point-set-topology

[ivory dragon]

the circumference of a circle is a line

the outer boundary of a sphere is a surface

etc

indeed, the outer boundary of a 4-ball would be 3-dimensional

so i was being a bit... informal when i said "n-sphere" earlier

[dawn geyser]

so the boundary of a glome(?) is a sphere?

[ivory dragon]

not necessarily a sphere, but 3 dimensional

[dawn geyser]

yeah okay

[ivory dragon]

anyway this is just a terminology/definition thing

[dawn geyser]

so a 4-ball has four dimensions, but a 4-sphere is 5d?

[ivory dragon]

the 4-sphere itself is 4-dimensional, but it's usually thought of as existing within 5-dimensional space as the boundary of a 5-dimensional ball

in the same way that a line is 1-dimensional, but if we wrap this line up into a circle, we view the circumference as existing within 2-dimensional space of a 2D circle

[dawn geyser]

oh okay, yeah. that makes sense

so if im using that formula to find the n-dimension volume of an 8-ball, it's just plugging k=4 into the forumla?

[ivory dragon]

yeah

and for the 9-ball, you'd plug in k = 4 also but use the other formula

the 2k+1 one

[dawn geyser]

okay i gotta read this all again, thank you so much!!

i'll send you my working when i'm done just to make sure i get it

[celest flare]

class

Does this mean topologies need not always be sets?

[midnight jewel]

since X is a set, its powerset is a set too, and all subsets of the powersets as well

so the topology is a set

[gritty widget]

why would they say its a class though?

[midnight jewel]

I mean it’s not wrong, maybe they wanted to emphasize that it’s a set of sets or sth. it’s weird tho, I agree

I’d have gone for sth like “family”

[gritty widget]

^^

[vocal wharf]

class means collection of sets

[midnight jewel]

not in set theory

[vocal wharf]

you think of proper class

[midnight jewel]

class = thing which is either a proper class or a set

“proper class” just means “class but not set”

I have no idea what the formal definition of a class is tho

[gritty widget]

maybe it helps in cat theory or sth

[midnight jewel]

never looked that deeply into it

[vocal wharf]

there is no formal definition of class in ZFC

[gritty widget]

because its not a thing in ZFC

right?

[midnight jewel]

yea cause zfc is about how sets behave

[gritty widget]

its defined in NBG

[vocal wharf]

yes, as a collection of sets

[gritty widget]

Sets are formulated in terms of classes. They are classes. The thing we refer to as a “proper class” is a class that is not a set like the universe of the ZFC.

yo N/U whats up mate

Not much kinda doing nothing atm

[marsh forge]

not here lol

[twin aurora]

usually people call something a "class" or a "collection" or a "family" when:

1. its elements are sets
2. they know that sometimes sets that contain other sets aren't actually sets themselves
3. but they don't know the set theory needed to know when that applies so they use a more non-committal term

[midnight jewel]

I call something a family when my intuition tells me I wanna call it a family cause “set” sounds too generic even though “family” literally is just a synonym

[twin aurora]

that's fair

a lot of times i use family when I mean like, a set of things but they have something big in common

e.g. a family of antiderivatives (only differing by a constant)

I've seen "collection" typically used there

like in the standard definition of a topology; e.g. in Munkres' "Topology" he calls it a set with a collection of subsets called open sets

[twin aurora]

from a cursory flip through that section I didn't see the word family once

[ivory dragon]

i generally use "family" to refer to a set composed of the relatives, ancestors, and descendants of an individual

[gritty widget]

Looking for help

U and V are given by the intersections with the open half-planes {y > -1/2} and {y < 1/2}

I tried to follow up the Mayer Vietoris sequence proof as the argument seems related but i have problem even with the start as I am not comfortable not even deciding if I shoul prove it directly or showing that P_ - P+ lays in the image of \partial and then use the exactness of the sequence

[mental spruce]

hey guys

@here Kevin Buzzard is doing his Algebraic geometry imperial class on twitch

https://www.twitch.tv/kbuzzard

thought you would like this

[delicate spire]

proof: idk the proof

[marsh forge]

Oh shit

Rn?

[mental spruce]

lecture over

[gritty widget]

Quick lecture

[mental spruce]

1 hour MSc lecture over livestream seems... fine

[gritty widget]

Oh I was going with the timestamps of your messages

[gritty widget]

Why is it that integrating any k-form over a not k surface is always 0?

is tht just by definition

[gritty widget]

Also another question

if I integrate a k-form over a k-surface that is only dimension less than k

is the reason why it gives 0 because when translating to classical integration, one of the variables dxi is integrated over an interval of length 0?

[bitter yoke]

<@&681259184582688842>

[river saffron]

If S is a subspace of the given metric space (l infinity d infinity), how can I show that there exists a Cauchy sequence in S which does not converge? I am trying to show that S is not complete.

(If nobody can help, might try again tomorrow when I wake up, and DMs are welcome ^_^ Thanks.)

[honest narwhal]

Depends a bit on S here

Well I mean so l^{infty} is a complete metric space so it boils down to showing S isn't closed

Subspace meaning vector here?

@river saffron

Topological subspace, basically subset

That's terminology which is used, and the fact that he said "subspace of the given metric space" suggests that a little bit

I guess we can give the vector subspace answer since that holds for both

The thing that worries me is more he's asking for techniques

Like he has some S in mind that he hasn't told us and he's asking the more vague "Let's say you find some S\subset l^{infty} in the wild, what are some techniques for showing it isn't closed?" like hmm

Anyway yeah I mean, I guess jan's business about finite support sequences is a non-closed vector subspace if that's what you're asking. If it's more "How do I go about solving these things?" the answer is, depends

If it's given to you explicitly there's usually some obvious candidate

Hmm I guess I'm reading a bit of chapter 2 of Silverman's elliptic curves, which is some general stuff about algebraic curves. If you're interested in that I can chat a bit in #groups-rings-fields

[gritty widget]

I tried to follow up the Mayer Vietoris sequence proof as the argument seems related but i have problem even with the start as I am not comfortable not even deciding if I shoul prove it directly or showing that P_ - P+ lays in the image of \partial and then use the exactness of the sequence
@gritty widget anyone can help?

[west spindle]

aight there's six problems here

which one do you wanna start with

uh huh

lemme just transcribe this here for my own sake

Let $X$ be a topological space, and let $A,B,C,D$ be subsets of $X$ such that $A \cap B, B \cap C$ and $C \cap D$ are all nonempty. Prove that $A \cup B \cup C \cup D$ is connected.

[gentle osprey]

Ann:

[west spindle]

okay so what have you got so far

[wanton marsh]

you forgot your quantifiers

[west spindle]

yeah that is not the definition at all

[wanton marsh]

it's also not the standard definition

[west spindle]

and ALSO you're clashing with the notation used in the problem

which isn't a good sign

[gritty widget]

yeah try to use different definiotion and proof by contradiction should work

it is, but have you heard about the one that says that X is connected if there arent any disjoint open subsets of X such that AuB =X?

[wanton marsh]

that's certainly more correct that the thing you wrote earlier

it's still wrong because using A=X and B = the empty set, you can show that every space is disconnected

it should be okay if you add that A and B have to be non-empty

[west spindle]

this certainly isn't a convenient definition to use here imo

[wanton marsh]

I don't think it's that different from the normal one

[marsh forge]

I think this will be about as hard with either defn

[gritty widget]

@gritty widget anyone can help?
@gritty widget any tips hint?

[lunar mural]

Is there a go to intro to topology group?

Trying to learn it over the next year or so for a physics project

[gritty widget]

Group?

[bitter yoke]

do you mean topological group?

[marsh forge]

No like a reading group i think

AMC im playing games rn but can take a look later tn

Remind me if i forget @gritty widget

[bitter yoke]

what games do you play?????

[austere dawn]

How do I do these?

[small obsidian]

@austere dawn
You just need two different sequences that go to (0,0) that have the same limit

Consider x = 0 and y = 0 as "the sequences"

[austere dawn]

so like, 1/n?

@small obsidian

[small obsidian]

You could call them
(1/n, 0)
And
(0, 1/n)

But, without fully justifying it, let x = 0 and instead let y = 0 to show that you can get two different answers

[austere dawn]

the answer would be that there is no limit right?

[small obsidian]

Yes, since you can get multiple answers

[austere dawn]

im assuming that if lim_x lim_y = lim_y lim_x is not true then the limit does not exist?

[small obsidian]

No matter how you approach (0,0), if the limit exists, you should get the same answer every time.

If you use two different paths, and get two different answers, the limit must not exist

[austere dawn]

thank you so much!

how about the one under it?

that one is confusing for me too

for that one, I think I should use the definition, right?

for multivariable differentiability; namely, lim |f(x+h)-f(x)-Ah|/||h||=0

?

[small obsidian]

Don't take my word for it, but a point is differentiable if both partial derivatives exist, right?

<@&681258879522177051>
A point R² → R is differentiable when?

[austere dawn]

Each entry of the jacobian is continuous?

[midnight jewel]

uh iirc both partial derivatives can exist but it’s still not differentiable. hang on we had a theorem about this lemme look in my analysis notes

yea okay it’s not true in general

they give this example as a function where the partial derivatives exist on all of ℝ² but it’s not differentiable at 0

however, this is true:
Let $U \subseteq \R^n$ be open, $f \colon U \to \R^m$ be a function. If for each $j \in {1, \dots, n}$ the partial derivative $\partial_j f$ exists on all of $U$ and defines a continuous function, then $f$ is differentiable on all of $U$

[gentle osprey]

Sascha Baer:

[midnight jewel]

so you just need the partial derivatives to be continuous and then it works

@small obsidian @austere dawn

[lunar mural]

@gritty widget @bitter yoke sorry I was so tired, I meant go to intro "book"

[grand monolith]

I had an idiot moment today. I tried to look up a bijective mapping between the Euclidean plane and a hyperbolic plane

[midnight jewel]

I mean that’s easy enough if you don’t want it to preserve any structure :P

just take the hyperbolic plane embedded in ℝ²’¹ and project upward from the xy-plane

[grand monolith]

Ok idiot moment number 2

Ty

Is it safe to assume that any of these projections between spherical or hyperbolic planes and such are continuous

Oh wait nope, boundaries at infinity

[grand monolith]

Are continuous bijective maps sufficient to show that a space filling curve in a Euclidean plane corresponds to a space filling curve in a hyperbolic plane?

Intuitively feels like yes, but that also implies that any space filling curve in hyperbolic space has a Euclidean analogue

Can't think of a reason why it wouldn't work

[chrome dew]

if it's bijective and space filling on one side, it's space filling on the other, otherwise it's not really bijective

[grand monolith]

That's what I thought

Ty

[river saffron]

Clearly (1, 1/2, 1/3, 1/n, 0, 0, ...) is a member of l-infinity

Could I then say that: (1, 1/3, 1/3, 1/n) is a member of the subspace of l-infinity?

[bitter yoke]

Uh, there are lots of subspaces of l infinity

So if that element is in your subspace depends a lot on what your subspace is

[river saffron]

I'm basically using the metric space l-infinity, with distance function d-infinity.

[bitter yoke]

There's definitely subspaces that contain that as an element

[river saffron]

We are asked to take a subset S of this space and find a cauchy sequence to prove that the subspace S is not complete.

Oooh ok, so if I find 2 sequences, and use d-infinity distance formula, I could perhaps somehow prove that S is not complete! Ty

[chrome dew]

you might want to find a sequence of sequences

converging to some sequence that's not in your space

[bitter yoke]

Okay yeah, they give you a specific subspace here

[river saffron]

The annoying bit is I fairly well understand what is being asked, but just cant make progress aaa

[bitter yoke]

Meros solution is good

Just play around with sequences and try to get a sequence of vectors in your subspace to converge to something outside of the subspace

[fervent citrus]

Your original idea of looking at (1, 1/2, ..., 1/n, 0, 0 ,...) was fine

this sequence (which nth term is (1, 1/2,..., 1/n, 0,...)) converges in l-infty but you don't need to know that to show it doesn't converge in S

[midnight jewel]

There's definitely subspaces that contain that as an element
I would argue that (1,⅓,⅓,1/n) is not an element of l^\infty since it’s not an infinite sequence

[round saddle]

I'll head to the math help channel 🙂

[winged viper]

Topology noob here. I've been trying to show that given any knot (closed 1-surface), you can construct a 2-surface that is oriented where the boundary of the 2-surface is the knot. Is this logic correct?

• any knot must be oriented, so it is homeomorphic with the unit circle
• construct a disc with the boundary as the unit circle and apply the homeomorphism to make it into the knot again

(and the disk is oriented and a homeomorphism can't change that)

again i really dont know anything about topology and im trying to show this for a calc class so a lot of the assumptions i make are pulled out of my ass

[marsh forge]

This would not really constitute a proof

as of course any map S^1->R^n

extends to a map D^2->R^n

But you haven't shown that this image is a manifold

and of course you have to choose the correct on as its not too hard to build one that isn't

[winged viper]

well i wouldnt need to extend the map from s^1 to D^2 right? the knot is already in 3 dimensions

so i would just need to extend this mapping to be continuous on the disk?

[honest narwhal]

Lol

[dire warren]

Ohh

[midnight jewel]

that’s also what I’m thinking

I couldn’t think of any reasonable counterexamples for the if direction, anyway

[rugged swan]

R²\{a,b} and S1nS1

but S1 is an "open ring"

[midnight jewel]

an annulus?

[rugged swan]

yes

and connected with smth "open"

like a pipe

[dire warren]

Disrespect? Where lol

[rugged swan]

that"s a counter examble

but you can ask "if two open sets have the same fundamental group, are they homotopically equivalent ?"

[midnight jewel]

how do you show an annulus is not homeo to ℝ² with a point removed? I remember showing that there is no biholomorphic map bewteen the two (as subsets of ℂ) and that was decently difficult

(was an exam question)

[rugged swan]

hum

[midnight jewel]

homeo is less restrictive and I don’t think the proof would extend

oh actualyl yea

they are homeo

since (0,∞) and (1,2) are

and you can just apply that radially

[rugged swan]

that's sad x)

[midnight jewel]

and if it works for a single annulus I’m sure it also works for two connected ones

it’s just gonna be a whole lot uglier

[rugged swan]

yes

maybe

[midnight jewel]

anyway as for the infinitely generated case

consider an infinite “row” of such annuli

[rugged swan]

take a non connected open set

[midnight jewel]

I’m pretty sure that has the same fundamental group as an infinite “cross” of such annuli but also doubt they’re homeo

they specified connected

[rugged swan]

ok

I don't see how you would connect these infinites annulus but in an open way

[midnight jewel]

lemme sketch how I mean

take an open strip, along the x axis; cut out a circle around each (n,0) such that none of the circles touch each other or the boundary

for the cross, just take this thing and also rotate it 90 degrees around (0,0), and take the union

[rugged swan]

I think that would be homeomorphic to R²\Z points too

[midnight jewel]

it should be yes

oh right

yea

[dire warren]

You can show they’re not homeomorphic too

The cross and the row

[midnight jewel]

oh can you?

[dire warren]

The cross has 4 ends

Row has 2

[rugged swan]

the cross ?

[dire warren]

Geometric group theory saves the day

[rugged swan]

oh ok

[dire warren]

LOL

Well I mean I learnt it in geometric group theory for the first time

Rite haha

[midnight jewel]

what does “ends” mean in this context. I mean I can see it goes off in four directions obviously

[rugged swan]

have you any references about that ?

or a name ?

[dire warren]

Geometric group theory by kapovich drutu is where I saw this

[midnight jewel]

anyway this should be an example of two open, connected subsets of ℝ² that aren’t homeomorphic, with (probably) isomorphic but not finitely generated fundamental groups. no proof that they are isomorphic but I reckon if you label each generating loop with an integer in the strip, and an element in ℤ×{0,1} in the cross and then take a bijection between the two sets you prolly get an iso

I assume what I wrote works I just didn’t wanna actually think through it

[rugged swan]

thx

[midnight jewel]

yea that’s fair

wanna bet it relies on CH somehow

[dire warren]

CH in my topology

[midnight jewel]

I wouldn’t even be surprised but openness kinda removes a lot of the ugliness that could arise

right you can just remove single points

[dire warren]

I mean in point set yes, but in topology in general is it still the case

Damnit

[marsh forge]

Wait wdym jan

[midnight jewel]

I keep thinking of the problem in terms of gluing together nice open sets but obviously you can just remove arbitrarily ugly closed sets

[marsh forge]

I cant really think of an example?

[rugged swan]

wtf how is it possbile that I've ever heard about ends

[marsh forge]

What do they have to do w continuum hypothesis?

Oh wild

[dire warren]

@midnight jewel closed sets aren’t that ugly.. are they?

So like is CH true? What is the consensus

Yeah I mean like

What do people think should be taken

As the axiom

like how people say “choice is true” that kinda thing

[marsh forge]

I find it hard to like

Care about results

Dependant on adding axioms

Dependent*

Which is like

Tautologically nonsense

[dire warren]

It’s tautological, doesn’t mean it’s nonsense xD

[marsh forge]

But like if theorem T requires CH

I feel like idc about theorem T

Unless it like can be applied to show something without CH

But idk if thats a thing

[midnight jewel]

compromise: there is exactly one cardinal between aleph0 and c

[gritty widget]

https://cdn.discordapp.com/attachments/486841106298961930/691402074457636864/241958658294939648.png

[dire warren]

Assume CH -> it’s true, assume not CH -> it’s true

I feel there should be proofs that go like that

by definition N/U yes

Usually you take 1 tho

[marsh forge]

Assume CH, there are inf many primes

[dire warren]

Not zero

[marsh forge]

Assume not CH

There are inf many primes

[midnight jewel]

I mean he’s not wrong

[gritty widget]

Ok thanks

[marsh forge]

I mean that was intended to be a joke

I feel like everything i say comes off more chaotic w this pfp

[dire warren]

:))

It’s crazy

Why is it independent

Why don’t we have absolute truth

In our maths

Math is insane

It is no sign of sanity to be well adjusted to an insane world

(Math is the insane world btw)

[marsh forge]

I guess my main reason

Is that I feel like the big pictures that i appreciate

Require that i work in zfc

And not assume extra stuff

I expand them

Just in different directions

[dire warren]

The year is 1900

“The stuff I appreciate requires I work in peano axioms”

Without assuming extra stuff

[marsh forge]

I prefer to be agnostic

[dire warren]

All proof requires faith

As godel proved

Therefore god exists so u cannot be agnostic

[marsh forge]

Models dont matter if something is provable from the axioms right im not insane?

[gritty widget]

Assume X models ZFC

[marsh forge]

Like provable iff true in every model

[midnight jewel]

I have faith in naive set theory

[marsh forge]

So is it unreasonable to not care?

[dire warren]

How can u

[midnight jewel]

that one barber can just never shave I don’t care

[dire warren]

Look at an interesting looking question

And then once u find out it’s independent just not care

Shouldn’t you be more like

WHY ISNT THIS DECIDABLE

WITH MY FAITH

[midnight jewel]

22 you've gone insane

[dire warren]

What’s the weakest possible system

Is it Peano arithmetic

LOL

[gritty widget]

The empty set

[marsh forge]

Not even close

[midnight jewel]

Just logical tautologies
didn’t you just assume tautologies exist

[dire warren]

The weakest system such that such and such holds

That’s a field of math right

Reverse math or something

[marsh forge]

If you truly have no assumptions

Are there even tautologies

Dennis Hirschfeldt does it

And taught us some

Like RCA and Konnig

[dire warren]

What’s the weakest system in which I can shitpost analysis questions

[marsh forge]

Analysis depends heavily on like

ZF

Right?

[dire warren]

I see

[marsh forge]

Like idrk how youd formulate

[gritty widget]

ZFC though?

[marsh forge]

Most questions

[dire warren]

“There’s no point”

“Points don’t exist”

[marsh forge]

How strong of choice do you get in ZF

[midnight jewel]

only finite, right?

[dire warren]

#sleepswithcrazy

[marsh forge]

Well like some surjections

Will have sections right

Even w ZF?

So like, are finite collections of finite sets ok

[midnight jewel]

you can make infinitely many “nonarbitrary choices” too, ofc. like if you can just write down a formula for how you’re choosing things

but that’s not choice

that’s just writing down a formula

[ebon zephyr]

Is there a way to detect orientation of a 3D vector to respective axis ?

I am using the atan method with all axis to determine the rotations , seems to go haywire

https://www.youtube.com/watch?v=IziiCZXqjwM

[gritty widget]

Is it true that if in a space every subset is saturated, that this space is T1?
I’ve proved the converse but I don’t think this is true since for the converse it required to take possibly infinite intersections.
But I seem to be unable to come up with a counter example.

https://cdn.discordapp.com/attachments/504663858162696217/691450981342249010/241958658294939648.png

<@&681259184582688842>

[gritty widget]

Solved.

[gritty widget]

Those two statements are equivalent btw so T1 is equivalent to all subsets are saturated

[grim coyote]

hey can someone explain CW complexes to me in simpler/more fundamental terms

[loud scarab]

Hello, I'm trying to figure out why the highlighted part here is true, is it meant that all the neighborhoods contain x and thus they intersect A at x only, or can it be any other point of A? If its any other point of A, how to justify that?

[tepid totem]

You don't necessarily have that x is in A, so it has to be another point. But if x_n converges to x, then by definition any neighborhood of x must contain all the x_n for large enough n, and those are all in A.

[marsh forge]

@grim coyote

So, we like manifolds right?

because they are locally R^n

Now, it's too restrictive in some sense to require that this n be fixed

also, we might want to allow some small places where the space isn't locally R^n for any n

But we still want to keep a lot of the nice intuition

how do we do this?

Well, the next best thing is to try to patch together spaces with really nice pieces

Disks!

So we build CW complexes by taking disks and gluing their boundaries

starting with dimension 0 (points)

adding the skeleton (1-disks)

and then we add 2 disks and so on

gluing each to the lower skeleton

Does this make sense?

[grim coyote]

Ah I see

[marsh forge]

Also it's worth knowing that these guys can basically approximate any nice topological space

and every space is weakly homotopy equivalent to a CW complex

[grim coyote]

how about their relationship to simplicial complexes

[marsh forge]

Very closely related

but a little 'looser' in some sense

[grim coyote]

okay so simplicial complexes are kinda like CW complexes but with more rules

[marsh forge]

kinda

[grim coyote]

sry lol, a high schooler interested in discrete morse theory

[marsh forge]

np

[loud scarab]

@tepid totem oh got it, I seem to have forgotten the topological definition

[gritty widget]

https://cdn.discordapp.com/attachments/612450134567026688/691744533079261226/241958658294939648.png

How could I approach this?

[marsh forge]

its not topoloogical

you just prove it on the point set level

take a point in one, show its in the other

[gritty widget]

So should I ask in a questions channel?

I'm gonna take this as a yes

[chrome dew]

your definition of D doesn't include the pt (a,b)

[gritty widget]

See afterwards

[chrome dew]

what

[gritty widget]

(a,b) is defined afterwards

[chrome dew]

you define the radius in terms of a,b then make an open ball not containin the boundary

[gritty widget]

Ofc is (a,b) in D, it is by definition

The radius is 1

N/U yo

Yo

[marsh forge]

i mean n/u

it's just like

demonstrate that that inequality holds

for any R_(a,b)

like let x be in R_(a,b)

it gives you some inequalities for free

then you have to show that x is in the unit disk

Is it just working w the inequalities that is tripping you up?

[gritty widget]

It seems to get quite long and leading nowhere

So I thought I'm overlooking something

[marsh forge]

No probably not

you just gotta compute |x| for all x in R_(a,b)

and show its less than 1

or at least bound |x|

[gritty widget]

Hm ok thanks

[odd thistle]

Do the metrics $$d_{1}=\max\limits_{a\leq x \leq b}|f(x)-g(x)|$$ and $$d_{2}=\int_{a}^{b}|f(x)-g(x)|dx$$ determine the same metric topology?

[gentle osprey]

Tiamtum:

[odd thistle]

Oops, on the space $$C[a,b]$$ , continuous functions from $$[a,b]$$ to $$\mathbb{R}$$

[gentle osprey]

Tiamtum:

[vital sail]

no

[midnight jewel]

consider [a,b] = [0,1] and the functions xⁿ for n∈ℕ
consider dᵢ(xⁿ,0) and conclude that the d₂-ball of radius 0<ε<1 around the 0-function cannot be contained in any d₁-ball around 0

@odd thistle

btw, single dollar signs for inline latex

[midnight jewel]

wrong channel

[white oyster]

it is for plane geometry, could you please direct me to right channel

[midnight jewel]

#geometry-and-trigonometry

this is for higher university-level courses

mostly topology and differential geometry

[white oyster]

i understand , this is undergrad though

i will try there also

[midnight jewel]

quite a few people have calculus in university but we put it in pre-university anyway

[white oyster]

i see

[minor stag]

So I'm trying to show union of connected sets is connected if their intersection is non empty

I'm not sure if my proof is 100% right but i took the case for two sets which are both connected

call these sets A,B, then by definition we can find a continuous function f such that $f:A -> {0,1}$ is constant, and $f:B -> {0,1}$ is constant as well

[gentle osprey]

こ こ だ よ:

[minor stag]

Then since their intersection is nonempty we can take some element c in both A and B, which must be mapped by f to the same element

so f is constant on A union B, so A union B is connected

Is this ok

[odd thistle]

@midnight jewel thanks, I thought so. sorry for the display mode spam.

[knotty pasture]

Hello friend

I am an italian guy closed in quarantine who want to help me with a problem?

I have to find explicitly an embedding for klein bottle in IR^4

[lyric garnet]

statement: the closure of any set is closed

isnt that only true if the set has at least one limit point?

[bitter yoke]

no

[lyric garnet]

oh if a set has no limit points then its closed

[loud scarab]

Hello, in the intermediate value theorem statement in Munkres' topology, the theorem stated that if there exists a real number 'r' between f(a) and f(b) in the image, then there must exist a real number 'c' such that f(c) = r, it is natural to think that this 'c' lies in the interval [a,b], but it does not say that in the statement, is the number 'c' free to be any number in X provided that its image gives 'r'?

in short, can this intermediate value 'c' be somewhere other than the interval [a,b], at least in a topological statement?

[vital sail]

The "interval [a,b]" is not defined in the general setting because X doesn't necessarily have an order

[loud scarab]

Y has an order and f(b) > f(a)

X does not have an order

f(a) < r < f(b)

so here the number c that makes f(c) = r is free to be less than a, right?

[bitter yoke]

Again, "less than a" makes no sense because X does not have an order

[loud scarab]

Ok, thanks for clearing that up, but this is a generalization of the theorem, the theorem in calculus is a special case, how to transition into calculus and say that a<c<b ( we let X be an interval now )

I mean, this generalized theorem guarantees that there exists a number c such that f(c) gives r which lies between f(a) and f(b)

regardless of whether the domain has an order or not

so when we impose an order on the domain, like an interval in calculus

If f is continuous, c can be anywhere?

maybe there is a non injective function

so two values give the same image

so we can allow multiple c's?

[bitter yoke]

Intermediate value theorem guarantees the existence of such a 'c' in the interval [a,b]. It's possible that there could be other values of c outside the interval [a,b]

[loud scarab]

Intermediate value theorem guarantees the existence of such a 'c' in the interval [a,b].
@bitter yoke

in which context does it guarantee that?

because in the generealized theorem, there is no order

from what I understand, the c is guaranteed to exist. fullstop

but the theorem does not state where since there is no order on X, are there extra tools needed to transition to calculus or is there soemthing missing?

[bitter yoke]

Well, the usual intermediate value theorem in calculus guarantees that when X is an interval in R

[loud scarab]

Oh I see

I understood now, thank you.

[gentle osprey]

kxrider:

[tepid totem]

You didnt show that f^-1(V) is open for any open set V, you showed it only in the case V is the image of an open set, i.e V=f(U).
For example, take f: S^1 to [0,1) by sending a point to its angle divided by 2pi. Clearly bijective, and its also open. But the inverse image of [0,1/2) is not open, precisely because it is not the image of an open set of S^1.

@little hemlock

[little hemlock]

ah i see, thank you

[dim meadow]

@little hemlock incidentally if you think about it your question is equivalent to the question "why is a bijective continuous function not a homeo"

Because f bijective open or closed map is equivalent to f^{-1} bijective continuous

[tacit stratus]

Is there some algorithmic way to create a simplical complex homeomorphic to a given delta complex?

[marsh forge]

Yes

Iirc

Double barycenteic

Barycentric

Does the trick

[tacit stratus]

ohh right i remember doing that

unfortunately thats not really practical for my purposes because I need the number of simplicies to stay reasonably small

[marsh forge]

You can def refine this

By finding points where it fails

And then restricting the barycenter

[lean cedar]

I have a small question: Given a vector bundle $\pi:E\rightarrow B$ and some open set $U\subseteq B$ we say a function $s:U\rightarrow \pi^{-1}(U)$ is a section of $E$ over $U$ if $s$ is continuous and $\pi\circ s=\text{id}_U$. The last condition is of course that $s$ is injective when $U\ne\emptyset$. This seems to me to be the easier definition when first getting into the topic (like I am). How were you introduced to sections, with injectivity or with $\pi\circ s=\text{id}$?

[gentle osprey]

MrMonday:

[rugged swan]

Injectivity doesn't imply pi o s = id

[lean cedar]

feck im stupid

it only implies left inverse

I'll see myself out

[gentle osprey]

Token:

Token:

[lean cedar]

well given a point (a,b) you can walk along the non-rational component until it's rational and then switch around and then walk along the other component. This should suffice to walk to any point of the space. Like: (1,π)->(1,3)->(π,3) where every arrow is given by a path and thus the composition is a path aswell.

You also give a path from every point to (0,0). Then you can also find a path fro. ebeey point to every other and the space is pathconnected.

[gentle osprey]

Token:

[honest narwhal]

Actually computing the homology of a tetrahedron simplicially finally

And wow

I have to find SNF for these matrices and UGH

[lean cedar]

path seems good @humble obsidian now you have three other cases were different components are rational

[sleek thicket]

@honest narwhal are you doing weibel?

[honest narwhal]

Yeah

[sleek thicket]

I felt very proud of myself when I did that exercise

It's 90% of my understanding of homology

[west spindle]

does anyone here happen to be good at hyperbolic geometry

how would i go about figuring out the side lengths and the angles in this tiling of the hyperbolic plane by hexagons and heptagons

long shot but <@&286206848099549185> maybe?

[wanton marsh]

in hyperbolic geometry, aren't the geodesics little arcs instead of little line segments ?

[west spindle]

in the poincare model yes they are arcs

and the ones here are also arcs

i don't really see an issue

they just look like line segments bc they're short

[midnight jewel]

if you have a parametrization of the lines then it shouldn’t be too difficult I think? you’ll have to take the derivative and then integrate its norm (wrt to the poincaré metric ofc) from one corner to the next

there’s formulas given on wikipedia for distances between two points

so if you know the coordinates of some vertices then it’s easy enough

if not… then I have no idea either sorry

I dunno how you could construct them

[chrome dew]

it's a shot in the dark but this looks like it has a table on hyperbolic trig tables in section 9 http://web1.kcn.jp/hp28ah77/index.html

hard to read I think this incomplete and written by non native english speaker lol good luck

[honest narwhal]

@sleek thicket sucked it up and did the computation

[sleek thicket]

Yeah?

The tetrahedron one?

[dim meadow]

Isn't that in weibel chapter 1?

[honest narwhal]

Yeah at the time I was like eh I'll do it later. But nah I did it now

Really I'm on chapter 2 of Weibel modulo having skipped some bits of chapter 1. Though now with the online stuff I wanna try to finish up all of chapter 1, possibly doing all the problems

[small obsidian]

Reading over topology without tears again, making sure I take in all the details. There's a question in the Euclidean topology chapter:

Prove that {1/n, n ∈ N} U {0} is closed.
Prove that {1/n, n ∈ N} is neither.

The book hasn't introduced limit points yet. I really have no clue how this is supposed to be proven

[floral gust]

Closed where? R?

If so try arguing the fact that the complement is open.

[honest narwhal]

Yeah probably a good idea

Second one is obv neither open nor closed

[floral gust]

For the latter argue that the open sets of any (just showing one sufficies) of the point isn't contained (Can you find any open neighbourhood of 1 contained inside {1/n : n in N}? Any neighbourhood must contain an irrational? What can you say from there?)

To argue against the closure, complement is again a good idea. Specifically check what happens at the left side.

[wooden scarab]

If I have a connected topological manifold M and its universal cover C, is there a sense in which I can talk about the largest sets in M that are homeomorphic to some set in the universal cover? Clearly if a set is not simply connected then it's preimage for the projection map is not even the union of disjoint sets.
In other words there some necessary and sufficient topological properties that imply a set in M is homeomorphic to some set in the universal cover?

[small obsidian]

Oop, sorry I forgot I asked the question. I'm that ghost now.

Yes, the only way, right now, to prove a set is closed is to take the complement and instead prove open.

Clearly, you can just surround any point by (1/n, 1/n+1). The problem is the existence of {0}

Without the {0}, the set can't be closed, because you can't surround 0.

Huh. I think writing it out here, I just got it.

[sleek thicket]

I've decided to stan Hatcher

[tacit stratus]

This isn’t really what you asked, but for the preimage of projection to be a disjoint union of homeomorphic things, you need the image of the set’s fundamental group under inclusion into M’s fundamental group to be trivial. This isn’t necessary for it to be homeomorphic to some subset; consider a non-contractible loop on a torus

[honest narwhal]

@sleek thicket I've decided ur bad

[sleek thicket]

No I have a good reason for it

Someone is vagueing me on Twitter

And so I have to be contrarian

The proof in Hatcher is less than a page shorter than the one in Bredon! It doesn't skip all the messy details!

[honest narwhal]

Alright as long as it's not genuine. Still idk even for the sake of being contrarian there are certain positions that for the sake of your status as a rational being as such you just don't take

[chrome dew]

I disagree

for the sake of being contrarian ofc

[honest narwhal]

So I'm trying to show the business about maps factoring as B->im(f)->C in a general abelian category

In particular that B->im(f) is epi

This takes some work lol

[dim meadow]

Oh I did that problem, I remembered there being something off about the statement

[honest narwhal]

So it seems in an abelian cat, mono + epi = iso

But I'm not seeing it

I showed that mono <=> 0 kernel and epi <=> 0 cokernel

And I feel like it should be easy from here

So A->B has kernel and cokernel 0

We wanna manufacture a map B->A

I'm not sure what info can be extracted out of that statement. Is it to consider the 0 map B->A and mess with that?

Oh hold up that might work lemme see

Oh shit yo

Okay okay

That's slick

B->0 has the identity as the kernel

Or wait no this arrow goes the other way

Press F to pay respects

Well it goes the wrong way in the sense of "I didn't do shit"

Like B->0 has the identity as the kernel, because f:A->B is 0 there's a map A->B such that f factors as id circ f

But like

Lol

Replace "remember" with "learn"

because idk what an equalizer is

f:B->C, kernel is a map i:A->B which is initial

Initial wrt fi = 0 lol

Important detail

Oh okay actually no the trick I think is gonna use the fact that you're abelian, not just additive

Like in commutative rings this is false

Because Z->Q

So we need abelian cats

So the distinguishing factor for abelian cats is that monic = ker(coker) and epic = coker(ker)

So f:A->B is monic and epic

So cokernel is B->0

OH

A->B is the kernel of B->0

As is identity boom

Well the stronger statement that you're the kernel, not of anything but of your cokernel, is what's key here

Because id:B->B, then B->0

Can now lift to B->A

Woot

And reverse for the other composition

Yeah this is making me really appreciate R-modules

I like my elements

Well, small stuff, but I feel the property of being a kernel or cokernel is something you need the whole category for

Or I guess hmm

Or I guess if you take the abelian cat generated by stuff you want, then the 0 object of that guy should be unique? So in particular you can make some claim that "the kernel is unique downstairs"

But then the kernel of the large cat is a kernel of the small cat

Small meaning set of objects

Generated just like, add in everything needed to turn it into an abelian category

Wait then that could cause a problem maybe I need to think about this more carefully

Well no my concern is a bit different it's more like

What kinds of arguments you can Freyd-Mitchell over

Well, small ones according to Weibel at least

Yeah that's the thing that's making me worried

Like, some things allow you to just take the "category generated by a diagram"

Or abelian category generated by a diagram

In particular stuff being exact works out fine

But what worries me is that if you can construct a kernel in R-mod, that doesn't automatically Freyd-Mitchell over

Well so, here's the concern. You have a map in some abelian cat and you wanna show it's a kernel. You pass to a small abelian cat containing it and show that with respect to that smaller cat, this map is a kernel

By Freyd-Mitchell

Thing is, could a map be a kernel in this small abelian cat but fail to be a kernel in the whole thing?

Ah wait I guess you could play some game where you're like

Oh I wanna test whether this map factors right. Stick that on, push it across, your thing satisfies the property in this intermediate cat

So it's a kernel

So that map factors

And just do that to every map

Okay so if you have some theorem of the form "Maps satisfying blah are monic/epic" you can just prove this in R-mod

[honest narwhal]

Alright finally proved that factorization and wow it's a fuckin mess

[wooden scarab]

This isn’t really what you asked, but for the preimage of projection to be a disjoint union of homeomorphic things, you need the image of the set’s fundamental group under inclusion into M’s fundamental group to be trivial. This isn’t necessary for it to be homeomorphic to some subset; consider a non-contractible loop on a torus
@tacit stratus it's something to get a handle on it. Thanks

I didn't think about attempting to do stuff with the fundamental group

[gentle osprey]

Trichloromethane:

[rugged swan]

it does not cover [0,1]

[gritty widget]

https://math.stackexchange.com/questions/1221467/characterization-of-compactness-in-terms-of-closed-sets
I don't understand how this characterization doesn't require the proof that OP gave?

Are they just meaning that the proof is simple to show by properties of open and closed sets, or literally that its unnecessarily? What then is the more trivial proof?

wait nvm it just sunk in lol

union of a collection is the complement of the intersection of the collection with the complement of every set in the original, and its still equal to the whole compact metric space, take complements on both sides, intersection of collection of each set's complement is empty

Yeah and then instead of "every open collection, if it's an open cover -> it has finite subcover" just put "every closed collection, if its intersection is empty -> it has a finite intersection which is empty"

so you don't need that proof OP made

Is there a name for a collection whose intersection is empty? A dual to "cover", since for every such set you can make a cover from the complements of its elements

[mortal terrace]

i mean a collection of disjoint sets

[real notch]

Just because the intersection is empty doesn’t mean they’re all disjoint my dude

You could just say “A family of sets $A$ such that $\bigcap A=\varnothing$”

[gentle osprey]

Darkrifts:

[mortal terrace]

i mean but from any such set you can construct it easily into a collection of disjoint sets which has this complement is a cover condition

[gritty widget]

aw I guess there's no popular name for it then ;(
It would be nice to have a term to formulate compactness as not just "every open cover has a finite subcover" but also "every closed [term] has a finite sub[term]", where the term is that dual notion of cover

Which would absolutely be true

[mortal terrace]

but yeah you would usually just say empty intersection

[real notch]

co-cover

just \cap A = 0 it

[gritty widget]

Hey I’m not sure if this is too difficult but I really don’t know the answer to this problem and I’m struggling , it would mean a lot if someone could solve it , thank you : What is the riemannien curvature tensor if the minkowski metric g=dt^2-dx^2-dy^2-dz^2

[chrome dew]

what are you having trouble with?

should be purely formula plug and chugging

Riemann curvature tensor can be written in terms of Christoffel symbols which can be written in terms of the metric tensor

if you're doing too much work, you're doing it wrong btw

[gritty widget]

I’m currently in algebra II

what textbook u workin thru btw

this is a problem I need to solve for an application , it’s completely random and not from a textbook . For algebra II, I don’t have a textbook we just do workshoots

•worksheets

I have no idea how to do the problem and it’s way over my head

[bitter yoke]

Uh, algebra 2 like high school algebra 2?

[gritty widget]

High school algebra II

so I’m a little fucked I have no idea how to do it

[bitter yoke]

Uh, well, why do you need to do this

[gritty widget]

My sister is in college math but she hasn’t gotten to do this so she can’t help me

[dire warren]

Why are u doing differential geometry in high school

[gritty widget]

It’s for an application and nobody knows how to do it but I least want to answer it

[dire warren]

What do you mean an application

[gritty widget]

I’m trying to get leadership for college and a friend is willing to teach me how to program in java and work with spigot for Minecraft but I have to apply for it

[sleek thicket]

lmao

[gritty widget]

it’s basically trying to program for a Minecraft server

[dire warren]

You have to apply for your friends mentorship?

[chrome dew]

lol what the hell

[gritty widget]

yUP

[dire warren]

Lol

[gritty widget]

it’s a little ridiculous

[sleek thicket]

This is so funny

[gritty widget]

But I just need an answer at this point I’m desperate

[dire warren]

And he set this question?

[gritty widget]

yup

[dire warren]

.

[chrome dew]

ok the answer is it's 0 cause minkowski space is flat ok there you go

[gritty widget]

There are three problems I’m hoping I answered the other two right

[sleek thicket]

I gotta play this minecraft mod

[gritty widget]

Wait is it seriously 0

[chrome dew]

yep

[gritty widget]

Thank you so much

This really means a lot

[chrome dew]

I'm sure

it means nothing to you

[sleek thicket]

Can somebody help me? I'm trying to get to the end but the portal isn't lighting

I'm in creative mode btw

[bitter yoke]

can someone actually teach me how to play minecraft tho

just bought it a few days ago

no meme

[gritty widget]

sure I’m down

[sleek thicket]

It's great dude

Do you want an invite to a server?

It's modded

[bitter yoke]

lmao im mostly joking

part of the fun is figuring out all the stuff myself

at least for me

[sleek thicket]

Yeah 100%

If you want more stuff to figure out though, this server has 180 mods

[bitter yoke]

jesus christ

what type of mods are there

[gritty widget]

I love modded MC but usually my PC can’t handle it

[sleek thicket]

All of them

Mods that add machines and power

Mods that add different magic systems

Pipes and giant storage setups for automation

Jetpacks and powersuits

This is what I've been doing for the past three weeks

[bitter yoke]

jesus christ, im definitely going to spend too much time playing mineceraft already

[sleek thicket]

I've played 4 full days

Since I got to California

Probably more, that's just what it was the last time I checked

[bitter yoke]

holy fuck yeah

I'm scared of getting super addicted lmao

[sleek thicket]

Yeah I'm gonna have to set hard limits

Since classes start Monday

I played so much minecraft in middle school

[bitter yoke]

haha I wish I had started earlier now

[gritty widget]

this channel just turned full meme I love it

I was crazy addicted to Minecraft in middle school

also how were you gonna apply the curvature tensor to minecraft lmao

NO CLUE

The only problem related to it was how do you use algebra to find a stronghold using only 2 ended pearls

That one wasn’t too difficult to answer so I was able to do it

do you look at the java directly?

I have a friend who breaks stuff in mc like that and was mentioned by antvenom because he had a breakthrough solution recently

are you in the monkeys server?

I’m not sure what you mean but I just got coordinates from the game and substituted them in for y-y1= m(x-x1)

I don’t know how to program

[chrome dew]

you're in a good position to start learning if you wanted, you can easily make some goals of things you want to make.

[ivory dragon]

yeah thats just an application of trigonometry

standard triangulation

[gritty widget]

I’m interested in learning more advanced math but I’ve struggled in the past so I’m not really sure if it’s a good idea

[ivory dragon]

one valid question, though, is - considering that there's some "error" in the angle measurement of an ender pearl throw, where is the optimal spot to stand for a follow-up third throw to adjust for this error?

this is also just a trigonometry question

but a more interesting one

/sidetangent

[gritty widget]

I guess throw one horizontally , one vertically , and one up and down

I don’t really know though

I just put it so for he first coordinate you throw it as you’re standing on the ground and the second coordinate is from from throwing it onto the ground and recording the coordinates on the specific block it fell on

Tbh my answer is probably not that accurate but I tried my best and I’m hoping I got it right 😳

[ivory dragon]

its conventional triangulation

[sharp frost]

What is a neighborhood of a set

Does it have to do with open sets or a topology on a set

[ivory dragon]

yes.

a neighbourhood of a subset is a set containing an open set containing that subset

the more common definition is

[vocal wharf]

goddamn edit, I can't correct you

[ivory dragon]

i consider myself an expert at being eventually right.

[vocal wharf]

tbh i don't think I ever considered a neighborhood that is not open, so not sure why it's defined that way

[timid mason]

Would you guys consider Topology to be one of the more abstract fields of math?

[ivory dragon]

introductory topology, not particularly

it has actual applications

more abstract than what 99% of people who take math 100 in university see, obviously

but relative to the rest of mathematics, not really

[gritty widget]

in metric spaces neighborhoods are always open but not necessarily in the context of general topology @ Loch

[little hemlock]

in metric spaces neighborhoods are always open but not necessarily in the context of general topology
how? take any subset of a metric space. Then the subset is a neighborhood of any point belonging to its interior.

[bitter yoke]

This is just a definition argument, people define neighborhood different ways

[gritty widget]

neighbourhood

/ˈneɪbəhʊd/

noun

a district or community within a town or city.

👀

[gritty widget]

Thank you N/U, very cool!

[floral gust]

N/𝔄*

[gritty widget]

thats what I said

[feral bay]

𝔄 is A, right?

[ivory dragon]

$\mathfrak{A}$

[gentle osprey]

Namington:

[ivory dragon]

yes.

[floral gust]

O.o

[grim coyote]

what exactly does it mean to put a function on a CW complex?

[rugged swan]

Hello, if I have a Lie group G acting properly and freely on a differential manifold M, one can state that if x is in M and S is a sub-variety of M such that T_x M = T_x G.x (+) T_x S, the function f : GxS -> M (g,s) -> g.s is a local diffeomorphism

it is possible to compute T_(e,x) f which is equal to [(g(t),s(t))] -> [g(t).x] + [s(t)]

but I don't understand why it is equal

for me it's because the action is locally the same as addition, am I right ?

[gritty widget]

u play dota?

[vocal wharf]

what is your definition of lie group, isn't it usually required that inversion is a smooth map

just from smooth multiplication?

well ok, then i dunno

[midnight jewel]

we had the definition that (g,h) ↦ gh⁻¹ was smooth iirc

(from which it follows that both multiplication and inverse are smooth on their own)

from the thing I have you conclude first that inverses are smooth (fix g=e)

and then (g,h) to gh is smooth too by composition of smooth maps

since it’s (g,h) ↦ (g,h⁻¹) ↦ (g,h)

any further assumptions on G?

can you post the whole problem statement?

I don’t think you can in general conclude smoothness of the inverse from just smoothness of multiplication (but I also don’t know of a counterexample - I just don’t know how you could prove it)

actually I wonder

for a fixed h, (g,h)↦gh is a bijective smooth map

what you want is a map f:G→G such that (g,f(g)) ↦ e under multiplication

this screams implicit function theorem to me

but I honestly can never remember the assumptions nor the statement of that theorem

in any form

I know it isn’t

I just can’t remember it

I’ve even understood a proof of it at one point

and found it nice, too

a manifold isn’t a vector space

there’s the inverse function theorem

but the map is from G×G to G, which is problematic here

since it requires an isomorphism of tangent spaces

a^i?

what are the a_i?

that doesn’t make sense

the group multiplication in G is in no way compatible with the multiplication of ℝ if you embed G in ℝⁿ

you can embed it, sure, but the multiplication is just some map

nothing to do with the ambient space

anyway I found a stackexchange post about this, and it’s relatively complicated

but the proof does use the inverse function theorem

https://math.stackexchange.com/questions/1301221/redundance-of-the-smoothness-of-the-inversion-map-in-the-definition-of-a-lie-gro
here (spoilers)

[sleek thicket]

Oh hey, this was a problem on my final last quarter

f(g, h) = (g, gh) is a smooth bijection

If you can show it's a local diffeomorphism, you get that it's a diffeomorphism and so inversion is smooth

It suffices to show that it's an iso on tangent spaces by rank theorem stuff

That's annoying but you can do it

It's helpful to use the identification T_(g, h) (G×G) ≈ T_g G (+) T_h G

@tall coral idk if you solved this already but feel free to ask my if you're still working on it

[gritty widget]

This is basically trivial... So much that I can't prove it! Any hint?

(Y<X) makes it useless but in the general case? Thanks for the help.

[sleek thicket]

Hey @tall coral, let's talk in here

It's kind of messy, sorry

I was writing it for a take home final so I didn't have a lot of time

Better crop

A map into a product is smooth iff it's smooth in each component

The projection map is smooth

The multiplication map is smooth

Yup

Basically if you write it down and it looks smooth

It's smooth

Maybe that's too far lol

I promise I actually check these kind of details

But this was on the final

I was being careful like the week we introduced products

You don't really have to be in a course either

I'm on the upper end of it from personal experience

Yeah I would suggest not doing that

Not to be rude

But that's a good way to not learn math but think you are learning it

Reading books you're not comfortable with and not being able to do the exercises

[gritty widget]

If X is a finite compact space and Y is a topological space, is the product topology compact?

[rugged swan]

a finite compact space ?

that just means that K is discrete

[sleek thicket]

All finite spaces are compact

[rugged swan]

yes not discrete*

[honest narwhal]

All spaces are T1 so 🙃

[gritty widget]

I know that all finite spaces are compact, but is this product compact?

[sleek thicket]

So you're asking if X×Y is compact?

[gritty widget]

yes

[sleek thicket]

Take X = a single point space

then X×Y is homeomorphic to Y

In general it's a finite disjoint union of Ys

So I think X×Y is compact for all finite X iff it's compact for some finite X iff Y is compact

Oh wait I lied

That's not true

It's true that this implies Y is compact

But X×Y might not be a disjoint union of Ys

[gritty widget]

hm ok thanks

Another question: is every subset of the space compact in the discrete topology?

[sleek thicket]

Are you asking if every subspace of a discrete space is compact?

This is equivalent to "is every discrete space compact"

Since a subspace of a discrete space is compact (and every discrete space is compact)

[gritty widget]

Okay then yes I am asking this

[sleek thicket]

Well let's think about it

Let X be a discrete space

Suppose X is compact

What does that imply about X?

Can you think of any particular open covers of X?

[rugged swan]

btw I'm not mad, the only finite compact spaces are discrete

[gritty widget]

For every set $I$ and every family of open sets $O_i$ with $i\in I$ such that $X\subseteq \bigcup_{i\in I} O_i$ there exists a finite subfamily $O_{i_1}..$ such that $X$ is a subset of those

[gentle osprey]

N/𝔄:

[sleek thicket]

@rugged swan I disagree

Let X be a finite space

Let {Ui} be an open cover of X

For each p in X, we have an i_p such that p in U_{i_p}

Then { U_{i_p} : p in X } is a finite subcover

@gritty widget sure, but what are the open sets in a discrete space?

[gritty widget]

Every subset

[sleek thicket]

Yes

[gritty widget]

internet problems rn, so messages may come through with a little delay

[sleek thicket]

So for any family of subsets whose union is X, there's a finite subset whose union is X

Np

[gritty widget]

hmm yeah that would make X compact, no?

[sleek thicket]

We're assuming X is compact

And discrete

[gritty widget]

I meant discrete implies compact

[sleek thicket]

And trying to classify X in some way, or find out other properties it must have

[dim meadow]

XxY is compact iff X and Y are compact

[sleek thicket]

@gritty widget no

Discrete doesn't imply compact in general

[dim meadow]

Discrete implies compact iff finite

[floral gust]

finite implies discrete and compact if I recall correclty which I guess what he was saying

yeah

[dim meadow]

No

[sleek thicket]

yes everybody, I'm aware

Finite does not imply discrete

[floral gust]

oh soz yes the other way around

[sleek thicket]

Take a space with two points, one open and one closed

[dim meadow]

Yes

No

Discrete compact implies finite

[gritty widget]

So if I were given some set X and I want to define a topology T on it, such that this space (X,T) is compact, can this be done for any set X?

[dim meadow]

Yes

[sleek thicket]

Yes, the indiscrete topology is always compact

[gritty widget]

ohhhh

[dim meadow]

Take X minus a pt

[sleek thicket]

It's easier to be compact if you have very few open subsets

[dim meadow]

Put any topology on it

Take the one pt compactification

Lol

[sleek thicket]

If X is a set, is there a ring whose spec is in bijection with X?

I assume so

But I can't think of one

I'm thinking of other ways to make it compact lol

[gritty widget]

thank you all very much guys, I guess the wording of my book just confused me

[rugged swan]

a compact space is normal @sleek thicket

by definition

[sleek thicket]

What? What's your definition of compact?

[rugged swan]

hausdorff*

[dim meadow]

Compact hausdorff

[rugged swan]

not normal mb

definition = every cover has a finite subcover + hausdorff

[sleek thicket]

I'm gonna become a guy who says quasicompact

I guess

[rugged swan]

¯_(ツ)_/¯

[sleek thicket]

I made this a while back when being mad at hartshorne

Zak, where did you learn math?

Not to be rude

I've heard that compact is used to mean quasicompact + hausdorff in French

But haven't ever heard it used that way in English

[rugged swan]

guess what

I'm french lol

[sleek thicket]

Yeah, that tracks

[rugged swan]

with quasicompactness we don't really have the geometrical intuition of what a compact space is

but yeah, it preserves the topological mean

[sleek thicket]

I mean, I don't have a geometrical intuition of a compact hausdorff space

[rugged swan]

wut

take any n-ball

or bounded closed submanifold of R^n

[sleek thicket]

Yeah I know examples of compact hausdorff spaces

I also know examples of (quasi)compact spaces

[rugged swan]

that's the geometrical intuition

[sleek thicket]

But I thought you were saying we don't have that for q compact spaces

I don't see how it's better for compact hausdorff spaces

Is what I'm saying

I know there's better theorems

[rugged swan]

no, there are non hausdorff quasicompact space

[sleek thicket]

Yes I'm aware

You're saying that there isn't geometrical intuition for q compact spaces, right?

[rugged swan]

wait wtf why I have the COVID role

[sleek thicket]

lol, it's infectious

[rugged swan]

yeah

lol

[sleek thicket]

I'm saying there also isn't geometrical intuition for compact hausdorff spaces

And the fact that there are nice geometrical examples doesn't change that

[rugged swan]

no there is

because if you're in a metric space you have all the good properties you wanted

[sleek thicket]

Well I mean I'm saying I don't have it

Not trying to prescribe

Compact hausdorff spaces aren't necessarily metrizable though

[rugged swan]

no but in the case of metric spaces the word compact (with the hausdorff axiom) is well used because a compact space is a generalization of Bolzano-Weierstrass which have a meaning only in hausdorff spaces

[lyric garnet]

just learning about open covers, they can be finite right?

like {(0, 1)} is an open cover of (0,1) right

although a noninteresting one

[bitter yoke]

yes

[lyric garnet]

cool

[wooden scarab]

are there any references on orbifolds?

[honest narwhal]

@sleek thicket well if they're second countable they are metrizable so...

[sleek thicket]

true

but my whole point is that compact hausdorff spaces can still be nasty

e.g. not second countable and not metrizable

[honest narwhal]

I mean I define topological spaces to be second countable

[sleek thicket]

oh lol

[honest narwhal]

I guess I didn't tell you that part

What's l^{infinity} again?

[sleek thicket]

and T1

[honest narwhal]

"and T1"
Corollary: finite spaces are discrete 😛

[sleek thicket]

a topological space is:

• compact
• hausdorff
• second countable
• a sphere

tori are on thin fucking ice

[honest narwhal]

Theorem: pi_1 is the trivial functor

Well actually idk T^2 is a Lie group which is nice

Sphere isn't even paralleizable

[sleek thicket]

hmm, good point

okay, take 2

a topological space is:

• compact
• the spectrum of a ring

wait aren't c* algebras like just compact hausdorff spaces?

are we back in a circle?

[gritty widget]

Any hint on the following?