point-set-topology

[iron bolt]

I feel like that's one big part of why R is so important

[opaque scroll]

Nice for your sequences to have limit points I guess

[iron bolt]

all of the constructions you mentioned on the other hand, like the suspensions, joins, cones and simplices are not just based on R but also more specifically the unit interval

so I feel like that goes back to "why is the unit interval the parameter space used in homotopies"

and that I don't really have a good answer to other than "it's the space of all things you go through to go from 0 to 1"... idk if that's helpful to you

[unreal stratus]

There is a good question here in that you can do homotopy theory (which was mentioned above in the list of stuff, stuff like spectra) purely combinatorially but can also use topological spaces to model stuff

[silent garnet]

yeah spectra and simplicial sets are part of the reason i ended up asking this question.
like R and [0, 1] are very good at modeling many different constructions that are important even when you dont think just about R.

[opaque scroll]

I guess flipping the question, how are you motivating simplicial sets without (geometric) simplicial complexes?

How are you motivating spectra without topological spaces / subsets of R^n?

How can one care about geometry without knowing what a line segment is?

[silent garnet]

isnt scheme theory something like that though

[crisp lintel]

well the order structure on R induces the topology

[silent garnet]

good question. maybe ill know the answer after i learn more homotopy theory and topos theory

[crisp lintel]

so those other reasons to care about R connect

[opaque scroll]

I wouldn't say so no.

This is still just taking subsets of R^n (and C^n) turning the geometry into algebra, then going full tilt with the algebra

[unreal stratus]

Why R^n and C^n lol

Okay sure I guess you mean that is where you start from with motivation

[opaque scroll]

Shapes are things in R^n, that's what it all comes down to

[silent garnet]

how are you motivating simplicial sets without (geometric) simplicial complexes?
maybe (weak) Kan complexes can be motivated via infty-groupoids (or infty-catergories).

[opaque scroll]

Well, then I'm very intrigued how you're motivating infinity-categories without homotopies

[unreal stratus]

Well homotopy feels a bit orthogonal as you can have homotopies in simplicial sets etc

[unreal stratus]

Idk how you can do this lol cause every model of oo-category i know invokves simplicial sets or topological spaces

[iron bolt]

in some sense yes. but you could argue that simplices are so important precisely because they're used as the domains of higher-dimensional homotopies

if you take the singular simplicial set of a topological space you're encoding precisely the information of all points, all paths (homotopies between points), all homotopies between paths etc. in it

[brazen condor]

It generally refers to ahem

A compact connected metric space.

Continuum theory is a whole area of study

But I view continuum as usually a broad intuitive notion in this context, how to codify the idea of a continuous space

And also R wouldnt be continuum ig because its not compact lol

But you can make it locally compact instead

[unreal stratus]

I mean you definitely don't need to explain this lol I do homotopy theory

[brazen condor]

Different definitions for different folks

[unreal stratus]

I definitely think you need topological spaces or smth to motivate stuff lol

[brazen condor]

?

Im not talking to you?

[unreal stratus]

Sorry was talking to undefined, I am also not talking to you lol

[brazen condor]

Ah sorry

[unreal stratus]

np

Sorry lmao

[warped helm]

[brazen condor]

Continuum theory and dimension theory are some of the older branches of point set topology and kind of related

but lemme think on this

I think a more specific question to ask first is why do we use intervals in homotopy theory, as that will then answer for instance why we define deformation retraction and path-connectedness the way we do

i found this answer that might be of interest that essentially says ther interval I is effectively the 'smallest' compact Hausdorff space to join two points.

https://math.stackexchange.com/a/1824082/400654

I've heard of a notion of something known as an 'interval object' in categories, maybe someone can shed some light on other intervals one could (usefully) use?

i think its a very good question that's not so easy to answer i mean obviously topology is gonna be abstraction of spaces we care about

could you imagine some other wacked out topologies of use though?

[silent garnet]

topological category? the goal is to avoid R, not to avoid topological spaces altogether

[unreal stratus]

Ah lol fair then

[iron bolt]

lol

[unreal stratus]

Though that is the worst model probably and definitely for getting to simplicial sets

[iron bolt]

and yeah, I definitely expect that you know all of this already :)
it just seemed relevant, and sometimes for questions like this reframing the basics in some particular way can already be half the answer ig

[queen prism]

https://math.stackexchange.com/questions/426507/topological-characterization-of-mathbb-r

[pallid swan]

i have no idea if it's related but i always found it amusing that you can define homotopies of discrete graphs purely using the graph that looks like * - *

[brazen condor]

its very tempting for me to speak about path connectedness or deformation retract properties but I worry of sounding a lil circular. Since, path connectedness in terms of I (it is conceivable to use I ∩ Q as your interval, then boom Q is path connected). One main reason not to do this is I ∩ Q isn't compact and limits of an increasing sequence need not exist.

If then we have justified this compact condition as important, then I is the smallest way of talking about path connectedness.

Then, it's somewhat immediate why lots of R things show up because R like things are 'continuous spaces' in the sense I can get from one point to another via a path.

Then stuff like fundamental groups fall out naturally. Deformation retract falls out naturally too when you consider it as 'family of paths' with suitable continuity conditions.

[iron bolt]

every space is path-connected with respect to I ∩ Q. you can send [0,1/√2) to one end and (1/√2,1] to the other

that's kind of why connectedness or R is so important. without your interval being connected, your notion of path-connectedness won't be related to standard connectedness

[brazen condor]

then, in many such 'R defined spaces' the pieces used to build them are locally contractible in the sense there exists a neighbor hood which is contractible

which, effectively means all 'topological properties' are going to be arising from the more global connections of the space which is a desirable property. basically it means are space can be 'locally' squishy and deformed.

[brazen condor]

true

great point

but then ig we are now asking why continuity is defined the way it is

[iron bolt]

I was going to just say "no" or "I'm not" in response to that :p - but actually, good point

[brazen condor]

like i don't see a way to do it without the reals hence why the reals were created but i think one should feel in a sense that the function y=x is continuous but the step function at sqrt(2) is not

[iron bolt]

local Lipschitz continuity maybe? that seems like it would work, though of course only in very specific contexts

gotta go

[unreal stratus]

(also lol sorry if my message came across as agressive oops (not that your reply looks like you are annoyed w me or anything))

[iron bolt]

no worries, it didn't read as agressive to me

[frank burrow]

Ill also be more active in this chat

[stray harbor]

Here

Seems like it's an entirely non-trivial thing to prove

[snow frigate]

i think there are a few constraints here

like i think we care that it’s hausdorff?

[stray harbor]

yes

[snow frigate]

so that at least stops us from gluing on topologically indistinguishable points

[stray harbor]

I don't care about constraints

All relevant topologies separate points anyway

or are the Zariski topology, which still separate points in a weaker way

[snow frigate]

zariski is still T1 iirc

[stray harbor]

Yeah, give two points, each has a neighborhood not containing the other

||which is essentially the same as saying there are varieties that pass one point and not the other||

[tranquil cosmos]

not the "good" zariski tbh

the good zariski is sober instead of T1

[snow frigate]

the Spec one?

[tranquil cosmos]

yeah

with generic points

[snow frigate]

is there an upper bound for cardinality of closure

in a reasonably well-behaved topology

[tranquil cosmos]

if by closure you mean SCC, it's always bounded by 2^2^|X| iirc

but i'm not that well versed with SCC so specialists could probably tell you more about it

[timber estuary]

gosh such big boi talks

[tranquil cosmos]

the SCC doesn't touch compact spaces so $\beta[0,1]$ is just $[0,1]$ itself but $\beta[0,1)$ is completely horrible 💀

[gentle osprey]

PKThoron

[snow frigate]

{1} holding up the world here

[timber estuary]

beta(N) is itself horrible to begin with

[snow frigate]

whats a sensible construction for the SCC

none of the ones i read make any sense

[tranquil cosmos]

i guess the {1} in [0,1] forces continuity of functions from [0,1) to behave in exactly one way

and if you get rid of that constraint it explodes?

[snow frigate]

also does munkres talk about any of this

[tranquil cosmos]

idk either

supposedly the "maps into the interval" construction is supposed to be the intuitive one but i haven't understood it yet

[stray harbor]

mfw you glue all compact surjections and get shit

[timber estuary]

it is intuitive in the sense that it lets you "accept" it exists

[snow frigate]

these compactifications look like magic to me

[tranquil cosmos]

i thiiiiiink the idea is that given a space X, you want all limits of sequences (or filters or nets) of X to be adjoined

so you ask a nice CHaus space K for help by mapping X into K and asking K what limit points it offers

like N into [0,1] might give you a formal limit via the sequence (1/n), but another one through a different, more oscillatory or pathological sequence which might need a different adjoined limit point (this is the part where i get a little unsure)

[timber estuary]

i honestly just went like "well i guess this is like the topology version of completion of algebraic structures" and moved on

[snow frigate]

wdym by “completion of algebraic structures”

[tranquil cosmos]

injective hull prolly

which is the same shit but in algebra

[snow frigate]

i guess this is sort of like making an ACF by adjoining roots of every polynomial

[timber estuary]

I've been speedrunning reviewing munkres

[tranquil cosmos]

let's go to R-Mod. an R-module I is called injective if its contravariant hom-functor Hom(_,I) preserves exact sequences

[snow frigate]

except not really because you can still make a bigger ACF

[timber estuary]

three weeks in and im at chapter 6

[tranquil cosmos]

so if 0 -> A -> B -> C -> 0 is exact, then 0 -> Hom(C, I) -> Hom (B, I) -> Hom(A, I) -> 0 is exact

[snow frigate]

i see

[stray harbor]

I hate saying it like this

[snow frigate]

i gtg so ill read all this later

[tranquil cosmos]

projective modules are usually not much bigger than the base ring (like all free modules R^n are projective), but injective modules are quite different

[stray harbor]

Like ok it's true, but give the intuition before on what that means

[tranquil cosmos]

over Z, stuff like Q and Q/Z is injective

[frank burrow]

Damn bro

[tranquil cosmos]

note that Q/Z looks a lot like the topological interval [0,1] and it will take its role here

[frank burrow]

I should get on that method

Im on chapter 2 rn

[tranquil cosmos]

and i think the injective hull of a module M checks all maps into Q/Z and takes a massive product yadda yadda

and it's like the SCC

so it's maybe vaguely more enlightening than the SCC but not much

[timber estuary]

ive had a two-semester class on topology (although that was like 3 years ago due to military shit) so i should already know the stuff

but alas my brain is not braining and id forgotten what compactness is

[timber estuary]

yaa i hate this thing

[warped shore]

really great website

thanks

[timber estuary]

oh wow this site is good

[snow frigate]

fun combinatorics problem for yall

[timber estuary]

?

combinatorics?

[snow frigate]

for $n \in \mathbb N$, how many non-homeomorphic $T_0$ spaces of order $n$ are there?

[gentle osprey]

lexi

[unreal stratus]

1 right lol

[timber estuary]

huh

{{},{a},{a,b}} and {{},{a},{b},{a,b}} are both T0

[unreal stratus]

Oh I read T0 as T1, I am stupid

[snow frigate]

theres only 1 T1 space

yeah

[unreal stratus]

"yeah you are stupid" jk

[snow frigate]

noooo

[unreal stratus]

i'm jokingg dw

[timber estuary]

isnt it equal to the number of non-isomorphic posets of n elements

[snow frigate]

i think you have to be careful with number of open sets but thats probably on the right track

and maybe replace poset with lattice?

[silent garnet]

what is this refering to? edit: found #math-discussion message

[snow frigate]

that there is an upper bound on the cardinality of a space a (well-behaved) space is dense in

yeah

[frank burrow]

Ill look at yhat, is there an algebriac topology website like that

[opaque scroll]

Say x < y if y is in the closure of {x} should give you a correspondence with posets

[timber estuary]

Let x <= y if every nbd of x contains y -- reflexivity is obvious, x <= y and y <= z implies x <= z (every nbd of x is a nbd of y, so is a nbd of z) and antisymmetry is given by T0. Suppose that a topology T generates a partial order <=. For each x, define U(x) = {y >= x ; y in X}. Since U(x) is the intersection of all nbds of x, it is open under T. We claim that T consists of all upper sets under <=. Indeed, any open set V is clearly an upper set, while if V is an upper set, V is the union of U(x) for x in V. Hence there is a one-to-one correspondence between partial orders and T0 topologies.

yeah thats what i was thinking

[frank burrow]

What is this from

[timber estuary]

what is what from

[frank burrow]

Whered you get the justification from

[timber estuary]

uh from my keyboard

[fierce mesa]

I'm genuinely lost for part b), could anyone explain what the question is asking for and some hints on how to prove it?

[gentle osprey]

Average Math Student

Average Math Student

Average Math Student

[fierce mesa]

Any thoughts are greatly appreciated 👍

^ I managed to prove that x_n(lambda) is cauchy, using the identity provided in part a) I guess.

wait

Okay I think I have a decent idea. The uniform continuous bit requires n0 to be independent of lambda, so I reckon that n0 should be somewhat related to M, and I can exploit the fact that it's a uniform contraction.

Okay I think I got it. Nvm!

[warped helm]

im lost at this bit, this is a sequence

what's written there is not uniform continuity but just the definition of x_n(lambda) converging to x(lambda)

[quasi forum]

its a sequence of functions that uniformly converge

[warped helm]

ah ok i see

yes since x_n(lambda) is a function of lambda

Ok cool

[warped helm]

the M here is going to play the role of this expression

[fierce mesa]

Yeah 👍

I basically said that

[warped helm]

pretty cool exercise

[fierce mesa]

d(x0, x1) has a maximum value

Well

d(x0(lambda), x1) is bounded by a finite value

So yeah then you have L^n/(1 - L) * M and then this converges to 0, so I think this is the general idea 👍

[warped helm]

yep

[quasi forum]

the proof should look basically the same

you just have to replace certain expressions and the argument goes through verbatim

the assumption does 90% of the work for you because the constants are independent of lambda

[fierce mesa]

Right

It took me a while to realise that M was important 😭

Thank you both!

[quasi forum]

np

[warped helm]

Thank you for posting that

[snow frigate]

are two points topologically indistinguishable if they have the same closure?

[prime elbow]

Topologically indistinguishable means?

You mean we can't separate them by disjoint open sets

[snow frigate]

i mean they have the same set of neighborhoods

so violating T0 axiom

[rancid umbra]

i believe so, yes.

[snow frigate]

i’m trying to come up with a proof

maybe showing closed nhoods are precisely the closed sets containing the closure?

and so the same for both points

[rancid umbra]

say that x and y have the same closure.

let U be an open nbhd of x. if y is in X - U, then the closure of y is contained in X - U, contradiction.

this is because the closure of y is the intersection of all closed sets containing y, and X - U is closed.

[snow frigate]

ahh ok

but wait doesnt this only imply the space is not T_1

ah wait no

silly me

[storm wadi]

Is the empty set considered connected? Is it a bit 'cheap' to use that fact to obtain a counterexample?

Obviously, by definition it is connected, but I was wondering if it's considered an edge case or something and shouldn't really be used to disprove something for general sets.

[novel acorn]

uh it isn't connected

because 0 = 0 cup 0

and 0 cap 0 = 0

i.e. you can write it as the disjoint union of two empty sets

[storm wadi]

disconnected requires the sets to be non-empty right?

disjoint union of two non-empty open sets

[novel acorn]

hm okay that's fair

[midnight umbra]

yes otherwise it would be a trivial notion

[storm wadi]

yeah true lol

[midnight umbra]

every space $X$ would be disconnected as $X=\varnothing\sqcup X$

[gentle osprey]

Convergant

[midnight umbra]

anyway to answer the question at hand

[midnight umbra]

it should surely be on the author of the theorem to exclude the empty set

[storm wadi]

Ideally yes. It's an exercise, stated very vaguely:

• Does the connectedness of A \cup B and A \cap B imply that of A and B?
  It's easy to construct a counterexample by just choosing A and B disjoint. But it feels cheap considering A \cap B = \emptyset as connected

[midnight umbra]

then i would suggest considering what happens in the case where A,B are not chosen to be disjoint

it isn't cheap as a counterexample but i think it would still be helpful

[storm wadi]

Oh right yeah it's just as easy to construct a counterexample: A = (0,1) \cup (1,2), B = [1,2), standard topology. Feels strange when you can just define A disconnected from the start.

I appreciate the help : )

[opaque scroll]

It's a matter of convention, but I think there are some good reasons it should not be considered connected.

One is the same reason 1 is not prime: "a locally connected space is uniquely a disjoint union of connected sets".

Another is this alternate characterization of connected: X is connected iff Hom(X, -) preserves disjoint unions. The empty set fails this.

[quartz horizon]

Mfw Hom(X, -) preserves coproducts

[urban zinc]

Neat, is this a rephrasing of the gluing lemma (in the case where the sets don't overlap)?

I've never heard this way of saying it before

I guess that's a silly question, the whole point of the gluing lemma is that the sets can overlap

[snow frigate]

where can i learn about finite topology?

[pallid swan]

like, topologies on finite sets?

they happen to be equivalent to preorders

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

this is a good summary

[snow frigate]

that’s what i’ve learned so far

[pallid swan]

huh, today I've learned that https://en.wikipedia.org/wiki/Finite_topological_space is also an article that exists

[snow frigate]

is there somewhere i can learn more, like about how continuous maps behave and that sort of thing

[pallid swan]

there's actually a category equivalence, so cts. maps are precisely monotonic ones

[snow frigate]

maybe there’s a good categorical way to view this idk

oh yep ok

[pallid swan]

tbh it's actually a good exercise

[snow frigate]

alright ill think about it some more

[pallid swan]

i / wikipedia can provide hints if needed

[snow frigate]

thank you

i kind of want to write this stuff into a computer program

[pallid swan]

oh yea, oeis probably has some fun factoids

nvm i just checked and it doesn't

but it does have references so idk you could dig through those maybe ?!

idk

https://oeis.org/A001930

[snow frigate]

alr im going to start taking notes and writing proofs

[opaque scroll]

It's essentially just that the image of a connected set is connected

[urban zinc]

Oh wait disjoint union in the codomain not the domain duh

[opaque cloud]

Omg Keith Conrad is on mathcord??!?

Awesome!

[alpine nest]

It's actually Keith Starmer

[pallid swan]

i wish

[opaque cloud]

UK prime minister Keir Starmer?

[alpine nest]

I mean, they didn't deny it

[river abyss]

you actually did it (I'm the instigator)

[main pulsar]

#help-10 message also fits here

[snow frigate]

awww

||isolated point?||

err wait

maybe not

yeah

i havent seen that notation before

[dusk magnet]

An isolated point of A is a point that belongs to A but exists in a punctured neighborhood where no other point of A is located.

[snow frigate]

yep

i hadnt seen it expressed in terms of a metric(?) before

[storm wadi]

Yeah what is that notation with the empty circle? I'm not familiar either

[snow frigate]

i think that's supposed to mean "open ball of radius delta"

[dusk magnet]

De-mind Neighborhood

i don't know is that correct in English

$U(x_0, \delta)={x\in S|d(x,x_0)\lt \delta}$

[gentle osprey]

ComradeKonata
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[dusk magnet]

what

[storm wadi]

Oh right, yeah so just the open ball of radius delta about x_0

[storm wadi]

might be \lt, I haven't seen that before

[dusk magnet]

yes, in Euclidian space

[snow frigate]

i think this works in any metric, no?

or maybe not in some pathological cases

[dusk magnet]

some space they don't say open ball, i don't know why either

[snow frigate]

im used to "isolated point" of a set as an adherent point of a set that is not a limit point

[tender halo]

a point in a metric space is isolated iff its character is 1

[snow frigate]

whats the character

i havent heard that term before

[tender halo]

character of a point is the lowest cardinality for a base at that point

[snow frigate]

ah

[dusk magnet]

me either

[tender halo]

character of a space is the supremum of the cardinalities of its points

so space is first countable iff its character is \leq \aleph_0

[snow frigate]

i see

[dusk magnet]

oh

[snow frigate]

and an isolated point has a character of 1 because it is clopen?

in the subspace topology of the set

[tender halo]

character of one means there is a smallest basic set

at that point

[snow frigate]

yea so the singleton of the point is open

[tender halo]

not necessarily

[snow frigate]

at least in a metric space

[tender halo]

yea

[dusk magnet]

[tender halo]

spaces in which all points have a smallest nbhd are called Alexandrov spaces and they are pretty fun

[snow frigate]

omg i love alexandrov spaces

aka pre-ordered sets

lol

posets if they happen to be T0

[dusk magnet]

i don't even know what it is

I'm new in point-set topology

[snow frigate]

alexandrov spaces are spaces where an arbitrary intersection of open sets is open

not just finite intersection

and so infinite union of closed sets is closed, as well

[dusk magnet]

Like an inductive set?

[snow frigate]

sort of maybe

all finite spaces are alexandrov

do you know what a T1 space is?

[dusk magnet]

wow

no

[snow frigate]

its a space where singleton sets are closed

all metric spaces are T1, for instance

[dusk magnet]

omg

[snow frigate]

a T1 alexandrov space is always discrete (every set is open)

[dusk magnet]

I've only just begun to understand metric spaces, normed spaces, and inner product spaces.

[snow frigate]

so T1 alexandrov spaces are not very interesting

[storm wadi]

how is this different from Hausdorff? Excuse me if thats a dumb question, I'm also new to PStop

[rancid umbra]

it’s slightly weaker than being Hausdorff

[storm wadi]

Ah wait so is T2 Hausdorff?

[frozen sparrow]

Yes.

[snow frigate]

yes

[storm wadi]

that's cool

[frozen sparrow]

T2 and Hausdorff refer to the same separation axiom.

[snow frigate]

T0 is "for any two points x,y there is a open set containing one that does not contain the other"

[dusk magnet]

oh hell man, I just want to understand mathematical analysis

[rancid umbra]

i believe the Zariski topology has closed points but isn’t Hausdorff (in general)

[dusk magnet]

😭

[snow frigate]

i think of T0 as a basic sanity check on topological spaces

[storm wadi]

Alright thanks I'll have a look

[snow frigate]

non-T0 spaces are nasty

[snow frigate]

a very good example of a T1 space that is not hausdorff is the cofinite topology

[snow frigate]

i think the SpecR incarnation of zariski can have non-closed points

[rancid umbra]

at least on R^n it holds

[snow frigate]

the 'normal' roots of polynomials one is just cofinite topology in the 1D case

[rancid umbra]

yea

[rancid umbra]

i’m not so familiar with spec stuff or algebraic geometry

[snow frigate]

theres the weird SpecR zariski where prime ideals that are non-maximal are non-closed points

me either

[rancid umbra]

weird

[snow frigate]

algebraic geometry is scary

dont put a topology on the set of ideals of a ring

its weird

quit zariski today

i like reading about topologies in between T0 and T1

or between T0 and T2

[rancid umbra]

you can certainly get some interesting ones

[snow frigate]

any poset gives you a free T0 alexandroff topology

[rancid umbra]

do you have a copy of Counterexamples in Topology?

[snow frigate]

no... i really want one

[rancid umbra]

ah, yea. it’s a fun one

[snow frigate]

non-metrizable things are cool

long line

[unreal stratus]

yeah, put it on the prime ideals instead

[pallid swan]

put it on the maximal ideals to ragebait modern ag purists

[kind egret]

Today I wrote in my memoire a part about Stone-Čech's compactifcation of N, which is a compact hausdorff space and separable but which doesn't have a countable basis of open sets. So you still don't have a copy of the book you wish but I hope you learnt something from my message 🙂

Oh, or another funny one : there are Banach space which have isometric duals but which are not even isomorphic to each other

[pallid swan]

in a sense, it's the maximal chaus separable space

[kind egret]

Like C(K) where K is countable metrizable compact

[kind egret]

Yes it is, so it shouldn't be metrizable

Life is painfull sometimes

[dense laurel]

i guess countability has something to do with metrizability. i wonder if there's some program/yoga to examine filtrations of non-metrizable things in the same way that some people are out to add axioms about cardinalities other than countable and uncountable.

[gaunt linden]

"Something to do with" is simple: metrizable spaces are always first countable.

[lament steppe]

If I have a Manifold M that has a boundary (dM) and an open subset of U of M, is it true that the subspace of U n dM of U is the same topology as U n dM that is a subspace of dM?

I want to say that:
1.) U n dM can be a subspace of U and U can be subspace of M
2.) U n dM can be a subspace of dM and dM can be a subspace of M

=> both "versions" of U n dM "agree" in terms of their topologies because they are just subspaces of M. Is this accurate?

[opaque scroll]

Yes, whenever you have subspaces U < V < W, the subspace topology on U is the same wether you think of it as a subspace of V or W

[iron bolt]

one of the many little topology lemmas like that that are used everywhere but rarely ever explicitly stated

[cerulean oriole]

Indeed, instead construct a locale whose points happen to be the prime ideals of the ring.

[vocal dawn]

Hey, can someone help me on something in DMs?

[ruby delta]

why not ask the question here

[vocal dawn]

See for yourself, you'll understand why.

I'm trying to make this make sense.

[snow frigate]

one day i'll understand locales

oh its just a complete lattice without infinitary infs

its just the uhh

poset lattice of open sets

and its bounded

"product of locales" is the categorical product?

limit of two-object discrete diagram

i mean the "finite product of locales" they describe is the product object in the category of locales they give on the first slide

ah, no

oh i forgot lattices are categories

oh wait they actually have a really clean definition

small, thin, skeletal category with all products and coproducts

[crisp lintel]

how does this multiplication make any sense

you define a times b to be a repeated operation applied to a

but how many times?

b times? b is an element of a topological space, that makes no sense

[vocal dawn]

Uhh, well, non-standard clearly isn't an excuse either, uhh, damn.

[crisp lintel]

also what's even the point of any of this

[rancid umbra]

did you write this?

[vocal dawn]

Yes, I'll show you where it leads me.

[crisp lintel]

also how do you define this delta operation, you say it isn't a measure but then what is it?

[vocal dawn]

I'll "try" to elaborate as far as my understanding goes, the Heine-Borel theorem, it states that a compact set is closed and bounded, think of it this way, the general definition of compactness recreates the effect of "boundedness" for general topological spaces, that's what I was leaning on.

Now look, forgive the innacuracy of terms for a moment, I'm still just beginning doing this.

[crisp lintel]

Sure but what actually is this delta

you're trying to multiply by it

[snow frigate]

i see

[crisp lintel]

A solution to all of these issues would be to have a measure on the space and consider real valued functions, but then that's just the Lebesgue integral of course

[snow frigate]

is this for pointless topology?

[vocal dawn]

Um, well measures break in exotic spaces, you get my point?

And I'm trying to recreate a more general Riemann.

Or Henstock-Kurzweil depending on how you think of it.

[crisp lintel]

They don't really break though

measures on locally compact Hausdorff spaces are quite well behaved

[vocal dawn]

What about non-hausdorff, like here.

[crisp lintel]

the bigger issue is that your construction just doesn't really make sense

multiple of the key components are not defined or ill defined

[vocal dawn]

I know, that's why I asked for help.

What are some of the key components one needs?

Fractals, overlapping sets, the long line, the comb space, black holes at singularities.

Um, I am explicitly trying to not use measures.

[crisp lintel]

idk to me it's not even clear how to salvage any of this

[vocal dawn]

Can it be developed? Or should I just break it down and rebuild from scratch?

[crisp lintel]

There isn't really anything to develop

[vocal dawn]

Integrate and differentiate.

I see now.

I see what you mean.

[crisp lintel]

It's basically impossible to make differentiation work on arbitrary topological spaces

at least in any sensible way

[glacial nymph]

https://cdn.discordapp.com/attachments/1111993907060949112/1325143268794564699/TUPW_2.0-2.gif

[vocal dawn]

Defining it as the inverse of the integral.

[glacial nymph]

what even is point set topology?

[vocal dawn]

No, that isn't it, that's not what I'm looking for.

I'll just grab my shit and leave.

[crisp lintel]

If you don't add additional structure to your space you won't be able to define anything sensible

Topological spaces aren't defined to be able to support such a theory

p much every sensible integration theory relies on the real numbers in some way as well, and that's for a good reason

[urban zinc]

general topology

[snow frigate]

it's pretty hard to do differentiation and integration without some linearity

[vocal dawn]

Linearity in what sense?

[snow frigate]

that's sort of what it means to be differentiable

at least in the usual sense

[vocal dawn]

I want to say when you have an integral you can differentiate it.

[crisp lintel]

Integrals and derivatives aren't really very cleanly opposites in general

so even if you have integrals you don't automatically have derivatives

for instance arbitrary measure spaces don't have derivatives

[vocal dawn]

That’s what I’m trying to do, but without Hausdorff, connectedness, or countability.

[snow frigate]

idk if you can get something that looks much like anything useful with such a general topological space

[warped helm]

vector space for example

[vocal dawn]

Oh I thought they were talking about linearity of functions, turns out no.

[warped helm]

you would like a norm as well

[vocal dawn]

No, I don’t want norms either.

Personally okay, I don’t want them here.

[snow frigate]

well the derivative operator is usually linear in some sense but i suppose you probably don't need that

same with integral

though i guess you can work with the ring of real-valued functions out of your space or something

you probably want some idea of distance

[dense laurel]

or stay and we can talk more about spaces that aren't Hausdorff, connected, or countable look like. perhaps looking at examples will help. i'm short on examples, although the countable-compliment topology may satisfy some of these conditions.

[vocal dawn]

The etale space.

Or, for example, being able to handle integrals and derivatives over disconnected domains, like black holes everywhere except exactly the singularity.

This would have PDEs hold on a punctured space.

[dense laurel]

okay, algebraic geometry (which i think is what sheaves are) is a little advanced for me. i tried some independent reading a few years ago, and i think i processed the definition of a sheaf, but i haven't retained it.

(this is not an invitation to explain algebraic geometry: i will probably loop back around to it later)

[dense laurel]

[insert comment about branched manifolds]

wrt to geometry and singularities

[vocal dawn]

In reality, what I am trying to do is to create a definition of absolute continuity for functions that are traditionally discontinuous.

That’s the point.

[dense laurel]

traditionally, absolute continuity implies continuity. is part of the joga to create a definition of continuity for functions that are traditionally discontinuous?

in other words, i don't see why you say absolute continuity instead of non-traditional continuity, but then again, idk Hartshorne at all.

😢

[vocal dawn]

That’s what I was trying to say.

You see, the thing is, I wished to say that that delta thing up top in the image was dependent on the space we were in, but generally, we can not compute it for the general plug and chug case, it a bit of a non-standard multiplication.

It’s meant to work for other elements.

[dense laurel]

i have saved the pdf and will hopefully be reminded to look at it once i get around the cleaning up my Downloads

[vocal dawn]

You like the idea?

[dense laurel]

i'm open!

[vocal dawn]

Uhh, what do you mean by that?

[dense laurel]

i guess it's just what some people say when they've had a lot of caffeine? i got the idea that it was important to be this way from somewhere..
..also, it could be used as in open-minded, which i generally am, even when i haven't had two red bulls at midnight.

[unborn hawk]

open in which topology

[vocal dawn]

“Weak”-differentiation, need I say more?

[storm wadi]

Question asks to show a circle is not homeomorphic to any subset of R. Since S1 is connected we must have A \subset R an interval. It's easy to derive a contradiction then by removing a point from A. However, for the singleton case, since removing the element from the singleton leaves you with the empty set which is itself connected (and pathwise), I'm not sure where the contradiction comes from.

Could you use compactness actually?

Empty set is compact but S1 with a point removed isn't(?)

[prime elbow]

for the singelton case you can use cardinality argument

[storm wadi]

Oh actually this is dumb right? A set with one element cannot have a bijection to S1

yeah, thanks

Does compactness also work? Although, it is obviously more work

[prime elbow]

but here what is the contradiction if it is not singelton?

because what if you remove end point?

[storm wadi]

You can just choose the midpoint right?

[prime elbow]

S1 is compact so interval has to be form of [a,b]

ah

yeah

my bad

[storm wadi]

If its the non-singleton case, it should always have a mid point

[prime elbow]

yes

[unreal stratus]

I think an amusing way to see this is how like

For any two points a, b of S^1, there is a homeomorphism from S^1 to itself which sends a to b

(Just rotate)

This is not true for [c,d] lol unless c = d where yes cardinality

Actually there is a funnier way

[storm wadi]

Wait sorry why can that not exist for [c,d]?

[unreal stratus]

E.g. Intermediate value theorem basically

[storm wadi]

ah right okay

[gaunt linden]

There's no homeomorphism that sends c to (c+d)/2, because only one of those has the property that removing it disconnects the space.

[unreal stratus]

You take the value c and d somewhere in [c,d] and between there you take the whole interval, so to be injective your only option is to send endpoints to endpoints

[storm wadi]

ahh yeah that makes sense

[unreal stratus]

In fact there is a funny variant of this argument: every map [c,d] -> [c,d] has a fixed point

But definitely not true for S^1

[storm wadi]

Haha that's quite neat, definitely a much quicker argument than what I wrote down

is every map on that space having a fixed point a topological property then?

[prime elbow]

i think yeah

you can prove it

[gaunt linden]

Sure, because if you have two homeomorphic spaces X and Y, and a map X->X with no fixed point, then the composition Y->X->X->Y would have no fixed point either.

[snow frigate]

this is kinda reminding me of “circle is a topological group but interval isnt”

[frail crypt]

would the boundary of A = {x in Rn : |x| = 1} be A itself by boundary meaning each open neighbourhood of some x in A contains points both in A and outside of A

[snow frigate]

you can also probably show its the boundary of the unit ball and use the fact that boundary is idempotent

[gaunt linden]

It's not idempotent, though. The boundary of [0,1] cap Q in R is [0,1], but the boundary of [0,1] is {0,1}.

[main pulsar]

I got this to the point where we just need to show $f(V(M)) = V(f(M))$, but how do we do that set-theoretically? We defined $$V(M) \coloneqq {\mathfrak p \subset A \mid M \subseteq \mathfrak p},$$ so we need to show $${\mathfrak p \subset A \mid f(M) \subseteq \mathfrak p} = f\left({\mathfrak p \subset A \mid M \subseteq \mathfrak p}\right),$$ but how would we do that?

[gentle osprey]

ILikeMathematics

[tranquil cosmos]

f(M) lives in the ring B, so your prime ideals on the left hand side need to sit in B

[main pulsar]

Dont we just need to replace M with f(M)?

[tranquil cosmos]

they live in different rings

M is a subset of A, so it can only be contained in prime ideals p in A
but f(M) is a subset of B, so it can only be contained in prime ideals q in B

so I'd write the set on the left hand side as ${\mathfrak{q} \subset B \mid f(M) \subseteq \mathfrak{q} }$, take an arbitrary element $\mathfrak{q}$ in that and show that it's in the right hand side

[gentle osprey]

PKThoron

[tranquil cosmos]

and vice versa

[dusk magnet]

$U(x_0, \delta)={x\in S|d(x,x_0)\lt \delta}$

[gentle osprey]

ComradeKonata

$U(x_0, \delta)=\{x\in S|d(x,x_0)\lt \delta\}$
```Compilation error:```! Undefined control sequence.
l.49 $U(x_0, \delta)=\{x\in S|d(x,x_0)\lt
                                          \delta\}$
The control sequence at the end of the top line
of your error message was never \def'ed. If you have
misspelled it (e.g., `\hobx'), type `I' and the correct
spelling (e.g., `I\hbox'). Otherwise just continue,
and I'll forget about whatever was undefined.

Preview: Tightpage -1310720 -1310720 1310720 1310720
[1{/usr/local/texlive/2023/texmf-var/fonts/map/pdftex/updmap/pdftex.map}{/usr/l```

[dusk magnet]

what's wrong

[trail charm]

better for #latex-help
anyway \lt is a mathjax thing, not a latex thing

in latex just use <

[dusk magnet]

i see

[trail charm]

if you insist on using \lt for < you can just add \newcommand{\lt}{<} to your preamble

[quartz horizon]

ok, i'm trying to understand some results regarding nowhere-dense subsets

as part of trying to understand BCT

this is the definition i'm using

is this proof correct?

here's another equivalent characterization

i'd be happy to hear more efficient proofs of these as well

[unreal stratus]

Maybe this is not what you want, you can simplify things by appealing to the analogues for usual density instead of nowhere density. Like it's standard that TFAE: 1) you have closure the whole space or 2) you have nonempty intersection with every nonempty open. then from that youhave like

S is nowhere dense = for all U, U \cap S dense in U <=> U \cap S \cap V nonempty for all V open in U, which proves (1)

and then (2) similarly since closures behave well with going to a subset

but yes your proofs all look good

[quartz horizon]

i see i see

is this proof fine?

i'm using this to show that the nowhere-dense subsets form an ideal

i can include this for completeness but it's very short

[quartz horizon]

also this is a cool result, it gives me some nice intuition for what classes of subsets are nowhere-dense

intuitively it's like $\partial U$ is of "codimension one", and so should be negligible

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[crisp lintel]

yeah exactly

nowhere density is also useful for proving a lot of "obvious" results that are actually slightly not obvious if you think about how to prove them rigorously

[quartz horizon]

oh like what?

[crisp lintel]

for instance showing that finite or countable unions of lines in R^2 can't contain an open ball

I actually needed a result like this for a group theory project once (specifically I needed the fact that a countable number of great circles could not cover the entire sphere)

for the countable case you need BCT (or you can use measure theory as well)

[quartz horizon]

oh hm

so could you prove it like

a line in R^2 is nowhere-dense

thus a finite union of lines is nowhere-dense

and so can't have nonempty interior?

[crisp lintel]

yeah

[quartz horizon]

interesting

so i'm still trying to understand the definition of baire space

so would statement 4 be analogous to "every nonempty open set has positive measure"?

[crisp lintel]

along with this analogy I always like to think about baire spaces as spaces where open sets are "full"

yeah exactly

[quartz horizon]

wdym by "full"?

[crisp lintel]

idk if I really have a precise meaning, getting a good analogy is tricky without just making the measure space connection

but for example in the rational numbers

open sets are kinda full of holes

so intuitively the open sets kinda lack structure

or mass or however you think about it

of course there's a strong connection with completeness from both the BCT for metric spaces as well as the one for locally compact hausdorff spaces

[quartz horizon]

right yeah

[crisp lintel]

since local compactness also gives you a sort of "fullness" to open sets

[quartz horizon]

so i should interpret the baire category theorem as telling you that

complete metric spaces have "full" open sets

and locally compact hausdorff spaces also have "full" open sets?

[crisp lintel]

yeah that's kinda how I think about it

[quartz horizon]

i see

[crisp lintel]

I think you mentioned the word negligible in a previous conversion, that's a good one as well. In baire spaces open sets cannot be (topologically) negligible

[quartz horizon]

yeah i got that from here:

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

nowhere-dense subsets form an ideal

and meagre subsets form a sigma-ideal

[crisp lintel]

yeah sets of measure zero are also a sigma-ideal so that tracks with the measure space connection as well

the distinction between why countable matters a lot more than just finite I think is maybe a bit trickier to motivate

though countability is far more robust than finiteness in general which I think partly explains whats going on

and countability is the correct cardinality for summation so obviously comes up a lot

[quartz horizon]

I mean I guess every nonempty open subset is dense in itself

So couldn’t be nowhere-dense

[crisp lintel]

I think I started to build more intuition the more problems I saw that used BCT

I had a cool problem on a recent assignment that was to show if $f:(0,\infty)\to\mathbb{R}$ is continuous and $\lim_{n\to\infty} f(nx)=0$ for all $x>0$ (here $n$ is a natural number), then $\lim_{x\to\infty} f(x)=0$

[gentle osprey]

Blake

[crisp lintel]

which uses BCT

in general whenever you have sort of pointwise condition you can use BCT to pass to a more uniform condition

[quartz horizon]

Yeah I need to figure out how that works at some point

[urban zinc]

Relevant thread that is quite fun to read through: https://math.stackexchange.com/questions/165696/your-favourite-application-of-the-baire-category-theorem

The Baire category theorem has always been weird to me because you usually use it in a proof by contradiction which is non-constructive... so even though you can prove that "most" objects satisfy X property, you can't produce a single one that has the property!

[midnight umbra]

uh... what?

[unreal stratus]

Lol

[quartz horizon]

yeah it's some kind of eventㅤ

but you can get around it if you're sufficiently smartㅤ

[midnight umbra]

she says when making use of a word with five letters

[quartz horizon]

*she

[midnight umbra]

my bad

[quartz horizon]

it's okkkkㅤ

it's fun to trollㅤ

[opaque scroll]

Some kind of eventㅤ is how I'm gonn'a describe aprilㅤ1st from now on

[quartz horizon]

canonㅤevent

[unreal stratus]

Lol

[quartz horizon]

it's stillㅤtrans day of visbility for me and i can't even say transㅤsmh

[midnight umbra]

I'm just gona type the message in a different server and send the screenshot here if I'm talking it is not worthy of the effort

[quartz horizon]

yeah you can do that or use texitㅤ

[midnight umbra]

it genuinely pisses me off

[quartz horizon]

tspmoㅤ

[midnight umbra]

stop flexing pls

[opaque scroll]

How do you do it with texitt?

[opaque scroll]

Like you send the bot a message?

[quartz horizon]

yes exactly, and then you can just forward it to the channel of your choiceㅤ

[pallid swan]

i dont get what everyone's issueㅤ is

[queen prism]

mostest skillest issuest thingest I'vest everest heardest

[pallid swan]

fuck i messed up the joke

[quartz horizon]

yeah it will work better if you can get it to happen at the end of the phraseㅤ

that way it's not as obvious what you're doingㅤ

[quartz horizon]

ooh yes i have seen this resultㅤ

it was on one of my old problem sheets i believe?

had no idea how to do it though, i didn't understand BCT that well back then

[unreal stratus]

Lol convergant is ur nick purposely spelled that way

[midnight umbra]

no. I was bad at spelling when I was thirteen and have not changed it

[unreal stratus]

Valid

[midnight umbra]

it does work in my favour sometimes. for example the correct spelling is in use as a username many a time on different websites, but my incorrect spelling is not

[unreal stratus]

Ah lol nice

[novel acorn]

Cheater

[cosmic mirage]

theyr arent going to allow using words which split penta times?

that was a stretch but also wait I'm confused was it supposed to give me problems for sending a 5 letter word lol

[zealous berry]

Would the results of a real analysis course on metric spaces be covered in a general topology book?

I assume yes right

[zealous berry]

Yeah it's a poor April fools joke. You can opt out of it by adorning a poop colored role in the channels & roles section

[cosmic mirage]

I assume many of the results in analysis are corollaries of generalizations in tolology

bruh the five letter thingamabob worked now

[zealous berry]

Maybe they got rid of it

[cosmic mirage]

but also topology has little to say abou't some stuffs that's more metric spaces specific

[warped helm]

uhhh not quite

there are parts where you do have to do actual analysis

[cosmic mirage]

well yes but theres many things that follow from topology

like im'age of compact is compact is a staple of real analysis and trivial in topology

but you should stil'l see the analysis proofs it is good for you

[cosmic mirage]

ah sorr'y I see what happened. I originally said "thin'k" as in this is what I am currently thinking but then changed it to assume bc of the word thin'g lmao. I realize it seem's I mean't assume without actually knowing topology LOL

okay I'm getting a poop role this is horrivle

[zealous berry]

do ,iam not very april

💩

[cosmic mirage]

I am now poop

okay anyways yeah there are some interesting metric space phenomena topology doesn't really see

[warped helm]

i think this is a great example of when topology should be used

[cosmic mirage]

but you should definitely see things first in metric spaces

[warped helm]

this book proves this theorem in the most horrible way

[cosmic mirage]

or else topology just looks like nonsense

[warped helm]

like this is just awful

[cosmic mirage]

also things like completeness are just invisible to topology

which is more or less the only reason to care about the real numbers, lol

[cosmic mirage]

well it is nonsense but it looks like nonsense too if you haven't seen analysis throughly

[zealous berry]

what about the idea that R is a perfect set?

is that not a topology thing

[cosmic mirage]

it is but what does this have to do with completeness

like completeness means cauchy sequences converge but a topological space has no notion of a cauchy sequence

[zealous berry]

R is more interesting than being complete, it contains all its limit points and contains only limit points"" is what i meant

[zealous berry]

i forgot this tho, cuz topological spaces dont necessarily have a metric

[cosmic mirage]

yeah that's the whole reason to like them bc metrics suck lmao

[zealous berry]

yeah the idea of closedness/openness/limit points/isolated points is thrown out the window without a metric

[queen prism]

blocked

[warped helm]

no

[queen prism]

ok fixed

[cosmic mirage]

wdym? topology still defined closed and open sets and limit poins and isolated ooints

[queen prism]

I've heard about like one application of perfect sets and I don't even remember what it is

[cosmic mirage]

idk how far away two random points in CP^67 × S^420 / some messed up equivalence relation are

[zealous berry]

im only familiar with those concepts as an assessment of how close elements are to eachother (e.g., a set is open if every element has infinitely many elements close to it roughly speaking)

wouldnt that require a metric

[warped helm]

no

also not really true about open sets in general topological spaces

[zealous berry]

ah does topology just define open sets more generally

[warped helm]

an open set in a topological space is just a member of the topology

a topology is very abtract, it's a collection of sets that satisfies 3 properties

[cosmic mirage]

I think in analysis one gets taught that metrics measure how far away two points are. And then in topology one is supposed to either believe that goes away or that you should think of inclusion of two points in the same open set as meaning they're close, in which case you're wondering how that makes any sense at all

[warped helm]

namely that:

1. empty set and X are in the topology
2. arbitrary unions of open sets are open
3. finite intersections of closed sets are closed

[cosmic mirage]

I would argue saying a metric measures distance is already nonsensical bc the metric is arbitrary

[zealous berry]

arbitrary unions of open sets are open
finite intersections of closed sets are closed
ah ok these were corollaries for the definition i learned in my analysis book

[warped helm]

x is a limit point of A if every open set containing x (neighborhood) intersects A

[zealous berry]

yep

[cosmic mirage]

yeah exactly topology just defines it to be that and it is a theorem of real analysis that open sets in metric spaces are open sets in the underlying topological space

[warped helm]

isolated if there exists a neighborhood that doesnt intersect A, etc...

[cosmic mirage]

like if I replace a metric d with d/(1+d) this is still a metric and has the same topology but these games make any physical interpretation of a metric impossible

[zealous berry]

over R, perfect implies closed implies complete, which is what i was getting at here. but i'd assume this isn't true in topology due to the general nature of these new definitions

[cosmic mirage]

well complete isn't a thing in topology

[zealous berry]

yeah makes sense

[cosmic mirage]

but perfect does imply closed... doesn't perfect require closed as a hypothesis? I'm forgetting

[warped helm]

topological spaces aren't apriori equipped to handle the notion of cauchy sequences

[zealous berry]

yeah a set is perfect if it's closed and contains only limit points (in my analysis book)

[warped helm]

you can steer further in the topology-ish direction and talk about uniform spaces

and cauchy nets

[quartz horizon]

Is there0 a name for the opposite of a nowhere-dense0 subset? One whose0 interior is dense0, I think0

Wait what happened to my green0colour

[tender halo]

not to my knowledge

[quartz horizon]

Oh hm I think0

Nowhere-dense0 subsets are equivalently subsets of closed sets with empty0 interior

So, their0 complements should be supersets of dense0 open sets?

[tender halo]

yea

[quartz horizon]

Interesting

[tender halo]

everywhere dense0 is the same as dense0, sadly0

[quartz horizon]

Yeah I realised

I know that sometimes open dense0 subsets reflect a “generic” thing0?

[tender halo]

alternatively nowhere dense9 are sets whose1 closures1 are co-dense1 (a clever way to confuse anyone who is listening)

[quartz horizon]

codense has a different meaning in category theory

[quartz horizon]

ok i think0 i have a better understanding of baire0 spaces

does anyone have any recommendations for sources that cover0 the proofs of BCT? should i just use the wikipedia links0?

[alpine nest]

I propose we rename them to "bear spaces" for word-length reasons

In my lectures I mostly followed the approach in Munkres, I believe.

I like that it's basically the same argument for complete metric spaces and for LCH spaces, just the particularities of why the decreasing family has nonempty intersection differ slightly.

(The 'following lemming' in question is this:)

(also, Munkres's assumptions are "compact Hausdorff or complete metric", but the argument only uses properties that hold in locally compact regular or completely pseudo-metrizable spaces as well)

[inner valley]

what are your thoughts on this ?

omg we can only communicate in non five lettered-wordles

[inner valley]

month

ty !

[tender halo]

engelking proves both at the same time smile1

i like cech-complete spaces its fun

[quartz horizon]

cech-complete?

[tender halo]

a (tychonoff) space1 is cech complete if it is a G_\delta1 in its stone1-cech compacticifcation

(or, equivalently, in any compactitication, or in all compactifications)

or, equivalently

[quartz horizon]

@.@

oh wait hang on

[tender halo]

locally compact spaces are cech complete because they have an alexandroff compactification in which1 they are trivially G_\delta1

[quartz horizon]

is this at all related?

this is from schechter's book

[tender halo]

nyo thats1 just completely metrizable

[tender halo]

tough1 question

for the same reason BCT is about1 countable unions i would1 say

but thats1 a non answer1

[alpine nest]

I'm not sure if it's related, but as a side remark: a subspace of a completely metrizable is itself completely metrizable if and only if it's G_delta

(also another formulation of BCT is "in a Bair espace, intersection of countably many dennse G_deltas is also a dennse G_delta)

Basically what I'm saying, the concept of G_delta shows up a lot

[alpine nest]

As a fun corollary, the standard topology on the set of irrational numbers is completely metrizable

(but on the set of rational numbers it's not)

[gaunt linden]

(Exercise: construct such a metric explicitly).

[opaque scroll]

Hmmm, not sure this works:
||Define d(r, s) to be 1/n for the smallest n with a/n an element of [r - (s-r) , s + (s-r)] (for s>r)||.

||Now d(r, s) >= |r-s|, so if a sequence is not Cauchy in R it's not Cauchy with this metric either||

||Say a sequence is Cauchy in R. Then it converges to some real number. Let's first consider when it converges to a rational a/n.||

||Then for any r in the sequence there will be a later s more than halfway closer to a/n, so d(r, s) >= 1/n. So the sequence is not Cauchy, which fits with it not converging||.

||Now if the sequence converges to an irrational, then for any 1/n we can pick epsilon less than a third of the distance to nearest a/m with m<n. Then the sequence eventually gets within that epsilon in the usual metric, and then d(r, s) < 1/n so the sequence is Cauchy in the new metric and converges to that irrational.||

[gaunt linden]

Hmm. I don't think that's actually a metric.
Let e be a sufficiently tiny irrational and consider the three points a=0.01+e, b=0.02+e, c=0.03+e.
Then d(a,c) = 1, but d(a,b) and d(b,c) are both <= 1/25. So the triangle inequality fails.

(For full disclosure, I don't have a solution ready myself).

[opaque scroll]

Yeah, figures.

I feel like it might be possible to modify this to work, but don't have the energy to think more about it now

[gaunt linden]

For what it's worth, here's an idea I'm trying to make work: Let $\bR^$ be $\bR\cup{\infty}$ and define $$f:\bR^\to\bR^*: f(x) = x - 1/x$$ (with "Riemann-sphere arithmetic"). For $x\in\bR$ consider the sequence $$\tilde{x} = (x,f(x),f(f(x)),\ldots,f^n(x),\ldots).$$ Then the set $$A={x\in\bR\mid \tilde{x}\text{ is eventually }\infty}$$ is countable, unbounded, and dense, and is therefore order isomorphic to $\bQ$. Thus $\bR\setminus A$ is homeomorphic to $\bR\setminus\bQ$ by standard techniques (though it takes a bit of violence to force that into the word "explicitly"). Now, I think a metric on $\bR\setminus A$ is given by
$$d(x,y) = \sum_{n\ge 0} \max(|\tilde{x}_n-\tilde{y}_n|, 2^{-n})$$
and my hunch is this can be proved to induce the subspace topology on $\bR\setminus A$ and be complete.

[gentle osprey]

Troposphere

[gaunt linden]

(And yes, that is super handwavy and full of inaccuracies).

[frank burrow]

Rly real analysis vibes

[cerulean oriole]

Is the projection X ⨯ Y → Y open for any topological spaces X and Y?

[gritty widget]

A basic open set of XxY is of the form UxV where V is open in Y, so the image of any basic open set is open

extend this to all open sets

[cerulean oriole]

That checks out. Thanks!

[gritty widget]

No problem!

[cerulean oriole]

Let f: X → Y be a continuous function. Show that f^{-1}(int(A)) = int(f^{-1}(A)) for all A ⊆ Y iff cl(f^{-1}(A)) = f^{-1}(cl(A)) for all A ⊆ Y iff f is open.

What's up with the "symmetry breaking" here? If we look at images then f(cl(B)) = cl(f(B)) for all B ⊆ X iff f is closed, while one direction of int(f(B)) = f(int(B)) for all B ⊆ X doesn't hold at all (in all other cases, the left-in-right inclusion held for any continuous function f) and the other holds if f is open. Which is also pretty asymmetric, actually.

[cerulean oriole]

OK, I think the resolution is that open and closed are really not self-dual (we didn't dualise image to the left adjoint ot preimage, and if we do open will dualise to itself and likewise for closed).

Note that on all subsets, preimage has both adjoints. By composing with inclusion-interior and closure-inclusion adjunctions, preimage has a left adjoint image-closure for closed sets and a right adjoint coimage-interior for open sets.

[cerulean oriole]

These two conditions are equivalent because preimage preserves complements. They say that the "unexpected" adjunctions exist: left adjoint (image) for open sets and right adjoint (coimage) for closed sets.

[cerulean oriole]

OTOH, the first condition here (map being closed) says that the expected adjoints work without using an additional closure/interior, which is rather different.

Another condition we could impose is that the "unexpected" adjoints exist without requiring them to circumvent the adjunction to inclusion.

Is there a continuous map f: X → Y such that for every open set of X, its image has a unique smallest open neighbourhood, but is not always open?

[gritty widget]

There's lots of trivial examples of these with things like indiscrete topologies or excluded point topologies(let X be a space and Y a set, let F:X to Y be constant, with F(x)=y. Define a subset of Y to be open iff it does not contain y.)

[gaunt linden]

Ah, this was really excessively complex. Instead, take the following metric on $\bR\setminus\bQ$:
$$d'(x,y) = |x-y| + \sum_{n=1}^\infty \max\Bigl(2^{-n}, |\cot(n\pi x)-\cot(n\pi y)|\Bigr)$$
Each term of this separately satisfies the triangle inequality, so it's a metric.

To see that this is equivalent to the standard metric, we show that each ordinary open ball in $\bR\setminus\bQ$ contains a $d'$-ball with the same center, and vice versa. The first direction is immediate since $d'(x,y)\ge|x-y|$. For "vice versa", consider a $d'$-ball with center $x$ and radius $\varepsilon$. Split the metric into two parts, where one has finitely many terms, enough that the infinite tail can never exceed $\varepsilon/2$ due to the $2^{-n}$ bound. Then the set of points where the finite part sums to at most $\varepsilon/2$ is open in the usual topology (because $x$ is irrational so $\cot(n\pi x)$ is always finite).

Finally, a sequence that converges to $p/q$ in $\bR$ cannot be $d'$-Cauchy, due to the term with $\cot(q/pi x)$.

[gentle osprey]

Troposphere

[cerulean oriole]

Hmm, this second example is non-trivial enough to convince me that it's more general than being open.

[cerulean oriole]

Is it true that an arbitrary directed family of non-empty open sets has non-empty intersection in a topological space which is compact and has a basis of compact open subsets?

[kind egret]

Which is by the way homeomorphic to $\omega^{\omega}$ for the product topology !

[gentle osprey]

La Chouette Aveugle

[kind egret]

in general, if it is open, then it is not compact (as long as the open set is not the space itself)

(or empty)

In every metric space, it is the case for example

[gaunt linden]

Consider [0,1] U [2,3] with the subset [0,1].

[kind egret]

But if you have compact basis of neighborhood, then, as long as the open sets are not empty(and decreasing), the intersection remains non empty

[kind egret]

Oh yeah, I should have said not clopen

If it is compact then, in a hausdorff space, it is closed, so for it to be open it has to be clopen

Thank you

[gaunt linden]

Ah right, via continued fractions.
And that gives a simpler (ultra)metrization too: d(a,b) = the reciprocal of the length of the common prefix of the continued fraction expansions of a and b.

[cerulean oriole]

Indeed, I am considering spaces which are not Hausdorff (the underlying topological spaces of (quasicompact) schemes).

[kind egret]

Haha yes ! But there are other ways to show they are homeomorphic, with the Lusin schemes for example

[kind egret]

Ok, just in case, what do you mean by "directed family" ?

[cerulean oriole]

For any U, V in the family, there is W in the family with W ⊆ U ∩ V.

[kind egret]

I think it is not empty

[alpine nest]

Yeah, it's wild

[kind egret]

Is there any other hyptohesis about the space or not ?

'cause if the space you are studying is quasi-compact then there might be something we can do with ultrafilters

[limber wyvern]

Hi! why when talking about topologic spaces with a metric, it is said that every open x, it has a open ball of radius r, why is this?

[quartz horizon]

Have you met the definition of the topology induced by a metric?

[limber wyvern]

This is why I'm asking it

I mean

Why a topology with a metric is a topology that has open balls?

I don't know if I'm explaining myself

So far I know that every metric space induces a topological space

[quartz horizon]

Ah, this is to do with the characterisation of continuous functions between metric spaces in terms of open subsets

Let $M$ and $N$ be metric spaces, and $f : M \to N$ a function. Then $f$ is metric-continuous iff $f$ is topology-continuous

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[limber wyvern]

I'm sorry but I haven't reached that level of topology yet

I don't know what metric continous is

I can imagine

but I don't formally know

[quartz horizon]

ah, metric-continuous is the epsilon-delta definition of continuity

topology-continuous is the "preimage of open set is open" definition

[limber wyvern]

I see

but I still don't know why it has to have open balls to induce a topologic space

[quartz horizon]

the point is that, if you use open balls, then the topology-continuous definition agrees with the metric-continuous one

you could define a topology from a metric in another way, of course

[cerulean oriole]

It's sober. I believe that that's it.

[quartz horizon]

but then the notion of continuity you get might be different to the one you started with

i would recommend showing that:

• if f is epsilon-delta continuous, then the preimage of every open set is open
• if the preimage of every open set is open, then f is epsilon-delta continuous

[limber wyvern]

I didn't see continuity in topological spaces

😅

this is my first chapter in topology

[quasi forum]

in a metric space the topology's base is given by open balls centered at any point with any radius, and the usual condition for open set follows by the definition of a base

unless specified otherwise this is always the topology for a metric space

[limber wyvern]

mmm

but my question is

Why it has to be open balls?

why not just simply points with no open balls?

why that definition?

[quartz horizon]

it will be clear once you see continuity in topological spaces

like that is the motivation for the open ball def

[warped helm]

its modeled after R and R^n

these are the primordial examples of metric spaces

[quartz horizon]

ok i want to understand pseudometric spaces a little better

so, if $X$ is a set, then a function $d : X \times X \to [0, \infty)$ is a pseudometric if it is symmetric and satisfies the triangle inequality? but we no longer require $d(x, y)= 0 \implies x= y$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

i assume the definitions of convergent sequence and cauchy sequence are the same, it's just that you might not have uniqueness of limits? like your space isn't guaranteed to be hausdorff

where i assume the topology is still generated by "open balls"

[quasi forum]

more or less

[quartz horizon]

and a complete pseudometric space is one where every cauchy sequence is convergent?

[quasi forum]

basically all the terminology carries over as usual from my understanding

[crisp lintel]

yeah it's more or less the same

[quasi forum]

the most common example that you get is seminormed spaces

[crisp lintel]

more useful than pseudometrics are topologies induced by a family of pseudometrics

since those can actually be Hausdorff

[quartz horizon]

gauge spaces, then?

[crisp lintel]

though you lose first countability in general so things get a bit more delicate

yeah I guess so, haven't heard that term before

they show up a lot in functional analysis since many natural function spaces are not described by a single norm but by a family of seminorms

[quartz horizon]

so hm

if i have a family of pseudometrics, what is meant by the topology induced by this family?

[quartz horizon]

and, could you give some examples?

[crisp lintel]

you take the open balls as a subbase

[quartz horizon]

oh, open in any pseudometric you mean?

in the family

[crisp lintel]

Yeah, so the base would be finite intersections of open balls

There's a lot of examples let me give a few

[quartz horizon]

ah, ok

hm

[quasi forum]

arent frechet spaces the most common example of this

or am i misremembering

[quartz horizon]

so if you have a single pseudometric, these form a base right? not just a subbase

but if you have multiple, i can see why the intersection of an open ball in pseudometric 1 with another in pseudometric 2 might not be a union of open balls

maybe

[crisp lintel]

1. the space of all sequences, where you have a seminorm $p_n(x)= |x_n|$ for each natural number N. This gives the product topology

2. the set of continuous functions on a locally compact space. For each compact set, you have a seminorm $p_K(f)=\sup_{x\in K}|f(x)|$. This gives the topology of uniform convergence on compact sets. This is particularly useful for spaces of holomorphic functions, where local uniform convergence is the correct notation

3. set of smooth functions from R to R. For each compact set K and $k\geq $0, define $p_K(f)=\sup_{x\in K}|f^{(k)}(x)|$. This is similar to the previous example but now you need local uniform convergence of all derivatives.

4. If E is any normed space, then the weak topology on E is described by seminorms $p_f(x)=|f(x)|$ for $f\in E^*$

[gentle osprey]

Blake

[crisp lintel]

(a seminorm induces a pseudometric in the obvious way)

[crisp lintel]

Frechet space is just a locally convex space described by a countable family of seminorms that is complete so yeah

[quasi forum]

yeah ok

theres another definition i know of but i forgot about that one lol

[crisp lintel]

It's equivalent to having a complete translation invariant metric and being locally convex

[quasi forum]

yeah

thats the one im familiar with

[crisp lintel]

but most frechet spaces are more naturally described by seminorms than metrics

[quasi forum]

makes sense

because if you have a countable family of seminorms then you can get the metric pretty easily

[crisp lintel]

yeah

but it's a bit of a not very natural metric

[quasi forum]

yeah its not super pretty

[quartz horizon]

oh those are very interesting examples

[quasi forum]

i think it makes sense personally but fair enough

its not that nice to work with

[gaunt linden]

Well, there are simply points. The open balls are not themselves elements of the space. They're just helper concepts that participate in the description of which topology it is we're talking about. But once that topology has been made, the open balls are not different from any other open sets of the topology.

[quartz horizon]

i see, so a countable family of seminorms can be "compressed" into a single one?

[crisp lintel]

more examples of locally convex spaces: the schwarz space (for any pair of natural numbers you get a seminorm $p_{n,m}(f)=\sup_{x\in\mathbb{R}}|x^nf^{(m)}(x)|$)

[gentle osprey]

Blake

[crisp lintel]

Not into a single seminorm, but into a metric

a finite family of seminorms can be combined into a single seminorm

[quartz horizon]

ah ok

[crisp lintel]

(just add them)

[quasi forum]

in general these spaces are not normable

[quartz horizon]

right right

[crisp lintel]

you can see this with spaces of differentiable functions, the space C^k for k<\infty has a natural norm that incorporates all derivatives, but C^\infty does not

[quasi forum]

they dont obey homogeneity, so balls dont scale in the way you might expect

very roughly speaking

[quartz horizon]

ahhh right

[crisp lintel]

in general these spaces are useful when you have an infinite number of features you want to incorporate into your topology, but none individually give you a norm

[quartz horizon]

fascinating, i see

i was just trying to learn about these for BCT but it seems they're useful in lots of analysis

[quasi forum]

oh yeah if youre doing functional analysis it shows up plenty

[quartz horizon]

specifically about how every complete pseudometric space is a baire space

[crisp lintel]

They're called locally convex spaces for reasons (it turns out the condition is equivalent to having a neighborhood base of 0 conssiting of absolutely convex open sets)

[quasi forum]

iirc the condition is that a TVS is normable iff the origin is contained in a bounded convex open set

[crisp lintel]

something like that yeah

[quartz horizon]

ah because you can use this to define the norm by scaling?

[crisp lintel]

every absolutely convex absorbing set gives you a seminorm via the Minkowski functional

[quasi forum]

it has something to do with minkowski functional if i remember how the proof from rudin goes

[crisp lintel]

so there's a correspondence between the geometry and seminorms

It's very hard to remember all of the different conditions for tvs's ngl but that's the basic idea

[quasi forum]

yeah theres a whole laundry list of results lol

the point is that any norm should have to respect those two properties in a "reasonable" way because of homogeneity

otherwise your open sets might scale in a way that look more "spikey" than nice convex sets

[quartz horizon]

oh that's a neat perspective

so you can think of the space of cauchy sequences of a metric space as a pseudometric space?

and then the completion is precisely the standard way to convert a pseudometric space into a metric space

[crisp lintel]

This also comes up with the L^p spaces

[quasi forum]

yeah in fact L^p is the de facto example

[quartz horizon]

yeah i was thinking

[quasi forum]

f = 0 lebesgue a.e. iff L^1 norm of f is zero

[crisp lintel]

and it comes up all the time when working with hilbert spaces, you often take a semi-inner product and form the quotient + completion to get a Hilbert space

[quasi forum]

so you modulo out by f ~ g iff f = g lebesgue a.e. to get a banach space

[quartz horizon]

interesting

[quasi forum]

anytime people talk about L^p this identification is implicitly present

[quartz horizon]

so pseudometric spaces help separate out the steps of completion, rather than doing it in one go?

[quasi forum]

this is why it is useless to talk about pointwise definitions of such functions as well

unless you have some condition like continuity that forces a unique representative from the equivalence class

[crisp lintel]

yeah that's one way they help

just added flexibility

[quartz horizon]

that's quite cool

[crisp lintel]

the main example for this kind of thing for Hilbert spaces is the GNS construction (and GNS-type constructions, which show up everywhere in the areas of math I'm interested in)

The idea is that if you have a *-algebra A and a linear functional $f:A\to\mathbb{C}$ that is positive in the sense that $f(a^*a)\geq 0$ for all $a\in A$, then you get a semi-inner product $(a,b)\mapsto f(b^*a)$

[gentle osprey]

Blake

[crisp lintel]

and from this you can form a hilbert space H, and a representation of the algebra on H

[quartz horizon]

oh ok

[crisp lintel]

(it is this construction that is used to show that every C*-algebra is isomorphic to a subalgebra of the bounded operators on some Hilbert space, you form this construction for all (normalized) positive functionals and take a direct sum)

[vital shale]

Ive that the speed of a curve γ at x on a metric space is the limit limε→0 d(γ(t+ε),γ(t))/|ε|=v(t). How can i show that the length of the curve on [a, b] is the integral of v(t) if its continuous?

[opaque scroll]

What do you mean by length exactly here? Just the difference between the star and end point?

If so that's the fundamental theorem of analysis

I guess there's some problems with the speed always being positive, so I guess that's not what you mean by length...(?)

[vital shale]

No, the sup over partitions

In a metric space

For a curve γ:[α, β]→Χ

[opaque scroll]

Isn't that basically the definition of the Riemann integral

I guess you want the intermediate value theorem to say that the derivative equals the rise over run

[vital shale]

Well yes but ive trouble cause in R we used MVT and know we are on an arbitrary metric space

[opaque scroll]

But you're not really though.

Like t |-> d(gamma(a), gamma(t)) is just a function [a, b] -> R

[vital shale]

Yes but i dont know smth about differentiability

[opaque scroll]

What don't you know?

[vital shale]

Can u pls explain yrself?

[opaque scroll]

You said "i don't know smth about differentiability"

I don't know what that means / what you don't know

[vital shale]

Ah i see. I mean, to use riemann sums and somehow show some derivative i need differentiability

[opaque scroll]

So what you want to show is that
sup of partitions is
the integral of v(t).

The Riemman integral of v(t) is also a sup over partitions, so the only question is whether
v(t)epsilon is a good approximation for d(gamma(t), gamma(t + epsilon))

[vital shale]

Exactly

[vital shale]

And for whole that u said i need some uniform δ

For every t in [a, b]

[opaque scroll]

Well continuous on closed interval implies uniformly continuous, so that's no problem

(comes from [a, b] being compact)

[vital shale]

Well yes for continuity of v. But i have to use also that v is limit of ε→0 d(γ(t+ε),γ(t))/|ε|

And there is where i need the uniform δ

[opaque scroll]

Well, same deal right.

For a given epsilon and delta consider the t where v(t) is well approximated. Then this gives open cover of [a, b]. Compactness boom boom.

[vital shale]

Ahh i see. Nice, thanks ill try. 🙏🏿🙏🏿

[misty parrot]

How is this?

[kind marlin]

when you write a generic e as belonging to the neighborhood around e', is e one of the n centers of the metric balls, or a generic element of E? you should indicate what e is right when you use it if that's what you mean

regardless, up to saying that if e is within M/2 of e', then e is not within M/2 of l, everything seems correct

past that it's not really clear how you concluded that there are missing points from the metric balls

you never seemed to use the fact that l is a limit point either

[warped helm]

the notation and formatting are very obtuse

if you are to use open covers, it will be easier to show that the complement is open

[kind marlin]

you can still finish your proof here since you've already done a lot of the work, you just need to construct a specific neighborhood of l and use the inequalities you've already shown

and i would just scratch that entire last line of symbols and write it in words to communicate more clearly

[misty parrot]

Well thats the definition of a limit point isnt it?

[kind marlin]

right, you should just justify that there are e's in that neighborhood because l is a limit point

you never said or implied this anywhere

[misty parrot]

Oops i didnt mean to reply to that

[kind marlin]

yes, it's just hard to follow your reasoning

like the point of a proof is to convey each logical step

some things are fine to skip but this is imo the really important part of the proof:

a) some neighborhoods of l are disjoint from the finite subcover of E

b) every neighborhood of l has another point in E

these two statements give you the contradiction, and you should just give a justification for both

i guess what i'm really saying is not just to justify them, but to present these statements as the contradiction

you don't acknowledge what the limit point is doing in your paragraph or in your symbols

[misty parrot]

Okay thanks.
Also my e's throughout the proof are defined by the preceding definition, theyre just generic elements of some stated set to prove something.

It gets redefined throughout the proof like you would if you were programming and redefining a variable. Is that a generally accepted thing to do? I figured coming up with new notation for unimportant elements would be distracting

[kind marlin]

it's fine to reuse a variable (albeit not super desirable, you should just prioritize clear communication)

when reusing a variable, especially when doing a "for all" statement, you should definitely state what the for all is referring to

your first implication is not that bad because you only need to worry about when e belongs to that e' neighborhood

the second implication gets confusing because it's not clear what e belongs to now

the first implication can be interpreted as e belonging to E, or just to the e' neighborhood, or anything really (since an implication still holds when the premise is false)

and that creates ambiguity for following statements

[misty parrot]

Oh okay I see

[misty parrot]

I find it difficult to read my own proofs to find errors or ambiguous things, since I already know what Im trying to say.

One more thing. Was it fine to leave out the part of the proof showing that there are no e's in the union of our finite subset of the open cover that are closer to l than the ones in our e' neighborhood, which follows from the stuff above that last paragraph.

Like if I was taking an analysis or topology class would I get points taken off for that?

[kind marlin]

you should justify that too yeah

it's kind of hard to explain what can be omitted and what can't, as a rule of thumb especially starting out in fundamental proof-based classes like these, you should err on explaining every step

this will also help you find logical errors

[misty parrot]

Okay fair enough yeah

Thank you