point-set-topology

[tender halo]

hmm

just minimum right

[lucid ocean]

yeah

should be, or product if you want I guess

[tender halo]

if you say "separated" instead of "completely separated" here you get normality of compact Hausdorff spaces

[lucid ocean]

yeah

[tender halo]

also it gives you complete separation of compact sets and closed sets in tychonoff spaces

[lucid ocean]

anyway, I was trying to find a version of normality that worked well with the ol' system I came up with

with the complete separation by a function approach, I need to quantify over all subsets with a property preserved under continuous maps

[tender halo]

there is uhh

collectionwise normality?

normality is really about covers of the space

maybe you can try thinking in that terms

i had a funny screenshot about different characterizations of normality

the book is "normal topological spaces" if you are curious

[lucid ocean]

maybe, would have to look harder because collectionwise normality isn't even hereditary

[tender halo]

you can always say "hereditarily collectionwise normal"

[tender halo]

hmm clearly normal spaces are a natural setting for topology, its only natural to ask that every countable locally finite open cover of X has a countable star-finite normal cozero-set star refinement

[lucid ocean]

there really needs to be a handbook of mathematical results

just the results alone, even for topology in comprehensive detail

[lucid ocean]

also anything with closed sets is suspicious because 1. in a finer space, there's greater separation but also more closed sets to separate, 2. they aren't preserved by continuous maps

[rain gyro]

Guys is this correct, I felt I made it too much of a overcomplication.. 🫣

[tender halo]

its not continuous because of the definition of continuous maps

youve proven that preimages of subbasic open sets are open

which in turn gives you continuity, but its a separate lemma

[uneven bronze]

If $X$ is a completely regular space, a subalgebra $\mathcal{A}$ of $BC(X)$ is called completely regular if (i) it is closed and contains the constant functions, and (ii) $\mathcal{A}\cap C(X,I)$ separates points and closed sets. E.g. if $(Y,e)$ is a Hausdorff compactification of $X$, $\mathcal{A}_{Y}={f\circ e:f\in C(Y)}$ is a completely regular subalgebra of $BC(X)$.\

Exercise: The Hausdorff compactifications of $X$ are in one-to-one correspondence with the completely regular subalgebras of $BC(X)$.\

Consider the map $[Y]\mapsto \mathcal{A}_Y$. I can show this map is injective and surjective, but I struggle showing it is well-defined. If $Y,Y'$ are equivalent, why would $\mathcal{A}Y=\mathcal{A}{Y'}$?

[gentle osprey]

psie

[uneven bronze]

Never mind. I figured it out I think.

[lucid ocean]

how about precisely separated by a function

(precisely separated: f^(-1) ({0}) = C, etc.)

should be stronger, a poor man's T6

[tender halo]

i mean that is exactly T6

[lucid ocean]

is it

like exactly so?

[tender halo]

a T1 space is T6 iff all closed sets are zero iff all open sets are cozero

[lucid ocean]

no but I only quantify over compact subsets

all disjoint compact subsets are to be precisely separated by functions

[tender halo]

you are asking when all compact sets are zero, which does give you complete hausdorfness, hmm

[lucid ocean]

does it?

if so, that's a kind of funny characterisation

[tender halo]

no im just thinking, it doesnt characterize it

[lucid ocean]

well I mean it'll give a funny condition to which it is equivalent to

but yeah that's less believable

the trick with padding out the zero sets to use compactness no longer works

[tender halo]

it also always true in perfect tychonoff spaces

i see no way to wrangle either from what we have though hmm

[lucid ocean]

it should be stronger

completely T2, completely T3, T4, T5 have no mention of precise separation

[tender halo]

sure

[lucid ocean]

so in principle surely one could concoct an example where it is impossible

but if there is a counterexample, it has to have a coarser topology than a T6 topology

hmm that's not saying much

take your favourite T5 but not T6 space like idk some LOTS

then maybe try to find some compact subsets thereof that can't be precisely separated, inspired by the closed sets that can't

hmm wait, the lex order on [0,1]^2 is T5 but not T6, and is compact

in compact Hausdorff space, compact <=> closed, so the space is completely Hausdorff but not "compactly T6"

[tender halo]

no i mean as i said, take niemytski plane

it is tychonoff and perfect so it gives you our condition

but is not normal

[lucid ocean]

sure, so its a bit stronger than complete Hausdorffness, excellent

[lucid ocean]

ah weird, conjunction of nonseparation properties yielding a separation property

[tender halo]

perfectness is kind of a separation property

seeing as T6 = T5 + perfect

[lucid ocean]

Well in the sense of my system

unless if it actually is, lemme check

yeah it probably isn't

[steady gorge]

since $A_n \cap A_{n+1} \neq \emptyset$, that means there exists no separations, right? would that be enough for $\bigcup A_n$ to be connected?

[gentle osprey]

ushygushytoes

[prime elbow]

Can you explain more? There exists no separations means ?

[steady gorge]

oh my bad. i mean separations as in a separation of X

i can give you the definition

let $X$ be a topological space. a separation of $X$ is a pair $U \text{, } V$ of disjoint nonempty open subsets of $X$ whose union is $X$. the space $X$ is said to be connected if there isn't a separation of $X$.

[gentle osprey]

ushygushytoes

[sonic crane]

Thats what u need to prove really

You need to prove there exists no separations in Union An

Like u need to give some sort of argument for that

[steady gorge]

ah ok

since each subspace in the sequence is connected, the union should've have any separations of X, no?

[sonic crane]

If A1 and A2 were already disjoint then A1 union A2 is not connected

So u should use that other condition somewhere in the proof

The fact An cap An+1 is nonzero

Maybe assume that U An is not connected (so there exists a separation) and then show a contradiction

[steady gorge]

ok i think i can try a contradiction proof

because if you assume it's not connected, then that means there must be a separation of $X$ between at least two of the sets. but since we know that $A_n \cap A_{n+1}$ is nonempty for all $n$, that would be the contradiction. so $\bigcup A_n$ is connected.

[gentle osprey]

ushygushytoes

[prime elbow]

I think you need a fact if B is connected subspace of A and A can be written as C u D, as separation then either B\subset C or B\subset D

[steady gorge]

are you saying that's where i could pull my contradiction from?

[prime elbow]

Yes

[plush folio]

I read in Lee that a covering map pi : E -> M is a fiber bundle whose model fiber is discrete. Could we take this as the definition of a covering map, or do we need some additional assumptions?

[unreal stratus]

I would say this depends on conventions. This is a good definition, but sometimes people add extra connectivity conditions

[plush folio]

You mean E should be connected, right?

[unreal stratus]

Sometimes yeah

[opaque scroll]

With this definition you would need the size of the fibers to be the same at every point.

This would be true if M is connected, but is otherwise usually not required of covering maps.

[plush folio]

Ah, so this definition is actually stricter when M is disconnected

[hidden abyss]

the definition of a fibre bundle I've seen doesn't explicitly require homeomorphic fibers

maybe thats just a weird convention my prof prefers

[opaque scroll]

I guess in most cases where one cares about fiber bundles the base space is path connected, so it doesn't matter, but I've usually seen the definition with a fixed fiber.

[chrome trout]

Suppose a net $(a_\alpha){\alpha\in I}$ is indexed by a directed set $I$ with a greatest element $M$. If the net is contained in a compact space, then there is a subnet $(a{\alpha_j}){j\in J}$ converging to a point $x\in X$, where $N$ is the greatest element of $J$ and $\alpha_N=M$. We then must have that $a_M=a{\alpha_N}=x$.

[gentle osprey]

qwerty

[chrome trout]

Is there no problem with the above observation?

It seems odd that if a net is indexed by a directed set with greatest element and also has a limit point, then the final element of the net should equal the limit point.

Particularly, if I index a net by the entourages of some uniform space by reverse set inclusion, and if the set containing the net is compact, it seems too easy to figure out the limit point of the subnet. The greatest element of the indexing directed set in this case would just be the diagonal.

[gaunt linden]

A net whose index set has a greatest element is -- for the vast majority of purposes nets are used for -- just a roundabout way to speak about that final point of the net.

(One can argue that a net is just a suggestive way to specify the filter generated by its tail sets, and when the index set has a greatest element, that is simply the principal ultrafilter generated by the limit point).

[steady gorge]

i'm thinking this problem should be solved similarly to the previous one. should i try to find a contradiction in assuming that it's disconnected?

[steady gorge]

oop wait nvm i got it now

[open maple]

Can I have a continuous map between 2 spaces with a finite number of points?

[queen prism]

you're gonna have to be more specific than that because the constant map is always continuous

[stoic herald]

I have to construct a subset of R so that there is a map from the subset to itself such that the map is bijective and continuous but not a homeomorphism...can I get some hints.

[scarlet turtle]

hint:
there is a continuous bijection from [0, 1] ∪ (2, 3] to [0, 1] by mapping [0, 1] ↦ [0, ½] and mapping (2, 3] ↦ (½, 1]. and this is not a homeomorphism.

[stoic herald]

I need the map to be from the subset to itself

[scarlet turtle]

this is just a hint

[stoic herald]

Oo ok

[scarlet turtle]

tweak this into something where you can "replace" the interval (2, 3] that we lost...

another hint is that there's still a lot of ℝ that we have not yet used

[stoic herald]

Uh can we do something like this :- like considering this subset [0,1] U (2, infinity) and then mapping [0,1] U (2,3] as you have and then mapping (3, infinity) to (2, infinity)

@scarlet turtle

[scarlet turtle]

that breaks continuity, because (3, ∞) is connected, but (½, 1] ∪ (2, ∞) is disconnected

[scarlet turtle]

i'll give another example of this sort to push you in the direction of what i have in mind

there is a continuous bijection from [0, 1] ∪ (2, 3] ∪ (4, 5] to [0, 1] ∪ (2, 3] by mapping [0, 1] ↦ [0, ½], mapping (2, 3] ↦ (½, 1], and mapping (4, 5] ↦ (2, 3]. and this is not a homeomorphism.

[stoic herald]

Ok then I consider the set S = [0,1] U(n ,n+1] , n >= 2 and then map S to S as you have ...the mapping is continuous and bijective but the same is not true for the inverse

[scarlet turtle]

YES! but be careful with the indexing. you want to do stuff like (2, 3] ∪ (4, 5] ∪ (6, 7] ∪ ...

but saying (n, n+1] for n ≥ 2 will give you (2, 3] ∪ (3, 4] ∪ (4, 5] ∪ ... = (2, ∞), which is not what you want

[stoic herald]

Ah ok

[scarlet turtle]

but otherwise, yes, you've got it!

[stoic herald]

Sorry for that

[scarlet turtle]

np, it's an easy indexing mistake to make.

[stoic herald]

Thanks for the help

[limber dune]

To show two spaces (X and Y) are not homeomorphic, is it enough to give a bijection from a proper subset (A) of basis of topology (B) of one set to another. As in if there are basis elements in B that are missed by this bijection, then the topologies are different.

Context: want to show that the growing wedge of circles in R^2, each centered at (n,0) with radius n, (X) is not homeomorphic to the CW complex that is the countable wedge of circles (Y).

I know that the basis elements of X at 0 consist of equal length arcs of length 2\epsilon (\epsilon <1/2 say). But the open sets containing the wedge point in Y can have arcs of arbitrary length.

How do I make this precise?

[quartz horizon]

(note - same Q was posted in #alg-top-geo-top )

[limber dune]

Yes, because it is from AT, but I didn’t know whether my question was more appropriate for point set.

[quartz horizon]

yeah it’s not an issue

just best to make sure discussions don’t get too duplicated is all

[rancid umbra]

for a covering space p : E —> B to be considered a fiber bundle, B needs to be connected, right? otherwise, the number of sheets could vary on each component of B, and so the fiber won't necessarily be fixed.

or, is there a notion of a non-homogenous fiber bundle where this makes sense, where like, the fiber “varies continuously”?

[unreal stratus]

Well you somewhat tautologically just need the fibres to be of fixed cardinality

And yes one way to guarantee that is adding that B should be connected

[rancid umbra]

so what would we classify covering spaces with varying fibers as?

do we just not care about them?

or do they not have a nice generalization?

[unreal stratus]

Hm I do not know what the best terminology is, but people do still care about them. There is a similar thing with vector bundles where you can have varying ranks

[rancid umbra]

interesting. is there a name for the vector bundle case?

or maybe a reference?

[unreal stratus]

I would still call them vector bundles and just maybe indicate that they need not be of fixed rank

[rancid umbra]

oh. you don't need any like, continuous switching conditions or anything?

idrk what that would mean, but it seems like, locally, in the total space, you would want to transition nicely when you change fibers

[unreal stratus]

I guess for vector bundles or covering spaces you cannot really change fiber here except if you change connected components so it is not much of a concern

[rancid umbra]

true

but um. what about for smooth fiber bundles

[opaque scroll]

When I first learned about this stuff my lecture said, if your space isn't connected you don't have a topological space, you have two.

And that's basically how it goes here, a covering space just covers each component.

[quartz horizon]

Wouldn’t they just be covering spaces

A covering space is a locally trivial bundle of discrete spaces

[gaunt linden]

RIP Q.

[rancid umbra]

sure, but i guess i meant like a nice generalization like a bundle. i wasn't being super precise about what i meant, sorry

[rancid umbra]

that's another line of thought that i had. for a covering space p : E -> B, is p^{-1}(B_i) -> B_i a covering space of each component B_i of B?

[quartz horizon]

Sounds like the pullback bundle?

[opaque scroll]

I guess the better question is, assuming E -> B is such that p^-1(Bi) is a covering space for each component, is p aswell. In which case the answer is no unless B is locally connected

[rancid umbra]

okay, i have a couple of things to think about

thanks all

[unreal stratus]

Well i guess you need a generalised notion of fiber bundle here was the point

[quartz horizon]

Well yeah I’m not saying covering spaces are fiber bundles

They’re still what I said though

[unreal stratus]

Oh sure i guess i read fibre into there

[rancid umbra]

oh, so did i

wait, what is the more general notion of a bundle

[unreal stratus]

Though if i read that i would assume one meant fibre bundle aha

[quartz horizon]

Just a continuous function with codomain B

[unreal stratus]

The notion of a bundle over Bcan be generalised to the point where it justmeans E -> B

[rancid umbra]

and then a locally trivial bundle means that we don't have some fixed fiber, just that the local trivializations are homeomorphisms between p^{-1}(U_b) and U_b x p^{-1}(b) for b in B?

[unreal stratus]

This is one definition yes. I would personally be a lil careful as I do not think this is particularly standard terminology (though personally i like this now that I have heard it aha)

E.g. Husemoller in his standard textbook seems to say "locally trivial bundle" as an abbreviation for locally trivial fibre bundle (with some fixed fibre)

[rancid umbra]

off topic, but i wonder why the spelling difference: fiber vs fibre

[unreal stratus]

I am British

[rancid umbra]

it looks very classy lol

[unreal stratus]

And it seems I am slightly inconsistent lol

But at least in writing i always say fibre

[unreal stratus]

I guess probably because i am matching others a little

I guess one slightly funny thing to watch out for here is like what the data of the bundles are

Because for vector bundles you really need like the structure of vector spaces

So like idk i feel the theory of fibre bundles is a place where one should just be happy with various minor variants

[rancid umbra]

https://planetmath.org/locallytrivialbundle seems to have some information on locally trivial fiber bundles

[rancid umbra]

their def agrees with this one, but the diagram is all messed up

[unreal stratus]

There is also such a notion on nlab for example

I just mean to emphasise it like afaik is not too standard so i would be careful to make clear what one means but idk

[rancid umbra]

do you mind dropping the link?

[unreal stratus]

Can be found in the main article https://ncatlab.org/nlab/show/fiber+bundle

"One can also drop [...]"

[rancid umbra]

thanks. this will require some categorical language unwrapping on my part

actually, just that paragraph doesn't look so bad

[unreal stratus]

Dw it basically just means what you would expect, like there is a cover of B by some U_i so that it looks like F_i x U_i -> U_i for some F_i

Is there anything you are doing w fibre bundles heh

[rancid umbra]

the reason i was kind of asking about this whole mess was because i remember proving that every finite sheeted covering map is closed.
this is a special case - when the base is connected - of the more general statement that fiber bundle projections are closed whenever the model fiber is compact.
now i am wondering if it holds for locally trivial bundles in the sense described on the nlab, that if the fibers of a locally trivial bundle are compact, then the projection is closed

[unreal stratus]

Ah

[unreal stratus]

Isnt that also false without some more hypotheses

Maybe I am silly

[rancid umbra]

it is not

[unreal stratus]

Ah no it is probably okay like

[rancid umbra]

unintuitive, right?

i also remember thinking that you needed some extra condition like locally compact or something

but you don't

it just kind of works

[unreal stratus]

Presumably it is because the projections are closed maps by compactness and then you glue stuff

[rancid umbra]

yea, thats kind of the argument. you argue that p(p^{-1}(U) n A) = U n p(A) is closed in U for any closed A subset E

and this gets you an open neighborhood U - p(A) of a point b in B - p(A), where U is a locally triviallizing neighborhood of b

(btw, this is for the fiber bundle result. the covering space result requires a slightly different argument to apply when the base is not connected)

for the covering space argument, you can show that the intersection over the finitely many p(Ui \cap (E - A)) is an open nbhd of some point in B - p(A), where A is closed in E, and p^-1(U) is the disjoint union of the Ui

do these involve gluing in any sense?

[unreal stratus]

I was a bit sloppy there. I guess what I mean is that being a closed map should be "local on the base"
like suppose f: E -> B is a cts map, {U_i} is a cover of B and the maps f^{-1}(U_i) -> U_i are all closed maps. Then f should be closed

[rancid umbra]

yes, that is the argument used in the fiber bundle case, since the projection restricted to the preigmage of any locally trivializing nbhd is closed

[unreal stratus]

nice

Yeah and then you reduce to asking when X x F -> X is closed, and F is compact iff this holds for all X

The same argument should work for any locally trivial bundle where all the fibres are compact though

[rancid umbra]

interesting (i am leaning towards no because i can't quite see the connection between the covering space argument that i have the the argument for the fiber bundle case)

will investigate more tonight i think

[unreal stratus]

A fun exercise in this vein is like

If E -> B is a locally trivial fibre bundle with fibre F such that F and B are compact, then E is compact

(I forgot that covering spaces can have non-compact fibres lmao)

[rancid umbra]

yea, this one can be proven using the fact that fiber bundles with compact fiber are proper

[unreal stratus]

But by the same (start of the) argument, it is enough to show that E x F -> E is a closed map if F is discrete

and i think that should be easy

[unreal stratus]

Does this not need any Hausdorffness?

Well I guess I mean like you have to be careful cause proper has non-equivalent definitions in the non-Hausdorff case right

[rancid umbra]

not according to lee

ISM

[unreal stratus]

oh lol he sets it as an exercise

[rancid umbra]

he doesn't make any assumptions about the properties of the top spaces M,F, and E from what i can see

yea lol

[unreal stratus]

what does he mean by proper?

preimage of compact is compact?

[rancid umbra]

preimages of compact sets are compact, compact not having the requirement of being hausdorff

[unreal stratus]

ah yeah i guess you could probably do this by like restricting to covers by trivial opens

nice

[rancid umbra]

yea, i thought this set of exercises was kind of cool

[unreal stratus]

Though really these are just equivalent statements i guess aha

As in like "fibre bundle with compact fibres are proper" is super close to "fibre bundles with compact fibres and compact bases have compact total space"

[rancid umbra]

yea

[unreal stratus]

i now wish i had looked at this section of the book actually. seems a nice introduction to fibre bundles

[rancid umbra]

its only a brief section at the end of the chapter on vector bundles, but it is my intro to fiber bundles i guess

[unreal stratus]

I needed this statement a couple of years ago and couldn't find a proof online lol

[rancid umbra]

haha

yea, lee's books have a bunch of stuff in them

[unreal stratus]

ye they are cool!

[rancid umbra]

thanks for talking with me about this. i can work through this later tonight with comfort that i am not writing down gibberish

[unreal stratus]

np it is cool

i am like rusty on point set topology and this was fun stuff to think about

[rancid umbra]

i wonder if that is a common feeling among researchers

not saying that i am one

[unreal stratus]

i feel weird calling myself a researcher even though i guess i am now a baby researcher

lol

[rancid umbra]

lol

idk what i am rn

[unreal stratus]

c squared

[rancid umbra]

fax

[unreal stratus]

should change your nickname to 1

[rancid umbra]

just 1

[unreal stratus]

real

[rancid umbra]

nah, but i like c squared. my initials are c.c.

[unreal stratus]

lol complex conjugate nice

[fringe thorn]

off topic, but this is the first time I look in #point-set-topology and I find people talking about the exact problem (even the exact letter!) I'm currently working on

what a coincidence

[rancid umbra]

ha!

im skipping the smooth part because i forgot all about the smooth defs

[rancid umbra]

you were right, it does! its a very pretty argument

its at the right level of abstraction and has just the right tools to make the proof relatively straightforward.
the ad hoc proof that i had for covering spaces was kind of gross

because i wasn't really thinking about it in terms of local trivializations

[unreal stratus]

Noice glad to have helped :)

@rancid umbra im now confused though. So a space X is compact iff for all Y the projection X x Y -> Y is closed. So if X is an infinite discrete set then you can find Y such that this is not closed, but doesnt that contradict covering spaces being closed maps?

Presumably you want to assume you have finite coverings

[opaque scroll]

Just R -> S^1 is an easy example of a covering that is not closed I guess

[unreal stratus]

yeah

lol okay phew

[rain gyro]

Preimage operator preserves everything of the set property right under assumption of mapping is continuous? I kinda have this conjecture in mind.. I proved it preservs intersection and unions, also commute with complement operators.. And the preimage is order preserving too which is useful monotone lemma.. I kinda wonder what it doesn't preserves? I know that it wont preserves, I know commutation of preimage with closure operator hinges on continuity but I dont know other more counterintuitive examples

The closure wont be preserved is actually intuitive, since it takes continuous mapping to make the preimage also closed suppose the preimage admits some closure..

[quartz horizon]

Yes you can prove that preimage preserves any set-theoretical subset operation

This again is something category theory can help with

[rain gyro]

but will it be commute with different set operator given if its inverse is continuous?

[quartz horizon]

So if f is continuous, preimage should also preserve “topological” subset operations like interior and closure

[rain gyro]

I know my conjecture is right!!!

[quartz horizon]

Though I have to think about that

[rain gyro]

time to learn cat theory

[quartz horizon]

Currently unsure whether this is true

[rain gyro]

this is one problem on folland's chapter 4

[plush folio]

The preimage preserves any set theoretic operation as Pseudo said, but this doesn't rely on the function being continuous

[quartz horizon]

You might need f to be an open or closed map for this to work

For the topological ones

I can’t do this in my head unfortunately

[unreal stratus]

Consider the constant map R -> R at 0. {0} has empty interior but the preimage is all of R

[quartz horizon]

Yeah I imagined there’d be stuff like that

[rain gyro]

Yes, I mean commutation with different operators.. I am a bad typer 😭

[quartz horizon]

Ok so what I said earlier about “topological” subset operations only works if you have extra assumptions

Like being an open or closed maps

However for purely set-theoretic ones you should be fine

Stuff like union, intersection, complement, symmetric difference, …

[rain gyro]

if f, is continuous then inverse establishes two inclusion

though not necessarily giving equality.. I kinda did this exercise a couple day back, but couldnt really understand it or generalize it too

[quartz horizon]

So if f is an open map then preimage preserves interiors

And if f is a closed map then preimage preserves closures

Is my conjecture

[quartz horizon]

Again I’m head solving this so there may be technical details I’m missing

[rain gyro]

It gives you something like this… sorry my wording is a bit awkward I kinda want to redo this in streamlined way so will redo it in a week

[quartz horizon]

ok now i can type on a laptop

[rain gyro]

The corollary is absolutely true though

[quartz horizon]

so the interior of a set $A$ is defined as $U \subset \text{Int}(A) \iff U \subset A$ for $U$ an open set

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

then $U \subset f^{-1}(\text{Int}(A)) \iff f(U) \subset \text{Int}(A)$, and if $f$ is an open map then that's $\iff f(U) \subset A$, which is $\iff U \subset f^{-1}(A)$, which is $\iff U \subset \text{Int}(f^{-1}(A))$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

ok so if $f$ is open then preimage preserves interiors

[rain gyro]

Yes

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

now i need to check closures

closure of a set $A$ is defined by $\bar A \subset C \iff A \subset C$ for $C$ a closed subset

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

then $f^{-1}(\bar A) \subset C \iff \bar A \subset f(C^c)^c$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

if $f$ is an open map then $f(C^c)^c$ is closed

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

so $\bar A \subset f(C^c)^c \iff A \subset f(C^c)^c \iff f^{-1}(A) \subset C \iff \overline{f^{-1}(A) } \subset C$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[rain gyro]

open map, is like if C is open then f preserves the openness? I haven't been there yet but ill take note

[quartz horizon]

ok so if $f$ is an open map, then the preimage preserves both interiors and closures

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

neat

[rain gyro]

then is it possible to generalize for all operators satisfying these

[quartz horizon]

yes so this is an alternative axiomatisation of topology

in summary:

• if $f : A \to B$ is $\textit{any}$ function between sets, then $f^{-1} : P(B) \to P(A)$ commutes with any set-theoretic subset operation (like unions, intersections, complements)
• if $f : X \to Y$ is continuous and an open map, then $f^{-1} : P(Y) \to P(X)$ preserves both interiors and closures

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[rain gyro]

screenshoted, will prove this once I am at open maps

I wish I could have such clear mind ☺️☺️... gradually hopefully ❤️

[quartz horizon]

[granite crane]

wow

[quartz horizon]

hi Afzal!

[granite crane]

Hi pseudo

a bunch of unproductive days :c

i have dinner with my friends today, later i will start studying

[hazy hound]

Does 4 really mean for every subset

or does it mean for every closed subset?

[rain gyro]

Which part are you at afzal

[granite crane]

product top, same question

[rain gyro]

It’s actually very normal I feel like it’s making it much better to play some games I did it that way

[rain gyro]

Good old friend

[granite crane]

i played some games and end up being frsutrated lol

[rain gyro]

You can kinda think it a way that quotient topology is the finest topology that a mapping is continuous

[granite crane]

being unproductive sucks :c

[rain gyro]

And use this to relate to topology topology

[hidden abyss]

I think its correct the way its written

[hazy hound]

interesting okay :p

[rain gyro]

Quotient topology is usually meaningless for pointset topology but it gives you some picture why product topology is defined that way

[granite crane]

btw Emma sorry, i have to go for the dinner in another city. So i have to go righ now. I will reply when i will be free

[hazy hound]

I can prove it I just wanted to make sure it was correct before I jumped in

I'm a dummy lol I just realized why it makes sense for any arbitrary subset lol

[quartz horizon]

enjoy!!!

[rancid umbra]

the fibers need to be compact too. for discrete spaces, yea, this just means finite

[rancid umbra]

i think the generalized tube lemma 2 follows from transfinite induction + tychonoff's + generalized tube lemma 1?

tube lemma wiki page

or, maybe its simpler than that...

i think given an open cover of \prod_i A_i of basic opens contained in N you use tychnoff's to get a finite subcover of the A_i's.
all but finitely many indicies in this finite cover are trivial, i.e., equal to X_i. so we can just apply generalized tube lemma 1 + induction

would appreciate any feedback on how this sketch sounds

[tender halo]

yeah thats how its usually proven

[rancid umbra]

sweet

[tender halo]

its called Wallace's theorem

[rancid umbra]

cool, i have never heard of it referred to by that name

[tender halo]

its called that in the first line on the Wikipedia page lol

usually "Wallaces theorem" is specifically referring to the most general of the three

[frail bane]

I used induction but the solution seems to use double inclusion. When it comes to stuff about proving two sets are equal should I always do double inclusion?

this was my working out

I guess me assuming $B'{n+1} = B'{n} \cup A'_{n+1}$ could be a potenital mistake#

[gentle osprey]

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

[rancid umbra]

just skipped right over the statement of the lemma since i kind of already knew it lmao

[queen prism]

practically speaking you can just reduce this (and many other problems involving functions of two variables) to showing that cl(A u B) = cl(A) u cl(B), since then it holds for any finite family of subsets by associativity of \cup
anyway I would just explicitly check whether B_(n + 1)' = (B_n \cup A_(n + 1))' is actually equal to B_n' \cup A_(n + 1)' as you mention, since that seems to be the only nontrivial part here

[frail bane]

yeah got it thanks

also got another question about cantor sets which im curious about

why isnt the cantor set defined to be all possible endpoints?

e.g. {0,1,⅓,⅔,...}

isn't that equal to the cantor set?

[queen prism]

it can't be if I'm understanding your construction correctly, because it's a countable set (you can enumerate finitely many elements at each step and there are countably many steps)

that would also trivialize another special feature of the cantor set, which is that it's uncountable despite having zero "length" (called lebesgue measure)
and you can show that the cantor set has zero length using the fact that each step in its construction you remove a third of each component

[tender halo]

(note that being of measure zero is not a topological property, i.e. Cantor set having measure zero is property of a specific embedding of the Cantor set into the reals, not the Cantor set itself)

[rancid umbra]

https://ncatlab.org/nlab/show/closed-projection%2Bcharacterization%2Bof%2Bcompactness
thought that the proof of prop 1.3 on this page was cool. remark 1.4 is funny

this whole page is pretty cool actually

[unreal stratus]

I should also remark that one cool thing is this provides motivation for similar definitions in e.g. algebraic geometry

Like in (classical) algebraic geometry you have a product (which unfortunately does not agree with the "usual" product of spaces) but you can still contemplate spaces such that these projections are closed, and that is called completeness

[rancid umbra]

oh, interesting

[normal swift]

no way

https://tenor.com/view/stress-cabbage-beaver-eating-cabbages-gif-19336743

[rancid umbra]

bussy beaver, eating his math hw

[granite crane]

i havent started Quotient top yet

[granite crane]

thank you Pseudo

[normal swift]

dont

unless you want joy in your life

[rain gyro]

Don’t worry, he’s not doing algebraic topo yet so I don’t think there’s much occasion where he needs quotient topology to glue things together

It’s a relief to me too🤯🤯🤯🫠🫠🫠

[tiny obsidian]

classification of surfaces 🗿

[unreal stratus]

I mean this is not rly pointset right lol

Just was tacked onto our course for no reason

[rain gyro]

But classifications of surface is algebraic topology? I downloaded a note from Toronto uni

It says point set top has minimal content for quotient topology

[unreal stratus]

I guess there are not really nice lines but yes there is usually algebraic topology in the proof

[tiny obsidian]

I guess, but I think the surfaces course did the same (or at least similar) proof

[unreal stratus]

Sure ye

[tiny obsidian]

and really my point was more that defining and working with (regardless of classification) a lot of these surfaces is far easier by quotienting

e.g. klein bottle

[rain gyro]

I really hope it’s the case I was so tortured by product topology already

[rain gyro]

That’s algebraic topology right it said on the lec note

[unreal stratus]

Yeah this is what i assumed you meant

Products and quotients are like important things in pointset i would recommend getting quite familiar w

[rain gyro]

I felt happy a bit reading that quotient topology is minimally mentioned in ps top😭😭😭😭😭

😭😭

[tender halo]

quotient topology is pretty common

closed maps are extremely common and those are quotient

k-spaces (otherwise known as compactly generated spaces) are exactly the quotients of locally compact spaces for examples

[tender halo]

any (surjective) map from a compact space to a Hausdorff space is quotient

perfect maps (otherwise known as proper maps) are also a very important class of quotient maps

you rarely have theorems about quotient maps in general other than the basic properties

usually you study a class of quotient maps with particular properties

[frail bane]

My question is, isn't the cantor set going to be the set of all endpoints which is a subset of the rationals, hence countable?

i don't get the cantor set construction doesn't provide you with the set of endpoints which should be countable

[rancid umbra]

the cantor set is the set of all real numbers in [0,1] that only use 0 or 1 2 in their ternary expansion

this is an uncountable set

[alpine nest]

Nope, all the endpoints are going to be in the Cantor set, but there will be other points as well.

[queen prism]

you remove a lot of points, but not enough to change the cardinality

[rancid umbra]

to expand, the endpoints of intervals are the numbers whose ternary expansions using only 0 or 2 terminate

[frail bane]

can you give me an example point?

[alpine nest]

Well, c squared just outlined the idea

[frail bane]

because wouldn't these intervals eventually be points

[frail bane]

yeah

[alpine nest]

Another perspective (which is just the ternary argument in a different phrasing) is that elements of the Cantor set correspond to binary sequences, where 0 and 1 would correspond to whether you pick the left or the right interval at every stage, so for example the third interval in the bottom row would correspond to sequences starting with 010....

And there are uncountably many binary sequences

Endpoints would correspond to those sequences which are eventually constant

(of which there is indeed only countably many)

[frail bane]

can you write the infinite ternary expansion as a denary decimal expansion?

i can't think of an irrational point that would be in the set

oh wait

i think i can actually

[alpine nest]

$\sum_{n=1}^\infty\frac{2}{9^n}$

[gentle osprey]

Outsider

[frail bane]

it's kinda like cantor diagonal argument but this time you select a new interval to be in so you can always form an irrational number

[hidden abyss]

Shouldn't that be digits 0 and 2?

[rancid umbra]

yes sorry

oops

[frail bane]

isn't this just 1/4

[gaunt linden]

Yes, 1/4 is one of the non-endpoints in the Cantor set.

[alpine nest]

Happens to be rational, so fair enough if that was the issue

But just play with that

[frail bane]

that seems so counter intuitive

it is real analysis tbf

[alpine nest]

As c squared said, just do a ternary number with no 1's in the representation

[alpine nest]

Yes, the Cantor set is not intuitive, especially when your intuitions aren't well-developed yet

[frail bane]

😭

i never came across it in my first year

[gaunt linden]

Basically, there's none of the removed intervals that contains it, so it must be there in the final set too.

[frail bane]

yeahhh

i get that

[rancid umbra]

i think having the ternary expansion in the back of your mind helps a bunch. at least it did for me.

[frail bane]

yeah i get the ternary expansion

it just seems counter intuitive not considering that

but now i understand it

basic topology is so cool 😭

like wdym a set of nested compact sets is non empty

that is so sick

would picking "measure theory and lebesgue integration" as a module be a good choice if i like this stuff

[alpine nest]

I would say so

[frail bane]

Are there metric spaces where there exist "smallest" neighbourhoods of a point? (don't tell me what the metric space is if there exist any, I want to figure it out myself)

[opaque scroll]

Sure, though those neighborhoods won't be very exciting

[tender halo]

they are called Alexandroff spaces

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

[frail bane]

oooh okkk

[tender halo]

not metric, just any spaces though

[frail bane]

that can disprove my conjecture that every point of an open set is a limit point of the set

[tender halo]

T1 Alexandroff spaces are not particularly interesting

[frail bane]

because my proof messes up here

[opaque scroll]

Limit points of singletons can be a little tricky yeah

[frail bane]

singleton?

(i'm a noob at topology)

[opaque scroll]

Set with one element

[frail bane]

im doing rudin lmao

[frail bane]

ahhh

[frail bane]

wait this is so easy to construct 😭

if countable though

let me think about uncountable metric spaces

oooh perhaps i could use a cantor set

[unreal stratus]

Cantor set will not work

[frail bane]

it indeed does not

[tender halo]

i mean think about the smallest nbhd of a point and then shrink it in half

what happens then?

[normal swift]

in my topology book alg top is right after quotients.

[pulsar lagoon]

Given any subset of the power set the intersection of all topologies that include this subset is still a topology right?

[ruby delta]

The arbitrary intersection of subtopologies is a topology, yes

[gaunt linden]

Yes, that is the topology "generated by" your original set. The set you start with will be a "subbase" for the topology it generates.

[rain gyro]

This made me a bit scared😢 it seems I can never get out alive from product quotient top alive 😭

[rain gyro]

Product top is hard enough 😵‍💫😵‍💫😵‍💫

[rancid umbra]

quotients in topology are about cutting, gluing, and smashing. you can get some pretty cool stuff like this.

more generally, you work with quotients all the time. when you work with a space up to homeomorphism, or a group up to isomorphism, or the integers modulo n, for example. or whenever you want a slightly different notion of equality.

this happens all over the place

i believe the emerging field of HoTT is based on this principle (please somebody correct me if i’m wrong)

[rain gyro]

True but when it got formalized it’s just painful.. product topology was really hard already (managed to make sense of it by doing exercises) quotient topology doesn’t have much exercise on lec note i downloaded and it’s really scary looking

[quartz horizon]

i'd say HoTT is roughly about the idea of having different "levels" of equality, yeah

[rain gyro]

I should maybe not have moved to mapping and Hausdorff properties then? I will find more exercise to firm my understanding of quotient

[quartz horizon]

in my case i found that understanding the universal property of the quotient topology helped significantly

[rain gyro]

But do I need preliminary for learning uni properties?

And books

[quartz horizon]

to be honest i don't know

[rain gyro]

I would really want some tips

I heard cat theory is surprising self contained if you one learn a fraction to gain understanding

[quartz horizon]

it's spicy graph theory after all

[rancid umbra]

i feel like you can learn bits and pieces as you go

[quartz horizon]

and graph theory can often be fairly self-contained

[rain gyro]

But I still need a bit of references like a book or some preliminary stuff

[quartz horizon]

leinster's "basic category theory" may be helpful as a reference

[rain gyro]

Thanks I shall be getting it now

[rancid umbra]

the universal properties of most constructions in point set topology will be written down in most relatively modern texts, too

[quartz horizon]

but i would definitely recommend this

[rain gyro]

I am so tortured by quotients

[quartz horizon]

in the same way that you don't need to learn all of ZFC before starting math

you don't need to (and probably shouldn't) go very deep into category theory before learning other math

[rain gyro]

I just want some mental clarity like the la example you introduced to me

That was like really helpful and not heavy

[quartz horizon]

mhm, that's very understandable!

which la example did you mean?

[rain gyro]

The one you refers transformation as mapping

Like linear transformation and basis and they are isomorphism

[quartz horizon]

ah yes, viewing a basis as an isomorphism with $\mathbb{K}^n$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

i quite like that perspective

[rain gyro]

I really love example

[quartz horizon]

[rain gyro]

I just feel it’s gonna be wonderful if there’s something similar to this for topology

Like I had example of quotients how you glue stuff but there can even be directions which I feel my brain will explode

[quartz horizon]

mhm, i think there's many ways in which universal properties can be helpful in topology

[rain gyro]

I can imagine one glue the end points of closed interval and surfaces by equivalence relations but like you go from one dimension and twist it a certain way it’s just so hard and almost unmotivated too

[gaunt linden]

I find physical cardboard-and-glue construction to be a surprisingly apt mental image for topological quotients; if you haven't yet taken it seriously, now's a good time to start.

[rancid umbra]

i sometimes use yarn, paper, and tape too

[quartz horizon]

for me one of the hardest parts about getting started with topology was learning how to connect intuitive pictures to the formal math

there are enough pathological counterexamples that it was hard for me to figure out what intuitive arguments actually worked, and what intuitive arguments had a subtle flaw

[rancid umbra]

i have always wanted there to be some type of topological putty that somehow passes through itself, easily rips apart and sticks back together

[rain gyro]

Is this the physical cardboard and I try to play with it to build mental image? Or it’s a math concept

[gaunt linden]

Definitely have some experience with kindergarden-style physical construction.

[rain gyro]

I shall try those things 🥰🥰

I really hope I could understand, it’s quite difficult like really, but it seems cool too

I got pretty stuck after product top I tried to do exercises but quotients have been too challenging but it does seem fun though

[gaunt linden]

The good cardboard examples tend to be more examples of solid geometry ("print out this diagram, cut it out, fold along the lines and glue it together to become a dodecahedron" or whatever), but the physical gluing along cardboard flaps is pretty close to being the intuition that topological quotients try to capture.

[rain gyro]

That sounds super fun I love cardboard 🫣🫣🥰.. with patience eventually I’ll be able to understand it whole ❤️ definitely will finish quotient before moving onto different maps

I just kinda love playing those Lego style games.. lets goo🥰🥰🥰

[normal swift]

what? i thought it was alright & way better than quotients, guess im just allergic to these type of stuff, i hated ideals too.

[tender halo]

filters are way better than ideals

products are nice quotients are ass

[pulsar lagoon]

Yeah I was asking mainly as a sanity check cuz I couldn’t find this result in Munkres and was afraid that I was dreaming it up

[quick crane]

There should be a result that the topology generated by an arbitrary collection is the set of arbitrary unions of finite intersections of sets from the collection.

[rancid umbra]

keyword would be subbasis

[normal swift]

true

[normal swift]

havent studied filters yet so can decide

[rain gyro]

This thing really starts to get painful now

This seems to be a relatively trivial proposition, it turns out I have to write a lot to justify and I am not sure it’s correct

This is just partition topology and they should just change its name to that 😭😭😭

I wrote a page and a half of derivation

I should learn a bit universal property asap, otherwise my progress is literally not moving

[gaunt linden]

Is there a difference between [x] and pi(x)?

[rain gyro]

No my definition they are same I think

[unreal stratus]

This looks good

Just a bit wordy

[rain gyro]

I think it as exist x such that [x]=[y] where y in U

[pseudo marsh]

Is that union not just pi(U)?

Oh pi^-1 of that nvm

[rain gyro]

Exists x such that for every y in U [x]=[y]

That’s such a small detail

[gaunt linden]

It's pi^-1(pi(U)) I thjnk.

[rain gyro]

It’s really nice to see it as a mapping? And is my understanding of this correct?

I am so scared of confusing myself

[gaunt linden]

I would just say something like:

The union in the statement is pi^-1(pi(U)).
The map pi is open if, for every open U subset X, pi(U) is open in the quotient space.
But by definition of "open in the quotient" this is the same as saying, for every open U, pi^-1(pi(U)) is open. Which was to be proved, so we're done.

Your writeup obscures this structure somewhat by immediately splitting into separate "if" and "only if" cases, and doing the (arguably) interesting calculation of what the union is as something internal to the "if" branch.

[rain gyro]

Thank you I shall refine it further 🥰🥰 I wasn’t very familiar with the type of the proof so I kinda overcomplicated it too much 😭

[gaunt linden]

And you could probably use a final step in the calculation that removes x and gets you all the way to the union in the statement in the proposition -- since [x]=[y] when x~y anyway, you can just take the union of [y] for all y in U instead of taking the union of all x that are related to some y.

[rain gyro]

That’s such a nice argument year🥰🥰

I am editing already and thanks again 🥰🥰

[urban nimbus]

I am looking at part j) of this problem and I'm wondering if my understanding of this is correct. I see two ways to identify these points, one being "squishing" the sphere to make it so the poles touch from the inside (torus-shaped but with no hole), and the other would be "stretching" the sphere so that the poles touch from the outside (almost like a neck pillow, but with the ends touching each other). My question is, are these both correct solutions? Are they the same thing even though they look so different?

[ruby delta]

That does not sound right - observe that the image of the boundary in the quotient space, whatever it is, must be a single point

[urban nimbus]

What are you referring to with the boundary? Just where the poles map to?

[ruby delta]

Oh, wait you mean j not i lol

Let me reread this

The torus one looks right then

Not sure what you’re trying to describe with the second one

[pulsar lagoon]

The second one would be S2 V S2

[pulsar lagoon]

Nvm I misread what you wrote

[urban nimbus]

I’m imagining stretching the sphere out and meeting the ends with each other on the “outside” rather than squeezing it so they meet on the “inside”

[pulsar lagoon]

I think they’re the same cuz you just pull the squeezed points around

[urban nimbus]

Okay, that makes sense

Im fairly new to topology and only a few chapters in so it’s a bit strange to me that those two shapes are the same in this sense

[ruby delta]

The point of topology is being able to talk about two objects having the same “shape” even if they are different

[urban nimbus]

Okay, yeah that makes sense then

Thank you

[prime elbow]

i am going to start General topology by Engelking

[prime elbow]

Let (X,T) be a topological space, and assume B is a basis of (X,T), here B is basis means

1. For any B1 and B2 \in B, for all x in B1\cap B2 there exists B3 \in B such that x\in B3\subset B1\cap B2.

2. For any x in X, there is Bx\in B, x \in Bx.

From this information, can i conclude that all open sets in T is the union of elements of B?

for this i need for any open set V, for any x \in V, there exists Bx\in B such that x\in Bx\subset V.

I can guarantee that existence of Bx, but i don't know how can i guarantee the existence of Bx such that it is subset of V.

[rancid umbra]

how are you defining “B is a basis for the topology T on X”?

[prime elbow]

just 2 given properties

[prime elbow]

1 and 2

[rancid umbra]

but just those two properties say that B is a basis for some topology on X

[prime elbow]

yeah good point

[rancid umbra]

not necessarily T

[prime elbow]

now i see

thank you c squared

[tender halo]

that is not engelking says; B1 and B2 are just conditions that say that B is a basis

[prime elbow]

yes, this is my own question

actually in Engelking they give as converse

[tender halo]

without any relation to any existing topology

then if the topology that B generates is \Tau, then B is a basis for \Tau

[prime elbow]

so i thought converse hold?, i mean in Engelking they define basis B such that every open set is union of elements of B, and then by using the definition of basis, B has properties 1 and 2, i got this, but then i thought about reverse direction

[tender halo]

saying that "B is a basis of (X, \Tau)" is the same as saying that all open sets of \Tau are unions of elements of B

[prime elbow]

yes

[rancid umbra]

yea. to summarize, if B is a basis for a topology T on a set X, this typically means that any open set is the union of some collection of basis elements. any basis B for a given topology T will satisfy the two conditions you mentioned.

conversely, if a collection of subsets satisfies those two conditions, the the collection of all unions of elements in B is a topology on X, for which B is a basis.

[prime elbow]

yes

[rancid umbra]

i like that basis and subbasis are part of free constructions

[prime elbow]

so in Engelking they define the weight of topogical space, how is it useful?

[pulsar lagoon]

the only important one is countable weight i.e. 2nd countable

[prime elbow]

i see

one can easily check that....,

i don't want proof i don't get what does it mean?

[rancid umbra]

a basis for a topological space naturally gives rise to a neighborhood base of each of its points

[prime elbow]

so i have to take all such Bx \in B such that x\in Bx, the collection of such Bx will be B(x)

[rancid umbra]

yes. would avoid indexing by the point x tho.
rephrase this and write the neighborhood base for x as \mathcal{B}(x) = {B in \mathcal{B} : x in B}

[prime elbow]

Yes thank you

[tender halo]

its one of the more important cardinal charactersitics of spaces

in metric spaces, you will have useful theorems like density = weight = cellularity = \kappa-lindeloffness = extent

you can estimate the size of the space from the weight, for example any T0 space is size at most 2^weight

in general studying how various contininous functions preserve or don't preserve things like density and weight was one of the foundational objects of study in early general topology

[prime elbow]

i see

[prime elbow]

T0 space means every finite set is closed, right?

[tender halo]

it means that any two points can be separated by an open set

i.e. there is an open set that only contains one of them

[rancid umbra]

this property is also commonly rephrased by saying that the points of a T0 space are topologically distinguishable.

[prime elbow]

I see

[prime elbow]

Yes it is T1

[prime elbow]

Elobraote

[rancid umbra]

two distinct points in a topological space are topologically distinguished if there is an open set containing one of them but not the other.

[prime elbow]

So it is the definition

[rancid umbra]

yea. i was just giving another common way to say what bussy beaver said. i think that separated is kind of an overloaded term in topology

[granite crane]

What book is this? Can i know the name

[rain gyro]

Guys I revisited my proof done yesterday I didn’t omit the manipulation so later revision I might understand my write up. Is my new proof correct given the remark?

I used a slightly more formal method

[prime elbow]

General Topology by Engelking

[prime elbow]

So in this example they stated that O the family consisting of all subsets of X that don't contain x_0 and of all subsets of X that have finite complement.

So then How will X be open, i think it has to be or instead of and.

[granite crane]

Thank you <3

[tiny obsidian]

? X has a finite complement of size 0 so is open

O is {subsets not containing x 0} u {subsets with finite complement}

Which is the same as {subsets : (do not contain x0) or (have finite complement)}

[prime elbow]

actually i thought in the sense of and, U is open iff x_0 \not in U and complement of U is finite

[prime elbow]

how this basis is minimal basis?

i mean we don't need to take X\ {x_0} to be take in basis

[tiny obsidian]

I'm not seeing a claim its minimal

Minimal cardinality, yes

[prime elbow]

oh

how it has minimal cardinality

[prime elbow]

So why do we define locally finite? How is it helpful?

[normal swift]

like have you tried googling

[prime elbow]

no

[normal swift]

ok

[muted arrow]

Does the text not motivate this or use it more?

[prime elbow]

just define

[muted arrow]

strange. I don't know any statements which use this property off the top of my head, but it isn't hard to imagine such a thing since finiteness is a powerful condition. in this case, perhaps you can say useful things about a locally finite family with an atlas or more generally a sufficiently refined open cover on a topological space. this is a fairly powerful condition as it, in particular, implies there are finitely many sets in the family containing any given point. it's stronger than this, though.

[quartz horizon]

this helps for things like partitions of unity!

if you've heard of those

for example, if you have a collection of functions $f_s : A_s \to \mathbb{R}$, then it makes sense to define the sum $\sum_{s \in S} f_s : X \to \mathbb{R}$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

even though that sum is infinite, it is locally just a finite sum

more generally this is the sort of thing i've seen locally finite things used for - they let you define "infinitary operations" in a way that makes sense, because locally they just reduce to an ordinary finitary operation

[muted arrow]

yes pou is what I was trying to rember! ty ❤️

[quartz horizon]

[muted arrow]

I could remember the page number in Tu but not the name. weird how that works sometimes

[quartz horizon]

i haven't read Tu :P

[muted arrow]

Lee is better

[quartz horizon]

-# also haven't read that

[prime elbow]

is it correct, i used ai for latex

[prime elbow]

Since F = u f, so X\F = \bigcap X\f, since every X\f is closed so X\F closed implies F is open.

To show F is closed which comes from above result.

So I can say if every f's open then F is open.

Above result: 1.1.11

[tender halo]

local finiteness is a very powerful condition that ensures that your cover behaves nicely locally, paracompact spaces, that is, spaces where every cover has a locally finite refinement are very important objects in analysis for example

[prime elbow]

yes i see

can you verify my above arguments?

[gritty widget]

As I recall you can extend the pasting lemma to infinite collections of closed sets so long as they’re locally finite, too.

[unreal stratus]

Ye

[normal swift]

pasting lemma rules

[steady gorge]

i'm a little unsure of how to start proving that q is a quotient map here. any hints?

actually for surjectivity, since we're dealing with a projection map, we'd have any $q(x\times y)=x$ where $x$ would be in $\mathbb{R}$ already as an element of $A$, right?

[gentle osprey]

ushygushytoes

[rancid umbra]

not quite

this is fine for x >= 0

but you need to do something different for x < 0

[tender halo]

well by definition really

no tricks required

although if you had proven this theorem you can say that A is covered by sets (-inf, 0] x {0}; [0, inf) x R restrictions of f to which is quotient (on their respective images)

so it naturally follows that f itself is quotient

[rancid umbra]

my hint would be to draw A for showing surjectivity.

think about why this map is continuous (restrictions!)

then draw what q^{-1}(U) is for some open sets U of R

[steady gorge]

i thought x would be nonnegative since of how A is structured

[rancid umbra]

(-1,0) is in A, for example, since the second component is 0

[steady gorge]

ohhhh you're so right

i see i see

[rancid umbra]

yea, you should def draw this

[steady gorge]

so these would be bases for A essentially?

[steady gorge]

i shall try 🫡

[tender halo]

no, they are not bases

but if you have several compatible (meaning their combination is continuous) quotient maps such that their images are an open cover or a locally finite closed cover, then their combination is also quotient

[steady gorge]

should i also consider when x is nonnegative and y=0 as well? so the set [0, inf)x{0}? or does that complicate it a bit more

[tender halo]

i am unsure what for

[steady gorge]

it says that A consists of all points x x y for which either x is greater than or equal to 0, y=0, or both. so that's why i ask since you mentioned the sets that cover A

just wanted to make sure i don't have to account for when both conditions are true

[tender halo]

you dont

[steady gorge]

kk

[tender halo]

you can also just say that restricting f to the line y = 0 will give you a homeomorphism (and therefore quotient), that also lets you conclude that f is quotient

[rancid umbra]

you can just do this directly too. its a bit simpler imo.

if q^{-1}(U) is open in A, there are only a couple of cases to consider to show that U is open

like, for x in U, q^{-1}(x) = {x} x R if x >= 0, and {(x,0)} if x < 0. then you just have to think about what it means for q^{-1}(U) to be open in A.

[steady gorge]

kk

[steady gorge]

ok you said that what i originally had was better for when x>=0. so for x<0, it's a ray. how does that affect surjectivity?

[rancid umbra]

take a point x in R

x satisfies either x >= 0 or x < 0

if x >= 0, then what is a real number y such that (x,y) is in A and q(x,y) = x?

if x < 0, then what is a real number y such that (x,y) is in A and q(x,y) = x?

[steady gorge]

if x<0, then 0 is the real number such that (x,y) is in A and q(x,y)=x. if x>=0, then any real number y would satisfy?

[rancid umbra]

yes

[steady gorge]

ohhh i think i'm starting to see where my issue was. i wasn't paying attention to the restrictions on y

[rancid umbra]

yea

did you draw A?

[steady gorge]

mhm

ok i see what you're talking about now that i've said it lol

alright and then you said the restrictions are a hint to why q is continuous, no?

[rancid umbra]

yes

[steady gorge]

could you consider any neighborhoods? i know 0 is a bit tricky so i'm not sure how efficient this method is

[rancid umbra]

i think this method is pretty efficient.

U be a subset of R and suppose q^{-1}(U) is open in A.
Let x in U. There are three cases to consider:

• x < 0,
• x > 0,
• x = 0

Case (x < 0):
Since q^{-1}(U) is open in A and contains (x,0), then there is some open ball B_r(x,0) of radius r > 0 centered at (x,0) such that (x,0) ∈ B_r(x,0) ∩ A ⊆ q^{-1}(U).
Put s = min(-x,r) so that B_s(x,0) ∩ A = (x - s, x + s) x {0} ⊆ q^{-1}(U).
Then (x - s, x + s) is an open neighborhood of x contained in U.

[steady gorge]

and i should play with that for the other two cases? i imagine it wouldn't be too different from the first case

[rancid umbra]

Case (x > 0) is similar, yes

[steady gorge]

ah so case (x=0) is where it gets tricky

[rancid umbra]

you can draw a picture for all of these cases to guide you

[steady gorge]

i might need some help with a picture for neighborhoods

[rancid umbra]

here is the picture for Case (x < 0)

[steady gorge]

so when x=0, we can have any y in R. however we can't have as simple a neighborhood like the first two cases since of the restrictions on y, right?

could we examine a union of the two neighborhoods from the first two cases for 0?

[pulsar lagoon]

Is there any good resource to learn topology of function spaces better?

I dont wanna read munkres again

shit gave me ptsd for topology

[cosmic mirage]

what are you trying to learn about function spaces?

[pulsar lagoon]

just the common topologies used and some properties ig

like more on compact open topology etc.

[cosmic mirage]

gotcha. idk if this is helpful but any standard algtop textbook will talk about compact open probably

since algebraic topologists like to pretend only nice spaces exist

[pulsar lagoon]

I see lol

do you know if evans goes over topologies on function spaces?

cuz I was planning on reading that book too, so I might expedite it

[cosmic mirage]

no clue, this is probably searchable from the TOC tho

[pulsar lagoon]

bet

[cosmic mirage]

(hatcher has an appendix for it)

[prime elbow]

Is it true that if (X,T) is a topological space and B is base of T, and B has minimal cardinality which is finite, then for any basis S, B will be a subset of S.

I don't want any hint now

[tender halo]

yea

[prime elbow]

mse helped

[tribal palm]

do we assume the maps $e_\alpha$ to be embeddings ?

[gentle osprey]

red beet jen

[opaque scroll]

So the maps f_alpha are arbitrary, and the e_alpha maps agree with them on the boundary, so they need to even be injective.

But on the interior of D^n the maps are embeddings, but this is not an assumption. It's a consequence of the square being pushout

[tribal palm]

ohhh ok thanks i’ll see if i can work out the details myself

[prime elbow]

any hint for d?

[hexed steppe]

is the space of compact subsets of R^n equipped with Hausdorff metric separable?

nvm it is

[stoic herald]

Ok so I have this problem which says that f is a mapping from Rn to Rn and is also an isometry , I have to show that f is surjective...to do this I am using this result that if F is a mapping from an open subset A of nls V to V and F-T is lipchitz where T is an invertible linear map from V to V then F[A] is open now F is lipchitz on Rn and any invertible linear map is lipchitz on Rn and also I know that the set of all lipchitz maps from A to V forms a vector space therefore (in our case) F-T is lipchitz and F[Rn] is open since Rn is open now I will show that F[Rn] is also closed proof :-
Suppose some p elem of Rn is not in range then I can find an open ball around p such that there is no point of range(f) inside that ball (cuz if not then we will have a sequence of points from range(f) converging to p and their inverse images would be a cauchy sequence in Rn and therefore converges in Rn say to v and since f is continuous the image of v would be p) now for every point not in range(f) we can do this and therefore the union of these open balls would be open and range(f) is compliment of this so should be closed...now range(f) is both open and closed so clopen...the only clopen sets in Rn are Rn and phi and since range cannot be empty the range is Rn...can someone verify

[gaunt linden]

(Good god, use some full stops now and then.)

[prime elbow]

so to show Borel sets has given form, we have to show this family has all open set( which comes the fact every open set is F-sigma sets), and we have to show if A is in given family then A^c is in family and countable union of A_i is in family if A_i is is family, but i don't get this given family, please explain

[ruby delta]

it's just essentially repeating the fact that the borel algebra is generated by using union and intersection countably many times

[alpine nest]

You construct a collection of "auxiliary" families of sets, indexed by the countable ordinals; this is an inductive construction which defines a family for an ordinal alpha in terms of what's in the families for the "previous" ordinals (i.e. the ones less than alpha). Once you have all those auxiliary families, you take their union, and you argue that it's a sigma-algebra (and is equal to the Borel sigma-algebra).

If this is your first time seeing ordinals, then you should probably read up on them a bit (just on the general concept and on basics of transfinite induction/recursion)

This entire approach (and variants on it, because I don't really like the exact way it's done in 1.3.G; it seems it will make dealing with complements annoying) is described here: https://en.wikipedia.org/wiki/Borel_hierarchy

[prime elbow]

Okay sure

[alpine nest]

Engelking has a refresher on ordinal numbers in the introduction, in fact

[prime elbow]

Yes

[prime elbow]

yes so family F1 will be all countable union of sets of family F0, and F2 is countable intersection of sets from F1 u F0

so on...

so Fi \subset F{i+1}

[rancid umbra]

is it true that a map of bundles induces homeomorphisms on fibers?

[limpid kelp]

idts

[rancid umbra]

the map 𝛗 induces a map between fibers 𝛑_{E}^-1 (b) and 𝛑_{F}^-1(f(b))

[limpid kelp]

thats continuous

[rancid umbra]

yea

what about if the bundle morphisms are G-equivariant on fibers

[limpid kelp]

yes

would keep things clean

[rancid umbra]

okay

thanks

[limpid kelp]

got you

[unreal stratus]

What is G

The structure group?

[rancid umbra]

yea

[unreal stratus]

I feel like this still needs a bit more clarification on what you actually mean by bundle

What happens if you just take a fibre bundle F over a point, with trivial structure group

Then any map F -> F is probably as desired

If you mean G-principal bundles then that is different and it should be true

[rancid umbra]

hmm. im not sure i understand. lemme give some more context tho

[unreal stratus]

I mean like what about the example i gave

[rancid umbra]

okay

so ur saying just pick a constant map

[unreal stratus]

Well just take any map that isnt a homeo as your bundle map, and make the source a point

(More generally, you can give a trivial fibre bundle the structure group {1}

And so the equivariant bit dies and all you have are bundle maps and then a similar thing applies)

[rancid umbra]

okay, yea, then G-equivariant isn't enough to guarantee that the bundle maps induce homeomorphisms of fibers

[unreal stratus]

In fact like generally inducing a homeo on fibres is equivalent to the bundle map is an isomorphism

[rancid umbra]

im reading through this

and i was trying to see how i would define a morphism of bundles

and i came up with what the wiki page says, which is nice

but

i can't see how these are related

[unreal stratus]

This definition is odd to me lol, but perhaps I am missing context. In particular it says h is a map B -> B', but then it talks about the effect of h on fibres, which live in the total space

Which makes no sense to me

[rancid umbra]

this is from steenrod's paper, the topology of fibre bundles

[limpid kelp]

should have clarified - isn't a guarantee.

if we're talking about principal g bundles over the same base w g equivarant map covering the identity then yes

[unreal stratus]

Oh well i trust Steenrod a lot lol so now i am confused

I will look

[rancid umbra]

lol i haven't fully understood this definition yet

[unreal stratus]

Yeah i wonder if implicitly there is smth like this going on

[rancid umbra]

yea, im kind of trying to translate this into more modern terminology

[limpid kelp]

he's using it for both base + induced total space

[unreal stratus]

😭

So what is the difference between curly B and usual B

It would be helpful to know what these notations mean here

[rancid umbra]

i'll send pics

[unreal stratus]

Also like I would usually expect a map of bundles to be the identity on the base unless otherwise specified

(If they have the same base)

[unreal stratus]

Oh lol

Using B for the total space

😭

[rancid umbra]

yea

[limpid kelp]

the entire coordinate bundle (curly)

[rancid umbra]

i have been replacing B with E and X with B while taking notes

[unreal stratus]

I think it is more that this is so old that the conventions had not become standard

[limpid kelp]

i do not enjoy this

[rancid umbra]

and Y with F

[unreal stratus]

So i do not disrespect Steenrod at all

[rancid umbra]

I have also had this feeling

[unreal stratus]

Though tbh i imagine nowadays it is much more pleasant to learn from other texts lol

[rancid umbra]

im just trying to get through a few sections, but yea, agreed

[unreal stratus]

But okay this makds sense now

Thanks

Then okay it makes sense yeah

I think it is just a convention thing whether bundle maps are homeos on fibres

I checked and the (well-known) book "Characteristic classes" has the same convention

[rancid umbra]

hmm. so then what

what is a morphism of bundles lol

[unreal stratus]

Depends on convention

[rancid umbra]

thanks, i hate it

[unreal stratus]

But the usual one would be what wikipedia has

There are just many different notions of what you can mean by bundle ig

[rancid umbra]

is there one that is commonly used today? or are bundles just, shrouded in ambiguity

[unreal stratus]

When i wrote my ug thesis (on K-theory stuff) it seemed the usual convention is that fibre bundles should be locally trivial and a map E -> E' of bundles over B is just a map on total spaces which commutes with the projection to B. If you add extra data to your bundle — like a choice of trivialisations and sructure group G — then you may want to express compatibility with G

And then in some cases you may also wanna have isos on fibres lol

But i think i would just be careful with the text as well

[rancid umbra]

seems like a smoother read through after learning more about bundles from other sources then

[unreal stratus]

Like in this case you very explicitly have a G

I guess fortunately often you will just work with a special type of bundle aha

Like principal bundles or vector bundles

And then there is less ambiguity

@umbral panther do you think this is fair?

[umbral panther]

There are three main cases. Principal bundles, vector bundles and fiber bundles. If it’s a principal bundles, an equivariant map must be an isomorphism on fibers. If its vector bundles there are lots of options. I prefer maps preserving the linear structure, but they could have a kernel and the dimension of the kernel can vary, so the kernel and image aren’t guaranteed to be vector bundles. A general fiber bundle is just a map of spaces with special properties. I wouldn’t put any conditions on it

Lots of questions can be reduced to principal bundles. Asking that a map of bundles with fiber F is like asking for a map of principal bundles with group homeo(F)

[unreal stratus]

Agreed

[rancid umbra]

i don't understand the last sentence

asking that a map of bundles with fiber F (such that what?) is like asking for a map of principal bundles with group homeo(F)

[umbral panther]

For any F and E_i/B_i fiber bundles with fiber F, asking that a map E_1->E_2 send fibers to fibers by a homeomorphism is the same as a map of principal bundles with group homeo(F)

[rancid umbra]

oh, okay. thanks

[normal swift]

is this really point set topology.

[unreal stratus]

borderline lol

[normal swift]

i thought this was algebraic tbh

[unreal stratus]

I guess these are objects more part of alg top but there isn't anything alg top going on here either / the questions make sense for point set ig

[normal swift]

ok

[rancid umbra]

yea, i really didn’t know where to ask it

[rancid umbra]

this was my reasoning

[plain raven]

@rancid umbra Seems like the definition is engineered in order to obtain theorems 2.6 and 2.11. I wouldn't read into it too much more than that. The point of Theorem 2.6 is to illustrate that somehow we just care about the structure group and the way the fibers are glued together more than the fiber space itself, which is foreshadowing of some later theorems that establish that the classification problem for fiber bundles over B can be reduced to the classification problem for principal fiber bundles over B, where you take the fiber to just be G itself.

As others have said, any bundle map between principle fiber bundles is automatically fiberwise a homeomorphism, so in this case the distinction is not important. For the sake of the classification problem, it suffices to define a morphism to be an isomorphism, I suppose.

Who added this smh

I looked it up by IP address and it was someone at the MIT lab for nuclear science

note that if f : B -> B' is a map, and p : E' -> B' is a fiber bundle, then the induced pullback map p*f -> f is fiberwise a homeomorphism

so these are at least common

[pale plover]

is this problem related to number theory? because idk how to solve it if it is

[cosmic mirage]

not really, you don't need anything fancy to show its a metric space

well, for any part of the problem

[pale plover]

i just hear divisibility so i get intimidated 😔

[cosmic mirage]

haha you'll be alright !

as long as you know that prime factorizations Exist and how they work you can show the thing

[cosmic mirage]

huh i wonder why the nuclear scientists care about type theory lol

[prime elbow]

i don't get this

so what are the f_s in 1.4.A?, all possible mapping R-> {0,1} ?

[tender halo]

i mean it works but thats kinda cheating

any zero dimensional space can be generated that way

[prime elbow]

what is zero dimensional space?

[plain raven]

My guess is the IP thing doesn't distinguish between this and CSAIL, where I know for certain there are people who care about type theory, like Adam Chlipala and his students

[cosmic mirage]

yeah its also possible some random csail person happened to be connected to LNS wifi for whatever reason haha

i dont get the impression anyone in the math department cares about type theory there 💀

so probably csail

[rancid umbra]

okay, thanks. that feels like it clears up some of my confusion on this definition

[lofty orchid]

@thin scarab if you define divergent as the closed set of all non-convergent functions, then the functions we typically call "limit does not exist" would likely be a subset of divergent limits

That's the subtlety that I was thinking about

Bc in the extended real line you can say that functions that go to infinity converge to infinity, although in the ordinary real line they're divergent

I would personally not call that limit does not exist

Unlike oscillatory functions

[thin scarab]

This is true

I see what you're saying

[lofty orchid]

it's irrelevant in the topic that was being discussed there though lmao

Same difference

[thin scarab]

This is also not topology tho

I mean I guess you are adjoining two isolated points to the real line

[lofty orchid]

Initially I was thinking of holes that would make functions not converge

[thin scarab]

Not sure what you mean

[lofty orchid]

Consider the topological space X = R \ {0}

All functions that tend to 0 do not converge in this space (aka divergent) although a limit, does in fact exist

[thin scarab]

continuous functions X->X?

or real-valued?

Oh you just mean any function that takes values in X

[lofty orchid]

X to X

yes

the limit exists in R but not in X

[thin scarab]

Ok... I would not say that the limit exists

[lofty orchid]

so my first thought was in those cases divergent wouldn't necessarily mean limit dne, am I valid? Idk I just got off work and I'm dead tired 😭

[lofty orchid]

it exists in R

just not in the topology where the function is

[thin scarab]

But not in X, which is your codomain

[lofty orchid]

Yeah

[thin scarab]

So again I would not say that the limit exists because you are imagining X as embedded in another space

[lofty orchid]

Matter of nuance

[thin scarab]

No, not really. The limit point definitionally does not exist in the set which is the target of the map

Like yes it's natural to imagine a sequence heading away from the center of an open interval is heading "towards a boundary" but if there is no boundary then it just isn't

[thin scarab]

Okay, clearer way to make my case. (0,1) is homeomorphic to R, yeah? So a sequence "approaching 1" would be going to infinity under the image of such a homeomorphism. We already agreed that you wouldn't say that the limit exists if the sequence is going to infinity, unless you add in infinity via the extended real number line.

[lofty orchid]

I mean yeah a function f: (0,1) -> (0,1) that tends to 1 doesn't converge

Kinda the same idea, the limit point does not exist within the topology

It just makes sense to think that in this case there IS a limit point when you consider the topology to be embedded in a larger space

Kinda like the real line and the extended real line

As opposed to oscillatory functions like sine, for which you can't really say that

Don't get me wrong, I don't disagree with you

[thin scarab]

Okay but then you wouldn't consider your space to have a hole in it

In order for the limit to exist it has to be in your space

Like by definition

[lofty orchid]

I just think there are two kinds of diverging functions, and for one of them you can (I think?) almost always say that a limit does not exist no matter what space you're thinking about

[lofty orchid]

yeah yeah

Ik

😭

[thin scarab]

Sure, if your space is Hausdorff

[lofty orchid]

I didn't really think about T1 and T2 in this context

But yeah

[thin scarab]

well the trivial topology will do actually

[lofty orchid]

I mean now that I think about it

[thin scarab]

an alternating sequence converges to both points lol

[lofty orchid]

If f: R -> X such that X was non-Hausdorff you could make sin(x) have a limit too lol

well at least convergent into an open set

something like [-1,1]?

[thin scarab]

Huh? How are you defining sinx if you're mapping into a weird space

[lofty orchid]

Y = [-1,1] with

hmmmm

finite complement?

ugh that'll be so ugly I cant think off the top of my head if sin will even be continuous under this mapping

it won't be

[thin scarab]

idk i dont have an intuition for that space

[lofty orchid]

eitherway, good talk

Made me think a little bit about things I usually dont think about

are you a grad+ student

[thin scarab]

yeah

[lofty orchid]

makes sense lmao I'm just a senior undergrad and my experience with topology has so far been not very pleasant

[thin scarab]

point set topology is annoying but a necessarily prereq for basically anything interesting you want to do

[tender halo]

won't it? preimage of a single point is closed

so preimage of a closed set is a union of closed sets and therefore closed

[lofty orchid]

I know I took it so I could learn more about differential geometry

[lofty orchid]

I think originally I said that the map is sin: R -> Y

So preimage of a single point in Y should be open in R no?

[thin scarab]

no... the preimage of a single point in Y would be a collection of isolated points

[lofty orchid]

oh you're right

[thin scarab]

like, sinx=1 has discrete solutions

[lofty orchid]

I forgot it's a many to one mapping

[tender halo]

yea to have a map be continuous into a set with cofinite topology you just need the fibers to be closed

[lofty orchid]

singletons are closed too so yes I think it is actually closed you're right

I'm trying to think

Preimage of a closed set in Y

should be union of closed sets

Hmmmmm

damn that's actually crazy

😭

I find it so hard to believe but you're right

but is this a finite union

A closed set in finite complement topology should have all but finitely many points in Y right?

So if we're taking the preimage of this closed set, we're looking at the union of many many closed sets in R

Is that closed or open idk

[thin scarab]

open sets are those which have finite complement

[tender halo]

no? a closed set in a finite complement topology is finite

[thin scarab]

so all closed sets are finite

all finite sets are closed

[lofty orchid]

oh shucks im actually stupid lol

yes you're right

closed set is finite, and we have a finite union of closed sets in R

yeah that checks out

[tender halo]

honestly they should have just called it the finite set topology, nobody said you have to name your topology after the open sets instead of closed sets

[thin scarab]

i think it's a decent convention

[tender halo]

checking that finite union of finite sets is finite makes much more sense than checking that finite intersections of cofinite sets is cofinite

[thin scarab]

I think it would be confusing if some topologies were named after the open sets and some were named after the closed sets

[lofty orchid]

having a good intuition around finite complement, k-topology and all that for the first time last year was very weird

Took me a long time to get the hang of it

[tender halo]

or call it the Frechet filter topology