point-set-topology

[kind marlin]

ooh yeah and that’s just like the flip side of all of this

[kind marlin]

okay so the space needs to be disconnected, and specifically the union of two nonempty disjoint open sets that are each homeomorphic to the original space

this feels very fractally

let's say (0,1) but i remove all of the dyadic rationals? does that work

oh isnt this the same as baire space

[gritty widget]

baire space is omega^omega right

if so that makes sense

Or wait

idk where I was going with that

wait yeah

omega^omega times omega is the same as omega^omega

and generally X times omega(with the discrete topology) should have this property

[kind marlin]

right yeah

analogous to X^omega in the product setting

or i guess more loosely X^infinite set

[gritty widget]

Lets say coidempotent for now

For any infinite cardinal k, and any space X;

X^k is idempotent

X times k is coidempotent

The most interesting examples will probably, similar to the idempotent examples, not be of the form X times k

Though unlike before, there isn't a cardinality issue with countable X being unable to equal X^omega

Really we can generalize all of this to any sufficiently nice category right(e.g. a category with objects like cardinals, s.t. X^k makes sense and is equal to the product of X with itself k times, and similar case for X times k)

[kind marlin]

sorry im like a cat theory toddler

is X an object of the category

or i think ill just not press too deeply into that LOL i think that makes sense at face value

[gritty widget]

Yeah I mean in any category, it makes sense to define an object X to be idempotent, as long as we have objects like cardinals ig

[kind marlin]

wait okay i get that in the X times k setting bc you can define products with cardinals but im confused how to generalize X^k for arbitrarily infinite k in a category

oh wait i guess the universal property of products doesnt really rely on finiteness or anything

okay thats fine i think that makes sense

okay wikipedia confirms that great

[gritty widget]

A more categorical way of saying X is (co)idempotent is that Hom(_, X)xHom(_, X) is naturally isomorphic to Hom(_, X), so e.g. a finite collection of maps from Y to X correspond naturally to a single map from Y to X

[kind marlin]

is Hom(_,X) the collection of morphisms that point to X

like across the entire category

or is _ a placeholder for any particular object

[gritty widget]

Yep! The functor Hom(_, X)xHom(_, X) takes an object Y and maps it to the product of Hom(Y, X) with itself

It's from C^op to Set

[kind marlin]

ooh ok

[gritty widget]

In general, if Z is the product of X and Y, Hom(_, X)xHom(_, Y) is naturally isomorphic to Hom(_, Z)

Less formally, pairs of maps into X and Y are naturally maps into Z

[kind marlin]

ohh that makes sense

i think thats like taking the universal property of products in terms of the commuting morphism diagram and writing it in terms of the category of morphisms?

[gritty widget]

When X=Y=Z in the idempotent case, I guess an intuitive way of looking at it is that a finite collection of continuous functions into Q correspond to a single continuous function into Q

[gritty widget]

It's writing it in terms of representable functors more specifically

Almost all constructions in category theory can be interpreted as special cases of representable functors

or in terms of representable functors

[gritty widget]

Dually for a coidempotent object, a finite collection of maps out of it(with the same codomain) correspond to a single map out of it

[kind marlin]

ooh thats nice

[kind marlin]

simultaneously idempotent and coidempotent objects seem really nice then wow

[gritty widget]

Also to be clear idk if these have an actual name

I'm just using this for now

[kind marlin]

right LOL

i like these names though

[gritty widget]

Maybe you could extend it to like

k-idempotent for some cardinal k

E.g. omega-idempotent when the countably infinite power is equal to itself

Then idempotent would just be 2-idempotent

[kind marlin]

oh wait thats cool, you get a limit-type thing out of that

[gritty widget]

If a category has all small limits then there will be some k-idempotent objects for any cardinal k(simply take X^P(k))

[kind marlin]

whats P(k)?

oh wait i should look up small limits first

ive definitely heard of these before

[gritty widget]

Powerset of k

Really any cardinal > k works

But that in particular is always > k so

[gritty widget]

That's just the limit over a diagram with a set of morphisms, so no product with a proper class of indices

[kind marlin]

okay i think this might be going over my head but thats fine i just need to read more about limits later

i thought this was quirky: the cantor set is coidempotent since 2^N x 2 = 2^N, but it can't be written as X x K for infinite K because X x K isn't compact and the cantor set is compact

on the other hand, Q can be written as X x K (namely, Q x N)

so the cantor set became an "interesting" example here while Q stopped being one

[gritty widget]

That's neat!

I know the cantor set has a unique characterization kinda like Q does too

Could be related

Maybe the category of idempotent not-inf-power object could be neat or something

I would think the inclusion of the category of idempotent(including inf power objects) objects into the category should have an an adjoint, I'd think sending X to X^omega would be adjoint to the inclusion? Maybe not tho idk

[gritty widget]

Would a product of idempotent objects be idempotent?

[kind marlin]

i think so right? i was thinking about that as a better way to motivate why S idempotent * infinite discrete is also idempotent

then we just have 2 individually interesting idempotent topologies: Q and the discrete topology on infinite sets, which... probably aren't representable in terms of one another

well "just" as in "so far" 😭

well wait i guess the discrete topology not being representable as S^K for infinite K might require other things

[gritty widget]

Well

If X^2=X, Y^2=Y, (XY)^2=XYXY=XXYY=XY, so for finite products it should work

[kind marlin]

right, i didnt think about infinite products hm

but finite products seem reasonable

[gritty widget]

Generally S^K for infinite K will never be discrete(unless S is empty or a singleton)

[kind marlin]

oh wait so true

i was thinking about the continuum hypothesis for some reason 😭

okay thats great then

the infinite discrete topology is representable as S x K since we can just take D x D

i wonder if there's a topology that isn't representable by either S^K or S x K

(thats also idempotent / coidempotent)

[kind marlin]

what if we let X = disjoint union of N with R^N

obscene notational abuse but let disjoint union = + and product = x LOL

X = N + R^N

X^2 = (N + R^N)^2 = N x N + N x R^N + N x R^N + R^(2N) = N + N x R^N + R^N = N + N x R^N

i think these are not homeomorphic? (is this real math anymore idk)

then

X^3 = X * X^2 = (N + R^N) * (N + N x R^N) = N x N + N x N x R^N + N x R^N + N x R^N x R^N = N + N x R^N + N x R^N + N x R^N = N + N x R^N

so X^2 = X^3

i'm still not really sure if X = X^2 or not 😭 i think no? i'm also not super confident in any of these operations

my main reasoning why X != X^2 is because R^N is connected and N x R^N isn't but i'm not sure if the N + throws a wrench into anything

[gritty widget]

https://ncatlab.org/nlab/show/distributive+category#Examples according to nLab, Top is distributive

So (a+b)xc=axc+bxc

Take X=N+R^N

[kind marlin]

omg amazing

[gritty widget]

(N+R^N)•X=N•X+R^N•X

Uhh not sure where I was going with that honestly

Feel like I can get X^2=X out of this but not sure how

I’m tired

[kind marlin]

real 💔

[gritty widget]

Wait so N•(N+R^N)=N•N+N•R^N=N•R^N right

[kind marlin]

do we drop the N x N? can you do that with disjoint unions

[gritty widget]

Nah it’s because NxN=N

Wait

no I did that wrong yeah

N•N+N•R^N=N+N•R^N

[kind marlin]

right okay

[gritty widget]

So yeah I think X≠X^2, cause (N+R^N)^2=N+N•R^N+N•R^N+R^N

[gritty widget]

So yeah I think this reasoning is correct here

[kind marlin]

okay awesome

and was it valid for me to do N+N•R^N+N•R^N+R^N = N + N•R^N, just combining the right three terms?

[gritty widget]

I wonder if a simpler (but still nice) category would be better for this honestly

Like Grpd or something

[quartz horizon]

Categories mentioned

[kind marlin]

omg u missed so much cool stuff

[gritty widget]

Yeah I think so!

[quartz horizon]

What’d I miss

[kind marlin]

umm mostly just me geeking out over Q x Q being homeomorphic to Q

and then colimit did a lot of category theory things and i learned about disjoint unions in Top

[gritty widget]

Yeah we were discussing objects being homeomorphic to their own square in general and whatnot, pretty interesting

[kind marlin]

okay wait so we found a topology that eventually becomes idempotent omg

[quartz horizon]

I see

[kind marlin]

is there an easy way to construct higher powers

[gritty widget]

Since the fundamental groupoid preserves products, if X is idempotent(in Top), so is the fundamental groupoid of X(in Grpd), although it could potentially be the case the fundamental groupoid is of the form S^K in Grpd when X isn’t in Top, take e.g. the fundamental groupoid of Q which is just a countably infinite discrete groupoid, still idempotent

[kind marlin]

oh this is not hard i think

take n idempotent topologies that such that the pairwise products, 3-way products, etc are not homeomorphic to one another (i think this is generally not too difficult to do? since we have our S^K general construction)

then take the disjoint union of all of those

i think eventually some power of this will be idempotent, and the minimum power should grow with n (this feels like a very algebraic construct i should probably know)

im just treating this like a commutative ring now 😭

i think the only property of rings that fails here is additive inverses but that feels pretty important, mmmm i should double check this

maybe ill do that tomorrow

[rancid umbra]

i can show 1 <=> 4 and 2 <=> 3 but i can’t link the two. any hint?

2 => 1 is easy since if f is k-continuous, then in particular, f is k-continuous rel S, so f is continuous

but im having trouble showing that 1 => 2, i.e., selecting a set of CH spaces that determine the continuous ones out of X

i thought that maybe S should be the CH subsets of X, but i don’t think that works?

[quartz horizon]

maybe there's some kind of cardinality argument you could do?

[rancid umbra]

hmm. how do you mean?

[quartz horizon]

if you can use the cardinality of X to bound the cardinality of compact hausdorff spaces you need to consider

[rancid umbra]

oh, i hadn’t thought of that

what, like i only need C with |C| <= |X|? is that even a set?

[quartz horizon]

well if you mod out by homeomorphism then maybe?

[rancid umbra]

okay

[quartz horizon]

at the very least, up to homeomorphism, there is only a set of topological spaces with cardinality <= k

what you can do is

fix a set of cardinality k

consider all of its subsets, which range over all possible cardinalities <= k

and for each subset, consider all possible topologies on it (there's only a set's worth of these)

[rancid umbra]

makes sense, thanks

[quartz horizon]

Though, I’m not sure which cardinality of compact Hausdorff space you’d need

[umbral elm]

A problem from homework, X Hausdorff, X locally compact if for every U open and x in U exists an open V s.t. closure of V is contained in U and is compact, problem is to use this to get Baire category thm for locally compact Hausdorff X, but I can't see why compact has a thing to do with this. Supposedly choose a point x in X and a nbd, given a family of open dense sets Gn, nbd of x intersects all Gn and this gives Vn s.t. closure compact, intersection of all these Vn closures give still a closed compact set while intersection Vn is Gdelta, but what good does a finite open cover somewhere give an open dense set? Just a little hint to keep me going, or am I very close?

I guess the Gdelta set intersection Vn is the thick set i want? I don't even know if it is empty

[tender halo]

sets that are realizable as images of CHaus spaces in X

[umbral elm]

nvm I see, FIP

[oblique sentinel]

question about the urysohn-function from the classical proof. Let's assume we are in the special case of the real numbers as our normal topological space. Is the graph of the urysohn-function distinguishable from the identity function where:

The identity maps x to itself, whenever x is in between 0 or 1 and at the endpoints we extend it to 0 / 1

[rancid umbra]

okay, so my first guess was close

thank you

[tiny obsidian]

Why do you think it should be equal to the identity function at all?

The usual proof (of Urysohn's lemma I guess, and in your case for the specific closed sets {0} and {1}) doesn't say much about what the function you end up constructing actually is because you make a choice at the point you use normality. If your question is whether the function you gave can be constructed by following the usual proof, then yes

[oblique sentinel]

Oh Yes. I didn't mean it has to be the identity function. But it can be constructed in that way using the open sets U_p.
The closed sets I had in mind were more like [-a,0] and [1,b] for a>0 and b>1.

[trail wren]

I am new to the concept of complex projective spaces CP^n and study the topology on this space, i know that the topology on CP^n is the quotient topology given from the projection map, but somewhere i have seen that CP^n is metrizable via the Fubini--Study metric, but here the topology is induced from the coordinate charts

are these two topologies (quotient topology and the topology via coordinate charts) identical?

[gaunt linden]

Please do not post your question in three different channels (so far). It leads to wasted work and is inconsiderate towards people who may spend time writing a reply only to find that the points they had were already made in a different channel.

[trail wren]

Okay sorry, my bad

[pulsar lagoon]

I think they would be (they should at least be isometric as riemannian manifolds) since the fubini study metric also comes the projection as well

But im not too sure

[balmy nexus]

Maybe?
Note that in Uryshon you are choosing a set at each point of construction. The function may be quite wild, although it will be increasing.

you are effectively (70% sure this is correct) choosing Q inside R, and then taking its closure. There's a lot of Qs to choose inside R.

[chilly parrot]

it is always true that if you have a manifold with a Riemannian metric (which induces a metric space metric), its original topology agrees with the metric space topology

[trail wren]

i understand that the metric induces the topology compatible with the one induced from the atlas, but in my case i am interested in compatibility with the quotient topology (from the quotient projection map)

[chilly parrot]

the manifold topology is the quotient topology, since to get a manifold you start with some topological space and then give it charts

[trail wren]

so if i understand it right: we start with the quotient topology and then define charts to match this topology, hence they coincide

[cosmic mirage]

you can also get a topology on a set by putting a manifold structure on the set - i.e. declare which subsets are (hom/diff)eomorphic to specific open sets in R^n -- then you get more open sets for free by taking the preimage from atlases of open subsets. then intersection data gives you all the open sets. is your question whether the topology on CP^n induced by this combinatorial information is the same as the usual quotient topology?

(im pretty sure this is true... correct me if i am wrong)

[trail wren]

i think yes, this is the question, everywhere i see that the fubini study metric is canonical for CP^n, so i guess it must be true, but i am skeptical since it is new to me

[gritty widget]

By the way, I've been thinking some more about it, and, similarly to the case of a finite product of idempotent spaces being idempotent, if two spaces X, Y are both representable as S^K, H^K for some infinite discrete set K, then XY=(SH)^K is too, so an interesting space(e.g. idempotent, not of the form S^K) won't ever be a finite product of S^K spaces either! I think that's pretty neat. Haven't really found any super interesting results though

I'd expect a similar result for coidempotent spaces representable as S*K and coproducts of them

[kind marlin]

ooh yeah that makes sense

the closure properties definitely make sense

ive been tinkering with the semiring structure a little

[gritty widget]

I wouldn't expect all infinite products of idempotent spaces to be idempotent but I can't think of a counterexample

[kind marlin]

i thought it would maybe be possible to create a topology that is only finitely idempotent and finitely coidempotent by combining ones that are finitely one of them and infinitely the other

but there's not many options to choose from right now 😭

[kind marlin]

hmm i was thinking abt that too

i think they would be? If X_i are an indexed family of idempotent sets, and we take the product Y = prod_i X_i, then specifying finite entries in Y x Y should be analogous to specifying finite entries in Y? I think you can just group the products together by the common factors and define your map elementwise from X_i to X_i x X_i

[gritty widget]

According to wiki

"Isomorphism classes of objects in any distributive category, under coproduct and product operations, form a semiring known as a Burnside rig.[27] A Burnside rig is a ring if and only if the category is trivial."

I tried searching for more about burnside rigs but couldn't find any info

[kind marlin]

ooh

[gritty widget]

Oh yeah you're right, the order of the spaces doesn't matter, so say

AxBxCx...AxBxCx.... is equal to AxAxBxBxCxC... is equal to AxBxCx...

Neat!

I wonder if a product of idempotent spaces not representable as S^K are always not representable as S^K

[kind marlin]

i think no right, because you can just do Q^K

[gritty widget]

Oh yeah duh

What about finite products?

[kind marlin]

okay we do have that "representable as S^K" is equivalent to "infinitely idempotent" right

[gritty widget]

Man idk why but thinking about products and powers in Top is surprisingly interesting! I wonder if there are any very general theorems for things like when a space has a square root?

[gritty widget]

Yeah, S^omega=S so we can represent S as S^omega

but "representable as S^K" for some particular K gives us more information than just "for some infinite cardinal K"

[kind marlin]

hmm

[kind marlin]

right? i was thinking about that too

i actually made a math exchange post earlier today asking a lot of these questions LOL so ill see if anyone responds 😵‍💫

i find it super cool that you can manually let (x + y) be a square root of (x + y)^2

[gritty widget]

If not we should try to find some it doesn’t seem too absurdly complicated to do if we put effort into it

[kind marlin]

i would also be really interested in if there's any "power cycles"

like X -> Y -> X -> Y... etc

[hidden abyss]

I just found that post a few mins ago when trying to google QxQ = Q :D

[kind marlin]

that feels like something you can probably prove doesn't exist

omg LOL

[gritty widget]

Like XX=Y, YX=X? Or X^2=Y, Y^2=X?

[kind marlin]

the former but the latter sounds cool too :0

[gritty widget]

Well if X and XY are isomorphic, then X represents the functor Hom(_, X)xHom(_, Y), but idk where to go after that.

Hmmm

[kind marlin]

hmm that makes sense

like even if A is finitely idempotent and B is as well, A^N x B^N might equal A x B without A^N = A and B^N = B

it feels weird grasping what "finitely idempotent" means because the examples, Q and infinite discrete, seem to fail infinite idempotence for completely random reasons

or not random but... sporadic?

[gritty widget]

I think “most” the examples of only finitely idempotent spaces will fail infinite idempotency due to cardinality reasons

[kind marlin]

hmm that makes sense

[gritty widget]

Like Q^omega is, regardless of the topology, already too large to = Q

[kind marlin]

at least then, if we're multiplying finitely countably infinite topologies together, we know that won't be infinitely idempotent

[gritty widget]

Given any space $S$, $S^\omega$ is infinitely idempotent but won’t usually be $\aleph_1$ idempotent I think

[gentle osprey]

colimit

[kind marlin]

omg i was thinking abt this and forgot yeah there's fully different kinds of infinite idempotence here

[gritty widget]

Yeah that’s why I was sticking with “Representable as S^K” earlier even tho it’s longer

In my (so far very, very short) writing so far I’ve just decided to call a homeomorphism from S^K to X(for S not homeomorphic to X) for infinite discrete K a Boore representation, and a space is then Boore iff it admits a Boore representation. We should probably come up with a term for when it is a specific cardinal K, say “a Boore representation of degree/dimension/etc. K”

[kind marlin]

does a larger infinite idempotence imply every smaller infinite one

[gritty widget]

So then an interesting space would just be an idempotent, non-Boore space

[kind marlin]

oo that makes sense

[gritty widget]

I would think so, but I don’t actually know

Say K>J

S^K=S

Then S^J=(S^K)^J=S^(KJ)=S^K=S

I think this works

[kind marlin]

ooh yeah that checks out

nicee

and that contains the infinite -> finite logic too

[gritty widget]

Then a better way of phrasing something I posted earlier would be “If two spaces X, Y have Boore representations of equal dimension K, then their product XY has one of dimension K too”

[gritty widget]

yep! Neat

[kind marlin]

ahh thats nice

[gritty widget]

Generally I think a product of leq K many spaces with Boore representations of dimension K has a Boore representation of dimension K by the same argument

[kind marlin]

right that makes sense

[gritty widget]

I wonder if we can get some reducibility result like every idempotent space is a product of non-Boore idempotent spaces or something, that’d be neat

[kind marlin]

oh if we specify that S != X does that disallow some of our examples? idk for instance if erdos space has a boore representation

[gritty widget]

I don’t remember why I originally specified S!=X but I think the idea was that I’d want to allow spaces that are their own infinite power, but not equal to the infinite power of some different space, should be “interesting”

Like the reason 2^omega is relatively uninteresting is because it is “forced” to be infinitely idempotent because it is an infinite power of another space

[kind marlin]

right that makes sense

[gritty widget]

No clue which one will actually turn out to be more interesting though

[kind marlin]

maybe we could classify trivial and nontrivial boore representations

[gritty widget]

That’s probably a better idea yeah!

Then a Boore space would be K-idempotent iff there is a trivial Boore representation of it with dimension K

I think since most of this is in terms of the categorical products in Top and whatnot we won’t really need to do much topology, just category theory + some categorical properties of Top?

[kind marlin]

right that makes sense

the only thing that confuses me wrt that is that properties like compactness that we used to verify that the cantor set isn't an N-coidempotent boore space

does that still exist when we abstract away Top

or i guess are we moving away from characterizations of spaces

[gritty widget]

Yeah I guess we will still need some actual topology occasionally then

Interesting

I think some countability properties might interact with this in interesting ways too

If S has at least 2 points, and K is uncountable, then S^K is not second countable, so generally no Aleph_1-idempotent spaces will be second-countable

[kind marlin]

whoa that feels useful

[gritty widget]

I reckon most “nice” spaces won’t be k-idempotent for relatively large k due to cardinality issues

[kind marlin]

https://mathoverflow.net/questions/208960/space-x-such-that-x-lambda-cong-x-for-some-lambda

[gritty widget]

Woah that's neat!

[pearl valve]

What is the purpose of considering a larger ambient space X here? Having a seperation of a topological space Y seems independent of whatever ambient space Y lies in. So when using this theorem, why wouldn't we always just set X = Y?

[gritty widget]

I’m not sure, that seems a bit weird. Never seen connectedness phrased like that before, since it’s not a relative property.

[pearl valve]

its from munkres

[gritty widget]

Oh I guess I have seen it then? Weird I don’t remember that

[warped helm]

computing closures

[gritty widget]

Yeah idk why it’s phrased like that

[unreal stratus]

Munkres does connectedness weirdly imo

[gritty widget]

oh that makes sense

[warped helm]

i lowkey did not like that portion

[gritty widget]

Here is limit point synonymous with point of closure or something else?

[warped helm]

i also think munkres’s wordiness is very apparent in that section

[unreal stratus]

It shouldn't be

[hidden abyss]

Closure in Y or in X

[unreal stratus]

Unless Munkres has the wrong definition of limit point

[hidden abyss]

Should probably be in Y

[pearl valve]

but why not just take the closure in Y?

I guess I'm looking for an example when we might use this theorem in future and say "yeah its useful that we can consider limit points in X or something

actually I guess that makes sense

[warped helm]

the theorem is to justify this i would say

[unreal stratus]

To emphasise like in R, x is not a limit point of {x}

[pearl valve]

Maybe its convenient for example to say that two things share no limit points in R^2 and so then you don't hav to consider if there are overlapping limit points in the subspace topology?

[warped helm]

sure

(\operatorname{Cl}_Y (A) = \operatorname{Cl}_X (A) \cap Y )

[gentle osprey]

josemom2

[unreal stratus]

Real

[gritty widget]

Is a limit point of U a point x s.t. x is a point of closure of U\{x}

Or am I misremembering

[unreal stratus]

I think this works yes. Like "every neighbourhood of x intersects U in a point besides x"

[pearl valve]

Is this supposed to be capital N?

[warped helm]

seems like it

[umbral hamlet]

email munkres and tell him hes gotta lock in

[oblique sentinel]

Ok another question about urysohn lemma. If I want my normal topological space to be the real number line and A,B two closed intervalls. Are the U_p only between A and B on the number line? I mean by construction they have to include A. I am a bit puzzled by the x's away from A and B. Can someone explain to me the part where we define U_p for p>1 and p<0 when we are on the number line example

[tender halo]

they are gonna be just some open intervals with A in it

imagine just an open interval inflating from A not touching B

[gaunt linden]

Can you point to (or screenshot) the proof that confuses you? For example the proof sketch I can see at Wikipedia doesn't involve choosing any U(p) with p>1 or p<0, because it's constructing a function with codomain [0,1].

[cosmic zodiac]

So we're getting to topology in my intro to analysis course this semester and my textbook is talking about open and closed balls; is that just a way of conceptualizing neighborhoods?

[umbral hamlet]

Yes

Neighborhoods of a point are exactly the open sets that contain the point

[cosmic zodiac]

So is the ball a kind of R^2 way of thinking about it

?

[gaunt linden]

You're aware of the definition of "open ball", right? { x in X : d(x,p) < r } for some p in X and r > 0.

[cosmic zodiac]

Right yeah my kind of logic is that because it's describing a ball with a radius around a point instead of an interval around the point on a number line, then it's kind of considering two dimensions

I might be taking it too literally though I'm not sure

[gaunt linden]

Yeah, if the name is what you're asking about, then it's just chosen by analogy to what these sets look like in R³.

[unreal stratus]

Balls aren't just in R^2 or similar. They work in any R^n, or more generally any metric space (using that definition troposphere recalls)

[gaunt linden]

They don't need to look much like everyday balls in other metric spaces.

[cosmic zodiac]

Ahh i see

So this is just one example of what a ball might be (from my textbook)

Or actually it's two now that im seeing it

[gaunt linden]

Yes, those are just examples.

[cosmic zodiac]

Which actually answers my question lol

Thanks for the clarification though! Topology is def a little daunting lol

[gaunt linden]

One point that is often not made (because it is not necessary for reaching the conclusion of the lemma) is that every Urysohn function X -> [0,1] can arise from the construction in the standard proof.
In the case where X is the real line, you don't really need Urysohn -- it should be easy enough to imagine drawing some continuous function f: R -> [0,1] that is 0 on A and 1 on B and does whatever you please outside those two intervals. Then you can declare "this is the function I got out of Urysohn's lemma" simply by choosing each U_p to be the preimage of [0,p).
The U_p you get that way do not need to be intervals -- for example they can include parts on the other side of B if your chosen function starts dipping away from 1 there. But "normal space" doesn't guarantee they will be intervals, just the they will be open. And the preimage of the (relatively) open set [0,p) will be open as long as your f is continuous.

[oblique sentinel]

I am not sure what you mean by "every Urysohn function".

I am not interested in constructing a cts function on the real line with here stated properties, since that is an easy task.

I am more interested in visualizing the Urysohn Function (f = inf...) from R to [0,1] and thinking about how the U_p look like.

So I don't really have a question anymore right now except for the one above, but ill surely come back to this when I have the time and maybe ping you?

[gaunt linden]

I mean every possible continuous function R -> [0,1] that satisfies f(x)=0 when x in A and f(x)=1 when x in B.

Every such function is something the procedure in the standard proof can end up with.

There's no single "the" Urysohn function.

[oblique sentinel]

and how would they differ? Are you talking about the enumeration of the rational p's and the way they are counted? I don't get the point really

[gaunt linden]

How would what differ?

[tender halo]

proving the uryhsohn function exists needs like DC

CC is not enough

so it is meaningfully non-constructive

[gaunt linden]

In general, yes, but for the special case of R as the domain there are plenty of ways to choose the U sets that don't need any form of choice.

[gaunt linden]

Depending on which of the infinitely many possible U_p sets you choose at each step, the "f = inf ..." function that results can be anything. There's no "the" output to visualize.

[oblique sentinel]

Well that is hard for me to grasp. In what way can the function be "anything"? The function is from R to [0,1] so it surely can be visualized and is even continuous. Can you try to explain to me the values a Urysohn function can take? I am confused.

For topological spaces in general I felt it was easier since I didn't care about visualizing it

[gaunt linden]

I feel I'm just repeating myself and running out of ways to say it. Every possible function f: R -> [0,1] that satisfies:

1. f is continuous
2. f(x) = 0 whenever x in A
3. f(x) = 1 whenever x in B
   actually appears as a result from the procedure in the proof of the Urysohn lemma, given the right choice of U_p sets.
   If you can draw it, then there's a sequence of legal U_p choices that would make the function you draw come out of the lemma.

[gaunt linden]

Here are some of the possibilities.

(Ignore the point where my mouse-drawn blue line tips over so steeply that it fails to be a function).

[oblique sentinel]

Thank you! That was what I needed. Appreciate the concern 🙂

[quartz horizon]

The product of two open maps is open, right?

By which i mean

Suppose $f : X \to Y$ and $g : Z \to W$ are open maps

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

Consider $f \times g : X \times Z \to Y \times W$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

I claim this is also open

It’s enough to check this for a basis of the topology of X x Z, which consists of products $U \times V$ for $U \subseteq X, V \subseteq Z$ open

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

Then $(f \times g)(U \times V) = f(U) \times g(V)$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

Since f and g are open maps, the factors are open, meaning their product is open in Y x W

So i think that should imply f x g is an open map?

[unreal stratus]

Yes

[tender halo]

product of a family of maps is open iff all of them are open and at most finite amount are not onto

[quartz horizon]

Hm i see

[kind marlin]

i think the coproduct of two open maps is also open, right? and actually the coproduct of two closed maps is closed too i think. is that true for infinite coproducts too?

[kind marlin]

i just realized something

the product of a K-coidempotent space with any other space is K-coidempotent

if X is K-coidempotent, then K x X x S = (K x X) x S = X x S for any topology S

this feels like a broader version of the "boring" coidempotent topologies we got from multiplying by discrete K, i think K-coidempotence is actually an ideal of the Top semiring?

[gritty widget]

interesting

idk what generic means here

[kind marlin]

oh just like

any

[gritty widget]

oh

[kind marlin]

this at least has nothing to do Top which is maybe good

(i think)

idk if theres anything easy to deduce from this bc semirings seem so annoying compared to rings 😭

are there any large semirings that are easy to visualize with idempotent and coidempotent elements

[rancid umbra]

put X = colim_j X_j with natural inclusions i_j : X_j -> X.

the inclusions are open: if U is an open subset of X_j, then the preimage of i_j(U) under i_j is U, and under i_k for k != j is empty.

a map f : X -> Z is open iff f o i_j : X_j -> Z is open for each j:
the forwards direction is clear since the inclusions are open.
for the converse, let U be an open subset of X. then f(U) is the union of f o i_j (i_j^{-1}(U)) over all j, so f(U) is open.

you can specialize this to the case when you have a family of maps f_j : X_j -> Y_j and you want to look at the coproduct map U_j f_j : X -> Y

the same argument goes through for closed maps because the inclusions are closed as well

[rancid umbra]

if you contrast this with the argument that bussy beaver suggested,
in the limit case relies on the fact that the image of a set can be expressed as an essentially finite intersection of its factors,
in the colimit case, the image of a set can be expressed as a union of its factors

[supple island]

Can anyone help me here? It is true that for a topological space X that X is compact iff. every net has a convergent subnet right? Now all sequences are nets and any subsequence of a sequence is a subnet too right? And sequential compactness by definition means that every sequence has a convergent subsequence. Therefore since sequence convergence and net convergence are the same for a sequence, compactness in X implies every sequence has a convergent subsequence i.e sequential compactness holds for X. Anything wrong here?

Surely something must be wrong because the conclusion is false

[quartz horizon]

The part that’s wrong is that a subnet of a sequence need not be a subsequence

It might help to think in terms of cluster points

[supple island]

Ahhhhhhh okay thank you very much. That makes sense then. Everything else is fine?

[quartz horizon]

Mhm

[supple island]

Thank you

[quartz horizon]

Being compact is equivalent to every net having a cluster point

And a net has a cluster point iff it has a convergent subnet

However, for a sequence, having a convergent subsequence is strictly stronger than having a cluster point

So in every compact space, every sequence still has a cluster point

It’s just that it might not have a subsequence converging to that cluster point

The issue turns out to be a lack of first-countability

If your cluster point doesn’t have a countable neighbourhood basis, it’s not clear how to produce a convergent subsequence

[supple island]

Thank you very much for your help

[quartz horizon]

[prime elbow]

Cluster of a sequence define as if I take any open set around that point it contains infinitely many points of sequence, right?

[quartz horizon]

Yep

[prime elbow]

So if sequence has convergent subsequence I can say it has cluster point.

But if it has cluster point I can't say it has convergent subsequence.
So if my space is first countable then I can say, right?

[quartz horizon]

Mhm

[gaunt linden]

Do you happen to know a counterexample?

[crisp lintel]

afaik the space of functions from the interval [0,1] to itself with the product topology has an example but I forget the details

as its compact but not sequentially compact

[gaunt linden]

Yeah, I suppose I meant an example that explicitly describes a sequence with a particularly stated cluster point that no subsequence converges to.

Or perhaps that's too much to expect, since it may require choice to prove [0,1]^c is compact.

[quartz horizon]

Yeah I know of compact non-sequentially compact spaces

So you could take a sequence there with no convergent subsequence, and then try to figure out what cluster points it has..

[gaunt linden]

Hmm, we can whittle the space down to {0,1}^R with the product topology, and the discrete topology on each copy of {0,1}.
Without loss of generality, we can decide that our cluster point should be the everywhere zero function R×{0}.
Then we only need to figure out a sequence that has R×{0} as a cluster point but has no convergent subsequence.
(In fact for the time being I'd be satisfied with a sequence that has R×{0} as a cluster point but no subsequence that converges to that).

I can make a sequence-without-convergent-subsequence by diagonalizing (using a bijection between R and the subsequences of 1,2,3,4...), but it doesn't seem to be obvious how to modify that construction to create an explicit cluster point.

[tender halo]

funniest counterexample is \betaN

for which the sequence of natural numbers has everything except N as a cluster point

but has no convergent subsequences whatsoever

but also uhh the good version of the arens fort space

for a more explicit example

lemme find the engelking page

[exotic crater]

Hello! I need some help with this problem please:

Let X be an infinite set and Y a finite set with two or more elements.

Consider the set A formed by the functions f : X → Y such that
there exists an element y ∈ Y satisfying that f^−1(z) is finite for all z ≠ y.

One of the following statements is correct:

a) The sets A and X are equipotent.

b) The sets A and P(X) are equipotent.

Prove it.

[gaunt linden]

When it says "Frechet space", does it mean what Wikipedia calls a Frechet-Urysohn space?

[tender halo]

yes

[gaunt linden]

I'm finding the explanation of how that differs from a "sequential space" very difficult to understand.

[tender halo]

sequential space is the one where you if take sequential closures over and over (potentially transfinite amount of times) you get the closure

frechet space are ones where this process finishes in one step

[tender halo]

(omega 1 times at most i guess)

[gaunt linden]

I wonder if there's something wrong with the definition in the article I linked. It seems to imply that Frechet-Urysohn spaces are exactly the spaces where "closed" and "sequentially closed" are the same subsets.

[tender halo]

no, "closure and sequential closure are the same"

[gaunt linden]

Not in general, right?

[tender halo]

"closed and sequentially closed are the same" are sequential spaces

[gaunt linden]

How can those be different properties?
If closure and sequential closure are the same, then sets with cl(A)=A are the same sets where scl(A)=A, just by substitutivity.

[tender halo]

because if you take a sequential closure it might not be sequentially closed :)

sequential closure is just adding all limits

which might introduce more limits

[gaunt linden]

So you're saying .... that sequential closure is not a closure operator?

[tender halo]

its not, no

[gaunt linden]

Wonderful.

[tender halo]

wikipedia calls it a "preclosure operator"

[gaunt linden]

Okay, by going to the Wikipedia article on the Arens-Fort space I see that it does sketch (well, assert, but I can fill in those gaps) a sequence that has a cluster point that no subsequence converges to.
(I don't love how the screenshot above goes out of its way to present the space as a subset of R when it doesn't have the subspace topology anyway).

[gaunt linden]

Thanks for the reference.

[tender halo]

yea arens fort is the space engelking outlines but without the 1/i points

i think its much more comprehensible the way engelking does it (minus making it the subset of the plane)

[storm night]

I have a question. I'm reading about the definition of a covering space on Wikipedia and I feel like I have simplified the definition of a covering space, but I'm likely missing something. Is the following definition equivalent to the standard definition?

[gentle osprey]

Nuclear Catapult

[storm night]

If the restriction ( \pi|{V_d} : V_d \to U_x ) is a homeomorphism for every ( d \in D_x ), then is it the case that
[\bigsqcup{d \in D_x} V_d \cong (D_x \times U_x)]
where ( V ) is a family of sets indexed by ( D_x )

[gentle osprey]

Nuclear Catapult

[desert vortex]

yes

[storm night]

Interesting. I don't see why we would use the convoluted coproduct definition.

The topology book by Stephen Willard also defines the covering space using the coproduct. Is there a good reason why we're using a coproduct? Does it make proofs more intuitive?

[opaque scroll]

Is disjoint union more convoluted than product spaces?

I think disjoint union is pretty intuitive

[quick delta]

Intuitively here, what we’re doing is fundamentally a disjoint union, and it’s a cute fact of topology that it can be written as that product over a discrete set

[storm night]

I agree. A disjoint union is just as intuitive as a product space. But a product space by itself is more intuitive than a disjoint union equipped with the requirement that the restriction of π to every member of the family of V is a homeomorphism. Or something like that.

[hidden abyss]

You need to assume the homeomorphisms commute with π though

[storm night]

@hidden abyss Are you referring to my definition or my last post?

[hidden abyss]

Your definition. The homeomorphisms $\pi^{-1}(U) \cong D\times U$ can't be arbitrary, you need to assume $\pi$ corresponds to $\text{proj}_U$ via it

[gentle osprey]

Jussari

[storm night]

@hidden abyss Would you agree with the following statement if we were to talk strictly about sets rather than topologies?

[gentle osprey]

Nuclear Catapult

[storm night]

Because this statement I'm convinced is true. I proved it in Cubical Agda, or at least to myself.

{-# OPTIONS --cubical --safe #-}

open import Agda.Primitive public
open import Cubical.Foundations.Prelude
open import Cubical.Foundations.Isomorphism renaming (Iso to _≅_)
open import Cubical.Data.Sigma

open _≅_

variable
 D U : Type

P1 : (V : D → Type)(π : ∀ d → V d ≅ U) → (Σ[ x ∈ D ] V x) ≅ (D × U)
-- Proof below
P1 V π =
 iso (λ(x , H) → x , π x .fun H) (λ(a , b) → a , (π a .inv b)) (λ(a , b) →
  let H = π a in
   (a , fun H (inv H b)) ≡⟨ cong (a ,_) (rightInv H b) ⟩
   a , b ∎)
  λ(x , H) →
   let G : inv (π x) (fun (π x) H) ≡ H
       G = leftInv (π x) H in
  x , inv (π x) (fun (π x) H) ≡⟨ (λ i → x , G i) ⟩
         x , H ∎

[hidden abyss]

Yeah this is true

What I was trying to say above is that you explicitly need to assume the corresponding of these diagrams to commute:

[gentle osprey]

Jussari

[unreal stratus]

To be honest there are already various non-equivalent definitions of covering space

[quartz horizon]

You can think of it as tensoring a topological space with a set

[gaunt linden]

Yeah, with the originally quoted definition, then the line with two origins would be "covered" by two lines, which we presumably don't want.

[solemn iris]

Can I get some hint on this?

[gaunt linden]

The problem seems to miss defining a topology on C(M).

[alpine nest]

It does; I suppose by default it would be the uniform convergence topology

[alpine nest]

I think (especially if we're talking uniform convergence topology on the C(M) spaces), Stone-Weierstrass theorem should help here

[solemn iris]

oh.. i see

thank you

[quartz horizon]

Oh cool, in a way this is showing $C(X) \otimes C(Y) \cong C(X \times Y)$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[warm kettle]

Is the unit sphere connected in an infinite dimensional normed vector space?

[tiny obsidian]

I think it should be path connected (go through 0)?

[alpine nest]

0 is not in the unit sphere; but the sphere is path connected nonetheless

(take two points, take the "straight line" path connecting them, and project that path onto the unit sphere)

[warm kettle]

OH. That makes so much sense

[alpine nest]

That won't work if the two points are antipodal (or generally linearly dependent if we're over C), but then you can pick an intermediate point that's linearly independent of them

[warm kettle]

Thanks, Outsider. So we don't even need to know much about the space

[tiny obsidian]

Oh the sphere

[alpine nest]

I mean, the ball is connected as well, but that's much easier because it's convex

[quartz horizon]

It is strange to me that the infinite dimensional sphere is contractible

[cosmic mirage]

its so useful though!

[quartz horizon]

Oh I didn’t know it was useful

[cosmic mirage]

yeah, for example if you quotient out the antipodal C_2 action you get RP^\infty, and this establishes this space as BC_2

because S^\infty is a C_2-space that is non-equivariantly contractible

[quartz horizon]

Uh

Idk what BC_2 is

[cosmic mirage]

it is the classifying space for the group C_2

[quartz horizon]

@.@

[cosmic mirage]

i dont know what things you know, but if you know homological algebra things then group (co)homology of G with constant coefficients will be the same as singular (co)homology of BG with those same coefficients

actually iirc you seem to be working towards principal bundles, so it is good to note that S^\infty --> BC_2 is the universal one

i.e. all other principal C_2 bundles (on nice spaces) will be pulled back from this along a homotopy class fo maps into BC_2

and this is true in general, another name for S^\infty is EC_2, and this story holds if i replace C_2 with any (to not get in trouble, let me say compact Lie) group

[unreal stratus]

Here is a funny observation. Consider the contractible groupoid C = { * -> *} (the "walking isomorphism"). Its nerve is a contractible simplicial set, and if you stare enough, you can see that its realisation is S^oo

[quartz horizon]

bruh

[quartz horizon]

I am only just finishing chapter 3 of rotman

Chapter 4 is when he starts singular homology

[quartz horizon]

Ohhhh

This makes more sense to me

[unreal stratus]

c.f. e.g. https://mathoverflow.net/a/417632/510603

[quartz horizon]

So a principal C_2-bundle over a space X is encoded in a map X -> RP^infty?

[cosmic mirage]

yes!

lets say for X paracompact

[quartz horizon]

How do you actually get that map

Mhm

[unreal stratus]

Also like you have presumably seen BG for the groupoid with one element * with Hom(*,*) = G. If you take this groupoid, take its nerve, and then its geometric realisation, then this is BG (the space)

[cosmic mirage]

there are many, many, many ways

[quartz horizon]

Only works for discrete groups or smth

But I have heard of this vaguely

[unreal stratus]

Yes indeed

But this is kinda forced if we are only using bare (1,1)-cats too

[quartz horizon]

mhm, right

[cosmic mirage]

here is A way, and I will assume the space X is compact: C_2 has another name, O_1 (R). so these principal bundles will correspond bijectively to line bundles over the space with isometric transition maps. i claim that i can embed any such line bundle isometrically into a high dimensional trivial bundle (i.e. just X x R^n for some large n; here is where i have used compactness). then, each point x in the space X has a fiber that is a line in R^n, so you just send x to the point in RP^\infty corresponding to that line in R^n \subset R^\infty

[quartz horizon]

Oh that’s pretty clever

[cosmic mirage]

also, if you know anything about homotopy colimits: BG (for finite G, I don't want to get in trouble) is also the homotopy colimit of the constant diagram BG --> Top that sends everything to a point, where by BG I mean the category (i.e. the diagram that encodes the point with trivial G action)

if you did this with a usual colimit you would get something boring, so you need it to be homotopy coherent. in this case, this means replacing the point with a homotopy equivalent thing with a free G-action, which is precisely what EG is, and then taking the usual colimit (i.e. quotient out the G action) to get BG

[cerulean oriole]

I have formed the vague impression (IDK how useful it is) that BG is how you take the quotient of a point * by the trivial action of G in a meaningful way (i.e., so that you somehow identify G-many things in a "neighbourhood" of the point). Naively taking the quotient would just give you the same point, so you replace the point with a (G-equivariantly) homotopy-equivalent space EG with an action of G, such that the action of G on EG is now free. As a result EG/G is "really" a quotient by G and you call it BG.

This is the same thing as the homotopy-colimit, I think.

[unreal stratus]

Yes, though formalising this in an entirely non-circular way is a bit hard here aha

Like the diagram you can taking a homotopy colimit over is BG

[cerulean oriole]

Is it not true? The simplicial set construction of EG for discrete G has a G-equivariant contracting homotopy (at least as a simplicial set) AFAIR.

Or maybe I messed up.

[unreal stratus]

I am pretty sure if this were true you would get a homotopy equivalence on orbits, which is false

Like the homotopies, being equivariant, pass to the quotients

Or, better: there isn't even a single G-equivariant map * -> EG.

[kind marlin]

can someone verify this for me?

let S = Q coprod (Q x 2^N) coprod 2^N.

Is S homeomorphic to S^2 and S coprod S?

and is S not homeomorphic to S^N, and not homeomorphic to S x N?

where N has the discrete topology

[kind marlin]

@gritty widget ^ might be a candidate for being only finitely idempotent and coidempotent :0 but it feels very "non-primitive" :( but I think we can more generally say that

if A is N-coidempotent and only finitely idempotent, and B is N-idempotent and only finitely coidempotent, then

A coprod (A x B) coprod B is only finitely both

"pf" (😵‍💫 ) squaring gets us

(A + AB + B)^2 = A^2 + A^2 B^2 + B^2 + A^2 B + AB + AB^2
= A + AB + B + AB + AB + AB
= A + B + 4 x AB = A + B + AB

and adding gives us

(A + AB + B) + (A + AB + B) = A x 2 + AB x 2 + B x 2 = A + AB + B

raising to N is where things get,,,, weird. i have thoughts, kind of like binomial theorem but where finite powers just get reduced and infinite powers get distributed amongst terms? i might have to think about if this can be more formalized. the heuristic idea:

(A + AB + B)^N =
A^N + A^N x AB + A^N x B + A^N x AB x B

• (AB)^N + (AB)^N x A + (AB)^N x B + (AB)^N x A x B
• B^N + B^N x A + B^N x AB + B^N x A x AB
• A^N x (AB)^N + A^N x (AB)^N x B
• A^N x B^N + A^N x B^N x AB
  etc

im basically "partitioning" N amongst the three terms such that the exponents "add to N", and finite exponents just reduce to 1 so our cases are just exponents of 0, 1 or N. I think you can then simplify a lot from the idempotence and coidempotence, and find contradicting properties or something, but im not really sure. before that even though, can this be made rigorous?

non-N coidempotence is a lot easier to verify I think

(A + AB + B) x N = A x N + AB x N + B x N = A + AB + B x N
and i think for coproducts, if n-1 arguments are homeomorphic and one isn't, then the overall coproducts cannot be homeomorphic, so this isn't N-coidempotent (but I might need fact-checking)

if that does work its kind of a quirky way to generate these things from families

it still doesnt suggest how to characterize only finite idempotence or only finite coidempotence in isolation though

if the ^N logic does work out, then idempotence/coidempotence lets us simplify down to

(A + AB + B)^N =

A^N + A^N x B

• (AB)^N
• B^N + B^N x A

using infinite idempotence of B:

= A^N + A^N x B + A^N x B + B + A x B
= A^N + A^N x B + B + AB

then, as long as A is not homeomorphic to A^N + A^N x B, we're good. i wanna say the fact that A and A^N are not homeomorphic should imply that? but im not entirely sure (and this is all predicated on iffy reasoning anyway)

[rancid umbra]

coming back around to this, why does a subset Z of X that is the image of a CH test map t : C —> X have to be Hausdorff? it’s obviously compact…

[tender halo]

it doesnt

[rancid umbra]

then i’m confused again. S should be a set of compact hausdorff spaces

am i misinterpreting what you meant?

[tender halo]

uh honestly i think i misread what the condition was

[rancid umbra]

oh okay lol

[tender halo]

and was answering a different question from what was asked lol

[rancid umbra]

thanks ha

i’ll keep thinking about it then

[tender halo]

i mean ok

if a k-continuous function fails to be continuous, then if you restrict the codomain to be the image it fails on a space of cardinality at most |X|

idk i really thing this is written badly

i think what it actually says is that you can take the testing spaces C in the definition of k-continuity to be from a set

instead of all CHaus spaces

that makes way more sense imo

[rancid umbra]

let \kappa be the least cardinal such that every point of X has a nbhd basis of cardinality at most \kappa (note that \kappa is bounded above by 2^|X|).
is it true that f : X -> Y is continuous if and only if for every convergent net x : \kappa -> X, fx : \kappa -> X is convergent and lim fx = f (lim x)?

this is sort of a generalization of the case when X is first-countable

[rancid umbra]

hoping to mimic the proof that a first-countable space is compactly generated to show 1 => 2 here

[tender halo]

that is called the character of the space

and i think its true yeah

or at least it does sound true

[rancid umbra]

if it is, then i think you just need to take S = {\kappa + 1}

or S = the compact ordinals in \kappa + 2 or something

like, for a first countable space X, f : X -> Y is continuous if and only if for every continuous function x : \omega + 1 -> X, fx : \omega + 1 -> Y is continuous

so you only need the test functions out of \omega + 1

[cerulean oriole]

Assume (1). For each topology t on X strictly finer than the given one, let f_t: C_t -> X be a map from a compact Hausdorff space to X which is continuous but not continuous if we put the topology t on X. The set of all (C_t, f_t)'s works (there are only set-many topologies).

(A simplified version of this is found in this mathse question.)

[thick vigil]

Is there a better way to set up this problem?

Also I meant z^2 not Z

[unreal stratus]

Yeah that seems fine

[thick vigil]

How is the rest of my solution? (If its blury i can send the whole file)

[thick vigil]

Ugh it blurred. Its problem 3

[rancid umbra]

will check this out when i can, thank you!

[drifting plaza]

Can somebody help me with Scewed Wilcoxon distribution please

[lusty trench]

To minimize $\Vert p-q+n \Vert$ is to minimize the distance between $n$ and $q-p$. So let $n$ be the result of rounding each component of $q-p$ toward the nearest integer. There may be up to 4 such $n$, but in any case, the minimum value is attained, i.e., your infimum is actually a minimum.

[gentle osprey]

Eduardo León

[lusty trench]

Also, the Euclidean distance on $\mathbb R^2$ isn't very convenient. It's better to use the distance induced by the norm $\Vert (x,y) \Vert = |x| + |y|$. You may recall from real analysis that all norms on $\mathbb R^n$ induce the same topology.

[gentle osprey]

Eduardo León

[lusty trench]

Notice that $\Vert p-q+n \Vert$ can also be interpreted as the distance between $p+n$ and $q$. For our optimal $n$, each component of $p-q+n$ is bounded between $-1/2$ and $1/2$. If we let $S(q)$ denote the square in $\mathbb R^2$ centered at $q$, whose sides have length $1$, then $p+n \in S(q)$.

So, when you prove the triangle inequality $d([p], [r]) \le d([p], [q]) + d([q], [r])$, you can assume that one of the following holds:
\begin{itemize}
\itemsep 0em
\item $p, q \in S(r)$
\item $p, r \in S(q)$
\item $q, r \in S(p)$
\end{itemize}
And clearly the third case is equivalent to the first, swapping the roles of $p$ and $r$, so you only need to consider the first and second cases.

[gentle osprey]

Eduardo León

[oblique sentinel]

What is the formal explanation for needing countably many sets U_p in the Urysohn lemma proof? My intuition is, that for finitely many U_p we have jumps that are not "smoothed out". But I would need that more rigourosly

[tender halo]

interpolating between finite amount of sets is the same as interpolating between two

so choosing a finite amount doesnt help

by interpolating i mean like constructing the relevant urysohn function which is kind of like interpolation if you squint your eyes at it

[oblique sentinel]

Ok I get that. But like if I had to prove that for a finite amount of U_p the function wasnt continuous in the topological sense, where does it go wrong

[crisp lintel]

It's going to involve connectedness basically

Basically if the domain is connected the image also must be connected

So if your image is a finite set of points in the reals, that isn't connected

so for some topological spaces you might get lucky and the finite amount will give you something continuous, but in particular for spaces that are at least a little bit connected it's unlikely

[tender halo]

it goes wrong in the moment where we prove the function is continuous because we rely on the property that between any two real r_1, r_2 there needs to be a U_p such that r_1 < p < r_2

we need p's to be dense

[oblique sentinel]

How does that make it go wrong? Can you explain a bit further? I tried "proving" it myself, but don't quite get it

[tender halo]

well to prove it is continuous you prove that preimages of (-inf; r) and (r; +inf) are open

we use density when proving continuity of the latter

in particular, we use that f^-1[(r; +inf)] is equal to \cup_{p > r} X \ cl(U_p)

that equality uses density, or more accurately it uses that p gets aribtrarily close to r

and it is here that we use that if p_1 < p_2 then cl(U_p_1) \subseteq U_p_2

[oblique sentinel]

and so the function isnt continuous because this set equality isn't true for finitely many p's. I really have big trouble getting to the conclusion here. Sorry

[kind marlin]

don't we need the indexing set to be dense in the interval

oh yeah that was already said

density is used a lot in the proof right, any finite set isn't going to be dense bc it's already closed

[tender halo]

i mean it might be continuous

it doesnt have to be discontinuous, we just cant prove that it is continuous

[tender halo]

i mean you should look at why this is true

and you will see where we use density

[oblique sentinel]

Ok I sadly have no time anymore today. I will try again tomorrow and hope I get it. So weird to me how some topics in math just are not easy for me and others are a joke

I guess everyone has their weaknesses

[oblique sentinel]

Well one last thought: I can imagime if we don't define the U_p for all rationals in [0,1], then there surely will be an open intervall of reals for which no x in X exists that gets mapped into it.

And so for that particular open set I guess we have no open preimage / not even a meaningful preimage so that we have something like a discontinuity?

[ripe creek]

spaces

[kind marlin]

You’ll never be able to find a general counter example because if X is discrete then we trivially have continuity

try picking a specific topology X and finding a counterexample for it

[ripe creek]

Topology is fun I think.

[alpine nest]

It is!

[ripe creek]

I started watching this playlist just now just for fun. Learning just for fun. I'm 12 minutes into the first video right now. It is very interesting.
https://youtube.com/playlist?list=PLOROtRhtegr7DmeMyFxfKxsljAVsAn_X4

I have heard of the Möbius strip, the torus and the Klein bottle before. It is interesting to learn that they all can be formed through certain ways of connecting the sides of a rectangle to each other.

[lusty trench]

Even better, you can parametrize them (as well as the real projective plane) as subsets of R^4 in a way that makes the idea obvious.

[umbral hamlet]

it might be tough to follow this without having pointset topology

[ripe creek]

I have learnt general topology before.

It was a few years ago.

[ripe creek]

wonky spaces

[umbral hamlet]

ah good

[prime elbow]

Let X be a metrizable space, I have to prove if every metric on X implies every continuous real valued function is bounded.

Since X is metrizable therefore there exists a metric function d such that it induce same topology.

Let f be the real valued continuous function.

Now I am defining new metric function d' such that, d'(x,y) = d(x,y) + | f(x) - f(y) |.

We can verify d' is a metric on X.

Since d and d' metric on X so it has to be bounded but if f is not bounded then d' is not bounded.

Is it correct? I didn't use the continuity of f.

[tiny obsidian]

What are you proving?

If every metric is bounded then every real valued continuous function is bounded?

Regardless, how do you know d' generates the same topology as d

[opaque scroll]

Did you mean to say that every metric on X is bounded or something like that?

You're sentence seem to cut off halfway there

[plush folio]

btw, for what kind of metrizable spaces are every metric bounded? It's atleast true for compact spaces, but are there non-compact examples?

[robust shadow]

Don’t think so, if the space isn’t compact then there is an unbounded continuous function(take a countably infinite set with no limit point {x_n} the set is closed f(x_n)=n can be extended to a cts function on the whole space by Tietze extension theorem)

[prime elbow]

Yeah sorry, yes every metric on X is bounded implies every continuous real valued function is bounded

[prime elbow]

No, d' not induce the same topology as d

[tiny obsidian]

That's where you go wrong. When we say metric on a metrizable space we would implicitly mean metrics that induce the same topology. You've found d' a metric on X as a set, but not on X as a top space, and so d' unbounded is fine

[plush folio]

Also the fact that you didn't use continuity of f is a sign that the proof must be wrong, because you can easily create an unbounded real-valued function on any (infinite) topological space if you don't care about continuity

[prime elbow]

Given : every metric on X is bounded so d' can't be unbounded

[prime elbow]

Maybe we don't need f to be continuous

[tiny obsidian]

"We would implicitly mean one that induces the same topology" - me just now

d' does not

We are not assuming it's bounded

[prime elbow]

Not same topology

[tiny obsidian]

Yes exactly

[prime elbow]

It is given that every metric on X is bounded

And X is metrizable

[plush folio]

Every metric on [0, 1] is bounded, but there are loads of discontinuous unbounded functions on [0, 1]

[tiny obsidian]

Every metric (that induces the same topology)

With the bit in brackets implicit

I agree that maybe this should be explicit

[tiny obsidian]

Wait it is even explicit

[prime elbow]

So what is the wrong in my proof?

[tiny obsidian]

Yeah so you found a metric d', it does not induce the same topology, and so (i) does not apply and mean d' is bounded

[plush folio]

I think that's what Edward is trying to explain

[prime elbow]

I see

[tiny obsidian]

d' unbounded is not a contradiction

[prime elbow]

My mistake 🥲

Any hint how to proceed?

I think i should show i implies iii.

Because iii to ii and ii to iii are done, also I showed iii to i.

And if X is not the limit point compact, then there exists an infinite set which is discrete subspace. I can define metric on that set which is unbounded but can we extend that metric on X?

[tender halo]

d' does induce the same topology though?

it induces the same limit operator, x_n converges to x in d iff it converges to x in d'

[prime elbow]

Yes and i think then we can say they induce the same open set

[tender halo]

if they induce the same convergence then they induce same closed sets then they induce the same topology

[prime elbow]

Yes

Thanks @tender halo you saved me

Here we used the continuity

[crystal vortex]

What does the notation R^n/(R^n \ 0) mean ? I know you can quotient a topological space using an equivalence relation, but I haven’t seen this?

The context is I am trying to understand the definition of relative cohomology? I am reading Milnor and Stacheff and they use H^i(E, E0, Z/2Z) which I am not familiar with

[prime elbow]

Any hint for part a?

[cloud kindle]

Start with the definition of a Hausdorff space, use the definition of a retract and kinda just follow your nose

In a sense theres only one way the proof can go

[prime elbow]

So, I have to show cl Y = Y.

For that, if I assume z \in cl Y but not in Y then r(z) and z are two distinct points and I can separate them by open sets, but how do I use continuity of r?

[tender halo]

the retract is the solution of the equation r(x) = x and is therefore closed smile

[prime elbow]

Sorry i don't get it

[tender halo]

if f and g are continuous functions into a Hausdorff space, then the set {x|f(x)=g(x)} is closed

[prime elbow]

Yes

Oh so r: Z -> Z and id : Z -> Z gives Y = { x | r(x) = x }.

Thank you bussy beaver

[unreal stratus]

I should say like I very much doubt that they mean R^n/(R^n \ 0) for this relative cohomology

[unreal stratus]

Good

[night crane]

jack

wassup man

/summon jack

[pearl valve]

When dealing with pointed spaces definition of homotopy, is there an easy to visualize geometric example of spaces A <= X where A is a deformation retract of X but not a strong deformation retract of X?

Online I can find examples of "deformation retracts" as functions that aren't strong deformation retracts, but I'm unsatisfied by these as Im rather looking for a defoirmation retract of spaces, for which there is no possible strong deformation retract

[unreal stratus]

I think the usual example is the comb space (pictured). The point {(0,1)} is a deformation retract but not an SDR.

[undone yacht]

Wouldn't something like the projection π: R^2 \ {(0, 1)} -> R x {0} work?

[unreal stratus]

Where does (2,1) go

[undone yacht]

to (2, 0)

let me check maybe I'm wrong

[unreal stratus]

Okay I was just confused why it wasn't written as R lol but fair

Anyway this is not a deformation retract because the source and target aren't homotopy equivalent

[pearl valve]

are these spaces homotopic? the left one has a hole adn the right does not?

oh yeah

[undone yacht]

my idea was that deformation retracts force homotopy equivalence, so just pick a retract from non-simply connected to simply connected space

and projections are easy examples

[unreal stratus]

That isn't what the question is about though

[undone yacht]

oh right I misread it

[unreal stratus]

Nws

[unreal stratus]

Anyway this seems to be an example of what you want

I think there is a limit on how "geometric" an example you can give – usually a bit gnarly like this

[pearl valve]

why wouldnt this be a strong deformation retract?

I think I saw an example similar to this in one of my classes, it had something to do with sequential definitions of continuity being violated?

is it similar to this argument?

[unreal stratus]

Ah yeah seems reasonable

[unreal stratus]

I believe so yes

Ig the argument here is the top left (0,1) is fixed, so all of the tops of the rays also have to be sent somewhere close to (0,1), which is bad cause they should travel down the thing?

I believe yes same argument for the comb space, just the rays are arranged slightly different

[pearl valve]

yeah this was for a slightly different argument about connectedness

but the contradiction had something to do with sequentail continuity being violated

[unreal stratus]

Yeah local path connectedness?

[pearl valve]

yeah

[unreal stratus]

But yeah exactly this sort of argument should work

[undone yacht]

lol

is the AI thinking of S^1 living in R^2?

[pearl valve]

probably

[unreal stratus]

Like basically in the example you wrote, if x_n is the top of the nth ray (with x_n getting closer to (0,1) as n gets better), then uh

Since x_n -> (0,1), for any SDR H : X x I -> X (X = ur space)you would have to like have H(x_n, t) -> x_n as n -> oo for any t. Then this is kind of bad if you use compactness as you can see the x_n have to stay on the same ray for all time and hence stay away from (0,1)

That kind of argument

Yikes that is terrible

Lol my advisor put smth into AI yesterday and found a super useful reference I wish I had known about like a year ago, so seems very variable

[undone yacht]

For tracking references it seems really good, like a search engine on steroids

[pearl valve]

hmm I guess I am still a bit confused by this example, to me it seems that all you need for a deformation retract in this caes is for (0,1) to stay at (0,1) during the whole shrinking process, I dont see how this comes with a contradictiojn

[queen prism]

this is like the one good application I've found

[unreal stratus]

Maybe I am mistaken and you do actually need the comb space

Actually no, I am fairly confident in this example

Problem is still yeah that there is nowhere for the tips of the rays to go

[gaunt linden]

Yeah, (0,1) has arbitrarily close neighbors that need to get all the way down to (0,0) in order to retract to (0,1).

[kind marlin]

locally compact + ~compact + locally connected + ~connected + locally metrizable + ~metrizable

pi base couldnt find an example or prove this impossible :0 does anyone an example/proof

0 motivation for this i was just playing with keywords

oh if i just drop ~connected theres lots of results, and i guess i can just do a disjoint union of two copies of something to get a disconnected one

[desert vortex]

I cant decipher what youre asking, what is ~

[queen prism]

not?

[desert vortex]

mmmmmmmmmmmmmmmmmmmmm I see

[queen prism]

fellow seer

[desert vortex]

I c

[young stone]

They were asking for a space that satisfies all those properties and ~ means negation

[opaque scroll]

Like the disjoint union of two long lines? Would that work

It's locally homeomorphic to R, so has all the nice local properties, but is not compact connected or metrizable

[opaque scroll]

Yeah locally connected is equivalent to being the disjoint union of connected stuff, so that's actually the only thing you can do

[prime elbow]

If X is a topological space and U is an open subspace of X, then can I imbed U into X × R by defining u -> (u, 0)?

This function is continuous, injective and inverse image U × {0} -> U, (u,0)-> u, is continuous, right?

[desert vortex]

yes you certainly can!

[prime elbow]

Then in c part why do they choose specific mapping?

[desert vortex]

because I think the idea is to embed U into X x R as a closed subset

[prime elbow]

But image is graph of \phi, right? which is already closed in X × R

[desert vortex]

not quite so, the graph of phi would be closed in U x R

but now phi : U -> R and we consider the graph of phi as a subset of X x R

pick a simple example: U = (0,1) X = R, d(x,y) = |x-y| and draw the graph of phi as a subset of X x R = R^2

[prime elbow]

Yeah if I take phi is Identity then graph is not closed in R^2

I see

[desert vortex]

I think the basic idea is this: we cant use the same metric that gives us the topology on X for U because U might not be complete under this metric, a sequence might approach the boundary of U and therefore has no limit inside U, so we take U and "stretch" U close to the boundary (this is from the particular embedding), so no cauchy sequence in this new stretched U can get close to the boundary

phi(x) is going to be big if x is close to the boundary of U and as x moves closer to the boundary of U then phi(x) will blow up to infinity, this is how to stretch U

[prime elbow]

Yes we can't use the same metric, for example (0,1) in usual topology on R

[desert vortex]

cool exercise

[prime elbow]

How do I show the image is closed in X × R?

[desert vortex]

put a metric d' on the graph like: d'((x, phi(x)), (y, phi(y))) = d(x,y) + |phi(x) - phi(y)| and prove it gives you the product topology

this is not pulled from thin air but in fact just this: if d_X is metric on X and d_Y is metric on Y then d'( (x,y), (x', y')) = d_X(x,x') + d_Y(y,y') is a metric on X x Y that gives the product topology

you might try and prove this lemma first

then show that this metric makes graph of phi complete, you dont actually need to show its closed

let me know how it goes!

or you could show its closed w.r.t the metric d' it would also work

[prime elbow]

Yes

I know this facf

[desert vortex]

great

then if a sequence (x_n, phi(x_n)) is cauchy in this d' then its cauchy in both the coordinates, so x_n cauchy w.r.t d in X and phi(x_n) cauchy w.r.t | | in R

so phi(x_n) has to be bounded from above and then d(x_n, X - U) is bounded from below, this is exactly what we were looking for because that means it stays away from the boundary

[prime elbow]

So it gives that limit of x_n is in U

[desert vortex]

yep but why?

[prime elbow]

Because, let's assume x is the limit of x_n if it is in X\U then d(x_n, x) ≥ d(x_n, X\U) > 1/k, where k is fixed integer, for all n in N, since d(x_n, X-U) is bounded from below so we can find such k.

k is a positive integer

So it contradicts that x_n converges to x

[desert vortex]

nice

[prime elbow]

Thank you, I got it

[iron bolt]

local properties are preserved under disjoint unions, the global properties you mentioned are not

so you can take any space with all the nice local properties you want and take a large enough disjoint union of copies of it to get a space that doesn't also have the global properties

actually nevermind, metrisability is also preserved under disjoint unions. but taking a disjoint union of spaces that have all the local properties and are not metrisable should work, like the example jagr already gave

[granite slate]

Show that a continuous function $\varphi: \mathbb{S}^1 \rightarrow \mathbb{S}^1$ extends to a continuous function $\phi: D^2 \rightarrow \mathbb{S}^1$ iff $\varphi$ has degree 0

[gentle osprey]

dompa

[granite slate]

i can prove that extension implies degree 0 but i just cant see the reverse

[paper wedge]

is it true that

[cosmic mirage]

here is a hint: use a homotopy between \psi and the constant map that sends everything to a point

[paper wedge]

if it has degree 0 then its nullhomotopic

if so then maybe you can use that to find ur disc

oh i was just giving him a hint haha 😄

[cosmic mirage]

ah nvm lol, deleted

[paper wedge]

😄

ig the purpose of this hint was

that i dont think its obvious (ig i mean it requires proof) that degree 0 --> null-homotopic

[prime elbow]

For the last part, we know h(X) is dense in Y, so if I show every cauchy sequence in h(X) converge in Y then we are done.

Let h(x_n) is a cauchy sequence in Y.

Since h(x_n) = [ (x_n, x_n,...) ] so for each eps > 0 there exists k in N such that D( h(x_n), h(x_m) ) < eps for all n,m≥ N.

So it implies lim d(x_n, x_m) < eps for all n,m ≥ N, here limit is over t, but since h(x_n) and h(x_m) is constant sequence implies d(x_n, x_m) < eps for all n,m≥N, implies x_n is cauchy sequence.

Now how do I show h(x_n) is convergent?

[tender halo]

well

i would start with figuring out what point it converges to

[prime elbow]

I got it, h(x_n) convergent to [x] = [(x1, x2,...) ] and [x] in Y because x_n is a cauchy sequence which we showed already

Thank you

[orchid perch]

How do u show that {x|f(x)>g(x)} is measurable if u know that f and g are lebesgue measurable from R^n to R?

[rancid umbra]

f - g : R^n -> R is measurable since ||(a,b) |-> a - b is measurable and the composition of measurable functions is measurable.||

use this to ||express the set in question as the preimage of a measurable set under a measurable function||

[rancid umbra]

you can view this as the pullback of f - g along the inclusion of the (0,oo) into R.

[young stone]

small pedantic comment: || f \circ g is Lebesgue measurable if f is Lebesgue measurable and g is borel ||

[frail jewel]

Just a quick ques.

If $A,B$ are subspaces of $X$ and $A \subseteq B \subseteq cl(A)$. If $A$ is connected then $B$ is also connected.

In general if we look at a topological property $P$ instead of just a specific property - openness in subspace topology(True), connectedness (True), separability (True), compactness (False), path connectedness (False). Do we know for which topological property can the above statement be true for?

[gentle osprey]

HandT

[frail jewel]

or just A being dense in B

[brisk peak]

are you asking if A, B with those inclusion relations inside X are there lemmas like "A has P implies B has P" for all the important stuff like connected, compact, complete and so on

[frail jewel]

Yes

[brisk peak]

or a characterisation of the ones where you force it to be true

[frail jewel]

Yes

[brisk peak]

since you listed the answer for the main lnes

ones*

[frail jewel]

Yes, it looked very similar to squeeze theorem so I was wondering what class of topological properties can I apply the theorem

[brisk peak]

for totally arbitrary topological space X?

[frail jewel]

I guess so.

Although if you can suggest some restrictions that can also be helpful

[brisk peak]

I am extremely cautious of ever saying anything about non Hausdorff spaces

It's counterexample city

[frail jewel]

Hmm, sounds reasonable.
I will try to restrict it to Hausdorff spaces to see if I can classify the topological properties

[brisk peak]

It's a good question

I don't want to answer it without reviewing specifics as I'm rusty

[frail jewel]

Sure, no problem.

[last hound]

What kind of function algebra structure can satisfy separability (separating points), non-vanishing (vanishes nowhere), and compactness all at the same time? How do you construct it?

[crisp lintel]

What in the chatgpt

I don't even know what the question is supposed to be

[red sage]

lol

“function algebra structure”

@last hound maybe you rephrase your question with adding more context to it.

[gaunt linden]

It sounds somewhat like they're asking for an example of the Stone-Weierstrass theorem.

[quartz horizon]

what's your preferred method to show that $U(n)$ is connected?

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[opaque scroll]

Writing down paths from the identity to diagonal matrices and then conjugating to get a path to every element is very explicit at least

Could also show the exponential map is surjective, but I guess the proof amounts to the same

[quartz horizon]

that uses the spectral theorem, right?

[quick delta]

Yes

[desert vortex]

another W for the humble spectral theorem

[unreal stratus]

The proof of Gram–Schmidt shows that GL_n(C) deformation retracts onto U(n), and the former is path connected by other arguments lol

[opaque scroll]

Do you also have an "other argument" in mind?

[young stone]

Just follows from row reduction right?

There's a path taking a matrix A to an elementary matrix times A (within GL_n(C)),
For adding rows you can do t maps to A + t times the row you are adding.
For scaling a row you can't just do a straight line path because it will pass through zero, take a path in C that goes around zero (This is also where this breaks for GL_n(R))

hm how do you swap rows though

[unreal stratus]

Like rotations or smth

[unreal stratus]

Uhh there's an argument with paths and stuff which I forget

[young stone]

The rows are linearly indepdendent so we can just do tR_1 + (1-t)R_2 i think

and vice versa for the other row

[quartz horizon]

what happens at t = 1/2 though

[young stone]

nothing you just get 1/2 R_1 + 1/2 R_2, this can't be zero because they are linearly independent

[quartz horizon]

but then isn't the other row also 1/2 R_1 + 1/2 R_2?

[young stone]

oh crap you are right

you'll have to use paths in C

to avoid intermediate value theorem

[quartz horizon]

yeah i imagine that's what you want

[unreal stratus]

Ues

*yes

[lament steppe]

Ive been thinking about the below problem:

Given: f : X -> Y where X is a first countable space, if x_n -> x implies f(x_n) -> x, show that f is continuous.

I have shown that the converse is true (if f is continuous and x_n -> x, show that f(x_n) -> x... where the first countable condition is not necessary.)

Im not exactly sure where/how I can apply the first countable condition on X to achieve this. I have been trying to show that x_n -> x implies f(x_n) -> x require f to be continuous, but it feels like there are always cases where that requirement isnt exactly necessary.

Any hints on this would be great (either what contrapositive to look at or how I might try to attempt to apply the first countable condition.) Thanks!

[gaunt linden]

Yes, proving the contrapositive seems to be the right way forward.

[crisp lintel]

you want to go by contrapositive here, if you've seen the proof for metric spaces its largely the same

it might be worth just doing it for metric spaces first and then modifying the proof to get the first countable case

[gaunt linden]

Assume there's a point x where f is not continuous, and show there must be a sequence converging to x whose image doesn't converge to f(x).

It may help to show as a lemma that x must have a neighborhood basis that's not only countable but consists of a sequence of nested neighborhoods $N_1 \supseteq N_2 \supseteq N_3 \supseteq \cdots$.

[gentle osprey]

Troposphere

[tender halo]

the most straightforward way to show it is to show that preimages of closed sets are sequentially closed

from which what you want flows immediately

[gaunt linden]

Hmm, I suppose that by "a point x where f is not continuous" I must mean something like there's an open A in Y, and a point x in f^-1(A) such that f^-1(A) doesn't contain any basic neighborhood of x.

[lament steppe]

Awesome. This gives me a lot to chew on, thanks a bunch everyone 😄

[waxen urchin]

hey guys, one question on how we define an open set in topology. yesterday i pondered a little over the resemblance of the definitions of an open set in a topological space and an equivalence relation. both definition provide nothing concrete, staying general, giving us that any set (or relation) is open (or that of equivalence) if three respective properties hold. I undersand that the way we define an open set allows us many different kinds of sets, and I just wanted to clarify if there's much more to an open set in a topological space than there's to a more concrete open set in a metric space, because it seems that there are many things that we could call open using these three. Example: Let $X$ be a set. Let $A, B$ be subsets of $X$, i.e. $A \subseteq X, B \subseteq X$. We call set $Y$ open if $Y \subseteq X$. Verifying the properties of an open set is left to the reader. So in this sense A and B are open sets, right? Just want to understand what we can make out of the concept of an open set, because in metric spaces it's all about metric, i guess? Sorry for a long text.

[gentle osprey]

nezhivoy

[flat plinth]

Is a second countable Hausdorff locally Euclidean space same as a metrizable locally Euclidean space?

I think one direction is doable cuz of Urysohn's theorem but I am not sure if a metrizable locally Euclidean space has to be 2nd countable

[gaunt linden]

The disjoint union of uncountably many spheres is metrizable and locally Euclidean, but not second countable.

[tacit drift]

im having trouble finding information on the cayley klein metric

it seems to be defined over projective spaces, but projective spaces over what fields? the notions im finding seem to be tied to C and R

[flat plinth]

Ah I see. Is there any reason why we care about 2nd countability? Partitions of unity still work cuz a metrizable space is paracompact. So what do we lose?

[neon breach]

I think you're describing the discrete topology there. In general I don't think open sets and equivalence relationships are analogous

[gaunt linden]

But the concepts of "an equivalence relation" and "a topology" are at similar levels of abstraction, in that simply saying "this is an equivalence relation" doesn't tell you anything about which elements are equivalent, and saying "this is a topology" doesn't tell you anything about which sets are open.

[tacit drift]

does "bien déterminée" here mean "well defined" in english

[waxen urchin]

i think so. and it seems you are asked to prove that there's such a function, right?

[gaunt linden]

I think excluding such large disjoint unions is more of an incidental side effect of requiring second countability than a goal.
I know of some weird examples of path-connected Hausdorff locally Euclidean spaces that are separable but not second countable -- but I can't immediately cite examples of theorems that depend on second countability.

[tacit drift]

no this is a paper by frechet

[waxen urchin]

thought that was en exercise

[tacit drift]

im trying to decipher it for a project

[waxen urchin]

voisinage lol

that's where neighborhood comes from

[tacit drift]

i don't understand why the function would be well defined

like so

in R a function would be f(x)=x/2 right

or f(x)=2x

one of those

and then in R^2 a function would be like f(x)=2x still actually

but because the implication only goes one way couldnt it also be

some other thing

[waxen urchin]

yes, that's the sense i was trying to put in the analogy, and that satisfying these three properties provides exactly that — satisfying these three properties, nothing more. no word on how these sets and their elements are. but i guess it makes sense in differential topology, where metric doesn't exist

but maybe all I've written above makes no sense and I'm wandering in the mist of ignorance

[flat plinth]

Isn't a separable metric space necessarily 2nd countable?

[gaunt linden]

No.

Wait, metric space -- perhaps yes.

The examples I had in mind are almost certainly not metrizable.

[quick delta]

Yes, can’t you just take all rational balls centred at some point of the dense subset?

[flat plinth]

but then they can't be hausdorff and locally euclidean no?

[gaunt linden]

They are. Let me try to remember how it goes.

[flat plinth]

If they are separable as well that would certainly contradict Uryhson's theorem

[vagrant stirrup]

Surprisingly, Bourbaki give quite a nice readable motivation of why open sets are useful and how one comes from metric spaces to topological spaces via a concept of "neighbourhood" that is useful for discussions of limits and continuity. The books has a couple of pages on that in the introduction, which I found quite illuminating when I was studying this

I also had some problems with grokking open sets -- like why the need to be so abstract?

[tacit drift]

why are we "grokking" anything

[alpine nest]

If X is a metric space, then second-countability, Lindelof property, and separability are equivalent

[vagrant stirrup]

supposedly because of this: "The term was first popularized by Robert A. Heinlein in his 1961 novel Stranger in a Strange Land, where it means to understand something deeply and intuitively."

[waxen urchin]

sounds cool, what book is it, if you don't mind?

[tacit drift]

am i going insane

[vagrant stirrup]

It's "Общая топология. Основные структуры" ( "Topologie générale", "Structures Topologiques", ch. 1) -- I've got a copy in Russian

I can send you photos of two pages if you want 🙂

[gaunt linden]

I think it went something like:
Take X={(x,y) in R² | x != 0}, and glue continuum many copies of R² to it by the following rule: point (x,y) in sheet s is glued to (x,s+xy) in X if x is nonzero, and isn't glued to anything otherwise.
This ought to be separable because the rational points in X are dense in the full space.
On the other hand, it prevents it from being second countable that the uncountably many (0,0) in the various sheets each has a neighborhood that doesn't contain any other of those points.
(So by Outsider's comment it can't be metrizable).

[waxen urchin]

i have it in english in a public library, I'll def take a look. having it in russian is real nice and rare, something i wish i had in my library too. thanks for the help!!

[vagrant stirrup]

Yeah, I value it too, I asked a friend to bring me a copy from Ukraine, it was published in 1968 and looks quite stylish. I only have one part though, not their whole treatise, just the book about general topology 🙂

[vagrant stirrup]

"public library"? do you mean a university library? Or just a regular public library has Bourbaki?

[waxen urchin]

ргб, москва (она же "ленинка")

[warped helm]

(it’s been a word for longer than the twitter ai has been a thing)

[kind marlin]

I always get whiplash when I see grokking now

[alpine nest]

Yeah, like many things it has been ruined in recent years

[tacit drift]

my bad gang

ive only read moon is a harsh mistress

[gaunt linden]

It's a better story than Stranger, tbh.

[analog yarrow]

Hey guys I’m new, is there a specific set theory channel?

[rancid umbra]

#foundations

[gaunt linden]

Though depending on which sort of set theory you want to talk about, #proofs-and-logic might be more appropriate.

(Proofs-and-logic if you want to use set theory for doing real mathematics, foundations if you want to dig into hairy axiomatic pedantry).

[sterile jungle]

ive seen that RP^n/RP^n-1 is homeomorphic to S^n? is that true?

if so, we can write RP^n as S^n/~ where ~ identifies x with -x. what is the explicit homeomorphism from S^n to (S^n/~)/(S^n-1/~) ?

[quick delta]

If you view S^n as just S^{n-1} x [0, 1]/~’ where ~’ is generated by (x, 0) ~ (y, 0) and (x, 1) ~ (y, 1), it is exactly given by “division by 2” on the second coordinate (where the fact that we quotient out by S^{n-1}/~ means that this is well defined)

[gaunt linden]

What exactly does RP^n/RP^(n-1) mean there? Which equivalence relation are you quotienting by?
It is true that RP^n is the same as S^n with every pair of antipodes identified.

[quick delta]

Quotient of a topological space by a subspace

[sterile jungle]

when viewing RP^n-1 as a subspace of RP^n, then identifying all points in RP^n-1 together

[quick delta]

(Ie the relation that identifies every point of the subspace to each other)

[gaunt linden]

Okay, yes, I can see now that actually does make sense in this case.

[sterile jungle]

im sorry but im not really sure whats meant by this. i get viewing S^n as S^n-1 x [0,1]/~' but not sure how its "division by 2" or how i can use that

[opaque scroll]

Division by 2 as in
(s, x) |-> (s, x/2)

[quick delta]

(And it’s an exercise to check that this is well defined, surjective and injective)

[opaque scroll]

And I guess note that RP^n-1 should correspond to the equator S^n-1 x {1/2}

[sterile jungle]

oh so literally division by 2. okay thank you all ill have a go

[rotund prism]

this is jumping the gun, right? Take a long plane, L^2. Now take a subset thats the cartesian product of L and a basic interval like [-2,2]

Poke circular holes into the centers of the uncountable intervals used in the Long Line's construction, along all segments of the subset. Theres uncountable circles. Uncountable holes. Or is it apparently common convention that any "surface" has to be paracompact

[queen prism]

paracompactness/second-countability is a pretty typical assumption when we're talking about manifolds yea

[rotund prism]

For a manifold sure, but... ok i was about to say who says a surface is a manifold but typing it out yea thats also a pretty common assumption

[queen prism]

I think serge lang assumes that all manifolds are hausdorff but treats paracompactness as a separate assumption

but that's scary and I don't want to go near it

[unreal stratus]

A surface is usually by dimension a 2D manifold

[queen prism]

by dimension or by definition?

[plush folio]

The definition of a surface is 2

[brittle cedar]

I'm not sure why I can't piece this together, it feels like it should be really straightforward.

our (only) criterion is that for any point x there is a compact subset of X containing some neighborhood of x

of course the hint lets us know we should consider the one point compactification, lets say Y = X U {infinity}. I think we'll need to use the fact that a neighborhood of infinity is open when it is {inf} U (X - K) for a compact K

[strong lantern]

Take U' a open set containing x contained in a compact set. Take V the intersection of U with U', then the clouse of V is contain in a compact set therefore is compact, V is contain in U and contains x

I don't see why there's a need for the one-point compactification of X

[gritty widget]

Hint:Any open subset of the one-point compactification not containing that one point is an open subset of the space in question

[gritty widget]

It’s not necessary but it can be used

[plush folio]

V is contained in U, but how do you know V-bar is contained in U?

[strong lantern]

Uhh That's why. Then yes indeed you have to use the one-point compactification

[gritty widget]

You could just use ||regularity no?||

[brittle cedar]

yes, I don't think we were supposed to have regularity for this problem but it's quick to do it in this case anyway, I think I got a solution. thanks guys 🙂

[strong lantern]

Yes, but it feels like cheating

[plush folio]

I was wondering what property guarantees that for any point x with neighbourhood U there is a neighbourhood V whose closure is contained in U; is that regularity? I think that's atleast sufficient, maybe necessary too?

[gritty widget]

I think that’s equivalent to regularity

Let X be closed, x not in X. Then, there is a neighborhood U of x such that the closure of U is disjoint from X. Then, the complement of the closure of U is the open set disjoint from U containing X.

Dually, if U is an open set containing x, U^c is a closed set not containing x, so there is an open set V containing x and an open set G containing U^c, disjoint from V. Ergo, the closure of V is contained in U.

[strong lantern]

I think this even works if the space is not T_0

[gritty widget]

Yeah so this is equivalent to regularity but not T3-ness I guess?

[crisp lintel]

you'd want to basically separate some set from infinity in the one point compactification

useful here I believe is the fact that compact Hausdorff spaces are normal

I did this proof recently for something and didn't go through the one point compactification but I think it might be a nicer proof to use it

The core of the idea is the fact that an open neighborhood of x that is disjoint from a neighborhood of infinity must necessarily have compact closure

(in the compactification)

[prime elbow]

I am thinking about (0,1) in usual topology is baire space or not

I think it is

Because it is topological complete space

[gritty widget]

(0, 1) is homeomorphic to R

[prime elbow]

Yes

But in general, every open set of complete metric space is baire, because every open set is topological complete

Now, is it true in general?

[gritty widget]

What is a topological complete set?

[tender halo]

they mean completely metrizable

[prime elbow]

Yes

[gritty widget]

Is completely metrizability hereditary for open subsets? Sounds true enough but it isn’t complete in the inherited metric so I’m not sure

[prime elbow]

I have to check R with lower limit is baire space or not, so I know compact Hausdorff space is baire space, and one point compactification of R with that topology is compact and Hausdorff.

So is it true that open set of baire space is baire? Any hint?

[prime elbow]

Yes, I proved it

[prime elbow]

.

[tender halo]

its hereditary to G_\delta sets

(so open sets too)

moreover, a completely metrizable space is a G_\delta in an metric space it is a subset of (the metrics dont have to agree of course)

and thats actually a complete characteristic, completely metrizable spaces are exactly those who are G_\delta in every metric space they are a subspace of

[prime elbow]

Yes, open subspace of baire space is baire space

[crisp lintel]

every locally compact Hausdorff space is a baire space, which can be proven p easily using the one point compactification

though I guess that follows immediately from the open subset thing

[tender halo]

more generally, lch spaces are cech-complete and therefore baire

[novel stump]

how do i go about finding an example of a function whose graph isnt closed? im very confused lol

[plush folio]

Do you want a function whose graph isn't closed? The question asks you to find a discontinuous function whose graph is closed

[novel stump]

oh wait sorry i misread the question

yes i want to find a graph who is closed

[plush folio]

Do you want any hints? I guess just start thinking about your favorite discontinuous functions, and see if their graph is closed

[novel stump]

i can only really think of like, f(x)=0 if x = 0 and f(x) = 1 if x > 0

[plush folio]

okay, is its graph closed? btw, that's a function on the non-negative reals, the question asks about a function on R, but I guess it doesn't matter that much

[novel stump]

i dont really know how to check if the graph is closed i'll be honest

should i equip each topological space with the discrete topology? or maybe the indiscrete topology?

[plush folio]

No, we're supposed to use the usual topology of R. So the important property of closedness we need to use here is that a closed subset contains all its limit points

[novel stump]

oh

i would think since this is a topological space we have that the complement of the graph is open

we havent emphasized that part of beign closed

[plush folio]

Yeah, if a subset is closed, then it contains all its limit points, which is called being sequentially closed. For many spaces these two are actually equivalent, ie. sequentially closed implies closed (in particular this is true in R^n)

[plush folio]

Just to explain the intuition, I imagine walking along the graph from the positive side, slowly approaching the point (0, 1). If the graph is closed, then (0, 1) should be in the graph, since it's clearly a limit point, but it isn't, so the graph isn't closed. You can prove it rigorously if you want, by defining a sequence along the graph and noting that the sequence approaches (0, 1) etc, but when you get more experience you learn to see whether something is open or closed mostly by vibes and handwaving

[gritty widget]

Baire Category Theorem also says locally compact regular(not necessarily Hausdorff) spaces are Baire, which iirc are not in general cech complete.

[austere flare]

hint: this function has a jump discontinuity, which "shouldn't" work as a counterexample for the reason sheddow discussed (sequential closedness; the limits x->1 from the left and right both exist in R but don't agree). but what about other types of discontinuities?

[novel stump]

sorry for the late response i was taking a midterm :]

[novel stump]

hm, so then then how would we make a graph thats closed but not continuous? clearly there is some trick

[novel stump]

would a corner work?

no

[warped helm]

theres not really a “trick”

[novel stump]

nvm

corner is nondifferentiable but not discontinuous lol

would an asymptotic discontinuity work?

but then that wouldnt make the graph closed

[warped helm]

try very easy functions

recall that finite sets are closed in R

[novel stump]

maybe 1/x?

am i overcomplicating this

[warped helm]

yes

[novel stump]

great

[austere flare]

that's the idea; just add f(0)=0

[novel stump]

so 1/x \cup 0?

[novel stump]

what would a simpiler solution be

[warped helm]

step functions, indicator functions, etc

[novel stump]

ohhhhhhhhhh

[warped helm]

take f(x) = 0 for x < 0 and f(x) = 1 for x \geq 0

[novel stump]

idk what an indicator function is but i understand the step function

[plush folio]

but the graph of this isn't closed

[warped helm]

oh then im tripping

[austere flare]

yeah i think f(x) = 1/x, f(0) = 0 is the simplest example you can get

[warped helm]

Lol true

[plush folio]

yep, that's the example I was thinking of too

[warped helm]

sorry, i initially thought of the graph as (\mathbb{R} \times {0, 1 }) but i realize that's not correct

[gentle osprey]

josemom2

[novel stump]

so this works correct?

[plush folio]

yeah, except the notation 1/x \cup 0 is a bit weird. You should specify more precisely which function you're talking about

and try to prove that the graph is closed if you can

[novel stump]

how so?

oh i suppose, how should i improve it

[warped helm]

you're taking the union of a formula with a number

which is not really standard

[novel stump]

oh wait i need to make it into cases

duh

[warped helm]

or defined to begin with

[novel stump]

LMAO sorry

[plush folio]

yeah, I can guess what 1/x \cup 0 means, but it's worth being a bit more precise

[novel stump]

i understand dw

[queen prism]

can you show your results when you're done

[novel stump]

yep! i will need help proving why its closed

this is what i have so far

[warped helm]

to prove that it's closed you should determine whether it contains all its limit points

for points c that aren't 0, you can take sequences x_n -> c and consider the limit of (x_n, 1/x_n) and see what you can say about this

and then treat the special case of 0

[novel stump]

we dont have that definition unfortunately, we have to show that that its complement is open LOL