point-set-topology

[prime elbow]

i can say d(x_n, y_n) is cauchy sequence

[opaque scroll]

And Cauchy sequences are bounded, so then it's finite

[prime elbow]

sorry, but are we using completeness of R here?

[opaque scroll]

To insure that the limit exists, sure

[prime elbow]

but if i want to construct by Q, by the same construction then how do i can say that metric is finite?

[opaque scroll]

Cauchy sequences are always finite, be they in Q or R

Finiteness is not an issue

[alpine nest]

Also if you want to construct R from Q via that theorem, then you can't speak of a metric, because a metric is a real-valued function, so you need to have reals as an existing object.

[prime elbow]

thank you, i got it

[quartz horizon]

This is the main thing

[cosmic zodiac]

Would you guys recommend learning about filters from Bourbaki or another book?

I guess I should be more specific that I want to learn about filters to better understand Stone-Cech compactification on topological spaces

[tender halo]

read gillman

"rings for continuous functions", that is

thats the best intro to stone cech

[cosmic zodiac]

So I was told, but I can't find a copy :(

[prime elbow]

I also want to understand it

[desert vortex]

do you know how to sail?

[urban zinc]

folland's analysis has a brief section on filters and compactification in the chapter on topology

I found it useful

[fair idol]

Which Guillemin talks about filters?

[fiery cypress]

Have I got this correct?

[prime elbow]

what is D?

[opaque scroll]

You seem to have switched the isolated and interior boundary point if I interpret the diagram correctly.

[fiery cypress]

subset of R^n

[fiery cypress]

oh yeah thanks

[desert vortex]

if Y is locally compact Hausdorff and K is compact Hausdorff and there is a homeomorphism Y -> K - {p} then is Y -> K the one point compactification?

ok I literally just found this on wikipedia: if X is compact hausdorff and p is not an isolated point of X then X is the one point compactification of X - {p}

[unborn hawk]

yes assuming Y is not compact

[desert vortex]

mmmmm right right

ok I got it

[odd gulch]

Om?!?

[fiery cypress]

yo

[cosmic zodiac]

I'm probably going to be in this channel a lot this semester 😭

[queen prism]

very nice

[cosmic zodiac]

Yeah I'm very excited but also it's a big jump for me, so we'll see

[viscid blade]

family of sets is set of sets ye

?

[ruby delta]

usually yes, but sometimes when someone says "family", the collection of sets might be too big to be a set

[viscid blade]

wdym ?

[opaque scroll]

The collection of all sets does not form a set for example (as a set cannot contain itself)

[viscid blade]

any example where the set doesnt include itself ?

[opaque scroll]

I mean, even the collection of all one element sets is not a set, or anything the same "size" as the collection of all sets.

But I'm sure if you work hard enough you can probably reduce the problem of any of these sets in some way to some set containing itself

[viscid blade]

thank you

[civic verge]

Could someone give me an idea of how to prove this?
I was thinking of something like this

$\overset{\circ}{A}\subset X \text{ exists an open subset } V \text{ such that } \overset{\circ}{A}\subset V\subset U\
\implies \bigcup W_{i}\subset U \
\text{therefore }\overset{\circ}{A}=\left{ x\in A: \exists U \text{ of } x, U\subset A\right}$

[gentle osprey]

Sheep Raider

[warped helm]

No

This isnt a proof

Define the sets that show up

Try letting x \in IntA

[quartz horizon]

What definition of interior are you using

[quartz horizon]

What you want to do here is show that the membership predicates are equivalent

[latent lily]

it's a dover book so it's pretty affordable

[quartz horizon]

This is essentially correct

[sly geyser]

Can someone show the proof for 0.10, and is it related in any way to 0.9 (I mean I don't think it is but I'm not sure)

Hmm yeah, doesn't seem to have anything to do with it

[rancid umbra]

what is ^2(^ww)?

$^2(^{\omega}\omega)$

[gentle osprey]

c squared

[sly geyser]

this goofy aah book writes set exponentiation backwards

this is related to the baire space btw

[rancid umbra]

lmao why r they trying to be different for no reason

[quartz horizon]

I thought it was tetration 💀

[sly geyser]

w to w tetration aaah

how would that even work

[quartz horizon]

Infinite power tower of w

[sly geyser]

hmm wait would this basically be like something along the lines of W_w

[rancid umbra]

w^w^w^w^…

[sly geyser]

this is set exponentiation though

not ordinal exponentiation

anyway basically

[quartz horizon]

I think it might be $\epsilon_0$?

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[sly geyser]

read the message above

[quartz horizon]

Oh

[sly geyser]

anyway to start off

the baire space is the open set created by the base sets which have all the sequences starting with a specific initial segment basically

you can get a sigma algebra out of that of course through countable unions and intersections and complements blah blah blah

the borel's measure is gotten by assigning , for each basic open set representing a specific initial sequence s, the measure that is equal to the product of 2^-(s(i)+1) for each s(i) in the initial sequence s

a subset of X (w^w) is null if it is a subset of a set of borel measure zero

and finally, a set is lebesgue measurable if it's symmetric difference with some borel set is null, in which case its lebesgue measure is equal to that borel set's borel measure

this is all from kanamori's book btw

[rancid umbra]

what is set exponentiation?

[sly geyser]

A^B is the set of functions from B to A

[rancid umbra]

oh

[sly geyser]

for example A^2 is the set of functions from {0, 1} to A, and it is basically the same as pairs of elements of A, so it is isomorphic to the other A^2 that is AxA

[rancid umbra]

yea, i just haven’t heard it by the name of set exponentiation

wondering what ^w w looks like now lol

[sly geyser]

bro it is just the set of sequences from N to N lmfao

[rancid umbra]

tetration

[sly geyser]

?

infinite tertration is kinda weird, for ordinal exponentiation it is just done as a limit of the finite tetrations and it would likely need to be the same here

more specifically it'd probably need to be a union of the finite tetrations

so it would have the natural numbers of N, the functions from N to N, the functions from functions from N to N to N, the functions from functions from N to N to N, to N and so on and so forth

You could also go the approach of taking its own type as input and giving out a value in N, but that doesn't make much sense sadly

You could get a variant of Russell's paradox if you do that

for example you could make a function so that the output from a specific functional input would be that input inputted into itself + 1

[quartz horizon]

Yeah you can’t actually have an $X$ with $X \cong \mathbb{N}^X$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[rancid umbra]

that is interesting. so some things can’t be defined as fixed points

[quartz horizon]

Mhm

[sly geyser]

there's also some natural correspondence between normal and set exponentation
for example, (a^b)^c is a^(b*c) and (A^B)^C is practically isomorphic to A^(BXC) (not technically the same but equivalent for all practical uses) where X is the cartesian product

[quartz horizon]

Tensor-hom

[sly geyser]

in fact funny enough what I just said is basically currying but from a set exponentation perspective

[quartz horizon]

Mhm

[sly geyser]

currying being that you can turn any function from pairs in BXC to A, to a function that takes a value in B as an input and gives you a function from C to A as an output (and the opposite way around being true too)

[rancid umbra]

this is the type signature of curried/uncurried functions

[civic verge]

[nimble portal]

Why are partitions of unity only admissible if your space is second countable?

I could intuit this with some definition crunching I think but I’d like to hear it from someone else

I believe it’s actually a weaker result, you need paracompactness, but I don’t remember what that means

not ready to look it up yet either :p

[nimble portal]

Apparently second countability implies paracompactness

[unreal stratus]

Given regularity

[nimble portal]

So how do you “use” a partition of unity in a proof?

Like

Define one in a proof

Let’s start there

Let A be a partition of unity over B?

or do you have to define the functions with it too

Let A = cup A_i be a partition of unity over B with respective functions phi_i

Stupid library’s not open yet so I can’t get started “officially”

Gee this’ll be my first proof in a year and a half

I’m trying to prove that every smooth manifold admits a Riemannian metric aaaaand this isn’t the right channel for that kekw

But ig the question I asked is still applicable here so I’ll keep it

Any example proofs to look over not related to this one?

[warped helm]

see munkres section 36

[nimble portal]

Will do!

You're beautiful

[sly geyser]

um btw, if {1} is an open set and so is the {0}, then are the support and kernel both clopen too?

is this like a union of i clopens?

[viscid blade]

Ok I want to prove that if we have E_tilda where E_tilda is the set of x adherent to E that E_tilda is closed

is this correct ?

[warped helm]

yeppers

[cosmic zodiac]

Does anyone know any resources to help study the homeomorphism between wallman and stone-cech compactifications in T4 spaces :D

Or at least a point in the right direction if i need to come to my own conclusions about it

[urban zinc]

the support and kernel of what?

is this related to a problem? if so can you post the full problem

[sly geyser]

it is right above my message

[urban zinc]

Huh, are you talking about partitions of unity? {1} and {0} aren't open sets though in [0,1]?

I'm confused

[sly geyser]

I won't pretend as if I'm fully clear about the thing myself, I'm asking for a reason

OHHHHHH

I'm dumb

it's [0, 1] not {0, 1}

I apologise

this .... changes things I guess

[urban zinc]

Haha no problem

[unreal stratus]

Kernel wouldn't rly be standard terminology though

That's only rly in algebraic situations

[urban zinc]

There would indeed be no way to define continuous functions into {0, 1} unless your domain were disconnected

[viscid blade]

yo

[sly geyser]

.... Is it correct to say that in the regions where U_i doesn't intersect with anything else phi_i has to be one while in the intersecting regions it can "split the value" among the other phi_j's for which U_j intersects in that region

[viscid blade]

when we say B is a basis for a topology

[urban zinc]

Yes

[viscid blade]

is that topology the topology generated by B ?

[urban zinc]

Hence why it's called a partition of unity

You split (partition) the value 1 (unity) between multiple functions phi_i

[sly geyser]

more specifically any open set can be gotten through union of the basic open sets

[urban zinc]

Yeah

[sly geyser]

generation is for a subbasis, though from what I understand they're the same in a lot of cases anyway

[unreal stratus]

You can say generation for a basis too

[viscid blade]

why is that true from this ?

It's quite confusing because it seems in the first they're saying a basis can be for any topology on X so long as it satisfies thee conditions

[urban zinc]

Have you tried proving it yourself (that B is a basis for the topology generated by B)

I think it's worth working through this in detail

[viscid blade]

yes

[urban zinc]

What have you tried so far

Oh wait, are you asking about uniqueness here? Like whether B can be a basis for multiple topologies?

[viscid blade]

i mean i think i got it, i don't think it's connected to my question at least

[viscid blade]

yea

[sly geyser]

Hmm so basically from what I see, I guess in this case you should just ignore what I said, as clearly in this case the intersection of two basic open sets can be gotten through a union of basic open sets, more specifically (B_1 ∩ B_2) being equal to U((B_3)_x)

I don't want to just give the answer but for the generation thing you need to use something similar with the union

[urban zinc]

Okay yeah this is confusingly worded. A better way to say it would be that B is a basis for T iff the sets in T are exactly the union of sets in B. The definition that they give for a basis is a criterion that holds iff there exists a topology T on X such that B is a basis for T.

If we define basis in this alternative way, it's clear that a basis can only correspond to one topology (the topology generated by the basis).

You should also check that one topology can have multiple bases.

Does that make sense?

[viscid blade]

well

[viscid blade]

they prove this in the next lemma

But when they use the fact that B is a basis for a topology T, they assume that T is the topology generated by B

what if T has basis B but isn't the topology generated by B

[urban zinc]

You can take it as a definition that "T has basis B" means "T is the topology generated by B"

[hexed steppe]

i dont think you should take that as a definition

[viscid blade]

so the first 2 is wrong definition for a basis for a topology, you need the third condition ?

[hexed steppe]

you can just prove that if it has basis B then it is the topology generated by B

[viscid blade]

if that is the case then why do they define it, isn't it a result

[hexed steppe]

topology generated by a collection of sets S = smallest topology containing S

[urban zinc]

The first two points (1) and (2) give the criteria for being a basis for a topology in general, but not for being the basis for a specific topology T

[hexed steppe]

clearly the topology generated by B contains all unions of sets in B. on the other hand, the collection of all unions of sets in B is a topology.

[sly geyser]

if the first two are true that the thing generated by the basis is a topology

[urban zinc]

If you know B satisfies (1) and (2), you know that B is a basis, but you don't know what topology B is a basis for

[sly geyser]

generation meaning the set of subsets created through the union of the bases

intersection not necessarily included in the generation, so it has to be included in the 2nd condition

[urban zinc]

Well "the topology generated by X" will always be a topology, no matter if X is a basis

[sly geyser]

my point is it works even if the generation is only union

[viscid blade]

ah hah that is super confusing

[hexed steppe]

i think that is a confusing distinction to make

[sly geyser]

oh, and the first point is to make sure that X (the whole set, like R) is an element of the topology

[urban zinc]

Yeah it's weird that they introduce it like this IMO.

[hexed steppe]

really? i think this is pretty standard

often in practice one wants to specify a topology by describing its basis

[sly geyser]

let me explain what I mean with 2

[urban zinc]

I think the wording here is confusing

[viscid blade]

aren't these the same

[hexed steppe]

oh sorry i see how they are defining "generated by"

[sly geyser]

so if A is a union of B_i's and C is a union of B_j's

for i in I and j in J (these are just indices who cares)

then the intersection A ∩ C

is the union of the intersections B_i ∩ B_j

[urban zinc]

Let's say you have a topology T in mind and a collection of sets B. If B satisfies (1) and (2), you know that it is a "basis for a topology" but you don't know how it relates to T. If you're saying "B is a basis for T", that means specifically that "T is the topology generated by B" (as in how they define it in your screenshot).

[sly geyser]

B_i ∩ B_j, because of the second point, is a union of B_k's (just choose such a B_k for each x and unify them across the members x of B_i ∩ B_j)

[viscid blade]

so T having a basis is defined differently as B being a basis for T

tf

[sly geyser]

and if unify all of these B_k's across all of the B_i ∩ B_j's you get A ∩ C

so the intersection rule works for the generated topology

[urban zinc]

Wdym by T having a basis?

I don't think I used those words

[sly geyser]

and union works because it is union generation anyway

so it's a topology

[sly geyser]

what do you mean by not knowing how it relates to T?

[hexed steppe]

maybe it would be helpful to use another letter

[viscid blade]

nvm i'm more confused now that i re-read that

[urban zinc]

B is a basis for some topoology, but not necessarily for the specific topology T (obviously, since T wasn't used at all in the criteria (1) and (2))

[hexed steppe]

replace T with T_0 in eric's message

[sly geyser]

(1) and (2) make it eligible for being a basis of a topology, in that the topology that has the union rule, inclusion of the whole set X in it and the intersection rule being generated by the basis only through union

[sly geyser]

........ what? Isn't there only one topology that can be generated by the basis

[urban zinc]

Yes

[viscid blade]

https://tenor.com/view/confused-confusion-meh-bunny-rabbit-gif-19003098

[hexed steppe]

there's a difference between 1) starting with a basis and asking what topology it generates, and 2) starting with a topology and asking for a basis that generates it

[viscid blade]

i don't know what you mean

[urban zinc]

Okay wait so starting over, your question was just what "B is a basis for T" means right? "B is a basis" means it satisfies (1) and (2), and "B is a basis for T" additionally means that T is the topology generated by B, in the sense given in your screenshot. I think we're overcomplicating this.

[sly geyser]

though one should know that T being the topology generated by B implies (1) and (2) anyway

[hexed steppe]

no

[sly geyser]

in the basis generation sense though, maybe I'm confusing things unnecessrily for swissmq

[viscid blade]

Ok so from my understanding here is what's going on.
You can have multiple bases for a topology.
If you have a basis B given, then the topology T with basis B is the topology T generated by B.
But if you have a topology T, since you can have multiple bases, a basis must just satisfy those conditions, but all of them would generate T

[hexed steppe]

B being a basis is what "implies" (1) and (2)

[hexed steppe]

yes

[urban zinc]

Correct

[hexed steppe]

it might help to have an example in mind

consider R^2 with the euclidean topology and B = {all open balls} and B' = {all open rectangles of the form (a,b)x(c,d)}. both of these are bases for the euclidean topology.

[sly geyser]

I think the confusion comes from the notion of generation here, are we talking about generation through union or through union and finitary intersection

[viscid blade]

so we can't have the same basis for 2 different topologies

or can we

[sly geyser]

no

[viscid blade]

i see so it is a bit strange to say that we define T_0 to be the topology generated by that basis since it is the only topology that can have that as a basis

[hexed steppe]

why is that strange?

[viscid blade]

Because the definition seems to imply you can have multiple different topologies with that share a basis

[hexed steppe]

why does it imply that

if that were true then this would be an ambiguous definition

"define T_0 to be the topology generated by that basis"

[viscid blade]

because by saying for a topology if you take some subset and they satisfy some conditions then it is a basis, it implies that it isn't limited to 1 necessarily

[hexed steppe]

no

[sly geyser]

Ok I'll just say my point straight
If we are talking about the kind of generation where you only generate through union (no intersection), not any set of subsets of X will (in the union sense) generate a topology, you need the first two conditions such that what B generates through the union of its elements (I don't think this is the same generation that they are talking about so don't confuse this with that I'm saying my own thing) is a topology at the end

[hexed steppe]

a given topology can have multiple bases

a given basis generates only one topology

these are different statements

[sly geyser]

because, in addition, to containing the union of its elements, a topology contains the whole set X (like R^2 for the 2d euclidian topology for example) and the intersections of its open sets

[viscid blade]

i feel like if i read this i'll just be more confused

ok well anyway i think i got it, thanks everyone

[sly geyser]

to make sure that what is generated by the basis through only union also satisfies the rule for X and the intersection rule, you need (1) and (2)

Did I ruin things here

[hexed steppe]

i didnt understand what you were getting at throughout the discussion

[sly geyser]

I was talking about generation as in the way a basis generates a topology, through just union, not the more general notion of a topology being generated by a set

[hexed steppe]

you seemed to be talking about distinctions between various notions of "generation" but that had already been defined in the initial post

[sly geyser]

should I have just said "union-generated" or something

would that have been more convenient

[hexed steppe]

i just dont know why it was relevant to talk about that at all

[sly geyser]

What I was trying to say is that for union-generation to also be the normal generation you need (1) and (2)

I should have just shut up though having two people/groups of people talking at once trying to explain wasn't the greatest idea

[hexed steppe]

sure that's a good point

maybe i didnt read your messages closely enough but i didnt really get that from what you were saying

[urban zinc]

nidlatam I think this is too many cooks in the kitchen

I would just let memorylessfunctor explain

[sly geyser]

I was thinking the same thing three messages above, I'm sorry lol

[hexed steppe]

i probably shouldnt have gotten involved either

[urban zinc]

Lol no worries

[hexed steppe]

also why is this notion of generation natural?

[sly geyser]

Honestly, I don't like how you guys said that (1) and (2) were "just the rules for a basis"

Being a basis is equivalent to being a basis for some topology T (more formally, there exists a topology T such that B is a basis of it)

[hexed steppe]

well the initial question was working from that as a definition of basis

[sly geyser]

Ah

[hexed steppe]

reformulating definitions isnt always helpful when clarifying basic concepts for first-time learners of a subject

since they are already accustomed to a particular definition

[urban zinc]

I think it's just a convenient definition to check

I don't think it's well-motivated here

[viscid blade]

is the second paragrpah connected to the lemma at all or

[rancid umbra]

are you asking or were you suggesting swissmq to think about it?

[rancid umbra]

yes

the first paragraph shows that \mathcal{C} is a basis for some topology \mathcal{T}' on X

the second paragraph shows that the original topology \mathcal{T} on X and the topology \mathcal{T}' generated by \mathcal{C} are equal, literally as sets

[urban zinc]

Yes, the first paragraph shows that C is a basis, and the second paragraphs shows that C is specifically a basis for the topology of X (i.e. the topology of X is the topology generated by C).

If we're being more explicit, technically you should say something like "Let (X, T) be a topological space" where X is the set and T is the collection of open sets, and then you'd want to prove that T is the topology generated by C. Here we're instead saying implicitly that X comes with a specified topology, and we're looking to prove that C generates that specific topology.

Oh c squared already answered, whoops

[viscid blade]

but if we show C is a basis and C is in X doesn't that mean C is a basis for X

😭

[viscid blade]

sure but where is this in the statement

[urban zinc]

No

C is only a basis for the topological space X if the topology of X is the one generated by C

[viscid blade]

https://tenor.com/view/the-rizzler-rizzler-black-panther-costco-guys-lebron-gif-13487563474626765619

[urban zinc]

If you only prove the conditions (1) and (2), that only shows that C is a basis for a possible topology that you could put on the set X, not the actual topology of X

[viscid blade]

I am so confused

Ok so

in the first paragraph we have shown that C is one of many possible bases of X

[urban zinc]

So when they say "topological space X" that comes with two pieces of information, the literal set X and the implied topology T on X

[viscid blade]

yes i agree (T,X)

[urban zinc]

If you show (1) and (2), that shows that C is a basis for some pair (T', X)

The second paragraph shows that T = T'

So C is actually a basis for the original pair (T, X)

[viscid blade]

ah hah now i got it

thank you

[viscid blade]

i used this reasoning when i first read the lemma idk why i forgot it

[rancid umbra]

its the last sentence of the statement

[hexed steppe]

asking

[hexed steppe]

sure

[viscid blade]

yo can you verify if i have the right idea ?

for 4b first part

my idea was to create the collection is the union of all T_alpha as well as all possible intersections among em

[hexed steppe]

doesnt sound correct

[rancid umbra]

this is sort of on the right track tho

[viscid blade]

oh sorry i meant using them as the basis

i forgot to specify that

lol

[rancid umbra]

yea

[viscid blade]

so then we finished fr

or i guess i need to prove it

okay

[rancid umbra]

yea. there is a nice characterization for this

that may be helpful

[viscid blade]

wait

how are we defining smallest ?

[rancid umbra]

yea

its by subset inclusion

[viscid blade]

that's what i thought but couldn't we in theory have topologies that aren't comparable

[rancid umbra]

so if you are using the union of the T_a's as a basis for this topology, then you are okay, right?

[viscid blade]

no

we need intersections

[rancid umbra]

sorry, yes, as a subbasis then

[viscid blade]

idk subbasis

like i haven't learn that

i'm not a topologist fr

[rancid umbra]

it is literally just a family of subsets of X lol

well, for now, just think of it like that

anyways, forget i said subbasis

[viscid blade]

okay

[rancid umbra]

yea, you want the smallest topology generated by the union of all T_a. but this topology is going to contain each T_a in that case, right?

whatever it ends up being

[viscid blade]

yes sir

[rancid umbra]

cool

so now you just need to figure out how to describe it

[viscid blade]

ok ok but can't we have two different topologies that contain these but are not comparable ?

[rancid umbra]

no, any topology that contains all of the T_a's will necessarily contain the smallest topology containing all of the T_a's

[viscid blade]

ok but that's assuming we have a definition of smallest

i don't get how subset can be used for that

if we have 2 topologies that are say incomparable

[rancid umbra]

let's forget about the family being a family of topologies for a second

given a family of subsets of X, how can i define a topology containing this family of subsets?

[viscid blade]

well it'd be the same stuff right

add all the intersections

then let all these subsets act as a basis

[rancid umbra]

yea, that is what i would call a bottom-up description. it deals with how the open sets are generated from your collection of subsets.
there is a top-down description that may be more useful. my hint would be to use 4(a)

[viscid blade]

is that just intersections of all topologies that contain that thingy ?

[rancid umbra]

yea!

so in our case, the thingy should be the union of all T_a

[viscid blade]

in that case we are immediately done right

[rancid umbra]

almost, yea

[viscid blade]

and i don't need ot actually specify more information like basis in other case

[rancid umbra]

mhm

[viscid blade]

mhm = yes ?

[rancid umbra]

you still have to show that like, each T_a is contained in this intersection, and that any topology containing all T_a necessarily contains the one we just defined

these should be relatively easy to show now though

[viscid blade]

why do i have to show first part

aren't we taking the intersection of all that contain each T_a

it's like showing the intersection contains X and empty set

[rancid umbra]

it should be (1) a topology containing all T_a and (2) the smallest among all topologies containing all T_a

[rancid umbra]

the intersection of all topologies containing T_a is T_a lol

so you would just be intersecting all the T_a

this is used for the second part of the question tho

the largest topology contained in each T_a

[viscid blade]

ye

you just intersect all of em fr

[rancid umbra]

but yea, just to be clear, for the first part of the question

you want the intersection of all topologies containing \bigcup_a T_a

for the second part, you want \bigcap_a T_a

[viscid blade]

is C just powerset

[rancid umbra]

what is C

oh part C

[viscid blade]

and the second of C is just intersection ofc

@north ore yo which section is this question from ?

[rancid umbra]

no. T1 U T2 is {0,X,{a},{a,b},{b,c}} and so you need to close this under intersections (it is already closed under unions)

this is the same as intersecting all topologies containing T1 U T2

[viscid blade]

yea i meant power set - b - c

i should specify that

[rancid umbra]

well, {b} should be in there since {b} = {a,b} \cap {b,c}

[viscid blade]

oh ye

i am sloppy

[rancid umbra]

lol all g

[north ore]

if you are asking about this then its in section 13

[viscid blade]

thank you

[north ore]

ask munkres

[rancid umbra]

yea, should be P({a,b,c}) - {{c},{a,c}}

[viscid blade]

wat

you mean c

not b

[rancid umbra]

yes

gracias

this is all packaged nicely by saying that there is a free functor F : PP(X) -> Topologies(X). it takes a family of subsets of X to the intersection of all topologies containing it.
if S \subseteq S' then F(S) \subseteq F(S'), and F(S) satisfies the universal property that for any topology on X containing S, F(S) is necessarily contained in it

[viscid blade]

you lost me at functor

or free

[rancid umbra]

lol

here, it just means that for two collections of subsets S and S' of X, if S \subseteq S' then F(S) \subseteq F(S'), so F is an order-preserving function

[viscid blade]

so it's almost like a homomorphism

or smth

idk

preserves structure

[rancid umbra]

yea, exactly!

it is an order homomorphism

[viscid blade]

bro knows all this math

i never heard of

[rancid umbra]

it's almost like i like math or something, idk lol

[urban zinc]

this is why I gave up a chapter into munkres when I learned topology lol, it feels so dry and unmotivated

[hexed steppe]

it’s fine for quickly getting acquainted with the important definitions

[rancid umbra]

i guess the way i see this is similar to the notion of freeness that is a few messages above this.

another way of phrasing this notion of generation is that a subset U of X is open iff it is a union of elements of \mathcal{B}.
if you look, condition two says that the intersection of two basis elements is a union of basis elements.
often we work with a collection of sets where we want unions of these sets to be our opens, for example,
the collection of open balls in a metric space, or the intervals (-oo,a) for the lower-limit topology, or charts for a manifold a la the smooth manifold chart lemma.

if we are going to have our opens be unions of basis elements, then this should be true of any sets generated by them (from unions or finite intersections).
this clearly holds for the union of any family of basis elements, but it had better hold for finite intersections as well.

maybe this answers your question? @hexed steppe

so like, if you are going to freely generate a topology by imposing that unions of basis elements be open, the family of basis elements needs to have this property for finite intersections

[hexed steppe]

i dont follow

i think it is more to do with the fact that a topology = data of convergent nets

so the topology “generated by B” should just be the minimal thing where convergence is governed by sets in B?

like x_a converges to x iff for all basis elts B containing x, the net x_a is in B eventually

so from this perspective it’s natural that open sets are just those where you can find a B around every point

idk that’s how i think about bases and topologies anyway

[rancid umbra]

oh, i don't really think about them in terms of nets, im not used to using them

but i think i see what you mean.

[rancid umbra]

i just mean like, if we are imposing that U is open if and only if it is the union of basic opens, then each basic open is going to be open, and if i have two of them, say B1 and B2, i need B1 \cap B2 to be open as well - B1 \cap B2 should also be a union of opens, hence the second condition.

[tropic copper]

hello, I have a basic topology question, would somebody be kind enough to help me?

I'm trying to understand the uniform topology on the countably infinite euclidean space. in particular how it differs from the box topology. I came across the example set X[i]=(-1/i,1/i) (where i is the natural number index), which is supposed to be open in box, but not uniform. it makes sense to me since there are points in X that do not have balls w/ the uniform metric that are subsets of X that contain that point

but then I considered these open sets:
let x&y be points in R^w where x[1]=0, x[i>1]=1/i-1, y[i]=-x[i]. let U be a ball w/ the uniform metric centered on x with a radius of 1 in the uniform topology. let V be the same but centered on y. doesn't the intersection of U and V equal X? I was made to believe that the finite intersection of any open sets is open, but that doesn't seem to be the case here?

is there an asterisk on that "finite intersection" business when considering metrics? am I doing the math wrong on that U/V intersection? do I not truly understand what an open set in the uniform topology is here?

obviously something in my understanding is seriously wrong, but I can't find out what exactly on my own. can someone please point out what I'm missing? thank you

[warped helm]

what is the set X here?

[tropic copper]

((-1,1), (-1/2,1/2), (-1/3,1/3), ...)

[warped helm]

the product of these intervals right?

[tropic copper]

sorry, tired

ya ya

[warped helm]

im not sure how you arrived at the conclusion that the intersection should be X

theres no asterisk about intersections

[tropic copper]

um I graphed it hold on

U

V

U intersect V

x axis is the index

[warped helm]

what exactly are you graphing?

[tropic copper]

so the space between the black lines is X. at x=1, for example, the bounds are (-1,1). for x=2, the bounds are (-1/2,1/2). we're taking the product of the points and so on

first picture also graphs U with the green bar being the upper limit, purple being the lower

V has red upper, blue lower

[warped helm]

thats not really a faithful visualization of those sets

[tropic copper]

how so?

[warped helm]

subsets of R^omega are just sets of sequences

read: sets of functions N -> R

[tropic copper]

ya? I mean I get there's a bit of extra info on the graphs, but I think the message gets across. if some point p is in X, then if you plot the dots for p on this graph for each index x, then the dots will be between the lines

I'm sorry I'm just not really following

[warped helm]

i just dont think this is a fruitful way of thinking about this

it doesnt really capture the construction of the topologies in question

[warped helm]

did you show formally that X is not open in the uniform topology?

[tropic copper]

no, I just found that on stackoverflow

[warped helm]

the justification you gave is general enough to where it would apply to any two topologies t \subset t’ on a set

right, i would advise proceeding formally

the uniform topology is a bit tricky at first and you wont understand it unless you get your hands dirty

[tropic copper]

sure

X is open iff X is a union of basis elements iff for all p in X and there exists some positive real s such that the ball Bs(p) is a subset of X
allow p=(0,0,0...), assume such an s exists for this element
let j=ceiling(1/s)+1
note that 1/j < s, so X[j]=(-1/j,1/j) is a proper subset of (-s,s)
let's define a point q where q[j]=1/j+(s-1/j)/2, q[i]=0
1/j < 1/j+(s-1/j)/2 < s
so q[j] is not an element of X[j], so q is not an element of X
d(p[i],q[i]) = d(0,0) = 0 < s
d(p[j],q[j]) = d(0,q[j]) = |1/j+(s-1/j)/2| < s
z(p,q) = |1/j+(s-1/j)/2| < s where z is our uniform metric
but q is an element of Bs(p)
so Bs(p) is not a subset of X, a contradiction
therefore X is not open in the uniform topology

im not confident in myself at such an hour, but I do think that's right. I'm still completely lost on my original question

[flat plinth]

A closed discrete set of a sigma compact space has to be at most countable.

What if i remove the condition of being closed?

I am trying to find a counterexample in R but i can't

Is there an uncountable discrete subset of R or Rn?

[tender halo]

no

closed discrete and (just) discrete subsets of metric spaces have the same bounds on them

because uhh

basically because metric spaces are perfect (in the sense of closed subsets being G_\delta)

[vagrant stirrup]

Reminded me about some article on Arxiv by Andrej Bauer about making reals countable:

https://arxiv.org/abs/2404.01256

But I don’t know how all that works, just found it inspiring :)

[tender halo]

there is this master theorem about it

which im surprised is not usually covered in pointset

[flat plinth]

as far as sigma compactness is concerned i dont see how the proof would work

I use the fact that a closed subset of a compact is compact

to conclude countability

[tender halo]

oh it wont

im was just talking about R

and that if you want a counterexample you need a non T6 space i think

if it exists, that is

[flat plinth]

what kinda monstrosity even is that

[tender halo]

well a lot of spaces are not T6

[inner valley]

is this the general topology channel ?

in any case, I have a simple question. \
So I'm considering the space $[-1,1]$ quotiented by the relation $x\sim -x$ for $x\in[0,1[$, and I have proved that ${1}$ and ${-1}$ are closed (but I'm not honestly convinced myself). If what I did is right, then I imagine this would be a counterexample for the converse of : $X$ is T2 $\implies$ every singleton is closed
edit : changed T1 to T2

[gentle osprey]

lunatiq

[inner valley]

I do think there is a simpler way to prove this with T1 and T2 spaces, but I wanted to play around with this specifical topological space
edit : T0 -> T1, T1->T2

okay so I'm more convinced now, $\pi^{-1}(\pm 1)={\pm 1}$ is closed in $[-1,1]$ so all is good !

[gentle osprey]

lunatiq

[inner valley]

where $\pi$ is the quotient map

[gentle osprey]

lunatiq

[urban zinc]

1 and -1 are identified by the quotient relation, so 1 and -1 are not elements of the quotient space

{1,-1} is an element of the quotient space

[inner valley]

no they aren't, i excluded $1$ and $-1$ from the identification

[gentle osprey]

lunatiq

[urban zinc]

Oh okay

I misread

[inner valley]

no worries

[urban zinc]

What is the definition of T1 that you are using?

Because every singleton being closed is in fact equivalent to being T1

[inner valley]

oh i think i meant T2 and T1

T2 as Haussdorff and T1 as Frechet

[urban zinc]

Oh okay

Yeah this quotient space is T1 but not T2, that's correct

[inner valley]

oh this is interesting

[urban zinc]

also yea this is the general topology channel

[inner valley]

yep i was unfamiliar with the nomenclature

[urban zinc]

Try proving it!

[inner valley]

it goes instantly with the equivalence : a set is open iff it is a neighbourhood of all its points

the switching perspective from opens to neighbourhoods is so insightful

sometimes the minimality of topology makes me feel that the theory isn't enough to have profound understanding, but it's actually not the case

really should look into the history of topology

[urban zinc]

I'd say point set topology is definitely a subject where the definitions are more interesting than the theorems

There are only a handful of "interesting" theorems in a point-set topology, like Urysohn's lemma and its many equivalent statements

Once you get further to manifold theory and algebraic topology, the results get so interesting

That is also very true

[urban zinc]

Oh yeah I think once you get to stuff like Stone–Čech compactification it gets very interesting too, I want to learn more about that :<

[inner valley]

yes i'm actually doing algebraic topology rn with cover maps, and i was studying that set as a counterexample to the finite cover map theorem

(basically that if u have a local homeo with fibers that are all finite of same cardinality, then ur local homeo is a cover map)

[urban zinc]

awesome!

[quartz horizon]

Adjoint functor theorem time

[inner valley]

the only thing our bourbakist geometry prof told us about the stone-cech compactifiaction is that he found it really horrid lmao

[unborn hawk]

it’s not horrid it’s just too universal to be relatable

joel merker

[inner valley]

thank god it was not joel merker

thank god i do not have to interact with joel merker anymore

[slender glen]

call a continuous function f: X -> Y locally injective if for every p in X there exists an open neighborhood U of p such that f is injective on U. Does f have to be a homeomorphism?

[warped helm]

i dont think so?

what if f is just injective on all of X and continuous

theres no guarantee of surjectivity

[slender glen]

oh right yeah I meant if f is also surjective

[reef mural]

I like the concept of nets

[slender glen]

answer is still no

take like, R -> S1 by f(x) = e^{2pi it}

I wonder if X and Y being simply connected would be enough

[strong lantern]

I mean, this fails even if you assume that f is injective and surjective globally.

[slender glen]

right

[strong lantern]

This case is even more obvious, since by your definition take X= Y times S with S some set with the discrete topology i.e. every set in S is open then X->Y the projection is a continous function which is "locally injective" and "locally surjective" but is nowhere near a homeomorphism.

Obvious is the wrong word here. Sorry if that sounds ofensive or something.

[slender glen]

No, you're right, I have a version of the question that is interesting in my head but I'm trying to formulate it properly.

[robust drum]

No eg covering maps

Or any immersion (smooth map where the derivative doesn’t vanish anywhere)

Such a map is injective in a neighborhood of every point

[kind marlin]

i dont know if this is what you're thinking of, but any continuous bijection from a compact space to a hausdorff space is a homeomorphism

e.g. any invertible continuous function on a closed subset of R with the standard topology is necessarily a homeomorphism

[tender halo]

the proper version of locally injective is closed map

and locally surjective are open maps

thats kind of the analog

[rancid umbra]

that doesn't use the topological inverse function theorem, does it?

[tender halo]

i mean it has to be monotone and surjective

which makes it open

[rancid umbra]

im just confused where you are using compactness

[tender halo]

which makes it a homeomorphism

or right hmm

[rancid umbra]

i don't think you do

[tender halo]

well its gives you that its a local homeomorphism maybe?

[kind marlin]

wait sorry im confused what youre saying

there's probably alternate ways to prove the R->R thing, it just is also a result that works when mapping from any compact space to any hausdorff space

[tender halo]

how are the two connected

[kind marlin]

are you saying u dont see why compactness is needed in this statement?

[tender halo]

no, they are asking how are you using that fact to prove the R -> R thing

[kind marlin]

oh sorry idk why i said R to R

i was thinking of R like a closed interval for some reason

[rancid umbra]

lol [-oo,oo]

[tender halo]

finally, we have it, the two point compactification

[tender halo]

its easier to see with open maps

its kind of intuitive why they are sort-of-surjective

also makes it easy to intuitively see why projection maps from a product are open but not closed

for closed maps its easier to see on the alternative definition of closedness, the one about fibers

that for every open nbhd of a fiber of p you can find an open nbhd of p whose preimage is in that nbhd

which tells you that the function is locally uhh "thin"? "small"?

that if you shrink the nbhd of a point its preimage shrinks accordingly

[rancid umbra]

actually nvm

[rancid umbra]

oh this is neat

[rancid umbra]

is there something like this to see that open maps are some sort of generalization of locally injective maps (in the strict sense)?

sorry closed

wait eff

[tender halo]

[rancid umbra]

closed is for injective, right?

[tender halo]

yea

its not really injectivness

i think of it of like

a smallness condition

and openness is a bigness condition

i know its like very vague, just something that came to mind when trying to wrap my mind around what closedness and openness actually mean

[rancid umbra]

i see

i hadn't really thought of finding a deeper meaning to openness and closedness before

[quartz horizon]

Can you explain a little more?

Hm

I wonder if this makes sense in terms of predicates

[tender halo]

well ok, i think its clear with open maps right, cuz you need to fill the whole nbhd around an image, taking the open nbhd of a point in the image is like a natural setting in which to test surjectivity

you can test it pointwise, image of a nbhd of x is a nbhd of f(x)

closedness you can also test pointwise but like in the opposite direction for whatever reason?

[tender halo]

via this definition

its a smallness condition in that if you shrink the nbhd of a fiber you shrink the nbhd of the image

so you will eventually separate yourself from all other points

(kinda, in sufficiently separated spaces)

[rancid umbra]

i guess fibers are a natural setting to test injectivity?

[tender halo]

it would seem so, im sure there is categorical mumbo jumbo pseudo can tell us

that makes that true

[rancid umbra]

have you looked at this in terms of nets or filters?

[quartz horizon]

Well it’s like

Open sets correspond to locally true predicates

The direct image of a predicate $p(x)$ is $q(y) := \exists x \in f^{-1}(y), p(x)$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

And the kernel image is $s(y) := \forall x \in f^{-1}(y), p(x)$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

A map is open iff whenever p is locally true, so is q

A map is closed iff whenever p is locally true, so is s

I talk about the direct image of a predicate in more detail in one of my articles

https://pseudonium.github.io/2026/01/21/Subset_Images_Categorically.html

[quartz horizon]

Injectivity <=> fibers have size <= 1

Surjectivity <=> fibers have size >= 1

[rancid umbra]

why call it the kernel image?

[quartz horizon]

Idk actually

[rancid umbra]

lmao

[quartz horizon]

When I’ve seen this given a name that’s what it’s been called

[rancid umbra]

this feels very topos-y

or sheafy

[rancid umbra]

and yea, i like this. it is a good way to think about injectivity and surjectivity in terms of fibers

[rancid umbra]

https://math.stackexchange.com/questions/1287019/characterization-of-open-maps-in-terms-of-nets
https://math.stackexchange.com/questions/1277615/characterization-of-closed-map-by-sequences-nets

anyways, these links seem related to what you are talking about with open sets corresponding to locally true predicates. the structure of the characterization of open and closed maps in terms of nets resembles that of direct image and kernel image

[tender halo]

oh thats really cool

[quartz horizon]

Injectivity means that any element has at most one preimage

[rancid umbra]

surjectivity means that any element has at least one preimage

[quartz horizon]

No, it’s to do with the size of the fibre

Yes

[rancid umbra]

you don’t need the phrase “on the level of sets”

[quartz horizon]

yes

[queen prism]

every fiber has cardinality at most/at least 1

[viscid blade]

what about sqrt function

||(joke)||

[solemn iris]

How are we sure that {An} is nonempty? Like it says 'for each n for which it is possible', is it possible that there is no such n? I still haven't fully grasped the idea of this proof...

[tiny obsidian]

Each open set (except for Ø) contains a basis element as a subset

(General fact about bases)

So in particular each set A in curly A contains some basis element, and you'll be able to pick A_n for at least those basis sets

The "furthermore it covers X:" part shows this

[civic verge]

I recommend Stefan's Topology; the demonstration is more detailed.

[median peak]

I have just needed to use the compact-open topology on a diffeomorphism group Diff(X) to realize it as a topological group, but I am wondering why on earth we use the compact open topology. From my limited understanding, it generalizes the case of the uniform convergence topology when we map a compact space into a metric space, but I don't see why this is the "right" generalization (or whether there are other options which fail). Does anybody have any thoughts?

[cosmic mirage]

maybe not the most useful answer but it is the correct topology to put on mapping spaces to make (reasonable) spaces into a closed symmetric monoidal category with respect to the Cartesian product

if those words don't mean anything then the idea is that it satisfies some very desirable category theoretic properties that fully determine it

namely it allows you to curry continuous functions in the sense of https://en.wikipedia.org/wiki/Currying

[rancid umbra]

what is the right way to discover the compact open topology?

If X is locally compact, then X × − from the category of topological spaces always has a right adjoint Hom ( X , − ). This adjoint coincides with the compact-open topology and may be used to uniquely define it. The modification of the definition for compactly generated spaces may be viewed as taking the adjoint of the product in the category of compactly generated spaces instead of the category of topological spaces, which ensures that the right adjoint always exists.

wikipedia says this

but i can't quite figure out how to determine what the topology is on Hom(X,Y) from this

[cerulean oriole]

Yeah, I think it's pretty involved and doesn't even work for all spaces.

[rancid umbra]

im okay even if just working with locally compact spaces

but yea

[cerulean oriole]

For K compact and U open in Y, eval: Hom(K, Y) x K -> Y must be continuous, so V(U) := {(f, k) : f(k) in U} must be open, so because K is compact, by the tube lemma, W(U) := {f : {f} x K in V(U)} = {f : f(K) in U} mustbe open in Hom(K, Y).

Now if Hom(X, Y) exists for a given Y for X and all compact open subspaces of X and is functorial in the first argument for those spaces, then for any compact subspace K, restriction to K must be continuous, so W(K, U) := {f : f|_K in W(U)} = the usual thing must be open in Hom(X, Y).

[rancid umbra]

ah, this is a nice way of looking at it, thank you

[cerulean oriole]

This justifies that it needs to be at least that fine. I think the other direction would be more involved. (And then once you have shown this is (kind of) the only possibility, showing it works is hard too.)

[worn mortar]

I read this as well.

[crisp lintel]

not sure about diffeomorphism groups specifically, but in general its very useful to be able to reduce problems to compact sets b/c compactness makes the world go around

and the compact open topology is essentially the topology of uniform convergence on compact sets (although of course uniform doesn't really mean anything on arbitrary topological spaces, but still)

another fact that I'm fairly sure is true is that if you have a locally compact group G and consider the image of the cayley representation inside the homeomorphism group of G, then I believe the two are isomorphic

[inner valley]

In the product topology ExF, if U is an open set that contains {x}xV2 where x is in E and V2 is an open set of F, can we exhibit an open set V1 of E such that V1xV2 is included in U ?

[quartz horizon]

In general I don’t think so

[hidden abyss]

Let U be the red area in R² and take {0}×(0,∞)

[quartz horizon]

Mhm

[hidden abyss]

If E is compact then its true though

[quartz horizon]

Wait is it

The example I had in mind was

E = F = [-1, 1]

[hidden abyss]

Wait no it's not

[quartz horizon]

Take U to be the open disk centred at (0, 0) with radius 1

Then {0} x (0, 1) is contained in U

[tender halo]

v2 has to be compact, not E

for the tube lemma

[hidden abyss]

Right

[quartz horizon]

Ah interesting

[inner valley]

oh if that's true then my life is okay 'cause in the proof my V2 is [i/n,i+1/n]

where i is small enough for the interval to make sense

[tender halo]

yea thats just the tube lemma

[inner valley]

omgggg

tube lemma

new lemma acquired

im so happy

my prof is so lazy he just said compacity and let a whole thing be unjustified and i'm trying to wrestle with the details

[quartz horizon]

What’s the tube lemma actually

[inner valley]

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

[tender halo]

in general the following is true

tube lemma is just that applied to a binary product where A_1 is a point and A_2 is a compact subset

[inner valley]

is compact here haussdorf or just the borel-lebesgue property ?

[tender halo]

here compact can be understood as pseudo-compact

[inner valley]

aight ! i got it confirmed by the french wikipedia lol

also i need a second opinion on a matter

so basically let $h:Y\times[0,1] \to B$ a continuous map, and $(W_i)_i$ an open cover of $B$. \
Then is the following statement : \
\textit{By compactness of $[0,1]$, for all $y\in Y$, there exists an open nbhd $U_y$ of $y$ and $n=n_y\in\mathbb{N}$ non zero such that $h(U_y\times [(i-1)/n,(i+1)/n])$ is contained in an open nbhd of $h(y,i/n)$ contained in an element of the open cover of $B$.}\
not brushing too many details ?

[tender halo]

for some reason U is both a cover of B and of Y

[gentle osprey]

lunatiq

[inner valley]

sorry then it's just a notation problem on my part

(W_i) is basically opens that trivialise a cover map above B

[tender halo]

its very hard to read

like i can guess what you are going for

but its written backwards for some reason?

[inner valley]

it's basically in a proof of lifting homotopies (but i feel the argument is itself only topological)

[inner valley]

which is backwards ?

[tender halo]

wait where do you get i from

for all i from 1 to n - 1?

[inner valley]

yes basically

[tender halo]

you start with a cover of B and then pull it back, fix an y and then use compactness and then find n so that 1/n is less than the lebesgue number of the cover basically

but your claim is written in the opposite direction

i can only decipher it because its the obvious thing to try, i wouldnt call it a proof

[inner valley]

i see, i'll think it through tonight

[tender halo]

like its not wrong

[inner valley]

but won't the tube lemma be crcuial here

for the U_y

(also this is my prof's proof that i was trying to decipher lmao)

[tender halo]

but the steps of the proof are reversed which is hard to wrap one's head around

[tender halo]

well what you actually need is lebesgue number lemma

https://en.wikipedia.org/wiki/Lebesgue's_number_lemma

tube lemma doesnt really help i dont think

you need to find n so that your splitting of the interval is fine enough to fit into the cover

[rigid perch]

i have a question wrt smooth manifolds. let's say i have M be the set of points (x,|x|) for x real numbers, equipped with the subspace topology of R^2. Then, if i have a single chart that sends (x,|x|) to x, this is bijective and continuous with a continuous inverse, which makes M into a topological manifold. since i only have a single chart, this means that the atlas that consists of that single chart is in fact a smooth atlas, so this set of points a smooth manifold, by definition. how lol? it clearly has a sharp ridge at x=0.

[rancid umbra]

it is not a smooth manifold in the subspace topology

you are conflating two concepts

[queen prism]

it should be a smooth manifold with the subspace topology, no? the problem is the inclusion M -> R^2 fails to be a smooth embedding

[rigid perch]

!! this might be it

[rigid perch]

the definition of smooth manifold i'm working with is a topological manifold with a smooth atlas

and a smooth atlas is a collection of charts that are pairwise compatible

since my atlas consists of a single chart, there's no "pairs" to check

hence it's a smooth atlas

[rancid umbra]

what i mean is that it is not an embedded submanifold of R^2

that is sometimes how i use (and have seen others use) the phrase “in the subspace topology”

sorry if it caused any confusion

[rigid perch]

well i'm just specifying what topology i'm giving M

before specifying any charts

[rancid umbra]

this is an equivalent definition for an embedded manifold in R^n.
if you try to do this with your M, condition 3 fails at x = 0

so what i am trying to say is yes, M has a corner. this only matters if you try to view M as an embedded submanifold of R^2

conversely, if M is a manifold that is covered by a single chart, then any top space homeomorphic to M will also be a smooth manifold by transporting the structure through the homeomorphism

this applies in your case since {(x,|x|) : x in R} is homeomorphic to R

[rigid perch]

thanks !

[civic verge]

I made a forest of what should be happening. Is this correct? I assumed the subspace topology for A.

[warped helm]

the last implication doesnt make any sense

“implies empty” is meaningless

[civic verge]

sry, sry

[warped helm]

you need to find neighborhoods of x and y in A such that theyre disjoint

[civic verge]

Suppose two neighborhoods U_{x} and V_{y} in X where x,y belong to A.

[warped helm]

No

The larger space X being Hausdorff grants you the existence of neighborhoods U and V of x and y respectively that are disjoint

A set W is open in the subspace topology of A is open iff. W = A intersect U for some open set U from the larger space

[civic verge]

I understand that.

[warped helm]

You don’t want to consider just any two neighborhoods of x and y

[warped helm]

thats what i meant here

[ruby delta]

am i crazy or did you change your name

[warped helm]

i have not

[viscid blade]

it was josemom1

now it is josemom2

[civic verge]

I don't understand this

[warped helm]

Write what it means for A to be Hausdorff

[civic verge]

$U'{x}=U{x}\cap A$

[gentle osprey]

Sheep Raider

[civic verge]

this is an neighborhoods?

[warped helm]

thats what an open set in A looks like, assuming U_x is open in X here

[civic verge]

$ x\neq y, x,y\in A \left{ \begin{array}{cl}
U'{x}\subset A \
V'{x}\subset A
\end{array} \right. \rightarrow U'{x}\cap V'{x}= \emptyset $

[gentle osprey]

Sheep Raider

[warped helm]

Sure

But like write a proper proof

[civic verge]

I think it would be useful to use something about the topology of the subspace in A., no?

[warped helm]

.

Not just useful but required

Youre supposed to be proving that the space A with the subspace topology is Hausdorff

[civic verge]

I know that the proof comes from the definition, that's what the proof says, but I assumed that A comes from the subspace topology and then I assumed those open neighborhoods that when I intersect them, as you say, give an open set. Here comes the problem, right?

[warped helm]

There is no problem

The task is to produce two open sets in A, one containing x and the other containing y, that are disjoint

[civic verge]

What would this part be?

[warped helm]

i dont know what youre asking

this is why i suggest writing a proper proof

[warped helm]

Not quite

I'll just write the proof since it will be instructive

Let (a,b \in A) so that (a,b \in X). Since (X) is Hausdorff, there exist neighborhoods (U_x) and (U_y) of (x) and (y) such that (U_x) and (U_y) are disjoint ( (U_x \cap U_y = \emptyset )). By definition, (U_x \cap A) and (U_y \cap A) are neighborhoods of (x) and (y) in (A) with the subspace topology. Since (U_x \cap A \subset U_x) and (U_y \cap A \subset U_y), it follows that (U_x \cap A) and (U_y \cap A) are disjoint.

[gentle osprey]

josemom2

[civic verge]

mmmm

U_{a} and U_{b}?

What role do elements a and b play here?

[warped helm]

yes sorry mb

replace x and y with a and b throughout

[civic verge]

I understand now, ha ha ha.
That part of U_{x} intersection A is contained in U_{x}.

Thanks

[prime elbow]

How do I show if net is cauchy in complete metric space then net is convergent?

And net can be different, directed set can be different, right?

[ruby delta]

recall what the definition of a complete metric space is

[prime elbow]

I know definition

[ruby delta]

this seems like a relatively straightforward proof to me lol what are you stuck on

[prime elbow]

Okay let me do again

[crisp lintel]

It's not totally obvious I guess, but all you need to do is convert from a net to a sequence

Which you can do using balls of radius 1/n

[desert vortex]

is there some more elementary proof that the only continuous group homomorphisms R -> S^1 are the exponential functions ? I only managed to prove this using covering space theory

[hidden abyss]

Yeah you can show in general that for a Hausdorff topological group G, morphisms f: R->G are determined by f(1).
The idea is to first show that f(p/q) = p/q * f(1) and then you can argue via continuity

[desert vortex]

what is p/q * f(1) how can you multiply a rational number by an element of G

if G is divisible sure, or something along those lines

[hidden abyss]

Uhh wait

Yeah G needs to be uniquely divisible

Otherwise there wont be uniqueness

[desert vortex]

there also wont be existence

[hidden abyss]

I mean if G isnt divisible then the only group hom R->G is the zero hom

[hidden abyss]

But yeah G needs to have unique divisors for this to hold in full generality and thats not true for S¹ oops

I thought the method for solving the cauchy functional equation would fully generalize but I guess not

[desert vortex]

why ? the image f(R) will be a divisible group but unless f is surjective that wont be all of G so I dont see how to get a contradiction

[hidden abyss]

Oh true

[desert vortex]

in fact I think its not true if G = Z x R its not divisible

🤔 I appreciate your try though!!

im kinda stuck

[hidden abyss]

But I think you can still recover some ideas from the method for solving the cauchy equation?
f: R-> S¹ must still satisfy f(p) = f(1)^p, and by continuity, for q sufficiently large, f(1/q) should need to be f(1)^(1/q) in the first quadrant

Right?

[desert vortex]

Im not good at complex analysis

tbh

[hidden abyss]

Or I mean f(1)^(1/q) with the smallest positive argument

[desert vortex]

depends on how many times f winds around S^1 from f(0) to f(1) I think

[hidden abyss]

Yeah but we just consider q very large

[desert vortex]

mmmmmm

[hidden abyss]

i.e. for all q large: f(1/q) = exp(a*i/q) for some fixed a. Then multiplying by integers p we get f(p/q) = exp(api/q) for all p,q and by continuity f(x)=exp(aix)

[desert vortex]

what if f(1) = 1, then your approach would seem to imply that f = 1 constant

I guess maybe you can say ok if f = 1 constant function then we have nothing to prove but if f is not constant there is some r in R with f(r) not 1

that fixes the issue I spotted but Im not sure if the rest of the argument works like you say, I'd have to think about it some more later

maybe there is some elementary way to see all continuous f: R -> S^1 lift along the covering map R -> S^1

[desert vortex]

actually I think I might've slightly misinterpreted the exercise, it was asking this: To verify the statements made in Example 3.1.3

I dont know if that includes the isomorphism part or just checking the topology part

[opaque scroll]

Isn't that just the lifting property of covering maps?

[desert vortex]

my original question was how to do it without this since its not an algebraic topology book but an analysis book

its an exercise

[rotund badger]

if i took the intersection of (-1/n, 1/n) for n = 1, 2, ..... would in be left with {0}?
and {0} in R is closed since there is no r > 0 s.t B_r(0) is contained in {0}
or since like (-inf, 0) U (0, inf) is open, right?

would that be a valid example of an infinite intersection of open sets producing a closed set

[desert vortex]

right

yep

[rotund badger]

yay

thank you

[desert vortex]

nice!

(-inf, 0) U (0, inf) is open is what you want, not that B_r(0) is never contained in {0}

[rotund badger]

is there a more standard example or not really

[desert vortex]

that is a standard example 😄

welcome to math

glad to have you here

I guess you can intersect two non-overlapping open sets that gives you the empty set which is closed

[rotund badger]

oh? but i only need to show that it is not open right, not that it is closed
cause wouldnt (-inf, 0) U (0, inf) is open mean {0} is closed but not necessarily not open?

[rotund badger]

true but it is also open

oH my bad i said closed set earlier, i think they wanted one that wasnt open

[desert vortex]

aah makes sense

[rancid umbra]

{0} is not an open set

if im following correctly, that is the assertion you want

[rotund badger]

yes ! is stating that no ball around 0 would ever be contained in {0} be enough, or is there a more formal way to show that or can i just say it

[rancid umbra]

i guess the easiest way to say it more formally is that (a,b) is uncountable whenever a < b

in particular, it contains at least two points

[rotund badger]

i fear i haven reached this point in my course 😅 true (a,b) is uncountable, how does that help us?

oh because balls are intervals yes that makes sense

[rotund badger]

so it can NOT be contained in a singleton or finite (or counable?) set ! (yes?)

[rancid umbra]

yes

[rotund badger]

thank you ! <3

[rancid umbra]

i guess, here is a nice way to argue:
if (a,b) is a subset of {0} (with a < b), then (a + b)/2 = 0. this means that a = -b. since a < b, then a and b can't be 0, and a < 0 < b. since b is positive, then b/2 is also in (a,b), and so b/2 = 0. this is a contradiction.

the only open interval that is a subset of {0} is the empty interval.

[rancid umbra]

this avoids talking about uncountable sets if you haven't done that yet.

[rotund badger]

ooh thank you that makes sense

also would the union of (1/n, 1) from like n=2 to inf be an open cover of (0,1)
also if we take n=1 it still works right cause (1,1) = empty set?
and then any finite subcover {(1/n_i , 1) : 1 <= i <= m} would have union = (1/n_k , 1) where n_k = max{n_i : 1<= i <= m} which would not contain 1/n_k which is in (0,1), so (0,1) is not compact

its just my mark scheme uses this set, but it works either way right?

also i dont need to assume its compact right, i can just show any finite subcover would have at least one element of (0,1) which it doesnt contain

note we cannot use heine-borel here

[rancid umbra]

also would the union of (1/n,1) from like n=2 to inf be an open cover of (0,1)
yes

it still works with n = 1 for the reason you said.

your argument showing that any finite subfamily of {(1/n,1)} will not cover (0,1) is correct

[rancid umbra]

yes, your construction and theirs both demonstrate that (0,1) is not compact

[rotund badger]

thank you so much !!

[inner valley]

well for the n, sure, but you still need to have an open nbhd of y instead of just a slice

[tender halo]

oh for that yeah

i forgor about that part

[inner valley]

yeah i think the tube lemma is useful in this part

because the U_y will just be the intersection of finitely many opens that result from the tube lemma

i hate my professor

two important lemmas brushed under

[hasty thicket]

Hi, this might be a stupid qs but I was watching this video abt metric spaces and one of the conditions is d(x,y) = d(y,x). Is there any case of this property not holding? (Obviously not for a metric space)

[warped helm]

define d(x,y) = x - y

[hasty thicket]

Ah that’s pretty smart! Thank u :D

[viscid blade]

wb others

[crisp lintel]

real world distances

well ok maybe not distance but time

if you define d(x,y) to be the smallest amount of time it takes to travel from x to y, then it won't be symmetric since for instance you might have a hill that's faster to go down than up

[warped helm]

d(x,y) = x/y

like a lot of things wont satisfy this

[viscid blade]

hmmm

d(x,y) <= d(x,z) + d(y,z)
x/y <= x/z + y/z

why is this true

[queen prism]

I feel like the real intention is "satisfies every property of a metric but symmetry"

in which case b's example is good

[viscid blade]

does d(x,y) = x-y not satisfy allat

wait ofc not cuz it not positive

[finite token]

d(x,y) = |x-y| + (x-y)/2

[queen prism]

nvm it's good

[finite token]

We can generalize this: If d is a metric on X then d'(x,y) = d(x,y) + L(x) - L(y) is an asymmetric "metric" when L: X→R is K-Lipschitz with K<1

[hidden abyss]

Distances along the street map if you take one way streets into account

[finite token]

We can try the reverse too: if d'(x,y) is an asymmetric metric I imagine we can define a symmetric d(x,y) = (d'(x,y)+d'(y,x))/2

[kind marlin]

what properties does the asymmetric metric have to satisfy

[finite token]

Every property except symmetry

[kind marlin]

mainly asking because I think you can write the metric space properties in a way where nonnegativity is implied by the other properties including symmetry

but I’m also just used to that being part of the definition

I guess negative distances would be silly regardless

[finite token]

d(x,y) ≤ d(x,x) + d(x,y) implies d(x,x) ≥ 0

d(x,x) ≤ d(x,y) + d(y,x) with symmetry thus implies d(x,y)≥0

[quartz horizon]

You may be interested in lawvere metric spaces

[kind marlin]

right, but if we drop symmetry then an asymmetric metric probably could allow for negative distances)

[finite token]

Yeah

[kind marlin]

Mostly was referencing this

[finite token]

Ok then, for our version nonnegativity is explicit

[kind marlin]

Wow this did not sound like a category theoretic thing but the first link is very category theoretic

[finite token]

Though, we could maybe imagine distances being some sort of "energy cost" and some paths gaining energy

idk

[kind marlin]

ooh that sounds interesting

[finite token]

Never doubt the #1 cat theory fan

[finite token]

The google search preview text really did not know what to do with these sites

I was like, wtf is C and why is it being composed with st

turns out it was the word Cost

In retrospect I probably should have noticed

[kind marlin]

This actually does make intuitive sense in the setting where we drop symmetry and allow negative distances

going somewhere and back should exert nonnegative effort, no free lunch or something

This feels a lot like a line integral esque setting now ☹️

[finite token]

Imagine a d(x,y)+d(y,x)>0 axiom as replacement for d(x,y)>0

[finite token]

Oh, that reminds me

d(x,y)>0 is pretty much explicit already

I mean we showed that it could be rephrased as d(x,y)≠0 for a usual metric space but that's weird

Also I feel like a mathematician would just define it as a map to [0,infinity)

x≠y for the last like 10 statements I made I'm too lazy to say it

[kind marlin]

that’s fair but idk I’m kind of interested in negative metrics now

my favorite activity is inventing use cases for useless things

[finite token]

Show every negative metric is uniquely representable as d(x,y)+f(x)-f(y) where d is a metric and f is a function from X to R w/ f(0)=0 (doesn't work)

or something idk

[kind marlin]

that looks exactly like an algorithm I was thinking of but can’t remember the name

I think it computes shortest paths on directed graphs with negative edges or something? But not the one that does dp on path lengths, I remember it adding and subtracting weights at each vertex

and now that I think about it that is literally just a proxy for dealing with negative distances

[finite token]

wait there's no 0 in X

ok fix x_0 in X, same thing

lol

[finite token]

Oh that's cool

[kind marlin]

Johnson algorithm!

wow that’s really interesting

[finite token]

So we can treat them like normal graphs and do Djikstras or whatever

[kind marlin]

right

ha and the algorithm doesn’t really work if there’s a negative cycle

in the sense that if it finds one it just terminates since there is no shortest distance

Wow actually the d(x, y) + d(y, x) >= 0 is just a well-definedness check on distances

[finite token]

Yeah lol

Well that's the transitivity requirement

[kind marlin]

directed graphs seem like the perfect example of asymmetric metrics then

and negative distances are totally fine as long as you don’t have negative cycles, which implicitly agrees with what the axioms assume

[finite token]

Probably yeah

How hard would it be to do this another way though

Ig pretty annoying

Actually idk

I don't remember how Djikstras works so idk if it could be modified more easily

[kind marlin]

somehow I was an algos TA for 2 years and forgot everything

I really like this though :D that was such a neat rediscovery of Johnson

I should read through the graph algorithms again

[finite token]

It's cool seeing that it has a name lol

I kinda want to give this to my algorithm obsessed friends but I'm worried it's just djikstras

Maybe it's the opposite

We can't use my construction unless we already know the distances

So maybe Djikstras can be easily modified to find the distances, and that's used to compute the decomposition

[kind marlin]

hm like I know after the edge reweighings you just apply djikstra

you just don’t necessarily compute the same distances as in the original graph, so you have to add back the vertex weight offsets to get the “actual distance”

but the relative distances are all preserved

I’m not sure if that’s what u mean

[finite token]

Oh hmmmm

[kind marlin]

https://en.wikipedia.org/wiki/Johnson's_algorithm

im just gonna link this since im very fuzzy on the details

[finite token]

You choose something a little arbitrary for the vertex reweightings?

Ok

Yeah that would work if you only cared about turning it into a positive digraph and not a undirected graph

So I guess my version is that you can actually reweight the vertices to make it an undirected graph

[kind marlin]

oh interesting :0

as in, you’re enforcing symmetry on distances?

[finite token]

Yeah

If there's a directed edge from x to y in the old graph, the new graph has an edge between x and y with weight d*(x,y) = (d(x,y)+d(y,x))/2

Then, fix x_0 and let f(x) = (d(x,x_0)-d(x_0,x))/2

oh huh

this probably doesn't work lol

rip

[kind marlin]

wait I think I’m confused what are you trying to preserve / solve for

[red sage]

is it possible to find a nice metric that charaterizes pointwise convergence of sequence of real functions ?

for uniform convergence, we have sup norm with space of bounded functions.

I've just read a bit, is it equivalent to the question is the product topology on R^R metrisable ?

[unreal stratus]

There should be the problem that such a space would not be first countable, hence not metrisable

[red sage]

This is what we could get closer to what I said right ?

[quartz horizon]

Ooh this is a nice Q

[red sage]

This randomly popped on my mind, when encountered sup norm metric for uniform convergence.

[quartz horizon]

Do you know how to prove this result

[red sage]

I don't think so.

[red sage]

I've done some product topology, I might be able to read.

[quartz horizon]

Good luck

There’s a fun universal property proof of this

[red sage]

can you link it here ?