular old addition of an arbitrary function, besides orientation

[obtuse meteor]

I want to talk about the little n-disks operad with somebody but idk who

[dreamy smelt]

Oof

Well now I've got to go figure out what orientation means

thanks though!

[orchid forge]

each point in the set gets labelled with a + or -

when you integrate a function f over a point p, you get f(p) or -f(p) depending on the orientation

[obtuse meteor]

orientation for 0-manifolds is really boring (tm)

[dreamy smelt]

Is it like an abstraction/de-algebraization of the order structure on R or am I being too concrete/silly

[obtuse meteor]

0-manifolds are generally rather boring

[dreamy smelt]

Figures. 0-manifolds are boring

yeah

lol

[orchid forge]

orientation is a thing for manifolds in general

[obtuse meteor]

ye

[orchid forge]

pretty important to making sense of integration

[obtuse meteor]

orientation has to do with your charts matching up in a nice way

[orchid forge]

for example, why is the integral over [a,b] equal to F(b) - F(a) and not F(a) - F(b)

[obtuse meteor]

there are two orientations on a connected manifold

[dreamy smelt]

Right

[orchid forge]

just knowing the points in the set [a,b] isn't quite enough

[dreamy smelt]

Mfw the connected 0-manifolds are of literally 0 interest

[orchid forge]

to formalize integration to a level suitable for abstraction

[orchid forge]

there are only two of them lol

they are both very important manifolds, but not exactly complicated

[dreamy smelt]

Two?

I'm only aware of the singleton

what's the other

[orchid forge]

the empty set

[dreamy smelt]

Oh sure

[orchid forge]

although maybe your definition of manifold excludes it

[dreamy smelt]

Nah, just never really cared to think about \varnothing in topological contexts

it's still pretty counterintuitive to me, to think of things like functions into or out of it

[orchid forge]

as far as i know there's really nothing interesting you can say about the empty set topologically

[dreamy smelt]

Mhm lmao

Anyway, I've been trying to learn about cool calculus shit about manifolds n whatnot but I'm still lil dummy boi

I'm a freshman so crap like differential k-forms are a bit much atm

[gentle lark]

Should questions about differential topology go here or in the other channel?

Seems like a bit of an in-between area

[dreamy smelt]

0-manifolds aren't interesting but I don't specifically care about them. Just been thinking about classifications

[ivory dragon]

other channel

[orchid forge]

might be worthwhile to learn about differential forms and integration in R^n from baby rudin before tackling differential forms on manifolds

[dreamy smelt]

Right, I'll go crack it open

again, sorry for the specific ping

you just answered my previous question so I figured you'd be invested enough to answer again

much appreciated

[gritty widget]

rudin

[gentle lark]

If the diff top question overlaps heavily with alg top do I ask it here or the other channel still?

[dreamy smelt]

Sounds like something you can ask here, since they talk to each other

[gentle lark]

(I don't have any specific question in mind, but atm I'm reading up on the cup product and Poincaré duality and it feels like questions about these things could potentially go in either, so that's why I'm asking)

[dreamy smelt]

When I think of those terms, I specifically think of algebraic topology, so go ahead and ask here. If it's too differential topology, someone will direct you to #diff-geo-diff-top if they care enough; no problem.

[weak narwhal]

so quick question

we had already done this

does set minus work on finite direct products in the expected "piece-wise" way?

[tough imp]

No

Consider R x R and subtract the set (R\R- x R\R-), which is subtracting positive reals x positive reals

What you end up with is the plane minus the first quadrant

But what you get if you do R- x R- is only the third quadrant

[weak narwhal]

very good point

thank you

it felt a bit slimy

[rancid umbra]

do you mean everything but the third quadrant?

[tough imp]

No

Negative numbers times negative numbers is third quadrant

But if you did R x R ( R\R- x R\R-) you only excised the first quadrant

[rancid umbra]

thought u we’re doing R x R minus that set

[tough imp]

No

I was showing why the formula on this isn’t true

I set U1 = U2 = R-

The LHS is all but 1st quadrant and the RHS is only 3rd quadrant

[rancid umbra]

um. i agree with the formula (in the picture) tho

[tough imp]

But didn’t I show it’s false?

[rancid umbra]

it’s true in this case

[tough imp]

How?

[weak narwhal]

hes saying if i do it the way I wanted i get R- x R- which is Q3, but what we'd get from my operation is R- x R-= Q1,2,4

[tough imp]

Yuh

[weak narwhal]

i still think the idea is right

[rancid umbra]

oh gotcha gotcha

[weak narwhal]

i need a different set operation

[tough imp]

What are you trying to show?

[weak narwhal]

this

[tough imp]

That the product of open sets is open?

[weak narwhal]

yep

[tough imp]

Fucking hurbed

Rip

If you’re doing like general point set

The product topology is generated by a basis of products of opens

So to me this problem seems literally definitional

What’s your definition for the product topology?

[weak narwhal]

I kinda feel the same way

[tough imp]

Is it like in terms of open balls?

[weak narwhal]

i think this is all I have to work with for now

[tough imp]

I mean

This doesn’t define what sets are open tho

Haha

[weak narwhal]

oh oh oh

thats what u want

[tough imp]

Yuh

[weak narwhal]

in (X,d) A subset X is open if X\A is closed

[tough imp]

Bruhhhhhh

Wtf is a closed set then?!

Why are you defining a topology in terms of closed sets 😭😭

These definitions are WHACK

[weak narwhal]

i dont think this is atypical is it?

[tough imp]

Holy shit this is so painful

[rancid umbra]

bruh

[tough imp]

It definitely is

This is so trash tier

[rancid umbra]

just tell them what the open sets are holy shit lol

[tough imp]

Lmao

The point is like

Open sets are ones where you can wiggle a little bit and stay inside the set

Like if x in U

There’s some ball around x which stays in U

This is like the picture you should have in mind

But you totally don’t see it with these definitions

RIP

[weak narwhal]

well we have equivalently that for any a in A, there exists r>0 s.t. B(a,r) subset A

[tough imp]

BRO

PLS SAY THAT

Kekw

Yes

This is what I wanted

Okay so this is easy to do now

[rancid umbra]

lmao

[weak narwhal]

thats technically a theorem though?

[tough imp]

take a point x in U1 x … x Un

use it

[weak narwhal]

you first define closed sets

[rancid umbra]

no that should be definitional

[honest terrace]

no this is usually what's taken as a definition

[weak narwhal]

then use intrinsic defn for open

[tough imp]

Yeah that is ass-backwards from how like everyone else does it

Hahahahah

But this is the definition of open you want

[weak narwhal]

honestly I have no issue with it

[tough imp]

So the point is

That’s because

this is all you know

I assume

[weak narwhal]

no

[tough imp]

If you’ve seen and done more

Wut

[weak narwhal]

cause ive seen it in other courses and books

[tough imp]

BRUH WHAT

[weak narwhal]

idk, it seems fairly natural to me to define it in this way

[tough imp]

This book defines topology for metric spaces

In terms of closeds????

[honest terrace]

what

the actual

[rancid umbra]

i’m gonna hurl

[honest terrace]

fuck

[tough imp]

[rancid umbra]

kindly, please burn the book

[tough imp]

Okay I’m gonna tell you how to show product of open is open them I’m gonna gtfo cuz this is so crazy to me lmfao

[honest terrace]

You could not make things less intuitive than that, tbh, even if you actually tried

[tough imp]

If x is in U1 x … x Un

Then like consider what happens when you project x down to the i-th coordinate

This lands inside of Ui

[weak narwhal]

yea I know thats the typical way

[tough imp]

So there’s some ri such that B(ri,xi) < Ui

[weak narwhal]

but we havent talked about that yet so

[tough imp]

Then take the minimum among all ri

Call this r

And B(r,x)

If you have y inside of this

Then you need that d(yi,xi) <= r

So that yi is inside of B(ri,xi) < Ui

So this shows that every component of y is inside of Ui

So that y is in U1 x … x Un

So B(r,x) < U1 x … x Un

So U1 x … x Un is open

:D

Open ball good

[rancid umbra]

dude i still think this is right

[tough imp]

Bro it totally isn’t

Compute my example

You’re doing R x R minus (positive numbers x positive numbers)

That’s 3 quadrants

[weak narwhal]

either im also having an anyerism or theres no way that its true

[tough imp]

The RHS is negative numbers x negative numbers

[weak narwhal]

just replace R- as an arb. subset U and forget the open/closedness

wait

yea its way wrong

[tough imp]

Like as an extreme extreme example do U1 is empty U2 is everything

You get R x R = R x empty set = empty

[weak narwhal]

R\R- x R\R- is the 1st quadrant , so R x R\ (R\R- x R\R-) is Q 2,3,4 right

[tough imp]

That’s what I’m getting

[weak narwhal]

so you now get R\R- x R\R- = Q2,3,4 and R\R- x R\R- = Q1

[tough imp]

Which is why I gave it as an example

[weak narwhal]

contradiction

[weak narwhal]

[rancid umbra]

fuck why do some examples work and some don’t

[weak narwhal]

not sure about that example

[tough imp]

It’s true

R x R \ (R\R x R\ empty) = R x R \(empty x R) = R x R \ empty = R x R

[gritty widget]

[tough imp]

The other side is gonna just be R x empty = empty

[weak narwhal]

X x null= null

[tough imp]

Yeah

[honest terrace]

Yeah

that's literally what chmonkey is using lol

[tough imp]

Looking at pairs where the second coordinate is in empty set

[rancid umbra]

fuck i had the wrong pictures in my head

[tough imp]

Doesn’t wxist

[tough imp]

Lmao

[gritty widget]

wait what

[tough imp]

Also hi Shika:3

[gritty widget]

R x R/ empty x R isnt R?

[tough imp]

Well

It should be

(R x R)\ empty

I was lazy with parens

Oh wait no

[tough imp]

Yes cuz empty x R = empty

[gritty widget]

wut

what is R lol

[tough imp]

Take a pair (a,b) with a in empty set b in R

Any set

[gritty widget]

a topology?

[tough imp]

Legit doesn’t matter

[honest terrace]

Ok I like where this is going

[gritty widget]

oh

you mean empty set

not a point

[tough imp]

Yeah

[gritty widget]

ok yea

[tough imp]

Yehhhhhhhhh

[gritty widget]

lol

[weak narwhal]

im saying it happens to work for my false set subtraction thing

[tough imp]

No?

You end up with R x R on the left

And empty set on the right

That’s what my computation outlined

[gritty widget]

where is theirs example?

[gritty widget]

this it?

[honest terrace]

(I like how often general topology questions end up being discussions about elementary set theory. It happens so often )

[tough imp]

Cuz point set fucking sucks

Lol

[honest terrace]

true

[weak narwhal]

this is for an analysis class

[tough imp]

I encourage you to work through U1 = R, U2 = empty

[honest terrace]

Oh I'll happily say that analysis sucks too

[tough imp]

And see how tremendously bad it goes

[gritty widget]

R^2-(R^2)?

lol what

[tough imp]

Based

[gritty widget]

ur example is empty set dud

[tough imp]

No?

[gritty widget]

wait tbh nvm

i dont know what R- is

[tough imp]

No that’s a different one

[gritty widget]

oh he means reals

[tough imp]

Negative numbers or whatever

[weak narwhal]

i see your point

[tough imp]

Yes

[weak narwhal]

using the proper defn ud be subtracting the emtpyset from R x R which is allowed and u get R x R

[tough imp]

Yeah

[gritty widget]

yea

[weak narwhal]

cause empty set doesnt care about product lol

unless?

[tough imp]

Well

It destroys things in products

And turns them into nothing

It’s like how multiplying by 0 gives 0

[weak narwhal]

i should say

the empty set in Rn is the emtpy set in R right

[tough imp]

(Literally when you do cardinal arithmetic)

[rancid umbra]

okay, so the way to go about showing this with ur wonky definitions. let x_n be a convergent sequence in R^N cut the product of the N “open” sets. if the limit x that it converges to is in the the product of the N open sets, then each of the component sequences of x_n are converge in their respective set. but that’s impossible because each of the component sequences were in R but not in U_i.

[tough imp]

Idk what this means

You’re subtracting the empty set from R x R

Like the empty set is the empty set

You don’t have to think hard about this

[weak narwhal]

can you subtract R from R x R?

[tough imp]

No

[weak narwhal]

....

see my point

[tough imp]

I don’t

[weak narwhal]

alright

[tough imp]

The empty set is a subset of every set

[weak narwhal]

I know

[tough imp]

You’re doing

[gritty widget]

can you do (x,y) - x?

[tough imp]

(R x R) \ (empty set x R)

But empty set x R = empty set

It’s empty

Nothing there

[weak narwhal]

lmfao idk how to settle this loop

[gritty widget]

do you know definition of product of sets

A x B = {(a,b)| a in A and b in B}

[weak narwhal]

im saying your counter example was a good one because emtpy x R is well defined and R x R \ empty is well defined

so you dont need any other intuition to see its true

[tough imp]

I mean… idk what there is about empty x R being well defined

It’s a product of sets

Idk

[gritty widget]

oh i see what he is saying

or what he wants

[weak narwhal]

we have a definition for the direct product with the emtpy set

that we must satisfy

[rancid umbra]

u can subtract R from R^2. it’s just R^2 again. R^2\R = R^2 since there are no elements of R in R^2

[gritty widget]

yes it is the empty set

[weak narwhal]

and we have a definition for any set subtracting the emtpy set

[tough imp]

Yeah so I don’t see how well-definedness comes into play at any point?

[gritty widget]

it doesnt

[gritty widget]

ig if u abuse notation then sure

[rancid umbra]

what abuse of notation have i made lol

[gritty widget]

but if someone wrote R^2 - R id assume they are talking about removing a line of R2

[weak narwhal]

whats the abuse?

[rancid umbra]

nah

[gritty widget]

ive had it happen before

[rancid umbra]

R^2 - R in this context would def be R^2\R

[ivory dragon]

the one abusing notation isnt c squared

thats for sure

[gritty widget]

yea thats what i mean abuse of notation, you cant subtract not subsets usually

[ivory dragon]

?

[gritty widget]

but its fine for intution stuffs

[ivory dragon]

yes you can

[gritty widget]

what?

[ivory dragon]

A \ B = A iff B is not a subset of A

this is totally fine

A \ B = {a in A | a not in B}

[gritty widget]

yes ig

[ivory dragon]

nothing about this definition forces B to be a subset of A

and in fact, it would make a lot of topological definitions quite inconvenient if it did

[gritty widget]

yeah i should sleep

is my ezcuse

why are you guys not using -

so much better to use minus

[rancid umbra]

well u got it confused with R^2 removing a line, whatever that means. not a jab at u, just proving a point as to why that notation is bad

[weak narwhal]

cause its not regular subtraction

[gritty widget]

set minus