point-set-topology

[quartz horizon]

Just to check my understanding

Is a limit point a point where it makes sense to take limits and you’re guaranteed to get uniqueness of limits?

At least in metric space

[kind marlin]

a limit point p of a set X is such that every open neighborhood of p intersects X \ p

uniqueness of limits is given by any hausdorff space, and metric spaces are hausdorff, but i don't think there's anything special here wrt limit points

moreso for a limit point p of a set X, you can construct a sequence of points in X \ p that converges to p

[quartz horizon]

i think what i mean is the following

if $p$ is a non-limit point, then $\lim_{x \to p} f(x)$ can be whatever you want

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[balmy briar]

That's just because it satisfies the limit trivially right? If a is an isolated point of a set A, then lim x -> a f(x) = L for any L since lim x->a f(x) = L means for any nbhd of L there is a nbhd U of a such that f(x) in V for all x in U - {a}. Since a is an isolated point, there is a nbhd that contains no x at all so its always trivially satisfied

[quartz horizon]

yeah

[balmy briar]

yeah that works in any top space

[quartz horizon]

hm so maybe it'd be better to say

the definition of limit point is to make $\lim_{x \to p} f(x) = L$ a nontrivial condition?

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[balmy briar]

im not sure about the definition being to make lim x -> p f(x) = L nontrivial, we need p to be a limit point for it to be nontrivial yes but the definition of a limit point is just saying "points get arbitrarily close" i think

If you know nets then in topological spaces you have p is a limit point iff there exists a net converging to it

[quartz horizon]

where the elements of the net can't include p, right?

[balmy briar]

yeah

[quartz horizon]

it's $p \in \overline{(X \backslash {p})}$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[balmy briar]

i dont understand

[quartz horizon]

p is in the closure of X minus {p}

[balmy briar]

oh that's a limit point of X, i was a bit confused because i thought we were talking about a limit point of a subset A of X but yeah

[quartz horizon]

oh right sorry

i think that'd be $p \in \overline{(A \backslash {p})}$ right?

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[balmy briar]

yeah if p is a limit point then every nbhd of p intersects A - {p}, so in particular intersects the closure of (A - {p}). Conversely, if p in closure then any nbhd of p intersects A - {p} so it is a limit point.

[foggy quartz]

what are some ways to go about proving a topological space isnt C1?

[tender halo]

C1?

[foggy quartz]

First countable space

https://en.wikipedia.org/wiki/First-countable_space

[tender halo]

first time i see it called C1

uhh

well the easiest way is to find a point that doesnt have a countable base

if there is one

[foggy quartz]

yeah lol but like how do you properly proof that you need uncountable amount of sets to get a base?

[tender halo]

no, base at a point

it usually goes like "suppose there is a countable base \Tau_x at x, we can construct an open nbhd of x that does not lie inside any of the sets in \Tau_x"

[foggy quartz]

alrighty i will try that

[alpine nest]

Alternatively, show that the space doesn't have some kind of property that a first countable space should have.

Like finding a set whose sequential closure is different from topological closure

[crisp lintel]

Often you do this by doing something with sequences that doesn't work in general

like finding a function that is sequentially continuous but not continuous or what outsider said

[snow frigate]

compactness proofs are so hard

[quartz horizon]

Which ones are you trying

[snow frigate]

even trying to understand heine borel

or how the sequential definition matches with the open cover definition

(in a metric space)

it's easy to disprove compactness

it just feels like the definition is very difficult to work with

much more so than connectedness

if you want to prove something is connected, you suppose there is a clopen set and prove that it's either empty or the entire set

if you want to prove something is disconnected, you demonstrate a non-trivial clopen set

that's pretty easy

[frozen sparrow]

The open cover definition of compactness is hard to work with, yes, because you have to prove that every open cover has a finite subcover. There could be uncountably many such open covers, so for most cases, proper definition of compactness is never used. We usually use easier to prove equivalent definitions for it. This isn’t unique to compactness, really.

[quartz horizon]

Yeah it’s helpful to show

Compact <=> sequentially compact <=> complete and totally bounded

[snow frigate]

what are some easier definitions to use?

[quartz horizon]

I assume you mean the proof of heine-borel

[snow frigate]

yes

heine borel is easy to state

and intuitive

[quartz horizon]

Of compactness in metric spaces?

[snow frigate]

or in general topological spaces

or with some weaker condition like T2

[quartz horizon]

Ah, then

There’s the formulation of compactness in terms of closed sets

If you have a family of closed sets where every finite intersection is nonempty, then the infinite intersection must be nonempty

[snow frigate]

is it something about intersections?

ah yes

[quartz horizon]

There’s also compactness as an induction principle

And there’s also net-compactness which generalises sequential compactness

[foggy quartz]

Question2: Let $(X,T_x)$ and $(Y,Ty)$ be topological spaces. Let ${C\alpha}{\alpha \in A}$ be a collection of closed subsets of X such that their union is equal to $X$. Let $f: X \rightarrow Y$ be a function. Assume that $f|{C_\alpha}$ is continous for every $\alpha \in A$

[gentle osprey]

you_are_me

[foggy quartz]

I proved that f is continious if A is finite

We call a collection {C_alpha}_a(alpha in A) local finite if for every x in X there is an open neighborhood such that U intersect C_alpha for a finite amount of alpha.
prove that f is continious

iù trying to use some lemma that f ius continious if and only if or every x in X an for every open neighborhood V of f(x) there exists an open neighborhood U of x such that f(U) is a supspace of V

but imkinda not moving forward i dont rlly see anything

[warped helm]

work with the neighborhood local finiteness gives you

[warped helm]

and refer to this

[foggy quartz]

Yeah I took U an open neighborhood that fulfils the finite intersect requirement and then f|_U has to be continious due to that other thing i proved but then i get stuck because idk how to relate this to f being continious as a whole

[rancid umbra]

continuity is local, so you are done

[foggy quartz]

OMG PROPOSITION 2.1.14

oh damn

[rancid umbra]

if you know f|_U is continuous, then for an open V around f(x), its preimage under f|_U is U \cap f^{-1}(V) which is open, and then you can take unions

[foggy quartz]

this fixes everything

thanks

[rancid umbra]

just curious, what is this in your text lol

[foggy quartz]

Proposition 2.1.14 ("Continuity is local on the domain")

and then thing with an open covering and continous restrictions

[foggy quartz]

Ok one last question
So i proved that f is continious if A is finite but now they ask me to give a counterexample for if A is infinite
I can find a counterexample if A is uncountable infinite but im looking for one if it is countable infinite

I tried something with the closed intervals from [n,n+1] but that didn’t give me anything useful cuz you have that point of overlap and it doesnt work to make it discontinuous then

Well at least i fail to make it discontinuous

I was also thinking abt something with the discrete topology on Y but couldn’t find anything of that either

[gaunt linden]

Let X = Q and all the C_\alpha be singletons.

[foggy quartz]

but what topology do you define on Q?

[gaunt linden]

The usual one.

The one induced by the metric |x-y|.

(That is, the subspace topology it inherits from the usual topology on R).

[rotund prism]

So an extended metric satisfies the normal requirements to be a metric on a set, except that it can also map to infinity. And from what im reading, spaces with an extended metric are, at least for topological purposes, just as good as plain metric spaces.

Except clearly im not understanding some aspect of metric spaces or one of the related topological properties correctly.

Consider the long line. Its not metrizable. No metric exists on it, too long. Wikipedia and pi-base do both agree with my notion that the long line is locally metrizable.
So I consider that I should be able to equip it with an extended metric. Locally it agrees with the naive metric or whatever, and takes on infinity for two points separated by more than a countable number of intervals.

So like. Was that last sentence actually contradictory if i go through it formally? Cuz otherwise it seems people say that this extended metric should be equivalent to a plain metric, which cant exist for the long line.

and then, is there a different variation of metric I could reason about? A "extended metrizable" topological property that is distinct from normal metrizability?

[crisp lintel]

I think your intuition for extended metrics is not quite right

If X is an extended metric space, then you can define an equivalence relation by $x\sim y$ if $d(x,y)<\infty$. The equivalence classes are sometimes called "universes" and X breaks up as a topological disjoint union of each of the universes

in particular each universe is clopen

[gentle osprey]

Blake

[crisp lintel]

The long line doesn't really fit with this because although you have points that are "really far" from each other, it is still connected

[rotund prism]

wait, any two points with infinite distance according to an extended metric, cant be on a connected component of the topological space?

[crisp lintel]

yeah

[rotund prism]

...well dang thats unfortunate

[crisp lintel]

As you said there aren't any topological properties that separate metric spaces from extended metric spaces

Obviously there are metric properties because an extended metric takes on the value infinity, other than that I'm not sure I guess

[pallid swan]

i think the topology of metric spaces derives sorta from the "behavior near 0" in the sense that metric spaces are also equivalent to like. [0, 1]-valued stuff

well ok i guess what i said can't fully be correct cuz there's a difference between ultrametrics and metrics even in topology

but eh idk

[snow frigate]

wait doesnt that mean if you have a collection of metric spaces you can form an extended metric space

where each space of the collection embeds isometrically into a universe

[pallid swan]

i think quasimetrics are cool, they're what happens if you also drop the symmetry requirement, then you get a partially ordered set

of "reachability"

[pallid swan]

i.e. that d(x, y) = d(y, x)

[crisp lintel]

Yes exactly, that's just the disjoint union

one of the reasons for working with extended metrics is that you get a nice disjoint union

[snow frigate]

that’s cool

then the extended metric is uniquely determined by this too so it’s quite cool

[limber wyvern]

Hi!, In the definition of topology, why is the union arbitrary but the intersection has to be finite?

[gaunt linden]

Basically because that's how open sets in R^n behave, and those are what we're generalizing with the abstract concept of a topology.

[limber wyvern]

Thanks 🙂

[gaunt linden]

If we only required finite unions of open sets to be open, then the collection of closed sets in R^n would also be a topology, so we could only hope for the general theory to extend results from R^n that still hold after exchanging "open" and "closed" everywhere -- which are not very many.

[quartz horizon]

matches the notion of "open iff neighbourhood of all your points"

[kind egret]

if we allow countable intersection, and if the topology is metrizable, then you get (almost) all the borel subsets

and usually this is too big

and as it has been said, we can see that it works like that in R^n, so we tried to generalise what was understood from the usual spaces

but it was historically very difficult to decide the axioms of a topology

[kind egret]

actually im pretty sure you get all of them

[frozen sparrow]

It is to extend notions more familiar from R. Like, for example, allowing countable intersecions gives $\bigcap_{n = 1}^\infty(a-\frac1n, b +\frac1n) = [a,b]$, so you get like countable intersections of open intervals giving closed intervals, so infinite intersections need not preserve the "openness" even in R.

[gentle osprey]

Běta (YesHead)

[frozen sparrow]

Similar thing works for infinite unions of closed intervals. So, arbitrary intersections for open and arbitrary union for closed sets are just not desirable, really. At least, for being able to define continuity and other topological notions.

[austere dirge]

i think its this

they worked backwards

started from the more intuitive definition and noticed it could be made simpler

[austere dirge]

this is correct. clearly you end up with a set closed under countable union and intersection. to show closure under complements show that closed sets are G_delta then apply de Morgan's. this implies its a sigma algebra generated by open intervals, so it's the borel sets.

[lament steppe]

Lets say I have a collection of functions f_i : A_i -> Y that are continuous, where the collection of A_i's is an infinite closed cover of X (X and Y are topological spaces.) Also, Ai are all disjoint. I know that I can have a function from f : X -> Y where each f|A_i = f_i.

Something Im trying to wrap my head around is that since X is a topological space, then X is an element of the Topology for X. However, if I take the inverse image of Y with f, then this inverse image is the union of an infinite number of closed sets which not an element of the Topology of X... so its almost like I have a contradiction here.

The motivation here was to show that the gluing lemma doesnt hold for a infinite closed cover of X... so Im not sure if what Im thinking about is the root of this, or if Im misunderstanding something that is unrelated.

NOTE: Please dont spoil the counter example Im trying that I need to show that the gluing lemma fails in this case. Just want guidance on my current thought.

[ruby delta]

if you take the inverse image of the entirety of Y, as you observed the preimage is just X, which is certainly an open set and there is no issue there

you are on the right idea of there being an infinite union of closed sets somehow not being open, but you'll have to take the preimage of something smaller

also, are you trying to prove this in general? It would be much easier to just construct a counterexample

[lament steppe]

Yeah, Im just trying to provide a counter example 🙂

[ruby delta]

ah in that case it is much easier

[lament steppe]

Rereading what I stated above, I think I see something that is indeed incorrect:

"an infinite number of closed sets which not an element of the Topology of X"

This isnt true. For sure a union of a finite number of closed sets is an element of the topology, but I cant say an element of a topology isnt included if it is a infinite union of other elements.

[ruby delta]

or, actually, here is a more useful detail that I should point out:
if f is continuous from A_i to Y, the preimage of an open set of Y is open in A_i.

[quartz horizon]

yes so what is true is

in general, an infinite union of closed sets is not in the topology

[ruby delta]

but when your cover is infinite, that does not necessarily mean that that open set in A_i passes into a honest-to-god open set in X.
and indeed, this is the point of failure of the gluing lemma

[quartz horizon]

but it can be sometimes

[ruby delta]

also welcome back pseudo

[quartz horizon]

i've been here the whole time

[ruby delta]

so real

[ruby delta]

recall the proof of why, if the cover is finite, being open in A_i implies open in X

and now what I said about the infinite intersection of open sets not being open should make a lot more sense

[quartz horizon]

also it may be helpful to prove the following formula

let $S \subset A$ and $\iota : S \to A$ be the inclusion map. then for any $B \subset A$, we have $\iota^{-1}(B) = S \cap B$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

trying to make sure that i can prove part a

C is closed being the intersection of a bunch of closed sets

if x and y are in C, then you can find x < z < y with z not in C, by considering base-3 expansions and ensuring z has some 1s in its expansion

so C has empty interior, meaning it's nowhere-dense

also it's compact since it's bounded

in R the connected subsets are intervals, so the only connected subsets of C have to be single points, meaning C is totally disconnected

[warped helm]

easier than base 3 expansions is considering the width of each interval at every iteration, imo

[ruby delta]

yep all looks good

[quartz horizon]

and again by considering base-3 expansion, you can find points y in C arbitrarily close to some x in C with x not equal to y

so C doesn't have isolated points

[quartz horizon]

hm true

that's probably a better general strategy, thanks

[midnight umbra]

yeah that's fine. cheaty solution for empty interior is that any set with interior has positive lebesgue measure

so (b) technically gives it to you for free lol

[quartz horizon]

hm true, though that wouldn't work for generalised cantor sets

[midnight umbra]

like the fat cantor set?

yeah that guy has measure 1/2 it wouldnt work there

[quasi forum]

cantor set like constructions are beautifully cursed

i saw something like it used in my complex analysis textbook to construct a jordan curve with positive area

[quartz horizon]

why did that appear in complex analysis?

[quasi forum]

they prove jordan curve theorem there

you can get a proof through algebraic topology ofc, but cauchy's theorem already gives you all the necessary tools and in some way its just saying the same thing in a different language

[quartz horizon]

hm, i see..

[quasi forum]

so the fact you can get a proof via complex analysis is not terribly surprising

i wish to avoid saying the wrong things because my specialty is not algebra, but the essence of cauchy's theorem is that holomorphic functions on a simply connected domain have primitives, which i dont think is that far off from the idea of de rham cohomology

[quartz horizon]

ah, i'm trying to learn de rham cohomology atm

[quasi forum]

the notion of winding number can also be defined purely through complex analysis

[quartz horizon]

yes though i thought that was for C^1 curves..

[ruby delta]

ah I see, so the rough sketch is that you show that the connected components are exactly the regions where the winding numbers are constant?

i have no idea how to show that you can actually integrate over an arbitrary jordan curve, though

[quasi forum]

technically speaking you can sort of try to define this notion for any closed curve, although its pretty awful (see cartan for example)

[gaunt linden]

It's not clear to me that you can -- "arbitrary Jordan curves" allows curves that are not rectifiable (such as the graph of a Weierstrass function), and I don't think there's a canonical way to select a finite parameterization of them that you can integrate over.

[ruby delta]

but doesn't the statement of the jordan curve theorem hold for arbitrary paths in R^2?

[quasi forum]

it does

[ruby delta]

how the hell are you gonna use complex analysis to prove it then

or can you only prove it under the weaker rectificable assumption

[gaunt linden]

The Jordan curve theorem doesn't speak about integration, though.

[quasi forum]

you dont need rectifiability either

the proof is not what id call enlightening though

if you want the actual textbook i am pulling it from go to marshall's complex analysis

there is a proof given in chapter 12

too many details that i dont know enough about to recite it by heart

the point is that complex analysis gives you a way to identify "badly behaved" sets through cauchy's theorem

which is exactly what AT tries to do

they both give you ways to identify holes in your domain

which turns out to be the main thing you need to actually prove JCT

[rotund prism]

this reminds me of non-lebesgue measurable sets, which are pathological enough such that their existence relies on axiom of choice. Wonder if standard complex analysis bits like cauchy's theorem are by themselves able to try to describe a closed loop that spikes around a vitali set.

[kind egret]

If you admit the negation of the choice, then every subset of R is Lebesgue-mesurable

You can read more here if you're interested in (the document is available online)

[unreal stratus]

Shouldn't it more be that "there exists a model of ZF such that every subset of R is Lebesgue measurable"

[opaque scroll]

Yeah I don't think the existence of non-measurable sets is equivalent to choice. That seems a bit strong

[unreal stratus]

Sounds like you're saying ZF and negation of choice proves every subset of R is Lebesgue measurable, which should be false

(Like make the axiom of choice hold for enough sets that this theorem goes through but make AC fail for some other sets)

[kind egret]

Yes, sorry for that lack of precision, indeed it is that it is consistent with the negation of the choice

It is terribly badly said in my message

[unreal stratus]

Np np

[slender flame]

If I recall well, it’s worse than that: it’s in the metatheory “ZFC + there exists an inaccessible cardinal” that there exists a model of ZF + “all sets are Lebesgue measurable” (Solovay model). To the best of my knowledge, it’s still an open problem to show the existence of such a model with the metatheory/tools of ZFC only. The existence of an inaccessible cardinal cannot be shown in ZFC but "is thought" (so still open problem) to likely never be refutable (though in theory it could, if someone found a model of ZFC without one).

TLDR : Assuming consistency of ZFC, if you manage to construct a non Lebesgue measurable set without Choice, you would refute the existence of an inaccessible cardinal and likely collapse the hierarchy of large cardinals. But in essence, its not known if such a set exists but it is unlikely.

[gaunt linden]

The existence of an inaccessible cardinal cannot be shown in ZFC but "is thought" (so still open problem) to likely never be refutable (though in theory it could, if someone found a model of ZFC without one).
Since the existence of an inaccessible cardinal cannot be shown in ZFC (due to Gödel-Rosser), there must be a model of ZFC that doesn't have one.

Furthermore, if I recall correctly: if M is a model with an inaccessible cardinal, then that model's V_kappa where kappa is the smallest inaccessible cardinal in M, is a model of ZFC + "there are no inaccessible cardinals".

[slender flame]

Yep indeed, i didn't double check, thanks : i think, but to recheck, the constructible universe L doesn't have one, although your construction is very elegant.

I wanted to say that : its not known if the negation (nonexistence) can be shown in ZFC so there is a slight chance that they may not even exist. One possibility to show that is my TLDR.

[zealous berry]

so true

@slow widget

oops

<@&268886789983436800>

[midnight umbra]

I genuinely have to wonder who falls for these mrbeast scams

[urban zinc]

Discord is mostly 13 year olds

Also tbh even if you know it's a scam sometimes you get morbidly curious about how the scam works... that got me once

[queen prism]

I don't think it's really the mrbeast scam that does people in, they just install malware (because of 🏴‍☠️ for example) and then their accounts get hijacked and start spamming this kind of post

[midnight umbra]

it wouldn't be done if it didn't work

I'm pretty sure the scam is designed as a token stealer, so when you try to "claim your crypto" it steals all your active logins, both to take your crypto from you and then spam it with your discord account

[queen prism]

I mean yea if they make enough money from it then there's no reason not to do it, I just don't think that's how people usually get infected

[warm wasp]

I'm pretty sure Shelah proved that if there's a model of Dependent-Choice + All sets are measurable, then there's a model of ZFC with an inaccessible. So maybe we don't know if just "all sets are measurables" is stronger than ZFC (I'm not sure), adding DC is equiconsistent with an inaccessible so that can't be proven in ZFC

I don't have a reference in schlof though

[rotund prism]

so just to be sure, is any banach space locally simply connected and contractible? I know theyre path connected, baire, homogenous, and some others

oh and that they are locally compact IFF they are finite dimensional

[iron bolt]

every normed space is locally convex and hence locally contractible, so in particular locally simply connected

I'm not sure about general topological vector spaces

[crisp lintel]

should follow just from continuity of scalar multiplication

although actually I guess maybe not your open sets might be too big

[iron bolt]

right, I just realised. in a general TVS it still follows easily that every neighbourhood of the origin contains a balanced neigbourhood, and those are while not convex at least star-shaped

[crisp lintel]

yeah every neighborhood of 0 contains a balanced neighborhood then you can just define a deformation retract onto 0 in the obvious way

[prime elbow]

To say something is completion of metric space X it is enough to show that it is complete metric space and X is dense in it, right?

[digital sun]

Greetings. I'm trying to construct a counterexample for the statement $A \subseteq X,$ $X$ a topological space, $Int(A)$ connected, $Bd(A) $connected but $A$ disconnected. But I'm struggling. Is there any you know? Or is the statement true?

[gentle osprey]

Gol D Roger

[prime elbow]

What if you take A = Q in X = R ?

[digital sun]

So, $Int ( \mathbb{Q}) = \emptyset$?

[gentle osprey]

Gol D Roger

[prime elbow]

Yes

[digital sun]

Which is connected vacuously?

[prime elbow]

Yes

[digital sun]

Ok, thanks

🙂

[digital sun]

I would add isometry

That there exists an isometric embedding

For example if you change the usual metric for another one

[prime elbow]

I see

Thank you

[kind egret]

It is

Two different completions are always isometric, if the metric on X is fixed

[gaunt linden]

There are different conventions for whether the empty set is considered to be connected -- often it isn't. It would be a more robust solution to take, for example, A = Q union [42, infty).

[cloud kindle]

I hate to be the guy sharing an nlab article, but there’s genuinely very good discussion about this in here

https://ncatlab.org/nlab/show/empty+space

naively it does seem like it should be, but In practice it’s more convenient to exclude it, and this is what most people do in my experience (on the algebraic side of things)

[quartz horizon]

Nlab is inevitable

https://ncatlab.org/nlab/show/too+simple+to+be+simple

[cloud kindle]

It is, but the children should be shielded and protected for as long as possible

But yeah both of these articles are great

[jagged ridge]

Nlab is really annoying because 99% of its articles are completely garbage and then there’s the 1% that’s like genuinely insightful

So you can’t just disregard the entire website

The “too simple to be simple” article is in the 1%

[pallid swan]

truth supernova

i also love concept with an attitude

it feels like nlab has a lot of folklore articles

[jagged ridge]

Wow I’ve just read this for the first time and it is quite good

[pallid swan]

we should have a math folklore wiki

and then say it's objectively better than nlab

[cloud kindle]

Nlab is actually really good most people just don’t need it. I mostly just joke because the “basic” articles are ridiculous and it’s slightly silly/annoying when stuff like that gets sent to people who obviously aren’t ready for it

But I tend to find if you have an actual reason to be on there it’s very good. And the too simple to be simple stuff is just really really good for everyone

[jagged ridge]

I don’t quite think this is true. They have articles written on subjects that I am perfectly qualified in, understand everything they say about, and yet I still think those articles are bad. This is often due to a mixture of omitting important pieces, factual errors, or copy-pasted content between related articles

[pallid swan]

i do think some technical nlab articles are really nice, especially on niche subjects that other websites wouldn't cover

[jagged ridge]

Now admittedly I am an algebraic number theorist, which is not really the original intention of nlab, and many of the articles on that subject are sort of “unplanned bloat”. But even when I am reading an article on a category-theoretic concept I don’t think it’s very good

[pallid swan]

also it's a fine reference ime

[opaque scroll]

There's definitely a lot of bad articles on there. But 99% being garbage is perhaps an overstatement

[jagged ridge]

Perhaps

[rotund prism]

as a not professional in the slightest, i stinking love nlab

[jagged ridge]

Also their typesetting is a crime

This is far from the most important criticism but it annoys me

[cloud kindle]

This is a very different thing to 99% of the articles are garbage though

[cloud kindle]

This I agree with, and is apparently somewhat controversial. The category theory site should have better ways of displaying diagrams

[iron bolt]

heh. I actually have a whole collection of nlab screenshots like that

those three might be my favourites

[rotund prism]

wow i bet you'd be friends with the guy in the SOME server that i stole that image from

oh wait that wasnt actually specifically you

that was a different user and we simply both thought it was funny

[iron bolt]

lol. I've posted it a couple of times there, might well be someone who found it funny and reposted it

not that there's anything wrong with that of course, it's always nice to see good memes spread

[queen prism]

channel quality immediately gains a minus sign

[unreal stratus]

Idk lol like most homotopy theory people I talk to dunk on it

Including me lol

It was most useful for me when I started doing a lil cat theory ig

[cloud kindle]

Well ig this is just me lol

[quartz horizon]

Do you have some examples?

I occasionally edit nlab articles myself and would definitely be up for improving it

[jagged ridge]

i usually edit to fix the mistakes if they're pretty simple but this article https://ncatlab.org/nlab/show/ring+of+integers probably needs a complete rewrite

[quartz horizon]

Ah ok, I don’t know enough number theory to rewrite that

[jagged ridge]

it gives an excessively general definition of ring of integers that

1. no one has ever used
2. is wrong in significant amounts of cases, including ones mentioned in the article

oh and then after giving the wrong definition they mindlessly copy the function field analogy table with no additional context

[crisp lintel]

I'm confused why is the definition wrong

[jagged ridge]

it doesnt work for local fields

[crisp lintel]

What's the correct one

I'm not really an algebra person but I've always seen algebraic integer to mean root of a polynomial with coefficients in Z

[jagged ridge]

the definitzion they give in section 4, or to put it more in line with the language used in section 1, the integral closure of Z_p in the local field

[crisp lintel]

ah ok

so you'd want to take an algebraic element not over Z but over Z_p in that case

[jagged ridge]

whenever this definition is used, its probably restricted to algebraic extensions of Q, to avoid the conflict with language for local rings

but whoever wrote the nlab article tried to make it maximaly general, and in the process didnt realize it was causing problems

[crisp lintel]

Yeah that does seem to happen sometimes

[jagged ridge]

also, if I were to say "ring of integers in a global function field", number theorists would probably understand it to mean something different than whats in the nlab article

though many people would probably completely avoid that language

[opaque scroll]

Is there even a nice unifying definition for this?

Like maybe take the (topological) closure of the integral closure or something?

[jagged ridge]

Is there even a nice unifying definition for this?
imo probably not

topological closure of the integral closure of the integers works for characteristic 0, but gives the wrong answer for function fields

though maybe we dont care about function fields because #point-set-topology message

[opaque scroll]

So what's the right answer for function fields, and what do you get?

[jagged ridge]

Given a finite extension K/F_q(t), the "ring of integers" of K is the integral closure of F_q[t] in K

so replace Z with F_q[t]

[opaque scroll]

Hmm, maybe the right idea is to fix some PID that you do ring of integers relative to...

[jagged ridge]

but then at this point you might as well just write this definition:
"If (A,K) is a pair of a PID and its field of fractions, and L/K is a finite extension, then the ring of A-integers in L is the integral closure of A in L"

and then when you don't say A, there is a "standard" choice the reader is supposed to assume, whether that be Z, Z_p, or F_q[t]

this is what people actually use

I think whoever wrote the definition on nlab probably didnt like how ad-hoc this one seems. But there's a reason it is this way

[opaque scroll]

Seems like one can almost fix the article by adding a remark below the definition

But it examplafies the issue was your point I guess

[limber wyvern]

Hi!, I'm wondering what is the difference between the definition of interior point and a neighborhood, because, by formal definition they look the same to me.

Also, my book mentions that there is a relationship between the concept of interior and the concept of open, I don't clearly see this relationship, Could anyone clarify?

thanks in advance

Feel free to ping me

[prime elbow]

okay, so what is your definition of interior point, open set, and neighbourhoof?

[limber wyvern]

Second image is the definition of neighbourhood, in which "tal que" is the spanish traduction to "such that", and well, I don't have very clear if it's talking about open sets or just "opens", like the elements of the topology

I think it's talking about opens which definition is, it's open if G \in T

[quartz horizon]

The relationship is that x is an interior point of N iff N is a neighbourhood of x

[limber wyvern]

I always confuse between open sets and "opens", my teacher say they are not the same

[limber wyvern]

So interior point and neighbourhood are closely related

[quartz horizon]

Yes, the difference is just which one you focus on

[limber wyvern]

Okey, that's what I was thinking but I wanted validation xD

just in case

And what is the relationship between interior and open?

And, another question, does all elements of the topology have an infinite amout of neighbourhoods, or that's the case for metric spaces?

[quartz horizon]

x is an interior point of N iff there is an open subset U with x in U and U subset N

A subset U is open iff every x in U is an interior point of U

iff U is a neighbourhood of all its points

[limber wyvern]

So it's the same case as before, it depends in which one you focus?

[quartz horizon]

Yes

[limber wyvern]

mind blowing

It's amazing that you always have the answer

[quartz horizon]

That’s hasty generalisation

[quartz horizon]

Not necessarily, even in metric spaces

[limber wyvern]

Well, all the questions I've done you always answered them xDDD

[quartz horizon]

I have been learning some algtop recently so needed to brush up on gentop

[limber wyvern]

mmm I see, that may be more advanced that what I've seen so far, rn for me, in a metric space, the opens have infinite neighbourhoods

[quartz horizon]

Consider finite metric spaces though

[limber wyvern]

What is your recommendation in studying gentop? I'm struggling a lot with the exercises

[quartz horizon]

Draw lots of pictures as you’re working through Qs

[limber wyvern]

I'm spanish sorry, what's Qs?

[quartz horizon]

It’s also helpful to keep a standard list of topologies in mind

[quartz horizon]

Questions

[quartz horizon]

Both metrizable and non-metrizable examples

[limber wyvern]

So far I now know that the indiscrete topology is non metrizable

And that a topological space is metrizable if you can define open balls around x (which is an open) and they still in the topology

I don't know what galev is writing but I'm scared

[rough falcon]

Im currently trying to prove that a finite product of open maps is open.
Let $f: \prod_{i =1}^n X_i \to \prod_{i=1}^n Y_i$ and let $f_i: X_i \to Y_i$ be open maps. Then any open set $U \subset \prod_{i=1}^n X_i$ can be written as $U= \bigcup_{j \in J} \left( \prod_{i=1}^n U_{i,j} \right)$ where $\prod_{i=1}^n U_{i,j}$ are elements in the Basis $\mathcal{B}X$ for all $j \in J$. With basic set theory, we get that $f(U) = \bigcup{j \in J} \left( \prod_{i=1}^n f_i(U_{i,j}) \right)$ and $\prod_{i=1}^n f_i(U_{i,j})$ is an element in the Basis $\mathcal{B}Y$ of the product topology $\prod{i=1}^n Y_i \$.
But this proof should also work if we start with closed instead of open maps and subsets.
The problem is that $g: \mathbb{R} \to {0}$ and $f: \mathbb{R} \to \mathbb{R}$ with $f(x) = x$ are both closed maps but $f \times g$ isn't. So where did i go wrong?
I cant find any mistake in my proof above

[gentle osprey]

galev

[quartz horizon]

You can’t write a closed set as an arbitrary union like that

You can write it as a big intersection, but direct images don’t generally preserve intersections, only unions

[rough falcon]

yeah but let say we have a open set U, than we could get the closed set via X/U and now we could work with the intersection

[rough falcon]

Oh right

Im dumb

[quartz horizon]

Preimages preserve all subset operations however

[rough falcon]

Thanks!

But my proof should still be correct?

[quartz horizon]

Yes, finite products of open maps are always open

[rough falcon]

Thank you ❤️

[quartz horizon]

@limber wyvern I would also say topology has lots of equivalent ways of saying the same thing

It would do well for you to familiarise yourself with these, even make your own if that helps

For example, a subset S is dense iff the closure of S is the whole space

[limber wyvern]

It's almost confusing😅

[quartz horizon]

This turns out to be equivalence to

[limber wyvern]

I knew this

[rough falcon]

I really like how John Lee is visualizing a lot of his theorems in his book topological manifolds.

[quartz horizon]

For every nonempty open set U, S intersect U is nonempty

[quartz horizon]

I have heard good things about Lee, e.g. from @iron violet

[limber wyvern]

It's harder for me too see the same here that from the first definition

[quartz horizon]

It’s not obvious, but a good exercise to prove this

Working with these equivalent perspectives and knowing which one to choose goes a long way in topology

[rough falcon]

Yeah, he also has a lot of exercises so i often don't really know which ones i should do because i'm not in the mood to do 30 or more exercises each chapter xD

[quartz horizon]

Ah right, atm I’ve just been doing every exercise from folland

[limber wyvern]

I see

I'm sorry to ask this again but, in a topological space, the opens from the topology doesn't have always a neighbourhood?

By viewing the book recommended by galev I see this, and I see that they have a neighbourhoodness

or it's just to specify they are open sets?

[quartz horizon]

I would say this is a type error

Given a point p and a subset N, you can ask whether or not N is a neighbourhood of p

But “neighbourhood of an open set” isn’t really a sensible notion (at least not until later)

[rough falcon]

To be fair the picture is not about neighbourhood of sets

It's to visualize that a point is in Ext A iff it has a neighbourhood contained in X/A where A is any subset of X

[limber wyvern]

mmm, when I talk about opens it's not about open sets but about the elements of a topology

I don't know if it's the same

xD

[rough falcon]

And similiar for Int A and Par A

[limber wyvern]

My professor says it's not the same

[midnight umbra]

that's the same thing

literally the definition of an open set is "an element of the topology"

[rough falcon]

i think he means elements of the open set

[limber wyvern]

I mean, let me correct myself, he doesn't says it's not the same, he says that "open" is the name for the elements of the topology, but it's not like a open set

[rough falcon]

But this is the same

[midnight umbra]

I don't know what your professor is cooking, these are the same thing

[limber wyvern]

I mean, let X = {1,2} and a topology T = { \empty, X, {1} }, that {1} is what he calls "open"

[cloud kindle]

{1} is very much an open set in T

[midnight umbra]

... and he wouldn't call \empty and X open?

[cloud kindle]

Thats like, what an open set is

[limber wyvern]

yeah yeah, for sure

I just said 1 for taking an example

[midnight umbra]

ok

[limber wyvern]

so it's an open set

wow

it now makes sense xD

[midnight umbra]

at least in English there isn't a thing we call "an open," only "open sets"

[limber wyvern]

I see

So it can be treated literally like an open set

But it's not like a normal open set in R, because, how you interpret an open like {1,3,9}

[velvet salmon]

I mean I have definitely seen open used as a noun like that a lot

[midnight umbra]

really?

where

[cloud kindle]

An element of a topology is just called an open set. This is a generalisation of the thing you already knew to be open on R, this is where the terminology comes from

[velvet salmon]

Idk anywhere that talks about covers, the authors often say things like "covered by opens" or something

[cloud kindle]

I think this is quite common, ive definently heard like "A union of compact opens" or orwhatever

[midnight umbra]

i have never heard this in my life 😭

[velvet salmon]

It just gets super annoying to always write set

[midnight umbra]

i mean, what do you mean when you say this is "an open?"

[limber wyvern]

So, what do you mean is just called open set, it has something to do with the definition of open set in R?

or it's just the name

[velvet salmon]

You're conflating R as a set with R as a topological space with some topology

[limber wyvern]

It's an element of the topology

[midnight umbra]

which topology?

it's not an element of the eucliden topology

[velvet salmon]

An open set of R depends on the chosen topology, if you equip R with the discrete topology every subset is open

[limber wyvern]

I don't know, i just invented that example

[midnight umbra]

😭

[cloud kindle]

An open is is by definition any element of a topology. You maybe saw at other points sets like (0,1) in R be called open, but these are just the open sets in the usual (metric) topology on R

[midnight umbra]

so if its an element of a topology its an open set in that topology

[cloud kindle]

Theyre called open because it generalises that idea on R to other sets, and even just different topologies on R

[limber wyvern]

Wow

I see

So a point set like (0,1) it's an open for R, but it's a specific case

just for the usual topology

[cloud kindle]

(And when I say element of a topology I mean an element of T, not of the underlying space X)

[cloud kindle]

Yeah, this is the motivation for the terminology

[limber wyvern]

wow

i don't know what I would do if this discord didn't exists

[cloud kindle]

Closed sets in R look like [0,1], but they could look different in general, theyre just compliments of open sets in whatever topology you have

[limber wyvern]

I see

but

Just to wrap this up

and sorry to sound redundant, they are called open sets just because this idea comes from R, but it does not have nothing in common with being an open set for R like (0,1)

Sorry for asking this again but I've been studying topology a lot today and I have my brain in liquid state

[midnight umbra]

well, the open subsets of R in the metric sense also form a topology, so it's not that there's nothing in common

[limber wyvern]

Well, it's only common for R

[cloud kindle]

Yeah they need not have anything in common. Take for example any finite topological space, say the set X={0,1}, and put the topology {\empty,X,1} on it. Then {1} is open in this, but it doesnt really look like an interval on the real line.

The idea here is that open sets in some appropriate sense encode what it means for things to be close to each other, sort of like how they do in R, and hence why this generalises the idea. But yeah like (0,1) is an open set on R in the topological sense (with the usual topology)

[midnight umbra]

but the general definition "open set (in a topology)" is an element of said topology

[limber wyvern]

When I think about open sets I think about a set of numbers, which does not include the beginning nor the end: (0,1)

[cloud kindle]

Its not open under say the indiscrete topology

[cloud kindle]

Yeah this is the motivation, but you need to let yourself expand that picture a bit now if youre learning topology

[limber wyvern]

I'm sorry to ask more, but, what do you mean by "appropriate sense encode what it means for things to be close to each other" isn't this something exclusive to metric spaces?

I mean, the sense of close or far

[cloud kindle]

I dont mean anything too precise beyond that it is the generalisation of that idea. If you have a metric on your space that gives a natrual topology by making the open balls open sets. But of course in general for any topological space that idea might not make a lot of sense, look back on my example with {0,1}

Im just trying to motivate where these things come from. For any given topological space what it really means for things to be "close" probably isnt all that meaningful, we can just talk about open sets

[alpine nest]

Closed sets in R look like [0,1]
Yes, the Cantor set in particular

[limber wyvern]

The definition my books give for close, is:

A set is close if the complementary is open -_-

[cloud kindle]

Well I mean zoom in close enough....

[alpine nest]

... and you're still seeing the Cantor set

[cloud kindle]

This is the definition though, a set is closed if its compliment is open

[alpine nest]

To be fair, it's one of the definitions, closed sets are often defined in terms of limit points etc.

And a newcomer might not see it obvious why those are equivalent to the complement being open

[cloud kindle]

Its cantor all the way down, they just keep taking away the middle third of my set

[cloud kindle]

Yeah good point

[limber wyvern]

Just out of curiosity, in what year of your math degree did you coursed general topology?

[frozen sparrow]

Should be seeing it next semester, so in 4th year. But I have seen topological concepts in other courses such as Analysis on Rn, Metric Spaces, Geometric Topology and Calculus on Manifolds.

[cloud kindle]

But yeah I guess if theres something to take away from this its this: Your intiution about what open and closed mean in R is good to have in the back of your mind as a vauge motivation for this, but dont cling on too tightly because you will be studying spaces which actually look quite different to this. An open set is just an element of a topology, and theres many different topologies out there, and mnay of them can be quite whacky

[limber wyvern]

it's my first year in uni and I'm seeing topology

and I feel I lack a looooot of background

[frozen sparrow]

I've heard of some universities who substitutes full on Analysis for Calculus and General Topology for an Analysis course.

[limber wyvern]

So far I've just coursed Calculus I, and linear algebra xD

I mean, I've coursed statistics and physics too, but It doesn't help to study topology

[frozen sparrow]

I feel like a PS Topology course as a first year is somewhat insane. Especially if you haven't taken any Proofs class.

[limber wyvern]

yeah

its insane for me to confront any exercise

[frozen sparrow]

You're using Munkres?

[limber wyvern]

Well, I'm using a lot of books to try and understand topology, In my uni they have a custom book

But yeah, in my free time I'm with munkres

[frozen sparrow]

I see.

[limber wyvern]

Well, thank you everyone for the help and your time, you'll see me living in this channel xD

[willow stirrup]

topology without tears or munkres

for intro

[opaque scroll]

Second year for me.

[novel acorn]

We have topology of R^n in our second year but we don't have a course on general topology

[storm wadi]

second year also

[digital sun]

That's pretty much the standard case in most universities

[digital sun]

third semester, point set topology. Not fun that earilier.

I mean, nice if at least prof used one book, but he followed Viro Irodov. Not a sane experience

[digital sun]

Eh? Which uni is that?

Also, if someone can help me, in which topology scenario is $S^1 \cong [0, 2 \pi)$?

[gentle osprey]

Gol D Roger

[quartz horizon]

there is a continuous bijection $[0, 2\pi) \to S^1$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

however it doesn't have a continuous inverse

in fact this is the standard example of a continuous bijection that isn't a homeomorphism

[digital sun]

Hmm

$S^1$ as subspace?

[gentle osprey]

Gol D Roger

[digital sun]

Or as itself a space?

[quartz horizon]

either works

it's a good exercise to check if you haven't done so already that these give the same def

[digital sun]

$S^1$ as subspace and as itself a space?

[gentle osprey]

Gol D Roger

[quartz horizon]

yeah

[digital sun]

Oh well, thanks

Also, this might be interesting for you https://math.stackexchange.com/questions/4671250/making-0-2-pi-homeomorphic-to-s1-circle-in-a-plane

somehow with transport of structure they presented another topology.

I'm just so lazy to check it rn.

[quartz horizon]

oh yeah with transport of structure you can cheat

you can make $S^1 \cong S^{67}$ with that

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[digital sun]

😮

I'll think about it. For now I'm done, thanks for helping!

[unreal stratus]

Ah I like how you went for a prime number

[quartz horizon]

Wdym 1 isn’t prime

Silly potato

[limber wyvern]

Yeah, im having a pretty insane experience

Its my second quad-mester and I'm having topology without knowing nothing about analysis nor proofs

It's being pretty hard to follow

All my colleages are in the same position as me so :/

Not cool

[unreal stratus]

😭

[storm wadi]

topology before analysis is wild

[queen prism]

i think it's doable but probably not the best idea

topology without either analysis or proofs though...

[storm wadi]

everything must feel so unmotivated though. After all, what we now know as topology was first developed as a tool for analysis

[queen prism]

well historical accuracy isn't always pedagogically the best, but yea I wouldn't do it

I read a bit of topology before starting analysis and I didn't like it

though you can probably still get reasonably far

[storm wadi]

true, but in this case, at least from my experience, it is nice to have an understanding of particularly metric spaces to get a sense for why we are defining the things we are in topology

[queen prism]

yea metric spaces first would be a good idea

[digital sun]

Which uni is that?

[digital sun]

I would say most topological invariants are not that diff to digest

But getting to infinite products without knowing what a sequence is is wild

At least in the order Munkres deals with the topics

And Munkres is arguably the most pedagogic Topology book

[storm wadi]

I just picked up a copy of Munkres to go through some of the exercises

[urban zinc]

Maybe this is a hot take but Munkres is so... boring

I tried to read a chapter of it when I was learning topology and gave up

[quartz horizon]

Do you recommend any books for learning topology

[urban zinc]

I feel like I learned a lot of it from osmosis, but I remember someone on the server recommended John Lee's Introduction to Topological Manifolds to me, where the first three or four chapters do a reasonably quick intro to general topology

That was really helpful (I never read the rest of the book)

I know Hatcher also has a short intro to topology that people throw around, not sure if it's any good

[snow frigate]

i kind of liked willard

[snow frigate]

i did metric spaces first, got pretty far but now i'm struggling with compactness 😭

[cosmic mirage]

jacob lurie higher topos theory /j

[snow frigate]

munkres was alright and is very well-organized

[cloud kindle]

I would be surprised

[balmy briar]

the quotient topology notes there are good

[digital sun]

I don't think so, as a reference book, probably, like a dictionary. Not pedagogical

[digital sun]

Indeed

I mean, manifolds after analysis and some vector calculus course is a good introduction to most useful concepts un topology

[digital sun]

I thought that, but it depends on how you take a topology course for it. Self study with it surely is boring, but there are good profs who know how to extract the best of the book

At least that's how I passed Topology

After failing it at first using Viro Irodov as main book

But, I don't think Munkres is the best even for starters, there are better books, more challenging and pedagogical

And with colours

[kind marlin]

which book would you recommend?

wow thats long

wait so you dont recommend it 😭

is this the more challenging and pedagogical book with colors

[queen prism]

oh

well, another flowchart for my collection, I guess

[digital sun]

I do, nice book, awesome colours

[digital sun]

These

[kind marlin]

is there another?

[digital sun]

Nice book

[gaunt linden]

Are all those really ones you have permission to redistribute?
(Assuming they are not, I think the <@&268886789983436800> need to clean them up).

[digital sun]

Oh I see

If so I'll delete them

Please confirm

[urban zinc]

Yea that's fair

[digital sun]

I'll delete them anyway, I'll write the names only

[gaunt linden]

The #rules definitely state "We cannot allow posting links or files with pirated content."

[digital sun]

Thanks, I forgot about it

Indeed lol

I think I deleted them, if not, please do so.

[gaunt linden]

They seem to be gone alright.

[digital sun]

Thanks

So, the authors are (sorry If I misspell someone)

Alexandroff
Croom
Dugundji
Vipul Kakkar
Ronald Brown
George F Simmons
Lefschetz
Min Yan
Robert Messer
Emil Milewski
Andre
Gamelin
Ghirst
Herve
Kalajdzievski
Kuramesan
Kosniowski

Not in a particular order though

Those are the ones I know so far

Personally I like Kalajdzievski's Topology book a lot.

Is very complete and illustrative

[storm wadi]

Did you encounter compactness within metric spaces or topology first?

[snow frigate]

i saw in metric spaces but i didn’t absorb it

[storm wadi]

Yeah that's fair enough, I definitely did struggl (and still do) with compactness more in topology than metric spaces

[warped shore]

What about compactness

[urban zinc]

Terry Tao has an amazing explainer about compactness https://www.math.ucla.edu/~tao/preprints/compactness.pdf

[snow frigate]

just wrapping my head around the proofs of facts about it

like heine-borel

[polar storm]

Are compactness and limit point compactness equivalent in a second countable space?

I was able to prove this to be true if the space is T_1 but I don't know if this still holds for an arbitrary second countable space.

[zealous berry]

fuck you i tried that

the question was ill posed 🥸

greatest cope

[polar storm]

Here is my proof for $T_1$ case. Clearly compactness implies limit point compactness for any topological space. Now suppose $X$ is $T_1$, second countable, and limit point compact. Let $\mathcal{A}$ be an open cover of $X$. Since second countability implies Lindelofness, we may assume that $\mathcal{A} = {U_n}{n=1}^\infty$ is countable. Suppose for contradiction that ${U_n}{n=1}^\infty$ has no finite subcover. Then for every $n \in \mathbb{Z}^+$, we can choose
$$
x_n \in X \setminus \bigcup_{k=1}^n{U_k}.
$$
We claim that $E = {x_n : n \in \mathbb{Z}^+ }$ has no limit point. Indeed, if $x \in X$, choose $U_N$ that contains $x$, and we end up having $E \cap U_N$ being finite by construction. So $x$ cannot be a limit point of $E$, since $X$ is $T_1$. But this contradicts the fact that $X$ is limit point compact.

[gentle osprey]

Euiseok (Class of 2100)

[gritty widget]

any good arXiv papers on Alexandrov topology and Heyting algebras/interior operators?

[rotund prism]

This is more subjective, but are there any times when you would equip a banach space with a topology that is not metrizable?

i know the metric that comes with the definition of a banach space can induce a topology thats obviously metrizable (and completely metrizable at that) but idk how much i can just declare that the space itself is metrizable

[quasi forum]

calling something metrizable only makes sense if you have a topology in the first place

regardless if you have a banach space X then you can consider its weak topology, which is in general not metrizable

these come up frequently in analysis

the idea is that weaker topology => more compact sets => easier to extract convergent subsequences

[gaunt linden]

Is "weaker" synonymous with "coarser"?

[quasi forum]

in my book it is at least

by definition the weak topology is the coarsest topology on X for which all elements of X* are continuous, so in my head the connection is reasonable

[midnight umbra]

there are natural topological vector spaces that aren't metrisable

for example, the set of signed measures on a compact subset of R^n with the weak topology

[alpine nest]

Most vector spaces with the weak/weak* topology aren't metrizable.

[alpine nest]

Although it might be worth mentioning that the unit ball of this space (and consequently the set of probability measures on that compact subset) is metrizable.

[midnight umbra]

yep

[alpine nest]

I've always found it mildly unintuitive that the unit ball is metrizable but the whole space is not

[midnight umbra]

i assume by unit ball we mean those measures such that |m(K)| <= 1, since ofc the entire topology is not metrisable lol

[alpine nest]

Yeah, or equivalently |m|(K) \leq 1 (total variation)

[cerulean oriole]

I think it's because the topology is not the norm topology. So from the perspective of the topology, the "unit ball" is some random subset, albeit closed(?) and absorbing. But this doesn't help my intuition that much.

[digital sun]

I think of it as a subbasis though, like the weak topology induces a subbasis

For sequences in X*

But I can't remember it exactly

Though this summary from David Lecomte may be useful: https://perso.crans.org/lecomte/Math/WeakTopologies.pdf

[robust drum]

For measures I always wondered why ppl say weak instead of weakstar lol

[robust drum]

This is the same as the unit ball wrt the dual norm

Since the dual norm is sup of action on functions with infty norm <= 1

And that’s the same as the radon measure of the set

[midnight umbra]

the weak* topology is on the dual space right? if your perspective is that the set of measures is your base space then I think that's where the difference arises

[robust drum]

Well when people say weak wrt measures they usually mean weak wrt actions on some space of functions which is actually weak star

The weak would be dual space of space of measures which has too many opens

It’s too strong

[robust drum]

I mean you can do it but I’ve basically never seen it there’s not really a reason

[alpine nest]

Yeah, I agree with this, whenever I've seen this used, it was when treating the space of meaures as the dual of a continuous function space (via the Riesz-Markov-Kakutani representation theorem), so the topology considered would indeed be weak*

[kind egret]

You can proove that any infinite dimension Banach space is not metrizable for the weak or weak* topology (if it is a dual)

But as a nice exercise, you can show that R[X]* is metrizable for the weak* topology (for the II.II_1 on R[X])

[alpine nest]

I was hedging because I was dimly recalling Schur's theorem (in l^1, weak convergence is equivalent to norm convergence)

But I now remember that while convergence there is equivalent, the topology is still distinct

[kind egret]

Yep, the topology are the same only if they are metrizable

[kind egret]

This theorem is wild

[digital sun]

Double weak?

But then the original space V is also double weak, isn't it?

[robust drum]

No I mean that if you use the term weak consistently to refer to measures you’re not referring to duality with functions but duality with the dual space of the space of measures, which contains functions but is bigger

Since Eg L^1 or C^infty are not reflexive

In practice, when people say weak convergence of measures they’re almost surely talking about weak star

[urban zinc]

Yeah I agree, the star is just dropped because it's assumed

[limber wyvern]

Hi!, can someone clarify me, what is the local structure of a point in a open set?

[quartz horizon]

wdym by "local structure" exactly?

[limber wyvern]

tbh I don't know, in my book it talks about local structure in the points of a open set

but it does not explain what it is

[quartz horizon]

could you send a screenshot maybe

[limber wyvern]

it is in spanish

[quartz horizon]

might still help

[limber wyvern]

It also talks about local concepts

it explains that a local concept is a property that can be explain point by point

[limber wyvern]

Okay, by filtering the word I got down in the book and found the definition of it xD

is how the set behave around each of its points

[quartz horizon]

Well there is a notion of a property being true near a point p

[crisp lintel]

I would imagine this is referring to the neighborhood filter at a point

which is just the set of neighborhoods of the point

[limber wyvern]

I came back to see if my understanding on this concepts are correct:

Im trying to answer the question: "Why is Q dense in R".

And well, I know the answer in a Calculus I style but now that I'm with topology I want to do it in a topological way.

So, a set Q is dense if his clausure it's the whole space R. Knowing that an adjacent point is a point which is infinetly close to another point.

I try to answer this question by saying: Q is dense in R, because there is a infinetly close point to a point in Q which is in R

Is my answer right?

I mean, I know by Calculus 1 that is dense because every open set contains a rational number between the lower and upper bound

[quartz horizon]

that's not how i'd phrase density

[quartz horizon]

this is the correct argument

[limber wyvern]

How would you?

The general definition is that a set D is dense if $$ \overline{D} = X $$

[quartz horizon]

it's $\overline{D} = X$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

S0S4

[quartz horizon]

but we don't use "infinitely close"

[limber wyvern]

And how do you imagine an adjacent point?

I mean, I thinked of infinitely close by myself and I don't know if it's a correct way to think about it

[quartz horizon]

i mean my definition of closure is "smallest closed set containing our set"

[limber wyvern]

and you don't think of that like a set pretty close to the set?

[quartz horizon]

not really, no?

[limber wyvern]

but mmm, I think of it as points not as ser

Like, using your definition: "The smallest closed set containing our point"

[limber wyvern]

well, yeah, there is no close nor far in a topology without a metric but, if it's the smallest closed set

I imagine it as being close to the poing

[quartz horizon]

that's the closure of a singleton set

$\overline{D}$ is the smallest closed set containing all of $D$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[limber wyvern]

wow

[quartz horizon]

this just says $x \in \overline{A}$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[plush folio]

"infinitely close" is a bit imprecise, but I think it makes sense to say that if D is dense in X, then every point in X is arbitrarily close to a point in D

[limber wyvern]

Let $A \subset X$ and $x \in X$. We say that $x$ is adherent to $A$ if for all $G \in \mathcal{T}$ such that $x \in G$, we have $A \cap G \ne \emptyset$.

[gentle osprey]

S0S4

[limber wyvern]

this is my formal definition of adjacent point

[limber wyvern]

Yeah, I know it's imprecise because there is no such thing as distance in a topological space with no metric in it, but it's just a way to make it make sense in my head😅

[plush folio]

Yeah, but it's a good intuition IMO. I think almost all intuition/visualization you can have for topological spaces comes from metric spaces

[limber wyvern]

So, having that in mind, how would someone explain that Q is dense in R

knowing this

Is like, every real is arbitrarily close to a rational right?

[plush folio]

Yep 👍

[limber wyvern]

What do you think Pseudo?

[quartz horizon]

so you have to be a little careful with the quantifiers here

what is true is that for every $x \in \mathbb{R}$ and every $\epsilon > 0$ there exists a rational $q \in \mathbb{Q}$ with $|x - q| < \epsilon$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

so "there are rationals arbitrarily close to any real number"

however, what is not true is the following

for every $x \in \mathbb{R}$ there exists a rational $q \in \mathbb{Q}$ such that for every $\epsilon > 0$ we have $|x - q| < \epsilon$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

which is "for every real, there exists a rational number that is arbitrarily close to it"

do you see the difference?

[limber wyvern]

mmm

yeah

[quartz horizon]

so in particular, it is not true that "for every real number, there exists a rational number infinitely close to it"

[limber wyvern]

so it's like the opposite of what i said

Yeah, its most like, there are rationals arbitrarily close to any real number, but not for every real number there exists a rational number infinitely close to it

is it good?

[quartz horizon]

yeah

[limber wyvern]

wow

subtle difference

[quartz horizon]

this is why i don't really like the "infinitely close" language

because there are subtleties that are easy to miss

[limber wyvern]

okay, thanks to both of you for the help

[urban zinc]

well according to nonstandard analysis, ...

[sly geyser]

So when is homemorphic used and when is isomorphic used (cause I just saw homeomorphic used to describe equivalences of trees)

[zealous berry]

an isomorphism usually refers to two objects being identical in structure but perhaps different in name. for example, the set {1, 2, 3} is isomorphic to {4, 5, 6}, just with 1 relabelled as 4, 2 relabelled as 5, 3 relabelled as 6

in the same sense, two graphs are isomorphic if the structure of the graph is the same (same placement of nodes and same connections between them), but perhaps the names on the nodes are different or something

[crisp lintel]

A homeomorphism is just an isomorphism of topological spaces

[zealous berry]

not sure what a homeomorphism of graphs is

[sly geyser]

it's just an isomorphism in this case

[crisp lintel]

Any graph can be viewed as a topological space, so perhaps that

[sly geyser]

kruskal's tree theorem

[crisp lintel]

well, two graphs can be homeomorphic as topological space without being isomorphism as graphs

so it does depend what it means

for instance any cycle is homeomorphic to a circle

[sly geyser]

so all cycles are homeomorphic to each other

[crisp lintel]

If that's what they mean by homeomorphic then yeah, would have to see the rest of the paper to know

[sly geyser]

.... do you not know what kruskal's tree theorem is? (it's the TREE(3) thing)

[crisp lintel]

No I've never heard of that

[gaunt linden]

If I understand the tree theorem correctly, "homeomorphic embedding" is actually the topological concept if homeomorphism -- in particular, such an embedding is allowed to map an edge to an entire path in the larger tree, as long as the intermediate nodes are not used for anything else.

The same kinds of embeddings are used to characterize planar graphs in Kuratowski's theorem.

[autumn yacht]

How did they come this step?

[finite token]

The minimum of two things is whichever one is smaller

so either d(x,y) is smaller, meaning B(x,y) = d(x,y), or 1 is smaller, meaning B(x,y) = 1 < d(x,y). Either way, B(x,y) ≤ d(x,y).

[lament steppe]

As as sanity check, these are equal right?

X n ( Y \ Z) = (X n Y) \ Z = (X n Y) \ (X n Z)

[robust drum]

Yup, all follows from associativity of intersections

Since you can write C = X cup Y cup Z (just so they share a common ambient) then A\B = A cap C\B for A subset of C

[limber wyvern]

Hi!, what is the difference between a neighbourhood and a open ball?

[opaque scroll]

(in the context of metric spaces) a neighborhood is a set that contains an open ball.

[limber wyvern]

So a neighbourhood is more generic than a open ball?

[opaque scroll]

In the sense that an open ball is a neighbourhood, sure.

[limber wyvern]

Thanks

[fading meteor]

I need to prove that every metric space is a Hausdorff topological space. I have this so far:

Let $(M,d)$ be a metric space. Let $x \in M$ and $y \in M$ with $x \neq y$. Let $d(x,y) = r$. Then I need to prove that there exist neighbourhoods $V_x$ and $V_y$ for $x$ and $y$ respectively.

Let $k \in \mathbb{R}$ with $k < \frac{r}{2}$. Then $B(x,k) \cap B(y,k) = \emptyset$, because assume there exists $a \in B(x,k) \cap B(y,k)$, then $d(x,y) \leq d(a,x) + d(a,y) < 2k < r$, a contradiction.

[gentle osprey]

Nico

[fading meteor]

Is this a valid proof?

[fading meteor]

And here obv, $V_x = B(x,k)$ and $V_y = B(y,k)$

[gentle osprey]

Nico

[polar storm]

yeah, looks good.

maybe say k is positive tho this is obvious by context.

[fading meteor]

What is a basis and local basis for a topological space?

I don't quite understand that

[polar storm]

A basis is basically a building block for the topology (i.e. every open set). More specifically, it is a collection of open sets, called basis elements, such that every open set in X can be written as a union of basis elements (it can be an infinite union or finite union).

In a metric space (X,d), the collection of all open balls forms a basis of the "metric topology" on X i.e. you can use open balls to describe any open set in X as a union of open balls (which is obvious).

[polar storm]

For a local basis, here is the definition:

[polar storm]

You can think of a local basis at a point x as a collection of neighborhoods of x that captures all the ways of being “close” to x since any neighborhood of x is just a "larger version" of one of these local basis elements. This is why we use the word "local" basis.

One example where this is used is in the notion of a first countable space. In such spaces, sequences behave well enough that they capture all the important limiting behavior, so you can describe closures and continuity purely in terms of sequences like you did in metric spaces. In fact, first countable spaces generalize metric spaces, so you still retain these nice sequential properties in a more general setting.

[fading meteor]

And how do I prove:
Let $\mathscr{B} \subseteq \tau$. $\mathscr{B}$ is a basis for $\tau$ iff $(\forall x \in X)({U \in \mathscr{B} : x \in U }$ is a local basis for $x$?

[gentle osprey]

Nico

[polar storm]

$\mathcal{B}$ is a basis iff for every $x \in X$ , the collection $\mathcal{B}_x = { B \in \mathcal{B} : x \in B }$ is a local basis at $x$.

Proof.

The forward implication is immediate. For the converse, it is immediate that $X = \bigcup \mathcal{B}$, so it remains to check that if $x \in X$ and $B_1, B_2 \in \mathcal{B}$ such that $x \in B_1 \cap B_2$, then there exists $B_3 \in \mathcal{B}$ such that $x \in B_3 \subseteq B_1 \cap B_2$. Now note that $\mathcal{B}_x$ form a local basis at $x$ by hypothesis and that $B_1 \cap B_2$ is a neighborhood of $x$. This means that we can choose $B_3 \in \mathcal{B}_x$ such that $B_3 \subseteq B_1 \cap B_2$. This completes the proof.

[gentle osprey]

Euiseok (Class of 2100)

[polar storm]

@fading meteor I just found a typo in the definition (which came from my tex notes). In the condition (ii), it should be fixed so that the intersection of B_1 and B_2 contains x.

[fading meteor]

What's the best way to study topology?

I'm just struggling in general, actually

[zealous berry]

With drug

[polar storm]

I also struggled a lot (the motivation part)

I actually haven't taken any topology class yet.

[fading meteor]

Then how did you study it?

[polar storm]

But I found Munkres' Topology book really good

[polar storm]

I just self-studied.

[fading meteor]

Damn

[polar storm]

Point-set topology itself is really dry, and you just need to study more and more math to really see and understand the motivation. Many of the concepts introduced in the early part of Munkres' book are motivated by analysis (ofc not everything), at least that's how I understood. The proofs in Munkres' book are very detailed and not hard to read, and it has many exercises.

[fading meteor]

Yeah, my topology course also has a lot of motivation from analysis

[quartz horizon]

Lots of pictures help

I also think topology has a lot of equivalent characterisations of things

And it’s useful to know many different ways to say the same thing

[fading meteor]

Last year for example we saw a lot of theorems for multivariable calculus and now we're generalising stuff to topological spaces

[quartz horizon]

Because they can be useful in different contexts

[fading meteor]

Yeah, my syllabus doesn't really have that

[quartz horizon]

For example, a very useful characterisation of openness is “neighbourhood of all its points”

I’ve used this a lot in practice because it can be easier than checking a set is open directly

[fading meteor]

What do you mean by directly?

[quartz horizon]

Using the definition of the topology

[fading meteor]

If I need to check whether a set is open or not, I just check of all of its points are internal (is this the right word?)

[quartz horizon]

Interior, and yes that’s equivalent to what I said

[fading meteor]

How do I prove that if $X$ is a topological space and $U \subseteq Y \subseteq X$, then the closure of $U$ in $Y$ is $\bar{U} \cap Y$, where $\bar{U}$ is the closure of $U$ in $X$?

[gentle osprey]

Nico

[polar storm]

Before proving it, I'll use the notation A instead of U, cuz I always use U as an open set (it's just my habit).

[novel acorn]

note that every closed set in Y is the intersection of Y with a closed set in X :)

I think that's enough for a hint

[fading meteor]

Isn't this then just the definition of subspace?

[quartz horizon]

Well I think you should give a more explicit argument

[fading meteor]

What do you mean?

That the closure of U is closed?

[quartz horizon]

Like, can you explicitly provide your argument that the closure of U in Y equals the closure of U in X, intersect Y?

[polar storm]

We first show that $A \subseteq Y$ is closed in $Y$ if and only if $A = C \cap Y$ for some closed set $C$ in $X$.

Suppose $A$ is closed in $Y$. Then $Y \setminus A$ is open in $Y$, so $ Y \setminus A = U \cap Y$ for some open set $U$ in $X$ by the definition of subspace topology. Set $C = X \setminus U$, which is closed in $X$, and we have
$$
A = Y \setminus (U \cap Y) = Y \setminus U = C \cap Y.
$$

Conversely, suppose $A = C \cap Y$ for some closed set $C$ in $X$. Then
$$
Y \setminus A = Y \setminus (C \cap Y) = Y \setminus C = (X \setminus C) \cap Y,
$$
which is open in $Y$. Hence $A$ is closed in $Y$.

Now we prove the result. Clearly $\operatorname{cl}_Y(A) \subseteq \operatorname{cl}_X(A) \cap Y$. Conversely, let $C$ be a closed set in $X$ such that $\operatorname{cl}_Y(A) = C \cap Y$. Because $A \subseteq C$, we have $\operatorname{cl}_X(A) \subseteq C$. Hence $\operatorname{cl}_X(A) \cap Y \subseteq C \cap Y = \operatorname{cl}_Y(A)$.

[gentle osprey]

Euiseok (Class of 2100)

[novel acorn]

not really it still requires a bit of work to prove :)

I said the hint because the closure is defined as the intersection of all closed sets intersecting U...

[fading meteor]

Does every point in a first countable space have a local basis?

Yes, right?

[midnight umbra]

any point in any space has a local basis

first countable means every point has a countable local basis

[fading meteor]

Aha

I meant second countable

[polar storm]

second countable is stronger

[fading meteor]

Really?

[midnight umbra]

yes again, by taking every element of the countable basis containing your point

that gives a countable local basis

[midnight umbra]

yep, second countable implies first countable

[fading meteor]

How do I prove this: every FIP-sequence of closed subsets of $X$ has a non-empty intersection iff every countable open cover of $X$ has a finite subcover?

[gentle osprey]

Nico

[fading meteor]

Nvm

[midnight umbra]

you got it?

[fading meteor]

Yeah

[midnight umbra]

nice

[fading meteor]

I asked because I didn't understand the proof in my syllabus, but I got it after all

[midnight umbra]

i like the proof by contrapositive in both directions

[fading meteor]

I've only done 37 pages of my syllabus so far today, but it feels more like 100

I hate this subject

(No offense)

[rancid umbra]

take a break 😭

[storm wadi]

37 pages in one day 😭

that's like half the course for most of my classes

[fervent gull]

i am studying micro economics - and the concepts of open, closed, compact, bounded, convex and other such topics come up - any good online source to get a good understanding of these topics so that i can proceed with microeconomics study??

[unborn hawk]

micro economics??

[novel acorn]

Radu read about stochastic processes in economics with me to impress ur gf with ur money making abilities ;)

[unborn hawk]

I dunno what a stock option is I only know stacks x)

[iron bolt]

I prefer snacks, personally

my understanding of economics is that the more line go up, the more snacks I can afford

[urban zinc]

If you want videos, there are a lot of good real analysis playlists on YouTube, I would use one of those

https://www.youtube.com/playlist?list=PLBh2i93oe2quABbNq4I_-hyjhW8eOdgrO

[queen prism]

bright side

[fading meteor]

What's an example of an Alexandrov compactification?

It sounds so weird tbh

Also how do I prove this: If $(X, \tau)$ is a non compact, local compact Hausdorff space, then $X_\infty := X \sqcup \infty$ with the topology
[
\tau_\infty := \tau \cup {X_\infty \backslash K : K \text{is a compact subset of } X }
]
is a Hausdorff Alexandrov compactification of $X$

[gentle osprey]

Nico

[rancid umbra]

the circle S^1 is the one point compactification of R

[fading meteor]

How?

[rancid umbra]

take the real line

imagine a point sitting above it

drag -oo and oo to that point

you get a circle

[fading meteor]

Aha

Kinda like the Riemann surface for C?

[rancid umbra]

yea

[quaint spruce]

yes

[rancid umbra]

that’s just S^2

[fading meteor]

S^2?

[rancid umbra]

which is the one point compactification of C

S^2 is the unit sphere in R^3

[quaint spruce]

well you need to just check that the properties are satisfied

[fading meteor]

So I just need to check if it is Hausdorff and compact?

[quaint spruce]

and some more things:
X is a subspace of it
X is dense in it

[fading meteor]

Yeah ofc X is in it

[quaint spruce]

right, but X being a subspace means that the subspace topology inherited by X from X_\infty is the same as the topology on X. This may be obvious but its part of the requirement

(this amounts to the inclusion map being continuous)

[fading meteor]

What's that?

I'm just not familiar with most of the English terminology

[quaint spruce]

how did you define "Alexandrov Compactification"? just check that this construction satisfies those properties

[fading meteor]

A compactification X' of a topological space X is a one point compactification if X'\X contains exactly 1 point

[fading meteor]

So I thought I'd just check if X_\infty is a compactification of X, and that X_\infty \ X is a singleton

Right

[quaint spruce]

correct, and checking that X_\infty is a compactification is just doing this + checking that it is compact which you mentioned

[fading meteor]

Yeah

Now how do I do that tho

[quaint spruce]

well just keep unfolding the definitions

[fading meteor]

Isn't is just that we take the limit of all convergent series in X?

*sequences

Like, the set that contains the limits of all convergent sequences in X

Not even convergent

The set that contains the limits of all the sequences (a_n) where all a_i are in X

Isn't that \bar{X}?

And that needs to equal X_\infty

Or is this not the right approach?

Maybe not, considering the next chapter is about completeness

[quaint spruce]

here is what i recommend:
X_\infty \ X obviously only has one point
Now you need to check that X_\infty is compact, Hausdorff and X is a subspace of X_\infty, and X is dense in it.

start by showing that X_\infty is compact (you dont need any sequences for this, just follow the open cover definition)
next show that X is a subspace of X_\infty. This means that the subspace topology inherited by X as a subspace of X_\infty is the same as the topology of X
next show that X is dense in X_\infty, i.e that cl(X)=X_\infty
check that X_\infty is hausdorff

[fading meteor]

Compact means that every cover of X_\infty has a finite subcover, right?

Or smth like that

[quaint spruce]

"every open cover.." , but im assuming you mean that when you say "every cover"

[fading meteor]

Yes, we define cover by open subsets

Cover is just open cover and closed cover is cover with closed subsets

Or at least in my syllabus

Ok so here's what I got

X is locally compact, so that means that every point in x has a compact neighbourhood

So if x is in X, then we can just cover that up with some finite open cover for that neighbourhood

And if x is infty, then we can cover it with X_infty \ K, for any compact subset of X, because that's open and finite?

And then the union of all of these open covers is a finite cover for X_infty

Does that sound right?

[quaint spruce]

What you said may not even be finite because X may have infinitely many points

[fading meteor]

Okay

Then I was thinking about this theorem

X is compact iff ever FIP family of closed subsets of X has a non-empty intersection

But idk how to apply it to this one

[quaint spruce]

well you are kind of close

you dont need this FIP definition.

take any open cover of X_\infty. This means there is some open set containing infinity. What can you say about the complement of this open set? (look at how the open sets containing infinity are constructed)

[fading meteor]

It's just one of the sets that we added in tau_infty

[quaint spruce]

yes, but all those sets you added have a certain property

[fading meteor]

They are the complements of compact sets?

[quaint spruce]

correct, so their complement is a compact set in X

so the complement can be covered with finitely many open sets of the open cover

[fading meteor]

Okay, but taking the complement of that cover gives us a closed set

[quaint spruce]

it will be closed, but more importantly it is compact

[fading meteor]

Why

[quaint spruce]

because the complement of any open set containing infinity will be compact

[fading meteor]

So...

I'm really not good at this

[polar storm]

||Construct an open cover F of C in X using the open cover A of X^*. Then choose finite subcover of C from F. Here X^* \setminus C contain the inifnite point||

[quaint spruce]

ok, tell me if this makes sense
let U be open cover of X_\infty. we need to show there is a finite subcover. Because U is a cover, there is an open set V in U containing \infty. This means X_\infty \ V = K where K is a compact subset of X. Cover K with finitely many open sets U1,..,Un from U. Then V, U1,..,Un is a finite cover of X_\infty

[fading meteor]

Oh damn

Yeah now I get it

Also, what does this mean again: $\mathscr{S}(K,\mathbb{C})$?

[gentle osprey]

Nico

[fading meteor]

Idk if it's a topology thing, but it keeps appearing here and there in my syllabus

[polar storm]

looks like a continuous function space

C(K, \C)

[fading meteor]

That's a C!?

[polar storm]

set of all continuous functions from K to C

and I guess K is compact.

maybe ur syllabus is talking about Arzela-Ascoli theorem

[fading meteor]

Yes, but only very later on

[polar storm]

cuz it is about finding a compact subset of this function space

[fading meteor]

Ok not that much later, only 10 pages from where I am currently

[polar storm]

what class r u taking? Just curious lol.

[fading meteor]

It's just called topology and metric spaces

at Ghent University (Belgium)

[fading meteor]

I'm here now

And I don't understand it at all

[quartz horizon]

you can think of it as classifying the (pre)compact subsets of C(K)

[fading meteor]

What is precompact?

[quartz horizon]

a subset $S$ of a topological space $X$ is precompact if $\overline{S}$ is compact

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[fading meteor]

What's the difference with complete?

[quartz horizon]

completeness only makes sense for metric spaces, no?

[fading meteor]

Uhm

Right

[quartz horizon]

the main results i know for that are

let $M$ be a metric space. then if a subspace $S$ is complete, it is closed in $M$

[fading meteor]

I've not heard precompact before

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

conversely, let $M$ be a complete metric space. then if a subspace $S$ is closed, it is complete

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

it's a topological term

[fading meteor]

I saw this one, yeah

[quartz horizon]

it's equivalent to "total boundedness"

that is, let $M$ be a metric space. then a subspace $S$ is precompact if and only if it is totally bounded

[fading meteor]

Yes

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[fading meteor]

I know what that means

TOtally bounded

[quartz horizon]

so it's a kind of topological generalization of total boundedness

[fading meteor]

What's the difference between basis and subbasis?

[polar storm]

You can use a subbasis to build a basis.

[fading meteor]

I still don't see the difference with a normal basis

[quaint spruce]

well for a basis, you have an intersection condition at each point. If B1, B2 are basis elements containing x, there is B3 a basis element containing x and inside B1 intersection B2

[gaunt linden]

Looking at it from the other direction where we already have a topology and ask what would be a basis/subbasis for it:
For a basis you require that every open set is a (possibly infinite) union of basis sets.
A subbasis also allows finite intersections of them.

[fading meteor]

So if $S$ is a subbasis, then every open set $U$ is of the form $U =\bigcap_{i\in I} S_i$?

[gentle osprey]

Nico

[fading meteor]

And unions between them

[quaint spruce]

there are more general definitions of basis and subbasis, that dont depend on a topology (these are used when defining topologies generated by a basis, subbasis), so any notion of open doesnt exist yet.

Usually, in those definitions, a basis is defined as:
a collection of sets B such that U_{V\in B} V = X
for every x in X, B1,B2\in B containing X you have B3 in B with x \in B1\cap B2

a subbasis is defined as any collection of sets S such that U_{V\in S} V = X

[gaunt linden]

Yes. -- with the constraint that the I's must be finite.

[quaint spruce]

this makes the distinction very clear as well, between a basis and a subbasis

[gaunt linden]

For example the set of all open intervals of length 1 is a subbasis for the ordinary topology on R, but not a basis.

[fading meteor]

We defined subbasis as: $\mathscr{S} \subseteq \tau$ is a subbasis of a topology $\tau$ if ${ U_1 \cap \cdots \cap U_n : U_j \in \mathscr{S}, n \in \mathbb{N}}$ is a basis for $\tau$

[gentle osprey]

Nico

[fading meteor]

What does this mean?

Because this is what's in my syllabus and I don't really understand it

Also, can we have infinite intersections?

[gaunt linden]

What in it do you find unclear?

[gaunt linden]

Infinite intersections of open sets are not necessarily open, so that won't work.

[fading meteor]

Because you're defining a basis

[gaunt linden]

What?

[fading meteor]

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