point-set-topology

[umbral hamlet]

wait doesnt any neighborhood of x intersecting A automatically intersect A minus B

nope nvm

bc it could intersect at the parts A and B share

[kind marlin]

yeah 😭 theres a lot of confusing trains of thought

[umbral hamlet]

wait how do i know that U cap V will still intersect A

i mean i guess it intersects at the point x

[kind marlin]

right, so its a neighborhood of x

[umbral hamlet]

yeah

isnt there something about limit points having to be the neighborhood minus x intersecting

nah wait im just jumbling up definitions

[kind marlin]

dont overthink it

x is still in the closure of A

[umbral hamlet]

ur right

i guess the only thing now is showing that all neighborhoods of x that intersect A - B can be given as some U \cap V?

maybe thats kind of obvious though

[kind marlin]

mmm reframe that

we just want to show that x is in the closure of A-B

so every neighorhood of x contains a point in A-B

[umbral hamlet]

yeah meaning that any neighborhood of x intersects A - B

[kind marlin]

right

U was that arbitrary neighborhood

[umbral hamlet]

mmm i see

nah im still not seeing it

[kind marlin]

we know U cap V contains a point in A - B, right

[umbral hamlet]

yes

ah dude im so overcomplicating it

i see it now

U and U cap V both share x

U cap V and A - B both share x

so

no nvm i dont know that A - B has x

[kind marlin]

U cap V is a subset of U

thats all you need to finish this

[umbral hamlet]

ah right

idk why im thinking about this in terms of unions

ive been feeling as if my basic set theory is lacking

oh well im too deep into math now to go back to discrete math, eventually itll click

i took my discrete course at a college with barely any math before i transferred to my current college with pretty good math and it shows

[kind marlin]

to be honest i think this proof is just kind of hard (unless im just dumb)

for me a lot of the elements are used in ways that feel kind of indirect

we care about a point being in U, but we can't work with U, so we intersect with a set with a useful property to get a subset that is directly useful, but we only care about that subset because its a subset of our original set

[umbral hamlet]

yeah

[kind marlin]

its kind of sleight of handy

[umbral hamlet]

its just the fact that it took me that long to register that U cap V is a subset of U is a problem

[kind marlin]

thats fair

[umbral hamlet]

like im fully aware of that fact but for some reason it slipped my mind

honestly though i think i was just getting tied up with sets intersecting vs sets being subsets and that just scrambled my train of thought

so i wont worry too much about it

[rancid umbra]

there is a way to symbol bash this with demorgan’s i think lol

[umbral hamlet]

i was thinking about doing that but my profs tend to hate symbol spamming so i decided against it

yes okay good my next problem is easy

i need to find the point(s) that 1/n converges to in R under the cofinite topology

its just all of R

[umbral hamlet]

okay my professor barely went over quotient topologies so i have like nothing to deal with these

i hardly even understand what its asking me to do

so i know i have to show that q is continuous both ways since thats the definition of a quotient map

its forward continuous since its the projection of a continuous function

[kind marlin]

continuous both ways isnt really accurate, its moreso that a set is open if and only if its preimage is open

[warped helm]

@umbral hamlet some set theory comes in handy here. if g is the restriction of f to A, then g^{-1}(V) = f^{-1}(V) \cap A

[umbral hamlet]

yes i understand that

honestly the hardest part is showing that q inverse(U) open implies U open

U open implies preimage of U open is just showing that q is continuous

though i now understand why we cant say its exactly continuous both ways

[kind marlin]

idk if theres an easier way but i think you can kind of just do this directly

take an arbitrary set S in R, and consider the preimage S^-1. Suppose S^-1 is open. Then for any x in S, we want to show that we can form an interval around x that is contained in q(S^-1)

the main motivation is to find a preimage point of x in S^-1, and make an open neighborhood of the point that is contained in S^-1 and ideally maps to an open set in R, which will be contained in S and contains x. you just need to be careful with with preimage point you choose

[umbral hamlet]

i think i got it

frankly needed help from stack exchange but im not too sad bc im pretty clueless about this

[rancid umbra]

continuous both ways makes sense for open maps

which is kind of cute

[warped helm]

i'm confused, isn't the problem asking to show that q isn't open and also not closed?

[rancid umbra]

yea

[warped helm]

why would this be a goal then?

[rancid umbra]

to show that its a quotient map i suppose

[rancid umbra]

that is also part of the question ig

[warped helm]

ah ok

the target space has to have the quotient topology

[kind marlin]

oo open maps seem really convenient

If a function is continuous, open, and bijective, it is a homeomorphism.

If a function is continuous, open, and injective, it is an embedding.

If a function is continuous, open, and surjective, it is a quotient map.

so homeomorphisms are quotient map embeddings

[queen prism]

open maps sound nice until you take a look at a few examples

[kind marlin]

whenever a question involves proving a map is open i assume its going to be annoying :(

[umbral hamlet]

ok now i gotta show this thing isnt open or closed

tho its just two counterexamples so its whatever

pretty sure its equivalent to finding sets U that are open in the non negative reals but not on all reals

same for closed

[kind marlin]

https://math.stackexchange.com/questions/2065463/are-injective-continuous-functions-on-open-sets-open

this is... confusing me, what makes R^n special

oo theres a wikipedia page

[umbral hamlet]

wait wouldnt (-1, 1) restricted to A be open in the subspace topology on A but then [0,1) isnt open in R

so q cant be open

[kind marlin]

oh wait this is like super clear im just brainrotted from weird topology spaces

oh this is a thing that feels obvious but is actually hard to prove

[kind marlin]

are you referring to (-1, 1) x R

[umbral hamlet]

yeah but i realized thats not even a subset of A so

pretty sure just the set (R^+, 0) is open

and its image is R^+ which isnt open in R

[kind marlin]

well that’s okay we can take the intersection after but that set anyway doesn’t result in a non open set after projecting

[umbral hamlet]

yeah

but im pretty sure any set with the non negative reals in the first coordinate is open

[kind marlin]

are you referring to basically the positive x axis right now

[kind marlin]

Also this is open that’s just (0, inf)

[umbral hamlet]

yeah i meant [0, inf)

[kind marlin]

oh I see

but [0, inf) x 0 isn’t open in A?

[umbral hamlet]

how come

oh wait wait nvm

could i do like [0, inf) x U

where U is just some arbitrary open set

[kind marlin]

what open set in R^2 are you starting with to get these sets

[umbral hamlet]

(0, inf) x U

[kind marlin]

okay that’ll stay untouched in the A subspace topology, but that projects onto an open set so that’s no good

aiming for [0, inf) x U for some U is a good idea but you need to construct an R^2 set that’ll intersect to that, bc this isn’t an open set in R^2 by itself

[umbral hamlet]

no it wont

itll turn into [0, inf) x U

wait shit ur right

i would need like (-1, inf) x U

my bad

[kind marlin]

yeah but you also need to be a bit careful with U

[umbral hamlet]

how come

[kind marlin]

are we fine if U = (-1, 1)?

[umbral hamlet]

oh yeah ur right

could be that we take the negative chunk of the first coordinate and then all of U

and thats a subset of A

[kind marlin]

mm do you have an example im not sure what u mean

[umbral hamlet]

well like its either that the first coordinate is non negative or the other coordinate is 0

or both ig

so if we take the negative part of (-1, inf) then

oh the other part would have to be {0}

[kind marlin]

I still like this idea

[umbral hamlet]

i think U just has to not intersect 0

[kind marlin]

yup

[umbral hamlet]

dude this is terrible

ugh

anyways now i at least have my example

man idk why im struggling with this class so much though

i get its hard but i genuinely just have no idea whats going on

[kind marlin]

honestly I think some classes just click better than others for different people

[umbral hamlet]

yeah

algebra is what clicks best for me

i think im just completely unfamiliar with topology

and also the prof i have is apparently notoriously not good at teaching topology which isnt great

[warped helm]

tbh i've come to terms that i need to seek out more examples and motivation

since like, ok product and quotient topology cool! yay! (why am i being taught this)

[kind marlin]

topology is like the first pure math class that clicked for me lowkey, compared to algebra and analysis

[umbral hamlet]

same

ive also just like been completely unable to follow what my prof is doing in class

[kind marlin]

I feel like I would have had a much better time with my math major if I took topology first 😭

[warped helm]

like these constructions rose from the math that people were doing but of course that doesnt get mentioned

[umbral hamlet]

and she breezes over proofs

and she doesnt put her notes out

[warped helm]

my prof also says "its clear/obvious/trivial" for literally every step

and im like

why is the class being hosted then

[umbral hamlet]

yeah same

[kind marlin]

idk if you read the textbook and not to beat a dead horse but the textbook helps a lot :D I couldn’t really understand my prof’s lectures that much either but it made a lot more sense after reading through the relevant chapters

[warped helm]

yeah for sure

munkres comes in clutch

[kind marlin]

Flashback to us learning all of the algebraic topology chapters in 3 lectures 🥀

maybe learning is a strong word

and the backrests were so evil bc half of them were constructing coverings and stuff meanwhile half the class just learned what a group is 💔

[warped helm]

what are backrests?

[kind marlin]

Backtests*

like practice tests

[umbral hamlet]

im excited for algtop

i hope ill understand that better

geometric intuitions for group theory really clicks for me

so im hoping thatll be what happens for algtop

[umbral hamlet]

okay at the very least im good with the countability axioms

i think thats because i got good practice with them

so

maybe once i take my time with everything else ill be fine

[alpine nest]

I think the issue is that a lot of early general topology does consist of proofs where the individual steps are very simple, even though not necessarily combined in a way that's immediately obvious to a beginner; and also as a result this starts feeling really easy fairly soon. A lot of teachers have a hard time distinguishing between what is obvious to them and what is obvious to their students who are seeing all this for the first time, and I think that's particularly exacerbated in topology because the "things becoming obvious" happens fairly quickly compared to other fields (but still not instantly, so you can't treat it as obvious in lectures)

Tangentially related, here's a passage from Munkres which I found interesting:

I'm lecturing on topology this semester and I did sort of observe that, in the sense that up until this point my personal lecture notes consist of just definitions and statements of theorems/lemmas, and I know I'll be able to prove them on the fly in lectures.

[ruby delta]

ah, that's a nice characterization of a "deep" theorem

I like it a lot

[alpine nest]

But obviously I wouldn't expect the students to be able to just see how the proofs would work, especially on the fly.

They probably would given some time, which is also why I'm relegating some proofs to the interactive problem-solving sessions.

[alpine nest]

There's also a risk of flying too close to the sun here, because I've attended some lectures by really great mathematicians (and overall fairly good teachers) who misjujdged how much they'd be able to prove on the fly and got really stuck in lectures

[queen prism]

still haven't figured out how to pronounce Urysohn

[alpine nest]

The hardest part of that lemma, to be sure

A cursory google search resulted in me finding this https://www.nku.edu/~longa/classes/2009fall/mat305/resources/pronunciation.html

[queen prism]

oh god

[alpine nest]

What they give does jive with the Cyryllic spelling of his name (Урысо́н)

[queen prism]

oh the stress is on the ry, that's probably why I confused my prof

[prime elbow]

Outsider, do you maintain your lecture's notes or problems sheets?

[alpine nest]

Yes, but in Polish

[prime elbow]

😕

[tender halo]

yea good luck to english speakers to pronounce an ы

[alpine nest]

Well, what can you do, every language has its own phonemes

[sly geyser]

ok I have two questions

I'll ask my first

can you guys give me an example of a subset of R^n that is connected but not path-connected?

(I know that path-connectedness implies connectedness, but I just heard that it's not necessarily true the other way around)

ok I'll ask my second too

is there a specific name for a subset S of the underlying set of a topology T such that for all open sets X,Y, S ∩ X = S ∩ Y -> X = Y

basically meaning that the topology is the "same" (not necessarily homeomorphic but you get what I mean with the same structure and all so to speak) when restricted to the subset

[quick delta]

In R^2, take the graph of sin(1/x), combined with the line {x = 0, -1 <= y <= 1}

[sly geyser]

ok fair enough I guess

[sly geyser]

I had the idea when looking at the infinite product topology from the perspective of X^Y and the perspective of the set of elements of X^Y that have a cofinite kernel

[quartz horizon]

Such a subset has to be dense at least

You have an issue if singletons are closed

[sly geyser]

I noticed that too

it's admittedly a bit meaningless for R^n because of that

[quartz horizon]

Yeah if singletons are closed you need S to be the whole space

[sly geyser]

and meaningless for metric spaces in general I think

[opaque scroll]

If there is a closed point x not contained in S, then S will have the same intersection with X and X\{x}

So meaningless for any T1 space

[opaque scroll]

Which should include product spaces for X T1, not sure what you had in mind here

[sly geyser]

I meant the non-(box topology) version of the product topology?

is the non-(box topology) version still T1?

[opaque scroll]

Yeah, the usual product topology on a product of T1 spaces is T1

Let fi: Prod Xi -> Xi be the projection maps, then if {xi} in Xi is closed, so is the preimage.

Take the intersection of all those preimages and you get a singleton set

[sly geyser]

ok hmm yeah that works I guess

[opaque scroll]

Or perhaps simpler, if two points aren't equal, then there is a coordinate where their different. Then just take the basic open set that's only nontrivial at that coordinate

[sly geyser]

ok completely different question now

is R^2/{(0,0)} equivalent to R^2?

[quartz horizon]

No

[opaque scroll]

Not sure what you mean by equivalent, but the answer should be no in either case

[sly geyser]

I mean homeomorphic

[opaque scroll]

Still no

[quartz horizon]

Yeah they’re not homeomorphic

[sly geyser]

what's the condition for an open subset of R^n being homeomorphic to it?

[opaque scroll]

Being contractible is probably sufficient.

Appearantly not...

[unreal stratus]

What's a counterexample?

[opaque scroll]

Actually I'm not sure, there are open contractible manifolds that are not R^n. But I guess they probably don't embed as open subsets

[unreal stratus]

Apparently Whitehead manifold still works

Like it embeds as an open subset of R^3

[opaque scroll]

I see, yeah it's defined as an open subset of S^3 so that checks out

[unreal stratus]

This is v cool though lol glad you brought this up

Til smth lol

[sly geyser]

ok another question

why directed sets?

the elements are supposed to correspond to open sets (around a point), and upper bounds supposed to correspond to the intersections of those open sets right? but why pre-ordered instead of just partial ordered sets?

[crisp lintel]

I think it's just for generality sake

and yeah as you pointed out the main motivation is that the open neighborhoods of a point for a directed set

I don't think you actually need preorder to characterize all of the topological properties

Ah apparently it's useful to avoid the axiom of choice

since instead of choosing a net of open neighborhoods you can instead choose a net of pairs (y, U) where U is an open neighborhood of x and y is in U

(basically instead of choosing an element from every open neighborhood of a point which requires choice, you just consider all of the points)

[quartz horizon]

They correspond quite well to the notions of “limit” you come across in calculus

For example, you can consider the reals “directed towards a point”, and this isn’t partially ordered

Essentially, the antisymmetry requirement turns out to not really be that relevant for formalising the notion of a “direction” on a set

[alpine nest]

I think they're asking why preorders rather than partial orders (i.e. why no antisymmetry requirement), especially considering that inclusion is a partial order.

[quartz horizon]

Mhm, so I gave an example of a use case of directed sets where it’s not a partial order

[alpine nest]

Yeah, sorry, I didn't read the third message

Or the second one

Bad case of mansplaining, profound apologies

[quartz horizon]

[alpine nest]

What do you mean by this, though?

[quartz horizon]

I cribbed this from Wikipedia, actually

https://en.m.wikipedia.org/wiki/Directed_set#Directed_towards_a_point

[alpine nest]

Yeah, just found it; makes sense I suppose

I mean I can see how it's a preorder but not a partial order, and it's a fairly natural construction

[sly geyser]

wdym by the reals "directed towards a point"?

[quartz horizon]

See the link I posted

[gaunt linden]

Are you asking in the context of nets or something else?

[warped helm]

Tbf I think it's just his teaching philosophy and it's growing on me tbh

It's like he says those things to provoke you into actually going back and proving it yourself

[gritty widget]

This implies the only open set containing S is the whole space(trivially), and that S is dense(because SnT=Sn\emptyset implies T=\emptyset) , not sure what else

[sly geyser]

it also implies that the space ain't T1

cause if it was T1 it'd have every singleton be closed, making, for some x in X/S, SnX = Sn(X/{x}) -> X = X/{x} which can't be of course

the original place the premise came from in the first place was wrong

so ignore it smh smh

[gritty widget]

Ah

[umbral hamlet]

Ok I have to show that R omega isnt separable in box

Wouldn't the set of all possible countable products of open sets of the form (n, n +1) for n in Z show that there are uncountably many disjoint open sets

Yeah it would bc diagonal argument

I think that proves it then

[umbral hamlet]

This is a problem i did on my last hw that I apparently got wrong

Idk what it is i did that was wrong tho

Did I not prove that each coordinate has to be unbounded?

[unreal stratus]

"Some factor" doesn't quite make sense to me

Like firstly C needn't be of the form C1 x C2 x ...

And at the same time, the projection of C onto each coordinate could be all of R without C being all of R^oo - e.g. consider the set {(x, x, x,...)}

@umbral hamlet

[umbral hamlet]

Hmm

I mean i thought that I showed any coordinate being bounded in some direction causes a problem

[unreal stratus]

Sure but that doesn't seem to imply the conclusion

[umbral hamlet]

Ah well i suppose that doesn't make it all of R

[unreal stratus]

Well even if each coordinate were all of R you wouldn't be done

[unreal stratus]

By stuff like this

[umbral hamlet]

Well ig I was just logically lazy with this problem

Always next time I guess

[sly geyser]

I think I get why the first one is the way it is but why is the product topology on N^N and the irrationals equivalent?

oh hmm, I can kinda see it now I think

ok wait why can't it be ({0,1})^N too?

[iron bolt]

{0,1}^N is compact, the irrationals aren't

Tychonoff's theorem if you haven't heard about that in class yet

[sly geyser]

why are the irrationals homeomorphic to N^N?

[iron bolt]

I mean - your intuition for the Cantor set was probably that you can split it into two disjoint parts, then split each of those into two disjoint parts and so on forever, right?

it's a similar thing with the irrationals, just a little harder to see because they're not so spaced out

you can find a way to split them into countably many disjoint sets, each of those homeomorphic to the entire space of irrationals

[tender halo]

the easiest way to see is continued fractions

[iron bolt]

oh, interesting. that one I did not see coming

[tender halo]

irrational numbers correspond to infinite continued fractions

[sly geyser]

ok but like can't you also just constantly split the irrationals into two?

[tender halo]

that is to say, to infinite sequences of natural numbers

[sly geyser]

well fine there's a problem with the splitting thing

[iron bolt]

hm, right. looks like the details are a bit more subtle

you can keep splitting the irrationals into two and embed them into the Cantor set that way, but you'll find that the embedding is not surjective

[sly geyser]

when you do the splitting thing the infinite tuples basically represent how you approach the number through the split open sets, but this kind of construction would probably allow you to approach a rational too which would be a problem

[tender halo]

the main difference there is that a limit of irrational numbers need not be an irrational number

[iron bolt]

if you do the splitting carefully enough, I don't think so

[tender halo]

i.e. you have "gaps" in the space, where the rationals would be

[iron bolt]

you can probably arrange it so that every single rational number occurs as a split point once

[tender halo]

that is pretty much what happens with the continued fraction construction

every finite continued fraction is rational and a "split point"

[sly geyser]

or viewed in another way you would have to have one of the numbers "approach infinity" to get to a rational right?

[tender halo]

idk at this point this is stretching the metaphor to a point its innacurate lol

ok imagine N^N

every member has initial segments, N^{n}

the whole set is a tree, and its elements are the infinite branches of the tree

the rationals are the initial segments of these branches, the finite paths from the root

[sly geyser]

this is the idea right?

[tender halo]

yea

the blue points are the parents of all points to the right of them until the next blue point in line

the green points (direct children of blue ones) are parents of all points to the left (!) of them until the next green point

[sly geyser]

what my point was with "approaching infinity"

is that you don't touch a point by just getting closer and closer to it with intervals but never actually touching

and "the other side" gets "sealed" on the next iteration (the intervals approaching from 1/2 to 1 never "touch" 1)

such a process happens for every rational and so it's impossible to try to get to any rational by infinitely narrowing down the intervals

ok fine wait there's a bit of a mistake in what I'm saying

these open sets don't actually have that rational

by "approaching it through narrowing down the intervals infinitely" you technically just don't get anything as they technically "approach" a rational that isn't even inside them

[outer raptor]

What advantages are there in studying topology from the point of view of filters instead of open sets? It seems less natural to me. Is there any heuristic or practical reason to justify working with filters?

[crisp lintel]

usually the debate is filters vs nets not filters vs open sets

[outer raptor]

Let me phrase this another way, why add the overhead of any of these two systems?

[tender halo]

the system of.. open sets?

thats kinda the point of topology

[outer raptor]

No, that’s not what I mean. I mean, why go through the effort of introducing filters in topology? They seem like quite abstract, complex objects

In other words, why would one take the step from phrasing things in terms of open sets to phrasing things in terms of filters (or nets)

And ideally, I’d like an answer that goes further than just “to define convergence in bizarre spaces”

[tender halo]

the point of filters is to have a generalized the notion of limits

[crisp lintel]

well nets are often easier to work with than open sets for a lot of notions

and filters have advantages over nets

just as a random example if you want to show that the closure of a subspace is again a subspace (say of a topological vector space) you can prove it in a really quick and easy way with nets

not that it's difficult without but with nets it's just basically obvious

so phrasing things in terms of convergence is just really convenient, and filters are another way to do convergence which can be helpful

[tender halo]

ultrafilters have a lot of nice properties, such as every point having a canonical ultrafilter, ultrafilters corresponding to points in the stone cech compactification, images and preimages of ultrafilters being prime/ultra

[crisp lintel]

and filters show up in other places yeah

[tender halo]

they are the "right" abstraction when you think about what convergence is

the classical way to define an extension of a Hausdorff space is to define some lattice over certain subsets of the space and add all ultrafilters

stone-cech compactification, realcompactification, iliadis absolute all use this idea of generating a new space by taking all ultrafilters in a certain lattice associated with the space

dieudonne completion also? im unsure actually about that one, maybe not

a more palatable example is uhh

the duality between the ideals in the ring of continuous real-valued functions and ultrafilters on zero-sets of a space

(which is still stone-cech in disguise but w/e)

i think if you want to learn "why filters" then read "rings of continuous functiosn" by gillman

[gaunt linden]

How about "to define limits and convergence in nice everyday spaces"? Even in plain old $\bN$ and $\bR$, you can use filters for a uniform meaning of things like $\lim_{x\to c}$ vs $\lim_{x\to-\infty}$ vs $\lim_{x\to c^+}$, or for that matter $\lim_{n\to\infty}$, as well as "diverges to $+\infty$" versus "diverges to $\pm\infty$" versus "keeps oscillating", all without adding new points to the space just to have something to hang this up on.

[gentle osprey]

Troposphere

[quartz horizon]

Yeah filters are very based for defining convergence

They’re also a good way to model statements that hold “eventually”, or “forall things sufficiently [X]”

They’re closely connected to nets, which can be seen as a good topological generalisation of sequences

Sequences are good for probing metric spaces, while nets are good for probing arbitrary topological spaces

And any net defines a corresponding “eventuality filter”

Another nice thing about filters is that they give “canonical representatives” to any point in a topological space - namely, the neighbourhood filter

[small umbra]

not sure which topology chat this question belongs to, but ive had this problem about connectedness that ive been trying to understand for a while now (question is image).
I saw someone talking about two halfs of R^2 and then a bridge that connects them. This bridge is R x [n,infty) and then you increase n, etc.
what I do not get is how this ends up being disconnected as disconnected means a partition of two open sets if im not mistaken. these sets are both not open. any help would be appreciated!

realized my "closed" mistake

[alpine nest]

You can find two open disjoint subsets of R^2 which contain the respective halves; and also generally a disconnected set is one which is a union of two (nonempty) disjoint subsets open in relative topology

In fact those two subsets would be clopen, and existence of nontrivial clopen sets is equivalent to being disconnected

[small umbra]

okay so they just have to be nonempty disjoint open in a "relative" topology. and by relative you mean like with the same "rule" but restricted to a space?

[alpine nest]

Yes, in the case of metric spaces that just means you consider the subset (with the same metric) as a space in its own right, without any regard from points from the original space which aren't in your set.

As an example, if we consider [0,1] union (2,3) (in R with the Euclidean metric); [0,1] is an open subset of the relative topology, and (2,3) is closed in the relative topology.

[small umbra]

wait why is [0,1] open?

[quartz horizon]

It’s the intersection of an “ambient” open subset with your subspace

For example you could take (-1, 1.5) as the ambient open subset

Another way of thinking about this is that - you want the inclusion map from your subspace to R to be continuous

So the preimage under this inclusion map of any open subset of R should be open in your subspace

[alpine nest]

Also you can directly verify openness from devinition definition.

[quartz horizon]

So e.g. the preimage of (-1, 1.5) should be open

[alpine nest]

Every element of [0,1] is contained in [0,1] with an entire open neigbhorhood (remembering that points to the left side of 0 "don't exist" for us)

[quartz horizon]

Mhm

[gaunt linden]

By divination, this set is open.

[alpine nest]

Behold, an open crystal ball

In particular because points to the left of 0 don't exist in our space, a ball of radius 0.1 centered at 0 is the set [0,0.1)

Which is a subset of [0,1]

[small umbra]

ahh okay so its a union of open balls because we can just go (-0.5,1.5) intersect our space and then we have it

[gaunt linden]

Indeed.

[small umbra]

so then the two halves in the original answer will be open in the subspace because we can intersect it with for example (-infty, 0.5) x R and get itself so its open (similarly for the other side)

so just to make sure i have understood this discussion correctly would this line of argument be correct? still feels a bit handwavy to me. especially the last part

[supple island]

Is each V_n containing V_{n+1}?

[small umbra]

i see how i never actually checked that. id have to make sure of that yea thanks! is the last part right tho?

changed it to geq. should be good now

[smoky sigil]

This Lemma is from Munkres Topology. The part of this proof that I’m not getting is where it says “First note that h(Z+) cannot be contained in the finite set {1, …, c}, because h(Z+) is infinite (since h is injective)”. I understand that this is saying that h(Z+) is not a subset of {1, …, c}. What I’m not understanding is “since h is injective”. Can someone help me fill in the details to understand how they are justifying h(Z+) is not a subset of {1, …, c}? Thank you!

[young stone]

If h(Z+) is finite we can restrict the codomain of h to h(Z+) to get a bijection from Z+ to a finite set, but this can't happen because Z+ is not finite ( this should've been proved earlier in munkres)

[smoky sigil]

Ohhhh ok yeah I see it now. Thank you! Why was he using h is injective?

[young stone]

Otherwise the restriction won't be a bijection

[smoky sigil]

Ohhhhhhh right right that makes sense now. Thank you for your explanation!

[signal jackal]

So I'm looking at the cocountable topology. We can include the set A = reals minus pi in that topology, since its complement is a Singleton, right? And the rationals are a subset of A. But the rationals aren't in the topology since its complement isn't countable. I thought all subsets of open sets are also open.

What am I missing?

Nevermind I got it lol

[rancid umbra]

yea, not every subset of an open set is open.
if that were true, open sets would be kind of boring, since every set would be open in every top space

[signal jackal]

Yeah I got confused about the intersection rule. They have to be the intersection of open sets

[halcyon bone]

can anyone please make sure im not tweaking the fuck out

[rancid umbra]

Ross school of buisiness?

[halcyon bone]

ross math program

💀

basically the idea of compactness for convergence is:

convergence in ultrafilters is the same as cluster points, which is the same as closures of elements of the filter, e.g \bar{F}, but the filter base property tells us the \bar{F} satisfy the finite intersection property

but a collection of closed sets w finite intersection property having something in all of them (e.g nontrivial intersection) is the same as compactness just by taking complements of the closed sets

and then tychonoff thm is like trivial as a result

just by how convergence in products works

[balmy nexus]

tychonoff comes trivially from convergence in filters, it works for stuff where normal convergence doesn't work well.

[dawn frigate]

Is my proof for 4 correct? My profs proof looks a lot more detailed than this

[fallen canyon]

you can make this as detailed as ya wnt @dawn frigate like t1 non hausdorf

what was yr prof proof @dawn frigate

[prime elbow]

if X is infinite set with co-countable topology then only finite susbets are compact, right?

[stone kayak]

In a co-countable topology on an infinite set every finite subset is compact because any open cover can always be reduced to finitely many sets covering all its points however any infinite subset is not compact since you can create an open cover using complements of single points and no finite subcollection of these opens can cover the whole infinite set

[quartz horizon]

Finite sets are always compact!

[stone kayak]

Yeah in short

[gaunt linden]

Um, if you take complements of single points, then just two of those will cover everything.

Instead choose a countably infinite subset A and cover it with opens of the form (X\A) U {x}.

[stone kayak]

Ohh

I see

[alpine nest]

The other perspective would be to note that countable sets are closed in the co-countable topology, and consider the family of closed sets of the form A\{x} as x ranges over the elements of A.

It has the finite intersection property but its intersection is empty

[gaunt linden]

(Which is exactly the complementary view of what I said, btw).

[alpine nest]

Yeah, it's a different perspective on the same argument

[stone kayak]

Ah i get it now, mixed some stuff up

[gaunt linden]

((By the way, in the absence of (countable) choice, I think the right result is that the compact subspaces under the cocountable topology are exactly the Dedekind-finite subsets.))

((That is, picking a countably infinite subset A in the above argument is essential.))

[dark crypt]

i am confused about exercise 8: prove that an uncountable complete metric space has at least the cardinality of the continuum. — isnt this equivalent to CH, considering that any set can be equiped with a discrete metric, turning it into a complete metric space.

or am i missing something

[gaunt linden]

Hmm, perhaps they intended to assune connectedness?

[dark crypt]

connectedness hasnt been covered in the book yet. and the separable case is proven in that section as a theorem.

[gaunt linden]

Hmm, beats me. Your counterexample looks good to me, unless there's some hidden assumption somewhere.

[dark crypt]

well, hopefully i am not missing something stupid then

[gaunt linden]

Could it be something like "with no isolated points"?

No, that doesn't feel like it's enough.

[dark crypt]

i think it is? — take closed ball with diameter 1 around some point, find two points in the interior of that closed ball and take closed balls of diameter at most 1/2 around each of them. make the closed balls small enough to be disjoint. then repeat a similar procedure with each of the two balls and so on ad infinitum.

that way we can assign to each sequence in 2^ω a sequence of closed balls with diameters approaching zero and therefore a Cauchy sequence. and each of these Cauchy sequences should converge to a distinct point.

[gaunt linden]

Ah, the counterexample I thought I had in mind is not actually complete. :-)

[prime elbow]

What is Dedekind finite subsets?

[tender halo]

not in bijection with any of their subsets

(same thing) not having a countable subset

[prime elbow]

That's mean finite set?

[tender halo]

in absence of choice this is not equivalent to finiteness

a particular case of infinite Dedekind-infinite sets is amorphous sets - sets where if you split it in two, one part will be finite and the other infinite

[prime elbow]

Okay

I want non Hausdorff, non compact, locally compact space, i tried but I don't get it, any hint?

[rancid umbra]

it’s almost R, but with an additional point

[prime elbow]

Yes fixed point topology

Actually I have to show if X is locally compact then it is not necessarily that open subspace is locally compact

But in fixed point Topology it will be

And if X is Hausdorff then it will be true therefore I am taking non Hausdorff spaces

[rancid umbra]

not the particular point topology. i was thinking the line with two origins

[prime elbow]

What is line with two origin?

[opaque scroll]

Is your definition of locally compact just that every point has a compact neighborhood?

If so just take a space that isn't locally compact and take the one point compactification

[prime elbow]

Yes it is the definition which I am using

[rancid umbra]

they want a non-compact space tho

[opaque scroll]

Okay, then just take the disjoint union with whatever

[rancid umbra]

it is the quotient of R x {-1,1} under the equivalence relation that identifies (x,1) with (x,-1) for all x != 0.

[unreal stratus]

classic

[warped helm]

doing topology from munkres makes me appreciate that the modern development of the subject is essentially handed to you on a silver platter

can't imagine how frustrating it must have been to fumble around until hitting upon the right definitions for stuff

[prime elbow]

So Q is not locally compact, and if I take one point compactification of Q it will be compact by definition and Q is open subspace in this new topological space, thanks jagr ❤️

[prime elbow]

Seems interesting topology

[rancid umbra]

remember that you have to adjoin something non-hausdorff and non-compact

[rancid umbra]

it is

[prime elbow]

@rancid umbra Actually this was my original question, so therefore I tried with non Hausdorff and non compact space

Now i want to replace open subspace with closed subspace

[gaunt linden]

A closed subspace of a compact space is compact, so if A is closed and X is locally compact, then at any a in A you can just intersect a compact X-neighborhood of a with A.

[gaunt linden]

The one-point compactification of Q is already itself non-Hausdorff, though.

[rancid umbra]

oh dang

my b

[prime elbow]

But I think c squared addressed both non Hausdorff and non compact, but one point compactification of Q is compact, right?

[gaunt linden]

Yes, so if you did want a non-compact space you would indeed need something like take the disjoint union with something non-compact.

[prime elbow]

I see, thanks i have to spend more time on problems, because that's one easy

[gaunt linden]

(But "non-compact" is not half as weird as "non-Hausdorff").

[prime elbow]

One point compactification is a very good tool, if I want a Lindeolf space which has non Lindeolf subspace then if I take X with non Lindeolf space then its one point compactification will be Lindeolf, right?

[unreal stratus]

Meanwhile in algebraic geometry

[tender halo]

sure

i mean those are easy to come by

[prime elbow]

You react to your own message

[gaunt linden]

That sort of thing happens in algebraic geometry.

[prime elbow]

Algebraic geometry in my reading course, semester III

I don't know which one will be good for me?

[quick crane]

(11) Is most useful. (13) and (14) are also good. A solid course in (2) requires (13) as a prerequisite anyway. The rest are useless.

[prime elbow]

Lie algebra required Riemann geometry?

[prime elbow]

And why 11 is most useful?

[austere flare]

well that's a hot take

[austere flare]

well, what areas of math are you interested in?

[prime elbow]

I don't know, I like abstract algebra, analysis ( topology, trying to learn alg top, measure theory)

[austere flare]

hmm, if you're not sure/have broad interests, then it might be better to go with something that has more direct connections to different areas of math rather than something more specialized. if you're most interested in algebra and more topological flavors of analysis, then maybe intro to AG, lie algebras, riemann surfaces, fourier analysis, or rep theory?

[prime elbow]

Maybe sounds dumb, but I am trying to go in category theory because I found interesting

[austere flare]

sick! category theory is its own area of math; although these courses probably won't directly help you as a category theorist, some of these areas (like AG and homological algebra) have rich applications of category theory that you might find interesting

[prime elbow]

I am thinking of adding a category theory in reading course

[austere flare]

hell yeah

[valid patio]

less go

[quick crane]

yes, look at https://mtaylor.web.unc.edu/notes/lie-groups-and-representation-theory/

[quick crane]

Has the most broad applicability outside of pure mathematics, and also within pure mathematics. Look at the distribution of the areas math faculty in your university work in. Most likely, PDE is by far the field with the most faculty working in it.

[prime elbow]

I don't know where I have to ask this question,

Let L be a countably linearly ordered set. Show that there is a one to one, order preserving map f: L -> Q, where Q has usual order. Any hint? I don't know how to start

So this mapping does not need to be onto

[plush folio]

Self-reacts are common in algebraic geometry?

[dark crypt]

well, the set is countable, so you could build the map element-by-element.

[gaunt linden]

Since L is countable you have a bijection betwen N and L, so a natural place to start would be with element number 0 (or 1 depending on which naturals you like best).

[iron bolt]

so "one to one" here actually means "zero or one to one"?

that's why I like precise terms like bijective, injective and surjective better

[dark crypt]

one-to-one mapping usually means injective, one-to-one correspondence usually means bijection i think

[iron bolt]

that may be true but I'm still gonna skull-react it

especially because that convention means you can't use either term as an adjective (i.e. "f is one to one" meaning injective) but instead have to specify it like "f is a one to one mapping"

[gaunt linden]

I've definitely seen people write "f is 1-1". I don't like that terminology, but it's not like they've asked me for permission.

[prime elbow]

I know N to Q there is a natural one to one, order isomorphic mapping, are you suggesting me to find the one to one, order isomorphic L to N?

But I think it is not true that we can always find L to N such function, so maybe you are suggesting different

[gaunt linden]

"L is countable" means that there is a bijection betwen N and L. This bijection may not preserve order, but you don't need that. It just gives you a sequence in which you can pick images of each element of L in Q, one by one.

(There's no order isomorphism between N and Q. In the ordering of N, each element has an immediate successor, but no element of Q has an immediate successor under the usual orering).

[young stone]

Countable means there is an injection to N

[gaunt linden]

Ah poop, that word also has two related-but-distinct meanings. Some people consider "countable" to include finite sets, others don't.

[young stone]

Yeah unfortunate, this exercise is true for finite sets too so including finite sets makes more sense

[tender halo]

i like using countable to mean finite and countable and denumerable for with bijection to N

but ive seen it used the opposite way around lol

[outer raptor]

I like to imagine countable the following way. Let X be a set. If there’s an algorithm which “goes through X” that guarantees that you’ll eventually stumble on any given a\in X, then the set is countable (what you did was count up to a, for any a\in X)

It’s a matter of convention, sure, but this makes sense

I like the discussion in the book Logical Labyrinths about countability; it gives it that fun interpretation

Therefore finite counts as countable haha

[unreal stratus]

Rip when they are synonyms though

[rancid umbra]

i hate that enumerable and denumerable both meant countable

[queen prism]

{countable} = {finite} u {countably infinite} gang

[cosmic mirage]

yeah i use countable to mean countably infinite when it is clear that i would know if it were finite, because if i knew it were finite i would just say finite 💀

[crisp lintel]

technically no but if you want to learn it right then yes

lie algebras without the lie group/differential geometry stuff is really unmotivated

and the only real reason people care about lie algebras is because of the connections to lie groups

[dark crypt]

unless f is one-to-one means that f is injective, which i think it usually does

[gritty widget]

Maybe try by induction(or recursion? Not sure what would be the right term here)

Assuming it is infinite, pick some particular bijection to N. there is a 0-th element, say a_0, so pick idk 0 in Q and say P(a_0)=0. Then extend this map recursively by defining P(a_(n+1)) given P(a_n) in a way that is an embedding.

[prime elbow]

So to choose an image of a_1, if a_0 < a_1 then map a_1 to b > 0, otherwise map to b < 0.

To choose an image of a_2, if a_0 < a_1 < a_2 then map a_2 to c > b.

If a_0 < a_2 < a_1, then map a_2 to c, 0 < c < b.

Similarly for other possibilities.

Right? Thanks Eclipso, Troposphere

If there is one one order preserving mapping from A to B, and also from B to A, can we find order isomorphism mapping from A to B ?

[dark crypt]

does anyone have any idea what kaplansky means when he says that D(a, aₖ) can be made arbitrarily small? — i dont see how we would guarantee that aₖ ⟶ a.

[tender halo]

means the whole tail can be made arbitrarily small

which is basically the definition of convergence to a point

[dark crypt]

thats not what i meant. i am asking how you would prove that convergence.

[tender halo]

oh

[dark crypt]

he sort of implies that this is something relatively obvious

and i just cant think of any way to fill in that gap

[tender halo]

a_n and b_n are sequeces in a compact metric (and therefore sequentially compact) space, therefore they have (simultaneously) convergent subsequences which you can choose to converge towards a and b respectively

[dark crypt]

how do you choose them?

[tender halo]

you can do it because a and b are cluster points of a_n and b_n, otherwise (lets say we are talking about a) we can draw a circle of size epsilon around a where there are no points of the sequence, then every a_n has a circle of size epsilon where there are no other points of the sequence

and then a_n would have no convergent subsequence

but we already know that it does because the space is compact metric

[dark crypt]

why wouldnt it, it could have a subsequence that converges to some other limit right?

[tender halo]

it can't

if it has any cluster points, then one of them has to be a, and we know it has cluster points because the space is compact

[dark crypt]

in fact, assuming that f(a) ≠ a, i dont think you can have a as the sole limit point.

[tender halo]

(it can have other cluster points too, but we dont care about those)

[tender halo]

yea, f(a) would be a cluster point too

in fact every a_n will be a cluster point

[dark crypt]

ah right, i get what you mean now.

stupid

since f will expand open balls in that sequence

[tender halo]

yea also there is like an edge case if f(a) turns out to be periodic but thats easy to handle

[dark crypt]

thank you

i feel a bit stupid for failing to realize that for so long now, but that happens lol

[tender halo]

i mean its not a trivial observation, i had to think for a bit

[cosmic mirage]

not true in general, for example constant maps are always order preserving. if they are order-preserving and injective, then this is definitely true for finite sets but i havent had enough caffeine today to think about infinite sets and make no claims there

[tender halo]

Q + 1 and Q inject into each other but are not isomorphic

[cosmic mirage]

fantastic

[prime elbow]

yes

[tender halo]

there is a following theorem though

if A is order isomorphic to an initial segment of B and B is isomorphic to a final segment of A, then A and B are isomorphic

[prime elbow]

Proposition 3.3,

I proved one direction so now I have to prove net converge at most one point implies X is Hausdorff
Let I denote the set of all open neighborhood of x, and J denote the set of all open neighborhood of y.

They both are directed set with reverse inclusion

Assume X is not Hausdorff, so there exists two points x and y such that every open neighborhood of x intersect with every open neighborhood of y.

So for all (i,j) \in I × J[ we can make I × J directed set, (i_0, j_0 ) ≤ ( i_1, j_1) iff i_0≤ i_1 and j_0 ≤ j_1 ] , we can associate (i,j) to z_{ij}, where z in intersection of i and j

We can verify that Z: I × J -> X, (i,j) maps to z_{ij}, and it converges to both x and y.
That's a contradiction.

Is it correct?

[quartz horizon]

I think you should rename “Y” to “J”

Otherwise this seems fine to me?

Interesting that you’re working with nets

[prime elbow]

Thanks

So in this exercise I can show if that \phi is order homomorphism and its image is co final in I then it is cofinal function, but can we remove image is co final in I condition?

[humble schooner]

hey can someone help me pls ?

$\text{I took a sequence $(x_n)$ such that $x_n \in F_n$ for each $n$, and I showed that $(x_n)$ is a Cauchy sequence. \
Then I used the fact that $X$ is complete, and that the intersection of closed sets is closed, so the limit $x$ belongs to every $F_n$. \
Finally, I used the continuity of $f$.}$

[gentle osprey]

Ashraf10

[humble schooner]

\textbf{For the reverse inclusion.}

Let us show that
[
\bigcap_{n \in \mathbb{N}} f(F_n) \subset f\left(\bigcap_{n \in \mathbb{N}} F_n\right).
]
Suppose that ( y \in \bigcap_{n \in \mathbb{N}} f(F_n) ).
Then, for each ( n \in \mathbb{N} ), there exists ( x_n \in F_n ) such that ( f(x_n) = y ).
Since the sequence ( (F_n)_{n \in \mathbb{N}} ) is decreasing and the diameters of the ( F_n ) tend to ( 0 ),
the sequence ( (x_n) ) is Cauchy in ( X ).

Because ( X ) is complete, there exists ( x \in X ) such that ( x_n \to x ) as ( n \to \infty ).
Moreover, since each ( F_n ) is closed and ( x_n \in F_n ) for all ( n ),
we have ( x \in F_n ) for all ( n ).
Hence,
[
x \in \bigcap_{n \in \mathbb{N}} F_n.
]

Finally, by continuity of ( f ), we get
[
f(x_n) \to f(x).
]
But ( f(x_n) = y ) for all ( n ), so ( f(x) = y ).
Therefore,
[
y \in f\left(\bigcap_{n \in \mathbb{N}} F_n\right),
]
which proves the reverse inclusion.

[gentle osprey]

Ashraf10

[plush folio]

what do you need help with? that looks correct

[stuck geyser]

Any hints for (i) => (ii)

I can only really get as far as the diagonal in (S/~)^2 being compact but then I get stuck since S/~ isn’t a priori Hausdorff

[gritty widget]

Funnily enough someone just asked this exact question like a month ago!

Wait no

[stuck geyser]

Probably me because I took a hiatus for engineering

Then came back

[gritty widget]

Just similar

[stuck geyser]

And am still stuck

[gritty widget]

If ~ is closed in SxS, the complement is a union of sets of the form UxV where U, V are open in SxS. What can we know about U and V based on the fact ~ is an equivalence relation?

[stuck geyser]

By ~ do you mean the graph

[opaque scroll]

For any s in S the subspace
Sx{s} is homeomorphic to S

[gritty widget]

Yeah sorry

[stuck geyser]

thank you

[opaque scroll]

(I'm assuming your definition is that equivalence classes are closed in S)

[stuck geyser]

The defn for a closed relation here is that the projection map is closed

Equivalence classes are closed if the quotient is Hausdorff which is later in the problem

[opaque scroll]

I see, then I change my hint to SxF being closed for F closed. And that the projections SxS -> S are closed for compact S

[stuck geyser]

That latter part is what I should’ve realized

Let me try and prove that, thank you

[stuck geyser]

Just to make sure is the latter part true when S isn’t Hausdorff

[opaque scroll]

You don't need S to be hausdorff no

[stuck geyser]

Thank you I just didn’t want to go in circles

[stuck geyser]

I managed to prove it with an argument that was eerily similar to the proof that compact sets in hausdorff spaces are closed

is this related to the Tube Lemma

[gritty widget]

That method is in general very usefu!

[gritty widget]

The typical proofs that Hausdorff+Compact imply T_3 and T_4 respectively use practically the same method

[stuck geyser]

I just proved that compact “boxes” in an open set are contained in a open “box” in that subset instead because that was WAY more visually obvious

[umbral hamlet]

i get that i probably shouldve known this before doing topology but theres some things i missed out on ig

can this be proven using induction?

im not convinced since its not stating that the union of finitely many subspaces satisfying the properties is connected

i mean i know this doesnt require induction at all but im just curious

[kind marlin]

induction would only prove for finite sequences

you could probably use finite cases to generalize to infinite sequences though, you just need additional reasoning

[umbral hamlet]

yeah

though what i have so far suggests it isnt necessary

at least i dont think it does

i just have U and V being some separation of $A = \bigcup A_n$ and then showing that for any $k \in \mathbb{N}$, we have that $ U_k = U \cap A_k, V_k = V \cap A_k$ is a separation of $A_k$ and so WLOG $U_k = \emptyset$

[gentle osprey]

hiidostuff

[umbral hamlet]

actually wait im not sure WLOG U_k is empty

bc how can we be for sure that the chunk that comes from U isnt the entire A_k

[prime elbow]

How do I show a sequence which is universal net then it is eventually constant?

[kind marlin]

yeah, the argument kind of allows each Ak to belong to U or V, which doesn't help with showing one of them is empty

we already know that a connected subspace must belong to exactly one of two sets separating the space its contained it

we just need to show that all of our Ak sets must belong to the same one

[umbral hamlet]

ah but thats covered by the subsequent nonempty intersection thing im pretty sure

consider two subsequences of A_n that separate the sequence

there has to be at some point A j thats in U but then A {j + 1} is in V in that case

but those two elements arent disjoint

which contradicts that U and V are disjoint

[kind marlin]

mhm that works

[umbral hamlet]

i think that pretty much proves it

[kind marlin]

induction gives a neat contradictive proof too

by induction, any finite consecutive sequence of Ak must be connected. if we have a separation, then two Ai, Aj sets belong to the distinct separations. but Ai to Aj must be connected, which is a contradiction

its not that different but i kind of like it

[umbral hamlet]

oh i see

[prime elbow]

Anyone?

[prime elbow]

I want to know how nets idea comes, how Moore and Smith got intuition that if we develop this theory then it will help us in a complicated situation

And how do I continuously use them in my work?

[prime elbow]

Limit point compactness implies sequentially compactness, right? We don't need first countability here

I have filter F, say A is not in F.

Now I want to make ultrafilter F' such that F \subset F' and A not in F'.

So if I make F' = F u { X\B | B \subset A }.

Does it work?

No I have to add more elements

In my definition of filter they are taking collection of non empty subsets of X

Now i want to prove that every filter is intersection of all ultrafilter which contains that filter.

Say F is my filter such that A not in my F, then I have to make ultrafilter U containing F and not contain A.

i know for filter F there is ultrafilter U which contains F by Zorn's lemma, can we explicity describe that U?

[opaque scroll]

By Zorn's lemma there is a maximal filter containing F and not A, so you would just need to show that that is an ultrafilter

[opaque scroll]

Zorn is necessary to demonstrate that U exists, so there can't really be an explicit description.

[opaque scroll]

As a hint though:
||You might be able to give an explicit description of a filter that lies between F and U, with the property that no filter containing it contains A||

[prime elbow]

yes i got it

[prime elbow]

Exercise 5.2.4, I don't think it is true when X = {a,b}

[opaque scroll]

F1 = {{a}} and F2 = {{a}, X} works as long as X isn't just {a}

[prime elbow]

Ah

Yeah got it

Thanks jagr

[prime elbow]

In any topological space, compact space implies sequentially compactness, right?

[quartz horizon]

Nope

In metric spaces they are equivalent, but not in general topological spaces

One way I like to think about this is the following:

Compactness in general topological spaces is equivalent to “net-compactness”, which says every net has a convergent subnet

You can contrast this with every sequence having a convergent subsequence

[prime elbow]

But i think one way is true

[quartz horizon]

Sequential compactness is not strong enough to imply net compactness, because there are far more nets than sequences

However, net compactness does not imply sequential compactness!

The logic would go:

1. Every sequence is a net
2. By net compactness, it has a convergent subnet
3. Hence, every sequence has a convergent subsequence

But step 3 is actually false

A subnet of a sequence is not necessarily a subsequence!

[prime elbow]

Okay, but I think compactness implies sequentially compactness

[quartz horizon]

It doesn’t

There exist compact topological spaces which are not sequentially compact

Provably so

[prime elbow]

Can you give me a counterexample?

[quartz horizon]

There’s a nice construction you can do with [0,1]^{P(N)}

You use a kind of diagonal argument to construct a sequence with no convergent subsequence

However, by Tychonoff’s theorem, this space is compact

So the sequence you construct does have a convergent subnet, but no convergent subsequence

The way I like to think about this is that sequences are fundamentally limited by their “countability”

If you have a point that doesn’t have a countable neighbourhood basis, it’s a lot harder to get a sequence converging to it

[prime elbow]

What is wrong with this argument?

Let x_n be a sequence then take F_n = cl{ x_k | k≥n }.

This F_n are closed and they have finite intersection property so they have non empty intersection, say x.

My point is x will be the limit point of this sequence

[quartz horizon]

But neighbourhoods always form a directed set, so it’s a lot more straightforward to get a net converging to it

[quartz horizon]

So you’ve hit upon an important point

There is a distinction between “having a convergent subsequence” and “having a cluster point”

By a cluster point I mean the following

A point L such that, for every neighbourhood U of L, the sequence lands in U infinitely often

[prime elbow]

But if the sequence has limit point then it will have convergent subsequence, right? I am saying sequence has limit point not range of sequence has limit point

[quartz horizon]

This is the key point

[prime elbow]

Yes so why can't we make subsequence from here

Wait

[quartz horizon]

It is always true that, if a point is a limit of a subsequence, then it’s a limit point of the original sequence

But the reverse is not generally true

[quartz horizon]

And it’s precisely because of this issue

[prime elbow]

I see

So if it is first countable then we can do

[quartz horizon]

In general, your limit point L might have uncountably many neighbourhoods

Yep, exactly

[quartz horizon]

But you only have countably many terms of a subsequence to work with

[prime elbow]

Okay to say limit point compactness implies sequentially compactness we need first countability

[quartz horizon]

If there’s not a way to choose a good countable basis of these neighbourhoods, you won’t be able to actually make a subsequence

Mhm

So, what is true is that for nets

Having a limit point is equivalent to having a convergent subnet

For sequences, you only have convergent subsequence => limit point

And you need first countability to get the other direction

[prime elbow]

I see

[quartz horizon]

In this sense, “every sequence has a convergent subsequence” is strictly stronger than “every sequence has a limit point”

Whereas “every net has a convergent subnet” is equivalent to “every net has a limit point”

Metric spaces are always first-countable though, because you can take a basis of balls of rational radii

[prime elbow]

Yes

[prime elbow]

Yes

Thanks @quartz horizon

[quartz horizon]

I only learned about these characterisations of compactness recently, but I think they’re quite cool

[prime elbow]

Source?

[quartz horizon]

Uh mostly Wikipedia and nlab

[prime elbow]

Really?

[quartz horizon]

Mhm!

[prime elbow]

Are you a graduate student or PhD scholar?

[quartz horizon]

PhD student!

2nd year now

I wonder to what extent you need the full strength of sequential compactness, actually

I.e. when it is important to really have a convergent subsequence, as opposed to just a limit point?

[prime elbow]

And your research area?

[quartz horizon]

condensed matter physics

[prime elbow]

Really? You are doing your phd in physics and you are also doing mathematics

So why didn't you choose a PhD in mathematics?

[quartz horizon]

i like physics more

also i don't think i would be good enough at maths for a phd

[prime elbow]

Why?

[quartz horizon]

based on my track record in undergrad wrt pure math

[prime elbow]

Can I get an example of it?

[quartz horizon]

so you could take any compact space that isn't sequentially compact, like $[0, 1]^{P(\mathbb{N})}$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

in this case, by compactness, any sequence has a limit point

but you can explicitly construct a sequence that has no convergent subsequence by a diagonal argument

i think there's a MSE post that goes into the details, lemme find it

[prime elbow]

So here sequence has a limit point means there exists x such that any open set of x contains infinitely many terms of sequence x_n

[quartz horizon]

https://math.stackexchange.com/questions/1558448/product-space-that-is-compact-but-isnt-sequentially-compact/1558796#1558796

[quartz horizon]

yep

[prime elbow]

Why?

I know by compactness we have limit point compact

[quartz horizon]

compactness is equivalent to every net having a limit point

and any sequence is a net

[prime elbow]

I see

[quartz horizon]

in this sense compactness is a kind of "topological pigeonhole principle"

for finite sets, any infinite sequence must have a cluster point (meaning a point that the sequence hits infinitely often)

similarly, for compact topological spaces, any net must have a cluster point

[prime elbow]

Yes

Thanks

[quartz horizon]

i think this gels quite well with the everday meaning of "compact"

[prime elbow]

I am not sure about 5.7, i) => ii).

For refinement, can't I just take F' = F?

Oh ni

I got it

@quartz horizon sorry to ping, but I want to know how do you keep remember all theories, i think one way is working with them regularly

[quartz horizon]

what do you mean by "keep all theories"?

oh remember

hm well anki definitely helped in the past

but yeah working with them regularly helps

[prime elbow]

And how do you use it regularly?

[quartz horizon]

in part by talking on this server :P

[prime elbow]

Oh I see

I want to be used to nets and filters

[quartz horizon]

nets and filters are quite cool

[prime elbow]

How do I consistently use it? My professor has not introduced it, I am reading on my own so I want to keep it consistent

[quartz horizon]

hm that i'm not sure

maybe other people in this channel would have a better idea?

[tender halo]

filters are one of the central tools of set theoretic topology

so you can just read books/solve exercises on the topic

"the stone-cech compactification" by walker, "the theory of ultrafilers" by comfort&negrepontis, "rings of continuous functions" by gillman, the handbook - these are all the more important texts concerning the topic, more niche texts are:

"universal spaces and mappings" by iliadis, "algebra in the stone-cech compactification" by hindman, "continuous pseudometrics" by comfort&negrepontis, "extensions and absolutes of hausdorff spaces" by porter

todorcevic has something im sure, in "topics in topology" maybe?

[gritty widget]

You can define connectedness in terms of chains in open covers, e.g. a space X is connected iff, for any points x, y, and for any open cover U_a of X, there is a sequence {U_1, …, U_n} of elements of the cover such that x is in U_1, y is in U_n, and U_i intersects U_(i+1) non-trivially.

This definition may give a better perspective for proving this(and proving this is equivalent to connectedness isn’t too hard, just prove the set of all points reachable by chains from x in arbitrary open covers is clopen)

[umbral hamlet]

fortunately i did manage to prove it, but ill keep this definition in mind

the proof turned out to be pretty cool

[gritty widget]

Alright, I might look back at it in a sec then!

[umbral hamlet]

here it is so u dont have to scroll through the convo

[tiny obsidian]

another useful equivalent definition is "connected iff the only continuous functions to {0,1} (discrete) are constant" which allows you to do an induction like thing here

[gritty widget]

That’s a pretty neat proof!

[quartz horizon]

Yes this is great

-# generalised to “connected iff Hom(X, -) preserves coproducts”

[stuck geyser]

But I am still mega stuck because I can't figure out how to connect this to H/~

I've been stuck on this for a pathetically long time

[stuck geyser]

Jesus I'm a moron

def of the fucking quotient topology

[kind marlin]

hm i think i came up with a solution to i -> ii but i feel like im redoing work that i shouldnt have to do is there a clean way of simplifying the logic here

take a closed set A in S, and consider pi(A). we want to show pi(A) is closed. equivalently, pi(A)^c is open, and equivalently, the preimage of pi(A)^c under pi is open. that preimage is just the set of points P in S that don't relate to any point in A under ~, so we want to show that P is open.

Since the graph C is closed, the complement of C, containing ordered pairs of elements that don't relate to one another, is open, so for any two elements that don't relate under ~, we can find respective neighborhoods such that no pair of points belonging to the respective neighborhoods relate.

Take an arbitrary point p in P. For each point x in A, construct a neighborhood P_x of p and a neighborhood A_x of x such that no point in P_x relates to any point in A_x. The A_x sets are an open cover of A, and A is a closed subspace of a compact space, so A is compact, meaning we can find a finite subcover of the A_x sets. The intersection of the corresponding finite collection of P_x sets is an open neighborhood of p that doesn't relate to any point in A. -> P is open

specifically this feels super similar to the proof that a closed subspace of a compact space is compact, and i can't tell if there's redundant logic here

[stuck geyser]

now I'm stuck on ii) => iii) lo

I can't find a way to find closure of some set without knowing S is hausdorff or knowing open sets a priori

and the product of closed maps is not necessarily closed.

[tender halo]

take two points in the quotient, their preimages are closed and therefore can be separated therefore the points themselves can also be separated

(through the standard trick of taking an the complement of the image of the complement of an open set)

[kind marlin]

could you explain why the preimages of the points are closed?

[tender halo]

because T1 is preserved under closed mappings

and quotient is continuous

oh we dont get that S is T1?

hmm

do we get it? does compact have T1 in it here

isnt the exercise wrong if S is indiscrete and the equivalencr is trivial or something

[deft crow]

Let $(E,\mathcal{T})$ be a topological space and $F$ be a metric space (both non empty) and $f:E\to F$. How do I show that for all $n\in\mathbb{N}$ the following set is closed?

$$A_n={x\in E,\ \forall V\in\mathcal{V}(x),\ \exists y\in V,\ d(f(x),f(y))\geq\frac{1}{n+1}}$$

[gentle osprey]

Dealersgrip

[balmy nexus]

who is V and f please?

im reading this as:

A_n = all x such that there are points in every neighborhood for which d(f(x),f(y)) is larger than ...

If this reading is true, then I think you can show a non-closed example, with something with a lot of discontinuity points.

[quartz horizon]

Is f continuous?

Because if it is I don’t see how A_n could be anything other than an empty set

[balmy nexus]

Hey pseudo did you know that compact hausdorf topologies stop being so if you add/remove a open set

[quartz horizon]

Which I guess is closed lol

[balmy nexus]

made me think of you lol it sounds like a limit

[quartz horizon]

Since if y tends to x, d(f(x), f(y)) should tend to 0

[quartz horizon]

Yep! This is a consequence of the topological inverse function theorem

[balmy nexus]

oh cmon it's already codified

[quartz horizon]

Specifically, if T is a compact Hausdorff topology on a set X, then any coarser topology is non-Hausdorff, and any finer topology is non-compact

[quartz horizon]

By this I mean - a continuous bijection from a compact space to a Hausdorff space is a homeomorphism

[balmy nexus]

you know I know this one as the closed map theorem

but folland called it like you do

[quartz horizon]

I see

[quartz horizon]

Wdym

[balmy nexus]

it's maximal in the Haus category in minimal in the Comp category in some way

or the othe way around

I tried to see if a set with the above property is compact hausdorff (if the converse is true )but I gave up

[quartz horizon]

Well you have to be a little careful here

Because of the fact that there are topologies which are non-comparable

[balmy nexus]

well sure you need to wrap it up with the right safeguards, but it's still a minimal\maximal item in the fine/coarse order

[quartz horizon]

Like, there can be multiple compact Hausdorff topologies on the same set that are non-homeomorphic

[balmy nexus]

yeah

still it's pretty cool
a maximal topology in Comp is not necessarily hausdorff I think and neither is a minimal Haus topology compact.

but both at once? interesting to think... probably not but still

[quartz horizon]

Yeah it is cool

It also hints at how compact Hausdorff spaces are secretly “algebraic”

Because the first isomorphism theorem works for them

[cosmic mirage]

huh thats cool

[balmy nexus]

? something something qoutient maps work well on LCH spaces?

i dimly remember

[quartz horizon]

Specifically, the category of compact Hausdorff spaces is monadic over Set

Indeed, you can view them as algebras for the ultrafilter monad, iirc

[cosmic mirage]

i am going to pretend a topological space is a CW complex. then a compact hausdorff space is a finite complex. this is a vague question but is there some way in which i should then view infinite complexes in this language?

[quartz horizon]

Well if you try to run the first isomorphism theorem for general topological spaces, you get that

Every continuous map can be decomposed as a quotient map, followed by a continuous bijection, followed by a subspace inclusion

[balmy nexus]

like I know simplicial complexes don't know CW yet. you may be looking for LCH which is locally compact hausdorff

[quartz horizon]

But the issue is that continuous bijections need not be homeomorphisms

However if you do this for compact Hausdorff spaces, you’re fine

[balmy nexus]

ah i see

[quartz horizon]

This works cause a quotient of a compact space is compact

[balmy nexus]

you sure you can break it down to a quotient? they are so damn fiddly in Top

[quartz horizon]

And a subspace of a Hausdorff space is Hausdorff

[cosmic mirage]

i say pretend but as a homotopy theorist i actually mean "this is my definition of a space"

[balmy nexus]

yeah checks out

[deft crow]

yes this is it

and f isn' continuous

it;s driving me up th wall

[balmy nexus]

try to find a function f:R->R with Q as it's set of discontinuities

[deft crow]

I've got that already

Thomae function

[balmy nexus]

yep, so the set is not closed

[deft crow]

But I'm being told to show that it is...

[balmy nexus]

hmm that doesn't work actually

[deft crow]

must it be a mistake?

[balmy nexus]

A_n would be R

[deft crow]

I've shown that the set of points where f is continuous is the intersection of the complements of An

[balmy nexus]

interesting so maybe it is true...

let's look at a point in A_n^c

[deft crow]

But I can't show it's closed

[deft crow]

I've tried this but I can't seem to find a contradiction assuming it isn;t open

[balmy nexus]

they have a neighborhood where all points are close under f. So all those points are also in the complement, so the complement is open.

[deft crow]

not sufficient as the closeness is quite strict

take two points in the given neighbourhood

[balmy nexus]

let me be precise then

ah ok i see what you mean

[deft crow]

yeah the triangular inequality gives us a painful 2

[quartz horizon]

Ooh, ok

So this is supposed to be true for an arbitrary function f?

[deft crow]

yes

[quartz horizon]

Interesting

Yeah, then my gut reaction is to show the complement of A_n is open

[deft crow]

right but I've been trying for an hour now and I haven;t succeeded

keep in mind this is the first question on my exam paper...

[quartz horizon]

Damn

[balmy nexus]

try for a point in the closure?

[deft crow]

Haven't tried that yet...

and the sequential characterisation seemed messy

[balmy nexus]

oof

The Ans are getting smaller as n gets larger

in order to prove that A_n^c is open, look at A_2n. Then that should get rid of that painful 2

[quartz horizon]

Ok I’m going to go back to trying to find a counterexample

[balmy nexus]

you don't trust my idea 🙁

[quartz horizon]

Alright I have an idea for a counterexample

Define a function from R to R as follows

At 0 it’s equal to 0

If x is rational and nonzero, it’s 0.1

If x is irrational, it’s -0.1

[balmy nexus]

A_n would be R/empty depending on n

[quartz horizon]

Hang on

Isn’t 0 in the complement of A_8?

Or have I misunderstood

[balmy nexus]

oh i see

[quartz horizon]

However, I don’t think there’s any ball around 0 contained in the complement of A_8

So 0 isn’t an interior point of the complement of A_8, meaning A_8 cannot be closed

[balmy nexus]

yes this works

[quartz horizon]

In fact my suspicion is that A_8 is precisely the complement of {0}

[balmy nexus]

maybe the theorem is true for all largen enough n

[quartz horizon]

That I’m not sure of

[balmy nexus]

but your idea extends for something for all n, you can do a sum of R's with each R engineered to be annoying for a specific n

[quartz horizon]

Yeah a disjoint union of Rs should work

[balmy nexus]

yippie we have destroyed this question

[quartz horizon]

[balmy nexus]

Thats what you get for innovating in exams 🤣

[quartz horizon]

I think there is some result that like

The set of continuity points of a function is G_delta or smth

[deft crow]

So it's not true?

Sorry si has to briefly check out

[quartz horizon]

But you’d need to modify the definition of A_n a little

I think that’s what the question is going for

As stated, the intended result is false

[deft crow]

Right, that's a mistake on the exam then 😅

Thanks a lot

[quartz horizon]

I think the following adjustment might work, but I haven’t thought about this carefully:

$$A_n={x\in E,\ \forall V\in\mathcal{V}(x),\ \exists y, z \in V,\ d(f(y),f(z))\geq\frac{1}{n+1}}$$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

I’ve seen something vaguely similar to this when proving Lebesgue’s criterion for Riemann integrability

Since part of the proof involves showing the set of discontinuities of any function is a F_sigma set, and then showing each of the closed sets in the union has measure 0 (meaning that the set of discontinuities has measure 0)

[kind marlin]

this is equivalent to saying that all compact Hausdorff topologies on a space are either equal or non comparable, right

I think that’s the very first exercise munkres gives for the compactness intro chapter

[crisp lintel]

however interestingly there exist maximal-compact spaces that aren't Hausdorff and minimal-hausdorff spaces that aren't compact

[kind marlin]

i think that makes sense

[prime elbow]

What is maximal compact space?

[quartz horizon]

I guess so?

[kind marlin]

it didnt come out of nowhere 😭 thats how the exercise was stated

the statements feel strange in my head though, it's straightforward to show that a finer/coarser compact Hausdorff space is equivalent via an identity map, but the statement that adding an open set to a compact Hausdorff topology makes it noncompact feels more difficult to directly show/construct

[opaque scroll]

A subset of a compact hausdorff space is compact iff it is closed. So if you add a new open then it's complement won't be compact (as it wasn't closed before).

[crisp lintel]

compact space such that any stronger topology on the same set is not compact

[kind marlin]

that makes sense

mm is there a nice way to interpret “removing” an open set from a topology that gives a similar conclusion for the Hausdorff end

[opaque scroll]

I mean, basically the same argument works:

A coarser topology is still compact, so if it was hausdorff subspaces would be compact iff they are closed.

You've taken a set that was closed and made it not closed, but it's still compact

[kind marlin]

oh yeah that makes sense

i kind of stopped thinking about that lol, i was sort of wondering how you would construct a topology that's coarser, is there a process you can follow? adding a set feels clear since you can just make a subbasis and take finite intersections and arbitrary unions, but is there a way to construct a topology that is coarser than a given one, doesn't contain a specific set, and is finer than all other topologies that are coarser than the given and doesn't contain that set?

being finer than all others isnt really a requirement or anything but it sounds convenient

[opaque scroll]

I mean, you can order the coarser topologies not containing the set, then pick a maximal one by Zorn, but I doubt there is a finest one.

Like take {x, y, z} with the discrete topology and say we want to make {x, y} not open.

We could take the topology generated by {{x}, {z}} or we could take the one generated by {{y}, {z}}.

[kind marlin]

okay that makes sense, maximal sounds convenient enough

ty!

[wide kayak]

If X is a set, a topology on X is a certain collection of subsets satisfying some axioms, so it’s a subset of P(X) (the discrete topology is all of P(X)). Thus it’s a point of P(P(X)). Put a topology on P(P(X)); what is the fundamental group of P(P(X)) based at the discrete topology of X?

[rancid umbra]

-# r u okay?

[wide kayak]

I’ve been feeling a little funny ever since I fell into a chronosynclastic infundibulum

[rancid umbra]

i feel funny a little after reading chronosynclastic infundibulum

[unreal stratus]

"put a topology" makes this an almost meaningless question

[rancid umbra]

yea

[unreal stratus]

Take any topology on the set and then transfer it along a bijection to transfer the discrete topology to any other point

[wide kayak]

a good topology

[iron bolt]

"take a set of cardinality 2^2^|X| and put a topology on it. what is its fundamental group?"

^the question, basically

I'm not aware of any good topologies on the space of all topologies on a set, actually. maybe there are some with some nice properties, but I'm not aware of there being any canonical one

[rancid umbra]

i want to do something with continuous functions

[warped helm]

munkres has some strange writing at times

"here's a useful theorem"
"ok here's two non examples" (no examples of the theorem provided)

[fallen canyon]

@unreal stratus technically the q asked is just underspecified but not totally meaningless, they just need to suggest topolising how, coarse, continuous, etc

[wide kayak]

it was definitely a half baked “wowee” shower thought kind of thing, but it would be super interesting if there’s actually something to it

[balmy nexus]

Well, we do have a order on the set of topologies, although it's not a linear order.

we can have a topology generated by {x| a<x<b} for some comparable a,b

[topaz surge]

Hello everybody, I'm doing exercises on point set topology, im wondering whether this would be an acceptable proof for the closure of $A=(\sqrt{2},3)$ in the lower limit topology. I claim that the closure of A is $[\sqrt{2},3)$ bcs any basis B containing $\sqrt{2}$ would intersect A and $[3,4)$ is a basis element that contains 3 but doesnt intersect A. And any element less than $\sqrt{2}$ would not be in the closure of A for the same reason as 3.

[gentle osprey]

How Smart?

[alpine nest]

Yep, that's pretty much it; as for elements other than sqrt(2) and 3 you can also just make the general observation that elements of A itself have to be in the closure, meanwhile (since the lower limit topology is finer than the standard topology), the closure has to be a subset of the standard closure (so can't contain any points outside [sqrt(2),3]), which basically only leaves you with two points to consider.

(and your arguments for those two points are correct, you might be a bit more detailed with explaining why sqrt(2) is in the closure, but also the argument as you posted might be enough, that's probably up for debate)

[topaz surge]

Oh ok, thank you very much

[fickle tendon]

its a closure operator on P(S), right? that makes the poset of topologies on S a complete lattice at least

not sure about the properties - cant imagine it being algebraic lol

[sly geyser]

what is this demonic ahh sphere

[queen prism]

round

[sturdy fox]

one of my friends noticed "if you puncture a sphere it becomes a disk with 0 holes, so technically a sphere has -1 (1d) holes", is this something legit?

like is there some kind of formula/theorem thatd somehow make "negative" holes quantifiable and meaningful?

[sly geyser]

you can technically add a point p to R^2 such that every open set inside R^2 U {p} containing p must be a union of open sets in R^2 and sets of the form {d((x,y), (0,0)) > r} U {p} for some constant r

and get a topology on R^2 U {p} that is homeomorphic to the 2-sphere

thing is

there are multiple ways to add a point like this

this one corresponding to the missing point of the sphere isn't the only way

[sturdy fox]

you mean ones that arent manifolds?

[sturdy fox]

i think i figured out that its all about the euler characteristic,
χ(S^2) = 1 - 0 + 1
and puncturing turns it into R^2
χ(R^2) = 1 - 0 + 0
so χ goes down
and if you also define h = 1 - χ you get
h(S^2) = -1
h(R^2) = 0

h going up

and this h aligns with the more mainstream view of holes

[gritty widget]

manifold-ness is a topological invariant, so it’s still a manifold.

[sturdy fox]

well im asking that to see if compactifications into other well-behaved spaces are a thing, or would we have to abandon it being hausdorff etc

[balmy nexus]

in complex analysis there is 🙂

you can count zeros/poles

I think all of the compact 2-manifolds have that property you describe. You can poke a hole in a torus and the fundamental group would not change.

[opaque scroll]

The fundamental group goes from Z^2 to the free group on two generators, no?

[dawn frigate]

Do you know an example where the collection {A alpha} is countable and each set A alpha is closed but f is not continuous

[opaque scroll]

What is f?

Is f like a function defined on the union and continuous on each Aalpha?

If so you can take for example Q and have Aalpha just be individual points

[dawn frigate]

I already did induction proof for part a

For number 9

[balmy nexus]

oh right

never mind then...

[dawn frigate]

Thanks got it

[brittle wren]

today i was looking for an nlab reference for lindelof number property and got this instead

it is fascinating to me that people seem to think football is more popular than topology

[sturdy fox]

hmm wouldnt that be regular holes anyway?

also yeah the fundamental group changes, but the 1st homology group does not (i think?), the 2nd group does change which is why the euler characteristic also goes down

havent checked this enough to be sure

[fallen canyon]

@sturdy fox you're totally right a quantifyable measure would be the betti # or euler char

[sturdy fox]

yeah that makes sense now

[hidden crag]

What is fascinating about this

[alpine nest]

That anyone enjoys football

[cosmic mirage]

i have/am a counstructive disproof of this statement

and also prefer topology

[alpine nest]

Fascinating

[cosmic mirage]

wait no no I misread the message LOL

I read it as anyone likes football, and I do not

[umbral hamlet]

Let $p: X \to Y$ be a quotient map. Prove that if $p^{-1}(y)$ is connected for all $y \in Y$ and $Y$ is connected, then $X$ is connected

[gentle osprey]

hiidostuff

[umbral hamlet]

I thought i had a proof but im pretty sure its not right

[rancid umbra]

-# there is a really nice proof of this using chain connectedness, not sure how helpful that is for you though

[umbral hamlet]

I think i did prove chain connectedness

[rancid umbra]

oh, you know about it?

[umbral hamlet]

Like sequences of connected spaces that are sequentially intersecting has a connected union

Idk if thats what ur referring to

[rancid umbra]

ah, that is not quite what i am referring to