point-set-topology

[unreal stratus]

Well note any $Y \subseteq X$ can be written as $\bigcup_{x \in Y} {x}$

[gentle osprey]

potato

[unreal stratus]

That's how I'd write it

[wary furnace]

alright I get that

yeah it'll look much cleaner without those j's

so something like this right?

[unreal stratus]

ye although should quantify over all x in Y in the first or smth to be clearer I suppose

[wary furnace]

yeah a for all

[mint sage]

any idea?

[pearl holly]

a covering map is surjective and has the path homotopy lifting property I'm pretty sure or something like that

[coarse kestrel]

Well to lift a map $f:Y\to B$ to a map $Y\to E$ you need $f_(\pi_1(Y))\subseteq p_(\pi_1(E))$, so using $Y=S^n$ and the fact that $\pi_1(S^n)=0$ you have $p_*$ is surjective. Then injectivity follows from the fact that $S^n\times I$ also has trivial fundamental group

[gentle osprey]

Whoever

[mint sage]

I've done it, but what's next?

[coarse kestrel]

I mean that’s it

It’s already a homomorphism

And if it’s injective and surjective then it’s an isomorphism

[mint sage]

I got it. thx

[sharp frost]

Hi, I'm trying to show that the mapping cone of the Hopf map is homeomorphic to CP^2. Any hints?

[plain raven]

Hm. idk but the hopf map is a sphere bundle and the mapping cylinder (not mapping cone) of a sphere bundle can be understood as like, filling in all the spheres with solid matter

like

if p : E -> B is a sphere bundle with fiber S^n

then the mapping cylinder of p is also fibered over B with fiber the solid ball D^{n+1}

idk if that's helpful but it's the first thing that comes to mind.

[sharp frost]

I'm just struggling to figure out how exactly $S^2$ attaches to the cone of $S^3$ to make mapping cone

[sharp frost]

my friend is telling me that the Hopf map is a Lie group homomorphism but I don't believe him

[fair steeple]

and you shouldn't, since it lands in S^2 and that's not even a Lie group

[sharp frost]

aha thanks

[golden gust]

thanks

[empty grove]

Your friend Lied

[cursive flume]

are there more sigma algebras on a set, or more topologies?

like which structure is more „minimal“?

[gritty widget]

I'd suspect there'd be more sigma-fields, but I don't have precise numbers

[golden gust]

if the set is countable, then the sigma algebras are all topologies (since the union axioms agree, and the intersection & complement axioms are stronger)

but idk about uncountable

[rancid umbra]

is there a way to characterize density in terms of nets? there is a way to characterize density in terms of sequences in metric spaces, but i’m not sure if it breaks when you’re aren’t working in a metric space

[gritty widget]

https://en.wikipedia.org/wiki/Net_(mathematics)#Properties
yeah. It works the same way

a subset Y of X is dense iff every element of X is a limit of some net in Y

[rancid umbra]

thanks. would be interested if someone had an example where the sequential characterization broke in some top space. can’t think of one atm

[gritty widget]

Y = [0, omega1) is dense in X = [0, omega1], but every sequence in Y converges to an element in Y

[rancid umbra]

topology is the order topology here, right?

[gritty widget]

yes

[rancid umbra]

thanks

[rancid umbra]

surely not every sequence converges. what about 0,1,0,1,0,1,...? i could see how every non-decreasing sequence converges to some element in Y.

[gritty widget]

I didn't say that every sequence converges

[rancid umbra]

"every sequence in Y converges to an element in Y"

[gritty widget]

That's not math, that's just the way you speak. It might be confusing but it's correct

[rancid umbra]

?

[gritty widget]

What it means is that every convergent sequence in Y converges to an element in Y.

[rancid umbra]

that is different from what you said here

but thank you for clarifying

[gritty widget]

Like I said, it's an issue with English language, and it can be interpreted like I said that.

[rancid umbra]

i still disagree, but thats ok

[gritty widget]

c squared is right

there is no english language issue

"every sequence in Y converges to an element in Y" includes the phrase "every sequence in Y converges" which is wrong

but whatever everyone knows what was meant

that doesn't make sense, a sentence changes meaning with words missing

in general, sure. but in this case the missing words don't change the fact that you said every sequence in Y converges

which is the problem, since that's not true

I didn't say that

i am copy pasting from your messages

[rancid umbra]

lol

[gritty widget]

yes, I said that

[rancid umbra]

https://tenor.com/view/patrick-starfish-spongebob-its-not-my-gif-12227574

[gritty widget]

lmao

[rancid umbra]

https://tenor.com/view/thats-not-my-wallet-patrick-star-dense-the-incredibles-syndrome-gif-15425972

this one is better sry

[gritty widget]

I hope you realize that I realize what you all mean

I just completely disagree

[rancid umbra]

i did not know that you knew that we knew that you knew what we meant

[gritty widget]

"convergent" means "which converges" in this context

i think that everyone except you not understanding this should tell you that this meaning was not clear

even if it wasn't clear, that doesn't make it incorrect

[rancid umbra]

tbf, i have never seen this used before

[gritty widget]

and I agree that it wasn't completely clear

sure, if you interpret it the way you meant. but what you meant was not clear

please just use words to say what you mean

there is never any harm in writing more down

no one is omnipotent

[rancid umbra]

please just use words to say what you mean

lol

also, what?

[gritty widget]

I meant that I can't be careful all the time about people understanding what I said, everyone is prone to mistakes

such things will happen

admit your mistake when it's called out, then

[rancid umbra]

but what you said and what you meant are two diff things

[gritty widget]

I did agree

But I didn't agree that it's a wrong way to say it, and that I said it was convergent

[rancid umbra]

agree to disagree?

[gritty widget]

anyway, it's all offtopic

[rancid umbra]

right. so back to the example you gave, consider just listing the elements of Y in order, which you can do since Y is countable. this sequence converges to its supremum since it is non-decreasing, which is w_1

[gritty widget]

it's not a sequence though

[rancid umbra]

in what sense?

[gritty widget]

a sequence has to be ordered by natural numbers

[rancid umbra]

yes. there is a bijection N --> Y

[gritty widget]

every such sequence in [0, w_1) has a supremum which is < w1

this is because w1 is uncountable

[rancid umbra]

oh wait what is happening i wrote that down a few seconds ago

[gritty widget]

there is a notion of generalized sequence which are ordered by an ordinal, and spaces in which such sequences define the topology also have a name (which I forgot)

but not every topological space satisfies this

[rancid umbra]

wait so what is the supremum of {a : a < w_1}?

[gritty widget]

it's w_1

it just can't be reached by sequences

[rancid umbra]

im under the impression that this set is countable

am i correct in saying that?

[gritty widget]

it's uncountable

For example, consider the set {a : a < w0}

[rancid umbra]

right

[gritty widget]

this is countably infinite, but each of its members is a finite ordinal

[rancid umbra]

okay thanks, makes sense

[gritty widget]

{a : a < w_1} I think this is actually w1 by definition

so it's the same size as w1, trivially

[rancid umbra]

cant be, thats circular

[gritty widget]

oh you're right

[rancid umbra]

so Y is an open set, is sequentially closed, but not closed in X since its complement, {w_1} is not open in X. w_1 is a limit point of Y. interesting stuff going on here

[gritty widget]

what I meant is that if you look at the definition of ordinals by transfinite recursion then I think those should be the same by definition

[gritty widget]

yeah

[bright acorn]

Are you taking about nets?

[gritty widget]

No, I am talking about nets of the form $a:\alpha\to X$ for some ordinal $\alpha$

[gentle osprey]

Blitz

[bright acorn]

I mean, nets are indexed by a preordered set, not necessarily an ordinal.

[bright acorn]

Ok

[cosmic socket]

I’m reading Atiyah’s geometry and physics of knots and he is describing the axiomatic definition of a quantum field theory. He says that the axiom that $Z(\Sigma \times I)$ being the identity endomorphism of $Z(\Sigma)$ as well as functoriality is enough to give homotopy invariance. Can someone explain this.

[gentle osprey]

vivasvat

[gritty widget]

I know that I said generalized sequence which is also another name for a net, but I meant literally "a generalization of sequence", sorry for confusion

I just remembered, those are called pseudoradial spaces

[coarse kestrel]

In the definition of simplicial homology, is it important that the vertices of each simplex have the ordering of a simplex? Meaning can the vertices of a triangle be oriented v_1 -> v_2, v_2 -> v_3, v_3 -> v_1 and we keep the orientation consistent everywhere

Because the usual ordering is that v_i < v_j iff i<j

So the orientation is supposed to be v_1 -> v_2, v_1 -> v_3, v_2 -> v_3 and not v_3 -> v_1

[ancient zenith]

@coarse kestrel definitely! you can define whichever sign convention you want, so long as you preserve the property that d^2 = 0

the usual order v_1 < v_2 < ... < v_n is just the most convenient to state and generalize to all n

[coarse kestrel]

I see, and the resulting homology groups will still be the same? As in d^2=0 is the only thing we need?

[proper cobalt]

If I have isomorphisms for all cohomology groups, how can I show in general this induces a isomorphism of rings for the cohomology rings?

[fair steeple]

I don't think you can. There are examples of spaces with the same cohomology groups but different ring structures.

[narrow magnet]

i need some help with this

[proper cobalt]

@fair steeple Ah ok, specifically im trying to show the wedge sum isomorphism on cohomology rings from the isomorphisms on the groups. Does this just follow from the cup product satifying f^(a \smile b) = f^(a) \smile f^*(b)

[fair steeple]

I'm missing some context I think.. what do you mean by "the wedge sum isomorphism"?

[proper cobalt]

Have that \tilde{H}^{n}(X \vee Y) \cong \tilde{H}^{n}(X) \oplus \tilde{H}^{n}(Y) on the reduced cohomology groups where X, Y are path connected, strong retracts ect. Trying to extend to ring isomorphism for \tilde{H}^{*}

[fair steeple]

Oh I see. But doesn't H^*(X v Y) begin its life as a subring of the other one, by Mayer Vietoris?

[proper cobalt]

Hmm, only just been introduced to cohomology rings, it sounds true haha

[fair steeple]

I think it might be more beneficial to keep track of the entire ring structure from the very beginning, as you go through this argument

rather than just doing group and then trying to improve it to ring, for which there are counterexamples in general

when I say counterexamples I just mean spaces for which the groups match but the ring structures are different.. not counterexamples to the theorem you want, of course

[proper cobalt]

I see

[cursive flume]

i'm trying to understand why the comb space is not locally path connected, but it is path connected

I can't think in terms of pictures at all,just definitions

could anyone give some hint?

Let $E:=\left( {0} \times {0,1} \cup \left(\left{\frac{1}{n}|n \in N \right } \right) \cup [0,1] \right) \cup \left( [0,1] \times {0} \right)$

This is a subset of $\mathbb{R}^2$, which is equipped with the subspace topology of the standard topology.

I want to show, that for every $y,z \in E$, there exists a continuous path, which connects them.

[gentle osprey]

ProphetX

ProphetX

ProphetX

[haughty wave]

pretty sure you defined it incorrectly

should probably be {1/n} x [0, 1] and {0} x [0, 1]

but anyways if you take any point p on the first the hair of the comb (i.e. p = (0, y), where y ≠ 0), then consider any neighborhood basis

it necessarily contains a neighborhood that is a subset of B_y/2(p), so we examine that neighborhood

it doesn't reach all the way down to the base of the comb, but it does intersect the other hairs of the comb

so now you need to prove that there isn't a continuous arc connecting p and a point on another hair of the comb

[plain raven]

hey if i have a cohomology theory defined on a subcategory are there any interesting ways to try and extend this to the whole category

like if i have a cohomology theory defined on compact hausdorff spaces

is there a good way to extend this by formal nonsense to all compactly generated spaces?

maybe by some nonsense with like, localizations in a sheaf or presheaf category

like

there's probably some kind of grothendieck topology on the category CH such that like compactly generated spaces can be regarded as a subcategory of sheaves on that space

It wouldn't be the whole category because compactly generated spaces for sure don't form a topos. that would be nuts

(What would be the subobject classifier lol)

but yeah i'd be interested if anybody has a reference for a reference that allows you to formally extend a cohomology theory from a category C to some category of sheaves over C

but all i really care about is the CH -> CG case

[echo oyster]

Just to verify

A sequentially compact metric space is complete, right ?

But every compact metric space is not necessarily complete

[coarse kestrel]

This is a better question for #real-complex-analysis, but for metric spaces, compact and sequentially compact are the same, and a metric space is compact iff it’s complete and totally bounded

[echo oyster]

Ok sorry, thank you

[gritty widget]

I disagree. This is not about analysis

[coarse kestrel]

Oh I guess fair yeah

[wary furnace]

Ok this might sound pretty convoluted but lemme explain,
Basically I played around with X, the set of all functions from [0,1] to [0,1] and I gave it a topology induced by the basis described below
I listed a few examples of how such basis elements would look like
And that's when I noticed that this kind of topology would actually... behave in a similar way to the discrete topology on [0,1]
I started topology recently so I don't know some of the more advanced concepts, but is there a name for such a correspondence of topologies on different spaces? Does it have something to do with quotient spaces?

[gritty widget]

In this space, any two functions f, g which have the same zeros are topologically indistinguishable.
Identifying such elements, the topology becomes the same as the subspace of functions from [0, 1] to {0, 1}

[gaunt salmon]

Do anyone know any place to learn the cat theory needed for reading tom diecks algtop book? Would the first chapter of mac lanes book be good?

[gritty widget]

I've read it, you just need nlab really

Nothing too difficulty from category theory is there

At least, I didn't learn any category theory myself. But I guess I knew what it is. So maybe you do need something like Mac Lane

yeah, that'd probably be good

just read first chapter and supplement with nlab

[wary furnace]

and all functions from [0,1] to {0,1} have a cardinality of 2^|R|, same as P(R) or P([0,1]), so that's why there's that connection right?

[gritty widget]

It's not really discrete

[wary furnace]

oh?

[gritty widget]

No singleton except for the function f(x) = 0 for all x is open here

[tardy meadow]

sanity check: an "open basis at x" is a basis for the neighbourhoods of x right?

[gritty widget]

It's a family F of neighbourhoods of x such that any neighbourhood of x contains an element of F

[wary furnace]

I mean yeah but what I meant by "behave like" is that for example taking the intersection of Ua's corresponds to taking the intersections in the discrete topology on [0,1]

[tardy meadow]

cheers

[gritty widget]

But the unions don't correspond to unions

[wary furnace]

yeah that's true

this reminds me of how vector spaces behave

an intersection of 2 vector subspaces is another vector subspace

[gritty widget]

I am confused in show that two definitions of chain homotopy are equivalent

one being the standard definition where h is a chain homotopy between f and g if

f-g=hd+dh

im trying to show this is equivalent to a chain homotopy is a chain map H:C \otimes I to D

C, D being R-modules and I being the "interval object" or something like that?

C,D being chain complexes

chain complex, right, sorry

and yes I is the interval in chain complexes so its

I_0=R<x_0,x_1>

and I_1=R<e>

and the differential maps e to x_1-x_0

so like here you think of x_0 and x_1 being the boudary of the interval

e is the 1-simplex between them

then the differential is just the boundary map

where i am stuck is

oh never mind i left out a term

all good now?

sorry no i do have a mistake

H has the property that

H(x \otimes x_0)=f(x)

H(x \otimes x_1)=g(x)

H(x \otimes e)=h(x)

i want to show that requiring this to be a chain map, so dH=Hd is equivlanent to g-f=hd-dh

but if i do Hd(x\otimes e) i get

$H( dx\otimes e +(-1)^{|x|} x \otimes x_1 - (-1)^{|x|} x \otimes x_0)=h(dx)+ (-1)^{|x|} g(x) -(-1)^{|x|}f(x)$

[gentle osprey]

lime_soup

[gritty widget]

so putting this equal to dH(x\otimes e) we get

$hd(x)-dh(x)=(-1)^{|x|}(f(x)-g(x))$

[gentle osprey]

lime_soup

[gritty widget]

so there is this extra factor of (-1)^{|x|} that i don't understand

[empty grove]

I think these 2 definitions aren't exactly the same, but there's a canonical bijection between the homotopies defined in the 2 different ways

If you want on the nose agreement, you should take I ⊗ C → D instead of C ⊗ I → D

[gritty widget]

ah thank you very much

and this morphism is just the

natural iso that

makes the tensor product of chain complexes commutative up to natural iso

and the factor (-1)^degx was weird because

the element e has degree 1

and that why the f and g were inthe wrong order

thank you very very much

[empty grove]

Np ✓

[gritty widget]

so long story short

the morally right thing to do is always write homotopies as H:I times A

[empty grove]

Maybe

Or you could argue that the hd + dh = f - g equation is morally incorrect

It doesn't matter much when there's a natural isomorphism anyway

[broken light]

Hey guys!
I've just recently finished my linear algebra course, and our professor put down some bases for projective geometry. One of them was the notion of the projective line and projective plane. To display visually such thing, while accounting for points on such spherical surface (using this specific representation of the projective plane based on ℝ³) which are equal in the relation of equivalence defined by points being on the same line but on opposide sides (bozo notation "" -a=a ""), we reduced such surface to half of itself.
But then we still had to account for the line at infinity where technically we would have to bind all points on opposide sides.
He lacked time, so he didn't complete the weird transformation, but he smashed down the half sphere to a circle, cut it in 3, made a mobius strip out of the middle and yada yada.
I'm using yada yada because I guess that you know what I'm saying (hoping that I have explained myself to the fullest).
He advised us to look for a representation in 3d of the completed re-binded weirdly curved surface, and I was wondering if any of you knew what thing to look for online, because my luck has been bad :(

My course is in italian, so I suffer from "how the fuck do you translate this" and my explanations get really bad in english. Sorry if it's difficult to comprehend

[gaunt linden]

Sounds like you may be expected to invent the https://mathworld.wolfram.com/Cross-Cap.html

[broken light]

Thank you ❤️

[echo oyster]

Every compact subset of a matric space is always complete
Is this statement true ?

[stable kite]

(X,d) is compact if and only if (X,d) is complete and totally bounded. As such, your statement is true. 😇

[echo oyster]

Oh I see,thank you

[gritty widget]

I think you already asked that

[echo oyster]

Yes, i was still confused with the sequentially compact and compact

[formal tide]

I'm trying to show that R2/E has 3 connected components, where E is homeomorphic to "8" (two circles intersecting in 1 point)

could this be proven just using basic topology and jordan curve theorem?

[marsh forge]

yes

[formal tide]

thanks haha

any tips on how?

[marsh forge]

Hm the details of this are actually more annoying than I was thinking

Okay there should be some epsilon such that if you thicken the borders of both circles by epsilon they don’t touch

Then take the disks inside of these thickened boundaries

These will both have two connected components by JCT

Then you can show that two of those are the same in the larger space

Something like this ought to work

[formal tide]

I'm not sure I'm following this, they are already touching

[marsh forge]

Oh right I swapped the problem in my head (both are true)

You can probably adjust this same idea somehow

There also should be some (infinite) path in the plane separating the two and then you can solve it by splitting the plane and doing each circle on at a time

Maybe there’s an easier way

Let me think

Okay this is better I think

The JCT tells you that a circle splits the plane into an outside and inside component

And the outside components of both circles are the same

And the inside ones have to be disjoint

I think formalizing that to whatever level you need is the right strategy

Oh wait that’s not true

Oof

I’m at the airport and embarassing myself let me sit down and I’ll get back to this if someone doesn’t save me

[formal tide]

:p yep this is surprisingly harder than it seems, just like jct I guess

[marsh forge]

Okay what is true is that this is one of two cases

The other case is that one circle is inside the other

So here’s a better explanation

I place down one circle

I get two components

I place down a second circle that intersects the first at exactly one point

This has to be contained entirely in the inside or outside components

Both components are homeo to R^2

And placing a circle will separate it into two components

Giving 3 totale

I’m sweeping the point at which they meet under the rug, but it should be clear that this doesn’t make a difference

[sour geode]

I wonder if there is a way to construct explicit homeo from R2/“8” to space with 3 connected components

[marsh forge]

For sure not

Unless by R2/8 you mean a specific copy of 8

Like if you allow any embedding it won’t work

[formal tide]

ohhhh got it, cool proof!

[marsh forge]

This is only true locally***

But that’s a very easy thing to fix since circles are compact (bounded)

[gritty widget]

Embedd R^2 into S^2 then?

[marsh forge]

Oh that works nicely

[rancid umbra]

i dont see why that is true

R2/8 is basically just R2 minus a point along with 2 circles

[marsh forge]

If you allow an arbitrary embedding writing down an explicit homeomorphism is not going to really work

[rancid umbra]

can you give me any insight as to why it wont work?

[marsh forge]

I mean just try it hahaha

You don’t know enough information

[rancid umbra]

why cant i just take the disjoint union of the three spaces (R^2 minus a point and two copies of B^2) and then map R^2/8 into that space?

ig im just not sure what information im missing. also, not trying to argue with you, im just trying to see where my reasoning is breaking

[marsh forge]

I think we are having a linguistic difference

By explicit homeo I mean like

Writing down f(x,y)=…

The actual coordinate-wise shape of this general thing

Can be quite gross

[rancid umbra]

ah okay

[gritty widget]

I know the following statements are equivalent:

i) $X$ is Hausdorff, compact and completely disconnected
ii) $X$ is compact and completely separated
iii) $X$ is compact, $T_0$ and zero-dimensional
iv) $X$ is Hausdorff and coherent

My question is, is the condition v) ‘$X$ is Hausdorff, compact and extremally disconnected’ also equivalent?

[gentle osprey]

dgallego

[gritty widget]

I know these conditions define what it's called a 'Stone space' and I've found a theorem that (if I've understood it right) says (v) is equivalent, but the proof uses the Stone representation theorem for boolean algebras and I'd like a more fundamental way to see that, if it's possible

[plain raven]

Hi dgallego i'm researching this as well

and

I believe the answer is no.

https://ncatlab.org/nlab/show/complete+Boolean+algebra#SDGC

carefully reading section 4 here

there is a one to one correspondence between totally disconnected CH spaces and boolean algebras, up to isomorphism

and this restricts to a correspondence between extremally disconnected CH spaces and complete boolean algebras

Since not every Boolean algebra is complete, not every totally disconnected CH space is extremally disconnected

[rancid umbra]

this seems really interesting

[plain raven]

Yes, this is one of the first historical examples of an equivalence of categories.

If you have a couple minutes, I strongly recommend downloading Peter Johnstone's book on Stone spaces and reading the introduction

(about 10 pages or so)

[cursive flume]

there is the general procedure of: if you have an equivalence relation (say connectedness,path connectedness), then you have a functor from Top to Set, induced by this relation

now,I showed that being in the same quasi component defines an equivalence relation

does this mean, that this induces a functor?

[coarse night]

note that quasi components are preserved by continuous maps

[gritty widget]

Oh I see, I didn't realize the problem relies on the completeness of the Boolean algebra, thanks!

[plain raven]

Thanks for your question btw, the notion of extremally disconnected turned out to be exactly what i needed for a problem i'm working on

[gritty widget]

Cool!

I'm from Spain and here at the end of the degree it's mandatory to do some kind of 'research' (not producing new results but investigate about some topic and expose the existent results) and a big part of this assignment is about the correspondence between the categories of boolean algebras and stone spaces

if someone is interested in correspondences between spaces and algebra there's one very interesting between spectral spaces and spectrums of rings with the Zariski topology. It's a well known result from Hochster but I came across with a inductive process that allows you, given a poset, to find a ring whose spectrum is the poset you have when ordered by inclusion

in this case is not a duality or equivalence between categories since there are many rings that produce the same spectrum as a poset, but I think this inductive method it's quite interesting by it's own

[plain raven]

What is a spectral space?

[gritty widget]

The spaces that are T0, compact, coherent, sober and have the finite intersection property for open compacts (the intersection of two open compacts is compact)

The spectrum of a ring with the zariski topology is a spectral space, and the converse holds: every spectral space arises from the spectrum of a ring

[cursive flume]

https://math.stackexchange.com/questions/4124102/proving-a-map-that-admits-local-sections-is-a-quotient-map

can someone help me understanding the statement of the theorem? the title is "proving a map that admits local sections" is a quotient map

however,in the question, I do not see where existence of local sections is assumed. they assume there is an open cover?

this is the proposition I am trying to prove

[coarse kestrel]

They assumed that there is a cover V_j such that for each j, there is a map g_j such that f(g_j(y))=y. Notice how this is exactly the condition 2, where an open cover is just a quick way of saying for every y there is such open set

[cursive flume]

it is more than that

it assumes that the union of those open sets is the whole space

[coarse kestrel]

I said a cover

That's what it means to be a cover

[cursive flume]

but 2) never says that

2. only says there exists an open nbhd for each point

[coarse kestrel]

Right

[cursive flume]

it doesn't say union of all V is the space

[coarse kestrel]

So you get an open cover by taking all such neighborhood at each point

[cursive flume]

yes,this is what I fail to see.

Clearly,taking union of all those points will create a larger set than X

i.e. X subset union

why union subset X?

[coarse kestrel]

Each V is an open set

Meaning a subset of X

[cursive flume]

ah,then: union of subsets of X is subset of X

hence X=union for all x in X V_x, where V_x is open nbhd of x?

is this the argument?

[coarse kestrel]

Yes

[cursive flume]

also,then this statement is not minimal

cause in 2) it would need not require surjectivity

[cursive flume]

that follows

[coarse kestrel]

It's a very common situation: where "every point x has an open neighborhood V such that some statement holds" is the same as "there is an open cover where for each open set, a statement holds"

[coarse kestrel]

Well statement 2 implies surjectivity

[cursive flume]

that's my point

they assume it in the statement of 2

by assuming surjectivitiy of g

so it's not minimal,in this sense,even if true

[coarse kestrel]

I mean 1) needs surjectivity

So that's why it's included

[cursive flume]

yeah,but they could make 2 different theorems I think

like 2 propositions,instead of 1

to be minimal

[coarse kestrel]

I mean sure lol, that doesn't quite matter

[cursive flume]

for the univ property of quotients,why do we need to require constancy on fibers?

[coarse kestrel]

It's a necessary condition

[cursive flume]

why?

https://math.stackexchange.com/questions/2560556/f-continuous-leftrightarrow-f-circ-p-continuous-implies-p-quotient-map in this answer they seem not to use constancy on fibers

[coarse kestrel]

Say $\pi:X\to X/\sim$ is a quotient map, say $f:X\to Y$ is a map such that there exists a map $g:X/\sim\to Y$ with $f=g\circ\pi$, then if $x,y$ are both in the same fiber, meaning $x,y\in \pi\inv(a)$, then $f(x)=g(\pi(x))=g(a)=g(\pi(y))=f(y)$

[gentle osprey]

Whoever

[coarse kestrel]

Meaning if induced map exists then the original map has to be constant on the fibers

[cursive flume]

can you give a counterexample why this fails?

[coarse kestrel]

Why what fails

[cursive flume]

why is this theorem false?

I proved it

and I never used constancy on fibers

neither stated it

[coarse kestrel]

I mean $f=\bar{f}\circ\pi$ will imply $f$ is constant on the fibers as I literally just proved

[gentle osprey]

Whoever

[coarse kestrel]

You don't necessarily need to use f being constant on the fibers to prove the theorem

[cursive flume]

I'm not sure

that is,if f exists

I need to prove first,that the quotient topology exists,it satisfies the universal property

and after this, I get as corollary constancy on fibers?

[cursive flume]

or i'm unsure what are your assumptions here

[cursive flume]

the thing is you said this is a necessary condition

but i don't see where it needs be used per se

i'd say it's not necessary at all,it is a corollary(which i can't precisely see yet)

[gritty widget]

Because the function can't be defined otherwise.

This has nothing to do with topology, really

Just pure set theory

It's the same in algebra, for group, ring, module homomorphisms

If you want functions f, g, h such that h = fg then h(x) = h(y) whenever g(x) = g(y)

So the 'kernel' of g needs to be contained in the 'kernel' of h

It's not a corollary - you have to assume that ker(g) is contained in ker(h) in order for it to be possible to exist such f

For set theoretic reasons as above. Remarkable is that it's often enough to assume just that

At least in algebra, because in topology it only really works for quotient maps

Because if that criterion holds, we can define f uniquely (assuming g is surjective). But it turns out this f will often be a morphism in our desired category already

[peak crystal]

Is $\mathbb{C} - [-1, 1]$ is not dense in $\mathbb{C}$?

[gentle osprey]

ctx.author

[gritty widget]

It is dense

[peak crystal]

So if f is entire function whose range is C-[-1,1] then f must be constant how to prove this?

I think again i need to change the channel ☹️

[gritty widget]

It's just https://en.wikipedia.org/wiki/Picard_theorem

[peak crystal]

Oh yes, Thank you.

[wary furnace]

is this a good way of showing that an arbitrary intersection of open sets from the co-countable topology is not open?

or could this be written in a more compact and easy to understand way?

[golden gust]

the empty set is open

[wary furnace]

oh

yeah true lol

well I can just take away 1 set from the intersection and be left with a singleton right?

[golden gust]

yeah

that works and is clear enough

[mystic reef]

Let $X$ be a locally compact Hausdorff space, and let $X^+=X\cup{\infty}$ be its one-point compactification. Let $Y\subseteq X$ be a subset. Compute the closure of $Y$ in $X^+$ in terms of its closure in $X$. I dont really know how to go about this.

[gentle osprey]

Ursus1234

[grave maple]

Do you know how closures work in the one-point compactification?

[mystic reef]

no, not really. I have definition where if Y is compact Haursdorff space and X a proper subspace og Y whos closure equals Y, then Y is said to be a compactification of X

[grave maple]

OK, then I guess you'll have to work out some results about one-point compactifications using that definition. Suppose S is a closed set in the one-point compactification that does not contain oo (infinity). The what can you say about this set as a subspace of X?

[mystic reef]

hmm Im not sure. Wouldnt that mean it would be closed in X as well?

[grave maple]

Yes. Very good. But there is more you can say.

[mystic reef]

hmm ok, let me think 2 min

[mystic reef]

is it compact as well?

if Y is closed in X+ it follows that Y is compact?

[grave maple]

Yes. Very good.

In fact, this is an iff result.

closed in X+ iff closed and compact in X.

[mystic reef]

ahh yea ok

But Im not sure how I can use that to compute the closure...

[grave maple]

OK. We'll get there. Now suppose S is closed an contains infinity.

[mystic reef]

then its not a subspace of X right?

[grave maple]

Right, but what could you say about S - {oo} as a subspace of X?

[mystic reef]

closed and compact...

[grave maple]

Well, it can't be compact because of the iff result I just stated above.

[mystic reef]

but wasnt S-{oo} just the set we talked about before

[grave maple]

Yes. But in that case we were assuming that S - {oo} is closed in X+. Here we have assumed that S is closed in X+.

[mystic reef]

ok so now we are assuming that S a set containing infinity is closed in X+?

[grave maple]

Correct. We're considering two cases for S: it has oo and it doesn't.

[mystic reef]

if it doesnt its closed and compact

[grave maple]

We will have to consider these two cases when computing the closure too.

[mystic reef]

ok

[gritty widget]

Best to compute this by asking yourself question: What are closed sets in the one-point compactification of X?

[grave maple]

That's what we're doing in fact.

[mystic reef]

thats what Im having trouble with, still trying to understand one point compactification and closures

[gritty widget]

And those are 1) compact subsets of X and 2) closed subsets of X together with point at infinity

That's by definition of the one-point compactification

[grave maple]

Right. 2) is what we were working out right now.

[gritty widget]

So you have two cases 1) cl(Y) is compact and 2) cl(Y) is not compact

where cl(Y) means closure in X

If cl(Y) is compact, then this will be the same as closure in the one-point compactification

Because the one-point compactification is Hausdorff (since X was assumed to be loc. compact and Hausdorff)

[mystic reef]

and if cl(Y) is not compact?

[gritty widget]

If it's not compact in X, then cl(Y) can't be equal the closure of Y in the one-point compactification of X

but cl(Y) is the intersection of it with X

(that's how closures in subspaces work)

[mystic reef]

hmm ok, why is your notation for closure in X cl(Y)?

[grave maple]

cl_X(Y)

[gritty widget]

cl is short for closure

[grave maple]

That's the notation for closure of Y in X.

[mystic reef]

ok

[gritty widget]

I just didn't want to differ between those so I mentioned explicitly it's for X

so we know that cl(Y) is a proper subset of closure of Y in the one-point compactification, and that its intersection with X is precisely cl(Y)

and there is only one point we can really add to it

the point at infinity

so it must be precisely cl(Y) together with the point at infinity

[mystic reef]

hmm ok. Ill try wrapping my brain around it

need to get some lunch now

thank you both of you

[stable kite]

Hi there! I'm currently searching for examples of locally compact non-metrizable Hausdorff spaces.
One contender I came up with is the real product of [0,1], which is compact (Tychonoff) and Hausdorff, but not first countable, in particular non-metrizable, as far as I have made no mistake.
I'd love to know whether this is correct, and find other examples. :D

[gritty widget]

You mean [0, 1]^continuum ?

Any uncountable product of spaces is not metrizable

So any uncountable product of compact Hausdorff spaces is compact Hausdorff but not metrizable

And locally compact Hausdorff spaces are precisely compact Hausdorff spaces without a point

[fair steeple]

What if most spaces in your product consist of a single point?

[gritty widget]

I forgot to say non-trivial

[stable kite]

Can you give me a proof sketch for that? :D

[stable kite]

Oh nevermind, you can just show they're not first-countable just as you would do it for [0,1]^continuum I guess.

[stable kite]

How can I find a locally compact Hausdorff space X and a function f : X -> IR, such that f is the limit of a monotonous increasing net of continuous functions with compact support, but not the limit of a monotonous increasing sequence of continuous functions with compact support?

[coarse kestrel]

Isn't the constant sequence a monotonous increasing sequence of functions

[stable kite]

But only if f itself is continuous with compact support. :)

[coarse kestrel]

Ah that's what you meant ok

I didn't realize oops

[stable kite]

I know that X must be non-metrizable for that to exist, so I'm searching for some suitable f in like X={0;1}^continuum right now, but I have no idea how to prove some function is not a sequential monotonous limit of cont. f. with comp. supp.

[coarse kestrel]

Well I'm thinking maybe take the one point compactification X* and take a function f that is 0 at infinity and nonzero everywhere else. Then take a net indexed by the compact subsets of your X and functions that are supported in your compact set and agree with f in some region inside the compact set. Then if f is the limit of a monotonous sequence of compactly supported functions will imply X is sigma compact by taking the support of each function, so take X to be a non sigma compact space

But I'm trying to work out the details

Hmm doesn't make sense

[stable kite]

What doesn't make sense? :)

[coarse kestrel]

Well the existence of such f will require X to be sigma compact so it doesn't make sense to take X to be a non sigma compact space lmao

So my answer doesn't make sense nvm I tried

[stable kite]

Haha don't worry, thanks for your input regardless.

For context, I'll hold a seminar presentation on Bourbaki/Stone Integration Theory soon, which makes use of nets instead of sequences to remain correct for non-metrizable locally compact Hausdorff Spaces.
... But I have yet to find an example that legitimizes this decision. xD

[grave maple]

Any particular reason you are using nets and not filters?

[stable kite]

Uhh all the literature I have available uses nets, I honestly have never encountered the concept of a filter. 😅

[grave maple]

Wow, really? OK. For me, it's the other way around.

[opal pond]

hello, I really need help, I have an exam tomorrow and I am studying topology, sets, can anyone explain to me what an open, closed set is? and like... what's closure and interior point? I have to clarify that I am only a high school student in Georgia (Country) I am like in the 12th grade and This exam is going to be in an olympiad which if answered 30% correctly, I can get scholarship in my university

[gritty widget]

Metric or general topology?

[opal pond]

I think metric, like non-geometric 😄

It's all about sets

[gritty widget]

Were topological spaces defined, or were you guys working with metric spaces

[opal pond]

topological spaces are defined

like (X, t (tau))

[gritty widget]

alright, so a topology is a family of subsets of a given set X such that it's closed under arbitrary unions, finite intersections, and contains the empty set and X

elements of topology are called open sets

a complement of an open set is a closed set

[opal pond]

sorry if this is a dumb question... but What's a complement?

[gritty widget]

closure of a set A is the smallest closed set F such that A is contained in F

equivalently, closure of A is intersection of all closed sets containing A

[gritty widget]

If A is a set, then its complement (in X) is the set X\A, that is all points in X which aren't elements of A

[opal pond]

aah okay, got it

[gritty widget]

finally, the interior of a set A is the biggest open set contained in A

an interior point of A is a point in the interior of A

any questions?

[opal pond]

if X is {a, b, c} and tau is {empty set, {a}, {a, b}, X}

can you tell me, like which set is closed, open and which is the interior or closure?

[gritty widget]

we speak of interior and closure with respect to some subset A

open sets here are empty set, {a}, {a, b} and X

closed sets are empty set, {b, c}, {c} and X

Give me a set A and I can take closure/interior of it

[opal pond]

I am sorry, I am not used to this kind of theoritical explanations in algebra, I totally just do a lot of textbook problems 😅

[opal pond]

A {b, c} is that allowed?

[gritty widget]

It's not algebra, topology is borderline analysis and geometry (very roughly speaking)

[gritty widget]

Yes, any subset will do

[opal pond]

yeah, I know I just compared it's learning pattern to algebra's learning pattern

[gritty widget]

cl({b, c}) = {b, c} and int({b, c}) = empty set

cl means closure and int means interior here

[opal pond]

and one last question

what's the closure operator?

[gritty widget]

closure operator is a function, denoted cl, which to a subset A assigns its closure cl(A)

[opal pond]

okay, thank you so much ❤️ you were a great help ❤️

[gritty widget]

np

[opal pond]

∀

what does this mean?

[gritty widget]

for all

[opal pond]

and what's the basis of topology?

[gritty widget]

did you check wikipedia or any textbooks for the definition?

[gritty widget]

It's a collection of open sets which is closed under finite intersections

[opal pond]

yes, but I feel like it's kinda all over the place

[gritty widget]

@opal pond I said something wrong here, my bad

I must be tired or something

It a collection $\mathcal{B}$ of open sets such that every open set is union of elements of $\mathcal{B}$, or equivalently, for any $B_1, B_2\in \mathcal{B}$ and $x\in B_1\cap B_2$ there is $B_3\in\mathcal{B}$ such that $x\in B_3\subseteq B_1\cap B_2$

[gentle osprey]

Blitz

[gritty widget]

Motivation is like this: we often specify topology in terms of its basis.

For example, topology of a metric space, we define open sets as unions of open balls $B(x, r) = {y : d(x, y)<r}$ where $d$ is a given metric. Those balls then form a basis for topology.

[gentle osprey]

Blitz

[opal pond]

Thanks ❤️

[formal tide]

I'm back with another trivial looking result I can't deal with 🙂
If X in R2 is homeomorphic to the closed disk, show that R2\X is connected

seems like it should follow right from jordan curve theorem but only thing it tells me, as far as I can see, is that the interior of X is either in an unbounded component or in a bounded one

not sure what else to do from that

[gritty widget]

The property "compact subset of R^n separates it" is a topological invariant

This can be proven

So we can assume X is the unit disk

The proof uses some results such as no-retraction theorem, but it's elementary otherwise. It's not trivial to come up the proof with either

[formal tide]

what do you mean by "separates it"?

[gritty widget]

We say A separates Y if Y\A is disconnected

[gritty widget]

This result is how you can prove Jordan's theorem. At least in part, because it doesn't tell you about how many components does the subspace separates R^n into

You need further analysis using tools from algebraic topology for that

[formal tide]

I see, thanks

still wondering if there's just an easy way out with JCT, I think there should be...

[gritty widget]

It's doable

You can separate R^2 into two parts using the boundary of your disk

Main difficulty would be to prove that the interior of the disk is the bounded component

Things get a little complicated here

It's contained in it for sure, this only uses connectedness

[formal tide]

can you elaborate why? It could in principle be mapped outside, sort of like an annulus

(because of the fundamental group this is not the case, but other weird cases could hold (maybe?))

[gritty widget]

You're right, my bad

[shut moat]

isn't the connectedness of the two components part of the statement of the jordan curve theorem?

[formal tide]

yes

but that is not enough because of annulus-like images of the disk, I believe

[gritty widget]

Take a point in the bounded component such that the disk doesn't touch.

And treat the disk as a subset of punctured plane

[opal pond]

bruh wtf is this I am so confused

[gritty widget]

If we consider path around the boundary of that disk, we get a path in the disk which can't be contracted

This is a contradiction

[shut moat]

I could be brainfarting but given that X is homeomorphic to a closed disk, i don't think this can happen

[gritty widget]

So every point of the bounded component is a point from interior of the disk, from connectedness, it needs to be equal

[gritty widget]

It was just a hypothetical comment, I think

[gritty widget]

Hopefully you can formalize this

[formal tide]

I'm still processing it hmmm

[formal tide]

and yep, this shouldn't happen but my point was to proof that this does not happen

[formal tide]

I'm sorry I'm not getting it, loops in the punctured plane can't always be contracted

[gritty widget]

This is a loop in the disk

Which a contractible space

But at the same time, a non-trivial loop in the punctured plane

Hence contradiction

[formal tide]

ohhhhhhhhhhh

ok

[gritty widget]

Essential loop

[formal tide]

got it, wasn't understanding this

yup I think that works, thanks 🙂

[gritty widget]

You're welcome

[opal pond]

what's a natural topology?

[fair steeple]

One that, if polled, 9 out of 10 mathematicians will agree is the best one on a given set.

[opal pond]

whut?

[dry jolt]

Trying to understand a bit more about van Kampen, is it effectively just saying that if X = A \cup B and everything is path-connected (plus whatever other nice conditions) then the fundamental group functor preserves pushouts? In particular, X is the pushout of the inclusions A <- A \cap B -> B, and van Kampen is saying that \pi_1(X) is the pushout of \pi_1(A) <- \pi_1(A \cap B) -> \pi_1(B), right?

idk how to do diagrams with the latex bot here :/

[gritty widget]

Van Kampen is better thought of in terms of the fundamental groupoid functor

Then it's precisely a statement that this functor preserves push-outs

[cursive flume]

@coarse kestrel does this look legit now? the proof we discussed yesterday

@gritty widget sorry for not replying to your messages yesterday regarding the quotient topology,i'm still progressing it- will get back to you in like 30-45minutes when i TeX the proof

[gritty widget]

Natural topology on R^n is the one generated from the standard metric d(x, y) = |x-y| where |x| = sqrt(x_1^2+...+x_n^2)

[dry jolt]

it's not immediately obvious to me what pushouts are in the category of groupoids?

[opal pond]

$A^d$

[gentle osprey]

LukitoTheWise

[opal pond]

what is this?

is it the same as boundary?

[formal tide]

depends on the context

a lot of people use a lot of different notation

[opal pond]

what do you call this?

[formal tide]

I don't know

distance to A maybe?

idk

[gritty widget]

Do you need to know what they are? Computationally-wise, you'll still use the version with groups

[dry jolt]

but the picture I have in mind goes something like this: we can't apply normal van kampen to compute the fundamental group of the circle, but it feels like if you have two open arcs which cover the circle, you can fix two points (one in each intersection), and then use the fact that each arc is contractible, hence there's one homotopy class of paths between the points in each arc. Taking some sort of pushout here yields two distinct homotopy classes of paths between the points, and looking at the endomorphisms of one of the points in the fundamental groupoid shows that it's isomorphic to Z @gritty widget

[opal pond]

it says here that $if tau is a natural topology and A=(0;1)U{2}, then A^d=[0;1]$

bruh

[dry jolt]

ah yeah ig that's true

it was moreso about guaranteed existence but i feel okay taking that for granted

[gritty widget]

Existence of pushouts?

[dry jolt]

yeah

[formal tide]

I've been thinking about this, why would the image of the loop around the border of the disk be non contractible? It feels very obvious geometrically but I can't find any justification

if the point was in the disk, then sure, but by assumption it was not

[gritty widget]

Because of its winding number being non-zero

[formal tide]

was thinking about that too, but why it would be non-zero? which theorem is implying this?

I know that the winding number of a circle in the punctured plane is non-zero (1)

but is this preserved by homeorphic images? it surely is not by just continuous images (I could make it do two loops instead)

[cursive flume]

@gritty widget I got your point. the constancy on fibers has nothing to do with topology. it is pure set theoretic requirement for the map to exist, as yo usaid

[cursive flume]

found a counterexample to see why it is as you said

[opal pond]

Hey guys, I noticed that continuity in topology is not the same continuity in functions, do they differ greatly? Or is it the same thing, but just said topologicly?

Hey guys, I noticed that continuity in topology is not the same continuity in functions, do they differ greatly? Or is it the same thing, but just said topologicly?

[gritty widget]

the latter

[opal pond]

And what is homeomorphism?

[gritty widget]

a continuous map of topological spaces with a continuous inverse

[stable kite]

Can you give me an example of a non-sigma-compact locally-compact non-metrizable Hausdorff space?

[gritty widget]

I think [0, omega_1) is like this

@stable kite need me to prove it?

[stable kite]

What is omega_1?

[gritty widget]

first uncountable ordinal

[stable kite]

I haven't studied ordinals yet - any other manner of definining the space?

[gritty widget]

well, I could try but I don't see a reason to

ordinals correspond to well-orders

they are "equivalence classes" of them

ordinals can again be "well-ordered"

in particular we can consider what's the "smallest" uncountable well-order

this is omega_1

like, in my opinion, it's not necessary to study topology without knowledge of ordinals

but if you want examples, then it's good to have those

so imo, you should learn ordinals

and come back

[empty grove]

I feel like that's way too big a time investment just to construct examples

[gritty widget]

that's not an investment at all

[empty grove]

As long as you have some key properties of omega_1 you can prove these properties for the space

[gritty widget]

when you first learn about cardinals etc., you basically learn ordinals with them

[empty grove]

Well depends on how deep you go

[empty grove]

Assuming you study set theory properly

[gritty widget]

see Dugudnji's "Topology", first chapter

[empty grove]

Very often people just learn about basic cardinality arguments in an intro analysis course and never learn about ordinal arithmetic (like everyone at my college)

[stable kite]

@gritty widget Fair enough, I'll keep that in mind. :) In the end, I'm just searching for any way to answer my question linked here, which is probably an analysis question, but kind of lives off the topology of X, I guess.

[gritty widget]

[0, omega_1) isn't as scary as it might seem, it's an uncountable analogue of natural numbers

[grave maple]

If you go about finding counterexamples seriously, you will inexorably encounter ordinals/cardinals.

[gritty widget]

Yep. And if you want to dive deep enough into topology issues, there'll be some heavy set theory here and there.
Things like forcing and such... I don't know much about those things

[stable kite]

Basically, all I want to achieve is somehow legitimize that the Bourbaki Extension Process in Bourbaki Integration Theory uses nets instead of sequences.

[gritty widget]

Even when working with metric spaces - there are examples of subspaces of the real line for example, that are very cool but also set theoretical in nature (depend on the axiom of choice)

[gritty widget]

Working in general, you have to be prepared to work with some pretty general spaces

It's just how things are

[grave maple]

I know of an example of a non-metrizable topological space used in analysis: bounded [0,1] -> R functions. There is no metric that describes pointwise-convergence of these functions.

[stable kite]

Yeah, I get that absolutely. There's a lot to explore. ^^

[fervent citrus]

A little more elementary example would be an uncountable discrete space. Or more generally, you can take the sum (cf. Bourbaki TG I, §2, n°4) of an uncountable family of non empty compact spaces.

As for your actual goal, I think nets/filters are more canonical than sequences and it's actually sequences that lack legitimacy...

[stable kite]

A discrete space is sadly metrizable, I'll check out the ither example.
I know for sure nets are more legitimate than sequences, I just couldn't find a single example where monotonous nets of C_c(X) functions offer more limit functions than sequences of such functions 😅, but I will definitely need such an example to explain the entire approach in comparison to the Daniell-Stone net based approach on locally compact metric spaces we covered in class.

[opal pond]

Guys what's a card in topological mathematics?

[plain raven]

I haven't heard of this. Can you provide more context?

I couldn't find this concept on google.

[surreal lantern]

uh do you mean chart?

[fervent citrus]

oops didn"t read the whole original meassage xd

Ignoring metrizability (which isn't very important for the integral), you can see that if a function f on a locally compact space X is the upper envelope of a countable family of non negative compactly supported continuous functions, then the set of points x in X such that f(x)>0 is contained in the union of a countable family of compact sets.
So as long as X cannot be written as the union of a countable family of some of its compact subsets, then the constant function 1 on X is not the upper envelope of a countable family of non negative compactly supported continuous functions, but is for sure the upper envelope of all the non negative compactly supported continuous functions majorized by 1

[surreal lantern]

if you translate the word we use in german for chart literally you'd get card as well

[gritty widget]

I think they said they were Georgian

So might be a wrong lead there

[surreal lantern]

yeah true i just said that because there might be more languages who'd translate it like that

i'll just wait for them to provide more context

[mystic reef]

So im trying to show the following. Let $X$ be a topological space, $A\subseteq X$ a closed subset. Show that if $X$ is $T_3$, then the quotient space $X/A$ is Hausdorff. I can intuitively understand that its true, Im not really sure how to show it. I was wondering if instead of showing $p\Rightarrow q$ that I could show $\neg q\Rightarrow \neg p$. But I dont know if its better/easier...

[gritty widget]

well, T3 means you can separate points with closed sets

btw you wrote quotient wrong, it should be X/A

If you take two points x, y which are not in A, how would you separate them in X/A?

[gentle osprey]

Ursus1234

[mystic reef]

thank you

[mystic reef]

finding/showing a neighbourhood around each of those points...

I think

[gritty widget]

disjoint ones I presume

[mystic reef]

yes

[gritty widget]

yeah, we can always do that, but there's one thing, we don't want them to touch A

now if you project this into X/A, they should still be open and separate those points

so the only thing left is that if x is a point which is not in A, then we can separate x with A (assuming here that A is non-empty)

in X/A

And this is pretty much the same, you use that the space is T3 and then project onto X/A

[mystic reef]

ahh ok, I'll try doing that

thanks

[gritty widget]

np.

[stable kite]

Ohh, thank you, that's perfect! :D

[wary furnace]

Hey guys, I was able to prove that an infinite set has infinitely many topologies like this (using the well ordering theorem for any uncountable sets in a similar manner). How else could I prove it without relying on orderings?

[gritty widget]

Why not just take everything outside of a countable set to be open? Then you can assume your set is countable

[haughty wave]

to construct T0, just choose a point, which is always possible

To go from Tn to Tn+1, just choose a point outside of the largest open strict subset of Tn.

[wary furnace]

oh so we construct topologies based on the uncountable set we exclude?

[wary furnace]

also yeah that's a better way of expressing my idea, without orderings

[cursive flume]

how can one classify vector bundle structures on a fiber bundle with vector space fiber R^n?

[fair steeple]

Do you mean that the base is itself a fiber bundle?

I don't see what difference that would make, but it reads as if that's what you want.

Vector bundles are classified by homotopy classes of maps from the base into Grassmannians, but this is usually not very practical.

[marsh forge]

It is pretty practical in this case

[fair steeple]

What case?

[bitter smelt]

in the case in which you are trying to classify the things that you just gave a classification for lol

[fair steeple]

A classification theorem puts two things in bijection, the objects being classified and the "classifier" space, so to speak. In that sense any classification is great.

But that doesn't mean that it's easy to compute.. this is what I mean by practical.

[bitter smelt]

I know, I was being snide

[fair steeple]

If you classify something by natural numbers that's easy to understand, but I don't think anybody can claim to easily compute the cohomology of BO(n) for every n.

[marsh forge]

in the case that the space in question is contractible 🙂

OHHHH

IM so sorry

I misread prophet's message

hahaha

I thought they wanted to classify bundles over R^n

[fair steeple]

Yeah, the fiber is contractible.. which is always true.

[marsh forge]

yeah the representing thing isn't super useful most of the time i agree

[bitter smelt]

What does K(*,*) generally refer to?

Ive seen it several times and cannot google it

For example:

[gritty widget]

see "eilenberg-maclane space"

[bitter smelt]

https://en.wikipedia.org/wiki/Eilenberg–MacLane_space this?

oop

yep

thanks!

so for this, we're saying (using notation from the wiki page) that S=X, and G = $\pi_1(S)$?

[gentle osprey]

Migillope

[gritty widget]

yes

[bitter smelt]

And so then obviously $\pi_1(S) \cong G \coloneq \pi_1(S)$, and all other homotopy groups are trivial since the universal cover of S is contractible

[gritty widget]

No, S and X don't have to be the same, X = K(G, 1)

[gentle osprey]

Migillope

[bitter smelt]

I mean in the context of the screenshot, the S there is the X on the wiki page?

[gritty widget]

yes, sorry

[bitter smelt]

Basically checking that "is a K(*,*) space" means "is an Eilenberg-MacLane space of type K(*,*)"

[gritty widget]

you may be looking for $\coloneqq$

[gentle osprey]

TTerra

[gritty widget]

and you are correct on the homotopy groups

[bitter smelt]

I was, in #latex-testing. Who uses :- ? That's maniacal

Thank you!

[cursive flume]

I mean, i give you a fiber bdl with fiber R^n as input and ask how many inequivalent vector bundles can i build out of it

And which are those

[gritty widget]

at least in the smooth case, the multiplication by reals completely determines the vector bundle

that might mean something

[gritty widget]

this isn't a suggestive remark btw, i'm just throwing stuff out there

[shut moat]

all finite dimensional hasudorff TVSs of the same dimension are isomorphic

so i imagine the uniqueness of continuous vector space structures on R^n are enough to say that the vector bundle structure is unique for a fixed fiber bundle, but idk

[coarse night]

wait empty sets aren't supposed to be topological space

[gritty widget]

what's the book's definition of a topological space? i've never seen the requirement that they be non-empty

[coarse night]

I've seen non empty

[gritty widget]

the book's definition is the most important

[marsh forge]

Empty set is 100% a topological space.

[gritty widget]

just about every reference i can think of doesn't require non-emptyness

[marsh forge]

It is not a based topological space

[coarse night]

ig they don't need non-empty then

[marsh forge]

its not that big of a deal

[coarse night]

ya

[gritty widget]

at the pinnacle of mathematics are questions about signs and the empty set

[coarse night]

except idk if it messes with the Top category

I have no idea

[marsh forge]

Top as a category always has the empty set

The empty set is its initial object

[coarse night]

thing is, I thought singletons are

[marsh forge]

No

[coarse night]

I used to that is

[marsh forge]

They would not be even if you removed the empty set

[coarse night]

wait so what's the fundamental group of empty set?

[marsh forge]

You always take fundamental groups of based spaces

the empty set is not a based space

[coarse night]

ye

oh

I see where I messed up

[warped rover]

a cringe space if you will

[coarse night]

Mandela effect

[cursive flume]

Where can I find a proof for this?

[gritty widget]

https://arxiv.org/pdf/math/0702772.pdf

tl;dr: a map of total spaces of vector bundles is a vector bundle morphism if and only if it intertwines the scalar multiplications

[cursive flume]

Which theorem here?

[gritty widget]

around corollary 2.1

[cursive flume]

The language is so abstract

[gritty widget]

let me try to motivate the result. suppose $V, W$ are finite-dimensional real vector spaces and $\varphi\colon V \to W$ intertwines the scalar multiplications. picking bases, we can assume $V = \bR^n$ and $W = \bR$. fixing $x \in V$ and differentiating $f(tx) = tf(x)$ with respect to $t$ at $t = 0$, we have $$f(x) = \sum_{i=1}^n\frac{\partial f}{\partial x^i}(0) x^i,$$ which shows that $f$ is a linear map.

[gentle osprey]

TTerra

[gritty widget]

doing this consistently fiberwise is the more difficult part

maybe posting the whole paper just for this specific result was a little overkill, but hey there's some fun stuff in there so why not

[cursive flume]

Why W=R?

[gritty widget]

pick a basis of W and work componentwise

[gritty widget]

They definitely are. I just don't see "empty metric space" defined often

[gritty widget]

https://en.wikipedia.org/wiki/Metric_space
like on wikipedia here
once we had an argument, my professor was like "there is no empy metric space", and I give him counter arguments, that you can define a unique metric on it, and someone said that there's no empty metric space on wikipedia, lol. I gave up

[unreal stratus]

at my uni you lose a mark if you don't specify the metric space is non-empty when stating banach fixed point thm oof

[coarse night]

ok so metric spaces needed to be non empty but topological spaces aren't

[gritty widget]

Nah, it's all conventions. Things usually break for empty sets so people don't like to include them. But it's usually the same either way anyway, lol
Similarly in universal algebra you assume algebras are non-empty. But like, from category theory perspective, removing those means you remove initial objects from your categories.
In the end it doesn't change much

[stable kite]

Am I correct to assume that the Long Line is a locally compact non-metrizable non-sigma-compact Hausdorff Space?

[gritty widget]

yes

[stable kite]

Thanks!

[marsh forge]

The moral of the story re: emptyset is that the emptyset is included if and only if the theorems are true with it included

[shadow rampart]

Can you define a map between topological spaces by saying where the basic open sets of the domain space go to?

[wide kayak]

Open sets in a basis are a cover of the space, so yes. If you declare where basic open sets go, you're also specifying where all points go

[marsh forge]

There is a slight distinction to be made

One interperetation is that we are simply looking at the like

formal lattice of open sets

and describing a like ordered map between them

the other is to define a set map on all the open sets that can "glue" to a full map

the latter is equivalent to the data of a continuous map but the former is not unless you space is sober

[stable kite]

Theorem 1.1. Suppose U is open in a locally compact Hausdorff space X, K ⊂ U , and K is compact. Then there is an open set V with compact closure s.t. K ⊂ V ⊂ V- ⊂ U . [Rudin p.37]
I don't have a copy of Rudin - any tip on how to prove this? I remember a similar statement for normal spaces, and assuming this is not far from it, as we are locally normal.

[marsh forge]

You can find a pdf of rudin online, but this also should be doable? Don’t have time personally

[gritty widget]

yeah, you have lots of results for locally compact Hausdorff spaces analogical to those for normal spaces
Tietze and Urysohn theorem also get their own versions iirc

You prove this very easily actually, for every point of K take a compact neighborhood disjoint from U

I think

Or maybe like, if U^c is compact then we can separate K and U^c since the space is Hausdorff
Otherwise U^c union {infinity} is compact and disjoint from K
Take two disjoint neighbourhoods, then intersecting with X, they're still open

infinity is the point at infinity from the one point compactification

[stable kite]

Oh yeah, I forgot you can use one-point-compactification in such scenarios, gonna keep it in mind, thanks!

[gritty widget]

the one point compactification argument is cooler

[grave maple]

This result is similar to: In a locally compact space, every compact set is contained in the interior of a compact set.

I wonder how complicated the proof would be using filters.

[bronze lake]

Is there a difference between the product/trivial bundle and the tangent bundle other than how the local trivializations are constructed?

[spring geode]

i was told there were feet pics here

@gritty widget you lied

[odd flame]

why is munkres going through all this trouble

[marsh forge]

wdym

[odd flame]

like why define cartesian products with the idea of an m-tuple as a map

though if im being honest i dont really get how the m-tuple is a map from an index set to X

[marsh forge]

its the best way to do it formally

an m tuple is

a first element

a second element

...

an mth element

this is the same as a function from {1,...,m} to your set

[odd flame]

but (x1, x2, ..., xm) isn't an element of X

im probably being dense

[marsh forge]

x_1 is the image of 1 under this function

x_2 is the image of 2

and so on

so like if f:{1,...,m}-> X is a function

the associated tuple is

(f(1),f(2),...,f(m))

[odd flame]

TIL ryc is good at celeste

[odd flame]

this made more sense though thanks

so just to be clear that map he defines in the first line is just how we formally index the m-tuple?

[marsh forge]

yeah an m tuple is formalized this way

or at least this is one way, and probably the best one imo

the issue is that sets have no notion of order and can't contain duplicates

but functions out of ordered sets do have an order implicitly and also can have duplicates

[grave maple]

How does he formalize sequences?

[stable kite]

Don't know Munkres, but pretty universally an A-sequence is a function f : IN -> A.

[long grail]

What is the set X^{omega}? Is that standard notation? I can't find where it's previously mentioned in this book

This is munkres btw

[gritty widget]

omega = {0, 1, 2, ...}

X^omega is product of countable amount of copies of X

[rancid umbra]

it can be thought of as the set of all functions from N to {0,1}

[long grail]

OH

Ok

Thanks

That must be isomorphic to set of all infinite strings of 0's and 1's

f(n)=g_n(k), where g_n(k)=1 if n=k and 0 otherwise. Does that satisfy as an answer?

Also, concerning a "choice function," can I think of it as a function such that an element in the image has form ((A_1,x_1),(A_2, x_2), ...), in other words you input a set of sets, and it outputs a collection of ordered pairs consisting of an element and the set it lives in (from the set of sets).

Here x_n is in A_n

So, if the collection A is all nonempty subsets of the positive integers, a choice function could be "select the least integer in each"

[rancid umbra]

yea

[golden gust]

have you seen the notation B^A for the set of functions {f: A —> B} ? if so, then the space of m-tuples X^m (in the sense of a list) gets identified with functions {f: m —> X} where m is viewed as a set with m elements (as a von Neumann ordinal)

which is exactly the definition munkres has

maybe I'm weird but I think this is a cool example of good notation

[rancid umbra]

agreed

although, at this point, we need cartesian products to define functions

so there’s already a notion of a tuple

and it gets kind of hairy

[empty grove]

Functions only need you to define pairs

You define pairs, then functions, then arbitrary tuples

[rancid umbra]

but once i have pairs i can get triples

without needing functions

like, idk. there’s two diff tuples floating around

[empty grove]

Ye but then getting infinite tuples is not so nice

The standard thing in set theory is to define pairs to define Cartesian products and functions and then only use the function construction after that

Though they are in bijection so it won't matter much anyway

[long grail]

I'm confused. Doesn't the act of "picking" use the axiom of choice?

[marsh forge]

nope, your proof is correct

its kind of weird nuance to get used to but if you can specify some element uniquely for each choice then you arent using choice

For example, if i had an infinite set of nonempty well ordered sets

i could choose the minimal element from each

without using choice

[long grail]

oh

wait, so you wouldn't be using a choice function that selects each minimal element?

[marsh forge]

i mean ur using a "choice function" of some sort

this isnt a formal term

but you aren't invoking the axiom of choice to do it

[long grail]

oh i think i get it

when we define explicitly

we aren't using AoC

[plain raven]

Yeah.

[long grail]

AoC is for saying we can make arbitrary choices

[plain raven]

The AoC is needed exactly in the cases where it can't be defined explicitly

[long grail]

i get it now

Thanks a lot

[marsh forge]

oh thats a cute way to see that the well ordering theorem is eq. to choice

just well order all the sets and pick the minimal one

maybe thats how people usually prove it

[long grail]

ohhh

wow, all right. lightbulbs

[plain raven]

Generally if you only pick one time or you pick some finite number of times this doesn't require choice. this is just like, how we interpret the existential quantifier in logic

[long grail]

Yeah, I think I get it

[long grail]

But we have confidence in our arbitrary choice because if nothing else we can do what @marsh forge said

[marsh forge]

well well ordering theorem is similarly independent of ZFC

like choice and Zorn's lemma

but i think the point of this exercise is to help you distinguish choice from not choice

oh heres a fun fact

The statement that every surjection of sets has a left inverse is equivalent to the axiom of choice

[long grail]

That's literally the question I did right before lol

[marsh forge]

oh hahaha

makes sense

anyway thats not the real fun fact

the real fun fact is that you can axiomatize what it means for a category to be "similar to the category of sets" (this is called a topos)

and then in an arbitrary topos, even assuming choice, it won't always be true that every surjection has a left inverse

and so you can ask whether arbitrary toposes satisfy an internal form of the axiom of choice

and this is one reason you might care about whether a statement does or does not require choice

because if it doesn't you might be able to prove it for every topos

but if it does it won't be true

[long grail]

I don't know much about categories yet. Are you saying they aren't rooted in zfc?

[plain raven]

a topos is itself a model of set theory. you can define what a topos is using only first order logic without assuming set theory but the ones people care about in practice arise in the course of ordinary set-theoretic mathematics, and there are useful hypotheses and assumptions you might want to make about a topos which require set theory to state.

Even if you accept AOC you might be studying the topos for reasons related to like, applications in algebraic geometry and you might want to know whether an argument goes through in the model

[long grail]

Oh sweet

[empty grove]

That's circular because you end up having to choose a well order on each set

You would well order their union instead, then inherit that to each set in the family

[marsh forge]

Ah fair enough

[odd flame]

i came back to this and i just wanna make sure i got this: the m-tuple is really like an ordered tuple of the results of the indexing function? where the domain corresponds to the index of each element in the tuple

that's a weird way of phrasing it but i think that's it

he also says at the beginning of the section that this is a more general cartesian product but how

[marsh forge]

Yes that is it

[cursive flume]

I am on this proof for ages and tilted

Trying to Check any superset of an element in G is in G (filter axiom 3)

Can someone check my proof please

Let P cap U be in G, where P in F, U in N_x. We claim if V superset P cap U, then V in G. Indeed: P cup V superset P, and since P is in F, by axiom 3 of filter, P cup V in F. Moreover since U is a nbhd, we can enlarge it as much as we want: U cup V in N_x. We conclude: (P cup V) cap (U cup V) in G. We now use A cap (B cup C)= ( A cap B) cup (A cap C) for A=P cup V, B=U,C=V to obtain:
((P cup V) cap U) cup ((P cup V) cap V))
=((P cap U) cup (V cap U)) cup ((P cap V) cup (V cap V)).
Now note that this is equal to V, as P cap U subset V by assumption, V cap U subset V, P cap V subset V and V cap V=V. Qed

[grave maple]

Looks OK. This is basically the infimum of two filters.

[sly mural]

If $(\tilde{X}, p)$ is a covering space of $X$, with $x_1, x_2 \in \tilde{X}$ st $p(x_1) = p(x_2) = x$ then is it true for any loop $\alpha$ based at $x$, there exists a lift of $\alpha$ which is a path from $x_1$ to $x_2$? I know that for each element of the fibre of $x$ there is a unique lift starting at that point but I'm not sure if you can use path inverses to get a similar fact for the end points of the paths

[gentle osprey]

rustyjpeg

[dry jolt]

if your covering space is path connected, then this follows from the uniqueness of path lifting. given a path \gamma from x_1 to x_2, p o \gamma will be a loop at x in X. Then lifting p o \gamma to a path starting at x_1 yields the desired path

[sly mural]

great thank you 🙂

[sly mural]

i'm also trying to gain some intuition of regular covering spaces, is there any example or description that makes it more clear what a regular covering space "looks like" rather than just referring to normal subgroups

[sturdy notch]

There are some nice images of the different covering spaces of S1vS1 in hatcher iirc which give examples for essentially all the theory

[fickle umbra]

A homeomorphism is a type of a structure-preserving map. What kind of structure is it preserving?

[golden gust]

the topology

[fickle umbra]

as in the collection of open sets?

[gritty widget]

open sets = elements of the topology

[golden gust]

yeah, exactly. if f: X —> Y is a homeomorphism then the topology on Y is exactly the f-image of the topology on X

[gritty widget]

a homeomorphism is a bijection which induces a bijection on the topologies

So in a sense, it's just a relabeling of elements

such that after the relabeling, the way the family of open sets "lies" in your space, is the same as before

the family of sets wouldn't have to be a topology, really

we could generalize this

no information is lost, for sure

other than what we name the elements, similarly to how in graphs we could label them multiple ways

we could relabel them, but we still get the same graph

it's the same. It's simple, really

maybe not completely straightforward from the definition of a homeomorphism. But that's how they work

[urban panther]

i have a small question on cw complexes -- just collecting examples for myself; is there a cw complex structure on the cylinder S1 x [0,1]?

[quartz edge]

yeah

pretty easily

take 2 points

[urban panther]

yup

[quartz edge]

attach 3 1-cells to them, (2 are loops at each point, the third connects the 2 points)

attach a single 2-cell along all the 1-cells

done

[urban panther]

damn'

cool

thank you!