point-set-topology

[warped helm]

then as part of the problem prove that the two are equivalent

[novel stump]

before you ask, its my professors own notes that we're using as a textbook

[tender halo]

i dont think you have to, its like piecewise closed

and therefore closed

i guess you still need to prove that its piecewise closed and i dont think you can easily cheat there hmm

[warped helm]

you're saying to show like the part of the graph where x > 0 is closed, etc. and then take a finite union of closed sets?

[novel stump]

i was gonna say that the graph = R , and since Y is a topo space then R is closed

[tender halo]

yea

[warped helm]

the graph is not R

that's the range

the graph is specifically the set ({ ( x, f(x)) : x \in \mathbb{R}})

[gentle osprey]

josemom2

[warped helm]

yea that's the part that seems annoying to me

it seems plausible to show that you can fit open balls into the complement of the graph, but seems pretty involved geometrically

er i guess open rectangles would work too

actually not too bad just very annoying grunt work

[tender halo]

oh you can do hmm

you can cover the area of an arc with two closed rectangles

so that x > 0 is an embedding into a closed subspace

and therefore closed

[warped helm]

ye, i guess the difficulty is conserved by just having to check various details of any approach

[novel stump]

i dont think we need to exact;y prove it

we just need to explain why it is

[plush folio]

can't you just use the theorem mentioned here, applied to R^+ and R^- separately? Then the graph is the union of the three closed components and therefore closed

[warm kettle]

You could also define g(x, y) = xy. It's a continuous function and the preimage of {1} is exactly the set {(x, 1/x) for x != 0}. Then add the single point at zero (which forms a closed set) and it's done, I think

At least assuming the Euclidean topology

[warped helm]

dang these are all great solutions

[novel stump]

i don't understand any of them lmao

sorry for leaving again i had a topology midterm that i probably bombed LMAOOOO

[plush folio]

Thingoln's solution uses the fact that if f is continuous and S is closed then the preimage f^-1(S) is closed. Have you covered that property?

[novel stump]

we have used that

we really dont need to be this detailed about it though

[plush folio]

So the idea for all of these solutions is that we look at f on x > 0, x < 0 and x = 0 separately. Since 1/x is continuous on the positive or negative reals separately, we can use these properties

[warped helm]

is your class not proof based ?

[novel stump]

it is but this problem isnt

like specifically the example parts

[plush folio]

Tbh, I don't know how to do it without being this detailed. It's either this or just vague handwaving as I see it

[novel stump]

oh i just saw this, this seems like it would work

i have one last example that i need help on if thats okay with yall? i'm a bit confused about visualizing nonhausdorff spaces

i apologize for being unclear before

[plush folio]

Visualizing non-Hausdorff spaces is hard in general, I don't think anyone has any great tips there

[warped helm]

just have some examples

[plush folio]

Generally a space is Hausdorff if it has an abundance of open sets, so if you're looking for a counter example, you should look at a space with few open sets

[novel stump]

hm

[warped helm]

e.g X in the indiscrete topology

[novel stump]

true

[warped helm]

well X with at least two elements

[novel stump]

mhm

i just dont really know how to draw the picture too

like we have two topological spaces and we want to find two continuous maps such that f(x)!=g(x)

[warped helm]

try to look at where you use hausdorfness in your proof of (a)

[plush folio]

What do you know about indiscrete spaces? Like, what kind of maps into it are continuous?

[warped helm]

what can you not do without the assumption that Y is hausdorff

[novel stump]

i know nothing about indiscrete spaces lol

oh wow active, the first time i got that here in the years ive been here

[warped helm]

do you know the topology that defines it though?

{empty, X}

[novel stump]

yes

[warped helm]

if a map X -> Y is continuous then the preimages of opens in Y have to be open in X

or the preimages of closeds has to be closed

f^{-1}(U) either ought to be empty or the whole space in either case

[novel stump]

yeyeye

[warped helm]

this narrows down your class of continuous functions quite a bit

[novel stump]

i can only think of like. empty to empty or the whole space mapped to the other whole space

[plush folio]

You can take an arbitrary function f to the indiscrete space Y and check manually whether it is continuous

[warped helm]

well, the singleton sets {y} of Y can be closed

[novel stump]

how are we going to get like, the dense subset?

[plush folio]

Since you know what the open sets of Y are, it should be easy to check if f is continuous

[novel stump]

im so confused LMAOOOO

[plush folio]

What's the definition of continuity?

[novel stump]

for a topological space a function is continuous if for every open set V in the codomain, its preimage in the domain is also open

so we need two of those

[plush folio]

Yep, now what are the open sets in the indiscrete space Y?

[novel stump]

empty set and Y

[plush folio]

I want you to establish some properties of indiscrete spaces first, the problem becomes very easy once you've done this

[warped helm]

also i now realize i had the roles of the domain and codomain mixed up which made this slightly more confusing

[plush folio]

Yep, and what are their preimages under f?

[novel stump]

also open in X

idk if x is an indiscrete space or not

[plush folio]

It doesn't matter the topology of X. You can compute the preimages explicitly btw - what is f^-1(Ø) for example?

[novel stump]

the empty set?

[plush folio]

Yep

And the other preimage?

[novel stump]

like the preimage is the empty set

[novel stump]

no

[plush folio]

Yeah, I mean f^-1(Ø) is the empty set

[novel stump]

ye

[plush folio]

Dunno what you're saying no to, I think you misunderstood

What is the preimage of the other open set in Y?

[novel stump]

oh i did i misread i thought you said "any other preimage?

thats my bad

[plush folio]

Oh I see, no worries

[novel stump]

its all of X

[plush folio]

Yep, which is always open

[novel stump]

yep

[plush folio]

So the conclusion is that f is continuous, right?

[novel stump]

yep

[plush folio]

That is, every function f : X -> Y is continuous if Y is indiscrete, no matter the topology of X

There is a space with the dual property, every map out of it is continuous, I'll let you figure out what it is on your own

[novel stump]

that makes sense

dual property?

[plush folio]

Yeah, it vaguely just means "reverse the arrows" etc. There's a more rigorous meaning to it, but it's not important

[novel stump]

hrm

[plush folio]

Hmm, so I guess you want to find an appropriate X for your counter example now. Any suggestions?

[novel stump]

i was thinking just any X would work

maybe like X=R

[plush folio]

Hmm... The dense subsets there are a little too dense for my taste 😅 I'm sure it works, but I can't think of an example rn

I would pick a much smaller space, so it's easier to think

[novel stump]

maybe like X=[0,1]?

[plush folio]

Yep, good choice 👍 with what topology?

[novel stump]

can i say the usual topology

or do we need a specific one

[plush folio]

Oh wait, I thought you meant X = {0, 1}. I don't like [0, 1], it's still too big

[novel stump]

oh then we'll go X={0,1}

[plush folio]

With a finite space it's much easier to see what the dense subsets are etc.

What topology on {0, 1}?

[novel stump]

indiscrete maybe?

[plush folio]

Good, now what are the dense subsets of {0, 1} in the indiscrete topology?

[novel stump]

uhm

{0,1}?

im guessing i apologize

[plush folio]

Yeah, that's one of them, but remember we want to find two maps that agree on a dense subset and disagree on the rest

Do you remember the definition of density?

[novel stump]

that the closure equals the whole space

[plush folio]

Yep, and do you know how to find the closure?

[novel stump]

intersection of all the closed sets

[plush folio]

That's one way, another is that the closure of S is the smallest closed set containing S

[novel stump]

true

[plush folio]

So now it should be possible to enumerate the dense subsets of {0, 1}

[warped helm]

ig for a concrete example, compute the closure of {0} in X @novel stump

certainly its not closed since its complement {1} is not open in X

[novel stump]

would it be {{0},{1}}?

[plush folio]

Yeah, I guess you don't need to enumerate them, but just find one dense set other than X itself

[plush folio]

Exactly, plus X of course

[warped helm]

(and X itself)

[novel stump]

ah

[plush folio]

So now you should have all the ingredients to construct your counter example

[novel stump]

lovely :')

[plush folio]

Let me know if you need more guidance! But take a look at the problem and make sure you know what you want to construct, and what properties it must have etc. It's good to have it all in front of you

[strong lantern]

The clearest example for me is the line with the double 0. You can take two real lines and identify the every point except for zero. Then you have two different ceros 0, 0' whose open sets containing one always intersect the open sets containing the other.

[novel stump]

i can only think of like, let f send X to X i think??

and then we want G to be mapped to another open thing such that f(x) != g(x)

[novel stump]

why would that satisfy everything tho?

or well be a good counterexample

[strong lantern]

for a non-hausdorff spaces. But the graph of the identity is non-close

[novel stump]

hmm interesting

[plush folio]

what's the dense subset you're working with?

[novel stump]

X

[plush folio]

X as a dense subset of X?

[novel stump]

i thought X is a dense subset of X

[novel stump]

.\

[plush folio]

yeah it is

but you want f and g to be not equal. If f and g agree on X, then they are by definition equal

[novel stump]

oh

then i guess {{0},{1}} works?

[plush folio]

hm, that's a set of sets, not actually a subset of X

do you mean one of them?

[novel stump]

i thought that that was a dense subset of X

[plush folio]

{{0}, {1}} can't be a dense subset of X, because it's not a subset at all

[novel stump]

im so confused

did you not say it was previously?

sorry that sounded snarky

i mean i thought you said it was previously

[plush folio]

{0} and {1} are dense subsets, so I thought you meant the collection of those

[novel stump]

ohhhhhhhh

[plush folio]

but you see how {{0}, {1}} is not a subset of {0, 1}, right?

[novel stump]

not really but i kinda just want to be done w this assignment LOL

[plush folio]

hmm, I get that, but this is kinda important to learn (not just for topology). What are the subsets of {0, 1}?

[novel stump]

0 and 1

[plush folio]

(btw, set theory is often one of the hardest parts of point-set topology for beginners, so don't feel bad if that's what you struggle with!)

[novel stump]

oh i got a C- in set theory i know i struggle with it a lot unfortunately

[plush folio]

is 0 an element or a subset of {0, 1}? It's important to know the difference between element and subset

[novel stump]

i would think 0 is a subset of {0,1}

maybe {0} is the right notation?

[plush folio]

yes, {0} is a set, while 0 is an element. The elements are what is listed between the curly brackets: a, b, c are elements of {a, b, c}

[novel stump]

sorry

[plush folio]

no worries

so {0} is a subset of {0, 1}, since every element of {0} is an element of {0, 1}

does that make sense?

[novel stump]

yes

same thing happens with {1}

[plush folio]

yep

[novel stump]

so im not sure what our functions would be then... like i guess we can do {1} but that just gets mapped to the emptyset i think?

im not quite sure what im doing

[plush folio]

hmm, I feel like you're struggling a bit with the concept of a function? {1} cannot be mapped to the empty set, every function must output something for any input

[novel stump]

then idk how im supposed to make a continuous function from {0,1} to {X,\empty}

[plush folio]

wait, why is the codomain {X, \empty}?

[novel stump]

is that not what Y is

its an indiscrete space

so

[plush folio]

it's the topology of an indiscrete space, but it's not its actual elements

[novel stump]

so then what are the elements?

im so sorry 😭

[plush folio]

don't be sorry, this is the place for learning

[novel stump]

i cant believe im a senior struggling w this lmao

anyway enough moping

[plush folio]

So you understand the idea of a topology? We have a set of elements X, but to turn it into a topological space, we must also have a collection of subsets of X, call it T, satisfying certain rules. So to give a space X the indiscrete topology means that we give it the set of subsets T = {Ø, X}. This means that the only open sets of X are Ø are X itself

[novel stump]

yeye i understand all of that

i just didnt know what to use for our codomain

bc i thought we had our codomain to be those two sets

[plush folio]

so when I've been talking about an indiscrete space Y, I haven't been talking about a particular space, just any space with that topology

[novel stump]

oh

oh my lord im so sorry

[plush folio]

the topology is {Ø, Y}, not the space itself

[novel stump]

yeyeye

[plush folio]

so I haven't actually decided what the codomain should be, sorry if that was unclear 😅

[novel stump]

its okay!

[plush folio]

but a good choice is again to just pick the smallest space that is not completely trivial (and give it the indiscrete topology)

[novel stump]

so can we also use {0,1}?

[plush folio]

yep, excellent choice

[gaunt linden]

May I suggest {2,3} instead, just to keep things minimally confusing?

[novel stump]

sure!

[plush folio]

yeah, good idea

[novel stump]

i have 30 minutes until i need to leave

[plush folio]

so we want to find two functions f, g : {0, 1} -> {2, 3} both with the indiscrete topology, that agree on a dense subset, but are not equal

no worries, it won't take that long I think

[novel stump]

heck yeah

[novel stump]

yes

[plush folio]

and f and g must be continuous, but as we already established, they automatically are

[novel stump]

yes

[plush folio]

so what's the next step? Any thoughts?

[novel stump]

we need to map a dense subset to different values in Y

[plush folio]

hmm, we want them to agree on a dense subset, which means they should take the same values on that subset

[novel stump]

yes

so f : x + 1, g : x-1

or smth to that effect

maybe x + 2 and x + 1

[plush folio]

is x + 1 a well defined function from {0, 1} to {2, 3}?

[novel stump]

rats

no its not

[plush folio]

there are just 2 elements in the domain, so it may be easier to define each value instead of using a formula

[novel stump]

x+2 isnt either

i dont think

[novel stump]

so 1 -> 3 and 0 -> 2

[plush folio]

okay, and the other function?

[novel stump]

1 -> 2, 0 -> 3

[plush folio]

do they agree on a dense subset?

[novel stump]

i dont think so

i really dont know what im doing im sorry

[plush folio]

how can you make sure they do? (this is where it's useful to know exactly what the dense subsets of X are)

[novel stump]

well {1} is a dense subset and gets mapped to both 2 and 3

[plush folio]

or a different question: how can you make two functions that agree on any non-empty subset? Like, what does it mean for two functions to agree on a subset?

[novel stump]

i do not know what it means, i think it means theyre equal?

[plush folio]

yeah, they take on the same values on that subset, ie. f(x) = g(x) for each x in that subset

[novel stump]

i see

okay now im even more confused

bc i thought we just wanted to find two maps such that they dont agre

or do we want it to agree but f isnt g

[plush folio]

yeah, a counterexample here means that everything in the premise is true (except Y being Hausdorff), but the conclusion isn't. So they must agree on a dense subset, but f != g

[novel stump]

okay

so then we want {1} to be mapped to {2} but f isnt g

right?

[plush folio]

yeah basically

[novel stump]

hm

idk how we would do that

[plush folio]

what's stopping you from trying a few maps? There are only 2 elements in the domain and codomain, so there's not much to try

[novel stump]

i mean like. map 1 to 2 and 0 to 3. map 1 to 3 and 0 to 2, those are the only two maps i can think of

[plush folio]

there are more maps than that between {0, 1} and {2, 3}

[novel stump]

? really?

maybe like 1 maps to 2 and 3?

i didnt think that would work

[plush folio]

1 maps to both 2 and 3? That's not a function

[novel stump]

then im out of ideas lol

[plush folio]

you don't have to map to different elements in the codomain. We're not restricting ourselves only to bijections

[novel stump]

then i guess {0,1} mapped to 2 would work

[gaunt linden]

(Is X also indiscrete here?)

[novel stump]

yes

[plush folio]

Yeah, if you mean f(0) = 2 and f(1) = 2

[novel stump]

i do

[plush folio]

yeah, I wasn't sure if you were mixing up sets and elements, because it should be clear that mapping 1 to {2, 3} isn't a function

so anyways, you have one function, what should the other be?

[novel stump]

maybe 1 -> 2 and 0 -> 3?

[plush folio]

okay, do they agree on a dense subset?

[novel stump]

yeah bc 1 still gets mapped to 2

[plush folio]

nice 👍 I think that's it, it's clear that f is not equal to g. Does all this make sense?

[novel stump]

mostly, ty :]

[plush folio]

no problem, feel free to ask if anything is unclear

[brisk coral]

I think I sort of get what this question is going for? but I don't really understand why they define e like that, and then go on to say that phi is where x_j=0 for almost all j??

surely the closure of phi includes (1,0,0,0,0,0,...), (1,0,0,0,0,...) ... which isn't a null sequence?

[rancid umbra]

e_j is the sequence that is 1 at position j and 0 otherwise

phi is the set of all sequences such that x_j = 0 for all but finitely many indicies j, or in other words, it consists of sequences which are eventually 0

i think null sequence means a sequence which converges to 0

[brisk coral]

ah okay I think I was confused at the notation, I thought that x_j=0 meant that for a particular sequence, the jth element is 0, but now I'm thinking it means that the jth sequence is entirely made up of zeros

thank you!

[rancid umbra]

no, it’s not the jth sequence

x_j is the j-th coordinate or j-position of the sequence x

like

maybe the e_j is confusing you?

[queen prism]

"almost all" is evil terminology

[rancid umbra]

i think it’s pretty widely used

why do u think it’s evil

[queen prism]

it looks too much like measure-theoretic "almost all" and it's not equivalent over N iirc

[rancid umbra]

almost all in this context means almost all indicies

idk like

[brisk coral]

ah! okay so the idea here is that if I go far out enough into the sequence, I go past all of the non-zero elements, I'll have only zeros and so it converges to 0 ?

[rancid umbra]

i would use the notation e^j btw

so like

(e^j)_k = 1 if j = k, 0 otherwise

[brisk coral]

yeah okay, I guess its called e because it is like a basis?

[rancid umbra]

yea sort of

it’s not exactly a basis

not really super important to the problem

[brisk coral]

yeah it generally just confused me because it defines it, and then doesn't mention it for the actual problem 😭

yes

[rancid umbra]

oh shoot

okay

then yea

idk

why they even mention e_j

[brisk coral]

if you are interested btw its from Meise and Vogt "Introduction to Functional Analysis"

[rancid umbra]

here is my solution: ||let y be in the closure of phi and 𝜀 > 0. there is some x in phi with |y - x| < 𝜀 or in other words, there is a natural number N such that for all j > N, we have |y_j - x_j| = |y_j - 0| = |y_j| < 𝜀||

[brisk coral]

oh ty! and thank you for spoilering, i'll have a try and then compare 🙂

[balmy briar]

If we have a metric space X and a subset A of X such that we have this strictly decreasing chain $$ A \supset A' \supset A'' \supset ...$$ What is the maximum length of the sequence? Here ' is taking the limit points of the set.

[gentle osprey]

XDStar

[balmy briar]

I know it can be 1 or 2.. but i can't think of anything else. As soon as we get a discrete set in the sequence we then get the empty set after

I think also taking the set {1/n + 1/k | n,k in N} U {0} makes it 3? but I'm not sure

[wide kayak]

I think that idea could be extended to happen more times, and make the sequence arbitrarily long

[balmy briar]

that was my thought too but is this even true?

[plush folio]

Are you using a different definition of limit point than the one I'm familiar with? Like, aren't the limit points of for example [0, 1] exactly [0, 1]? Maybe you're thinking of isolated points

[wide kayak]

your example looks true to me, the first set of limit points is all the 1/n, and then those have limit point 0, right?

[balmy briar]

yeah that's the idea

[balmy briar]

its the standard definition i believe.. every nbhd of x in X intersects A in a point other than itself

[plush folio]

yeah, then [0, 1]' = [0, 1], which gives an infinite sequence AFAICT

[balmy briar]

it has to be strictly decreasing

[plush folio]

okay [0, 1] union arbitrary many isolated points?

like [0, 1] U {2} U {3} ...

[balmy briar]

the limit points will be [0,1] then the sequence is no longer strictly decreasing

it's constant

so doesn't work I think

[wide kayak]

actually what I don’t quite see about your 1/n + 1/k with 0 idea is why we don’t get all the limit points “in one step”

[balmy briar]

I don't understand

[plush folio]

hmm, aren't the only points in A' \ A the isolated points of A? Which disappear after one application of '

[wide kayak]

oh, nevermind. I see the first taking of limit points gives 0 and all the 1/n, and then the second just gives 0

so yeah, how about 0 with 1/n + 1/k + 1/j, etc?

[gaunt linden]

You can make as long a finite sequence as you want by starting from the right end of the sequence and making A^(n) a singleton. Then for each k, make A^(k) from A^(k+1) by adding a sequence converging to each isolated point in A^(k+1). With just a small bit of care that will make the derived set of the result the A^(k+1) you already have.

[balmy briar]

isn't the inclusion the other way around? We have A strict superset of A'

[balmy briar]

that makes sense yeah

could it be infinite?

[plush folio]

yeah, but I'm probably wrong, it's too late at night for topology rn

[alpine nest]

||Indeed it can: https://en.wikipedia.org/wiki/Derived_set_(mathematics)#Cantor–Bendixson_rank||

[gaunt linden]

Now take the first set in each of those finite sequences, squeeze it into (0,1) by something like arctan(x)/pi, and put all of these squeezed first sets next to each other. That will give you a sequence of length omega.

[alpine nest]

I'm always a bit hazy on the historical details, but I believe this is one of the earliest uses of ordinals/transfinite induction

[gaunt linden]

It would be neat if Wikipedia had stated how large the rank can be (or what is known about that question). Obviously it cannot be an ordinal with larger cardinality than continuum, but other than that?

[balmy briar]

I'm still trying to understand this 😅 but thank you @alpine nest @gaunt linden !

[opaque scroll]

Why is that obvious?

[alpine nest]

Apparently for Polish spaces it has to be countable: https://math.stackexchange.com/questions/4937051/the-countability-of-the-cantor-bendixson-rank-in-polish-spaces

[gaunt linden]

At least for a strictly decreasing sequence it's obvious, and I don't think a point can enter the sequence more than once.

[opaque scroll]

I guess I still don't see exactly why it's obvious

[gaunt linden]

Ah, that may be because I had somehow convinced myself we were working in R rather than an arbitrary metric space. 🤣

[opaque scroll]

Seems vaugely plausible, but idk what the argument would be exactly

[alpine nest]

My gut tells me that if you start with an ordinal space, you can get the rank as high as you want

[gaunt linden]

The sequence is decreasing, at least past the first step. Up until the rank is reached, each set in the sequence will be missing some point that was in all of the previous sets contain at least one point that will be gone from the next. Once there's been more than continuum steps, there can't be any more points left to remove.

[opaque scroll]

Assuming the set was continuum in size to begin with you mean

[gaunt linden]

Yes, but as I said, this was under the assumption that we were working in R.

[opaque scroll]

Yes then I agree it is obvious

Still though, being a metric space is also quite restrictive

[tender halo]

thats just the meat of the proof of cantor bendixon really

well one of the proofs

the cool one

(condensation points are not cool)

[tender halo]

yep

[tender halo]

the problem is that ordinal spaces are not metric

past omega 1 (included)

[tender halo]

there is this

[gritty widget]

What’s degree?

[tender halo]

oh yeah probably shouldve included that

uhh

[crisp lintel]

Let X be locally compact and Hausdorff and R an equivalence relation on X that is (1) a closed subset of the product X\times X and (2) each equivalence class is compact. Is the quotient map X -> X/R necessarily closed?

This has been annoying me I think it should be true but I can't quite see it

[kind marlin]

does this work as a counterexample? Let $X = [0, 1] \times [0, \infty)$, and identify each $(x, 0)$ point with $(x, \frac{1}{x})$ for $0 < x \leq 1$. \

$X$ is locally compact Hausdorff, and I'm pretty sure the equivalence relation is closed since its the union of a diagonal set in a hausdorff space and a "graph" set that's parametrized by 1 value. i think the only problematic area is if we have (0,0) as the first point in a product, but i think then bc 1/x is unbounded near 0, you can always force your neighborhoods to exclude the relation curve. this might be wrong so maybe check that :( \

now let $S = {(x, 1/x) : 0 < x \leq 1}$. by quotient map defn, $\pi(S)$ is closed iff $\pi^{-1} \pi(S)$ is closed. but the preimage of $\pi(S)$ is the union of (0, 1] with the (x, 1/x) curve, and this is not closed since (0,0) is in the closure

wow horrendous

[gentle osprey]

snowflake

[crisp lintel]

Yeah I think you're right

[crisp lintel]

Here's why I'm interested: Let X be locally compact Hausdorff and consider the space $C_0(X)$ of continuous complex valued functions on X vanishing at infinity. A subalgebra $A\subseteq C_0(X)$ is nondegenerate if there is no point of X where every function in A simultaneously vanishes. Given such a subalgebra, you can form the equivalence relation $R={(x.y):f(x)=f(y)\forall f\in A}$. If A is a closed *-subalgebra, then A should be equal to the set of all functions that obey the relation R. This has a very short and nice proof if the quotient $X/R$ is locally compact and Hausdorff. You can show that this holds if the quotient map $q:X\to X/R$ is proper (i.e., it's a closed map and each equivalence class is compact)

[gentle osprey]

Blake

[crisp lintel]

I can prove that if X is compact, the result holds. I also know by Gelfand duality that if A is isomorphic to $C_0(Y)$ and since it is nondegenerate the inclusion of A into $C_0(X)$ must come from a proper nap $X\to Y$. However, it's not entirely clear to me that this map has to be surjective. I also don't want to directly appeal to Gelfand duality.

[gentle osprey]

Blake

[crisp lintel]

Ok actually the specific result I mentioned you can actually show using the one-point compactification, however it doesn't quite give you a clean correspondence between "nice" equivalence relations and *-subalgebras which is more what I want

[crisp lintel]

I guess a precise formulation is the following: Which equivalence relations on a locally compact Hausdorff space X can arise from a nondegenerate *-subalgebra of C_0(X)?

[waxen urchin]

for $\mathfrak{B}$ to be a basis for topology $\tau$ on $X$, its required that for any element x of $X$ that's in the intersection of two elements $B_1, B_2 \in \mathfrak{B}$ there's another element $B_3 \in \mathfrak{B}$, such that $B_3 \subsetneq B_1 \cap B_2$ and $x \in B_3$. Well, if it's true, then we have another intersection of two elements of $\tau$, namely $B_3 \cap B_2$ (or $B_3 \cap B_1$) such that $x \in B_3 \cap B_2$. But then since its a basis we require that there's another element of topology, namely, $B_4$, such that $x \in B_4$ and $B_4 \subsetneq B_3 \cap B_2$ etc. So for any element $x \in X$ and a basis $\mathfrak{B}$ of topology $\tau$ in $X$, we always have a shrinking element $B_i \in \tau$, leading us to thinking that any element $x \in X$ ultimately is contained (among many other $B_n$) in $B_i$ such that $B_i={x}$. Is that true? The whole idea though is based on that we always have a proper subset contained in the intersection of two others, meaning it's always shrinking

[gentle osprey]

nezhivoy

[waxen urchin]

another distracted question on topic, I've seen a couple of different definitions of a topology basis, this one above (+ another condition) and that B is a topology if for any subset A of X we have that A is a union of elements of the chosen topology

i guess at some point we prove that they're equivalent? or are they like used in different settings or something?

[hidden abyss]

B_3 isn't required to be a proper subset

[tender halo]

even if it is, what you say need not be true

the property you talk about is true in the standard basis in R^n for example

except R itself i guess

[waxen urchin]

but i didnt use anything except the definition of a basis i guess

[tender halo]

definition of the basis just says subset not proper subset

[waxen urchin]

yeah it's just that notation is different and I've seen this definition somewhere used without the low bar

[waxen urchin]

yep. understood

[tender halo]

well i guess there are different standard bases, i mean idk open balls with rational coefficients

or something

without the low bar just means subset too

not proper subset

unless specified otherwise, only crossed bar means proper subset

is the usual convention

[hidden abyss]

Consider the open sets $B_n = (-1/n,1/n)$ in $\mathbb{R}$. Then $B_{i+1}$ is always strictly contained in $B_i$ and all contain 0, but no $B_i$ is equal to ${0}$

[gentle osprey]

Jussari

[waxen urchin]

btw following this definition we have that for example $A \subseteq X, A={x}$ (singleton), then we have that $A$ is a union of the elements of our topology. So we have that for every such A there's a single-element subset of topology. I might be missing something here

[gentle osprey]

nezhivoy

[tender halo]

actually this is not true, if there two balls with the same center and different radii then the property is not true

i guess maybe hmm

[hidden abyss]

You must have misread, only open subsets need (and can!) be unions of basis elements

[waxen urchin]

oh, seems true

[opaque scroll]

I'm not sure I'm following what you're saying, but if {x} is an open set, then would need to be one of the basis sets yes

[waxen urchin]

that's a nice example, thanks

[waxen urchin]

nah i misread indeed, thinking that any subset of X has this property

[tender halo]

oh i know how to make a base for R^2 with the needed property, take a point in each integer square and take open balls with radius 2 with centers in those points, then take half-integer squares, and pick balls with radius 1 with centers in (unpicked) points, etc etc

i think that works

[hidden abyss]

There's two ways to look at bases. Either you're given a topological space $(X, \tau)$ and want to find a nice description for their opens, i.e. a basis of $\tau$. Or you're given some \textbf{set} X and some collection of subsets ${ B_i}$ of $X$ and you want to find a (as small as possible) topology $\tau$ on $X$ s.t. all the $B_i$s are open in $(X, \tau)$, i.e. the topology generated by the $B_i$s

[gentle osprey]

Jussari

[waxen urchin]

that's kind of analogous to a basis in linalg

trying to find a subset that describes the whole vector space conveniently

[hidden abyss]

The first case corresponds to this definition: you choose some collection of opens and verify that every open is a union of them

[quartz horizon]

In a sense a basis for a topology is akin to a spanning set for a vector space

Afaik there’s not an analog of “linear independence” for topologies

[waxen urchin]

seems like that

[hidden abyss]

In the second case, if the B_is fulfill this definition, then you get a topology on X by declaring a set U to be open iff it can be written as a union of B_is.

[hidden abyss]

It's a good exercise to verify that this really is a topology on X

[waxen urchin]

you mean that these unions of B_is form a topology on X?

[hidden abyss]

Another important exercise is to check the following: Suppose you have a topological $(X, \tau)$, and some basis $\mathcal{B}={B_i}$ of $\tau$ (in the sense of the second definition you posted). Then $\mathcal{B}$ is a basis for a topology on $X$ (in the sense of the first definition you posted), and the topology generated by $\mathcal{B}$ is $\tau$

[gentle osprey]

Jussari

[hidden abyss]

Yeah

[tender halo]

disjointness, kinda

maximal disjoint families of open/basic subsets are important in pointset

[hidden abyss]

Most spaces dont have pairwise disjoint bases though

[quartz horizon]

I can’t think of bases that consist of disjoint subsets

[waxen urchin]

yeah that's what i was seeking. does that work the other way around? to show that it holds "iff"

[quartz horizon]

Unless you’re working in, say, a discrete space

In which case singletons suffice

[hidden abyss]

Yeah! You can check that $\mathcal{B}$ is a basis for the topology it generates!

[gentle osprey]

Jussari

[tender halo]

well not bases, but if you think about like boolean algebras associated with spaces, there independence makes sense

and if you pull it back you would be looking at disjoint families of open (or regular open) sets

[waxen urchin]

pullback hmhmhm

btw are there any nice books on general topology? I've seen a million good books on analysis, abstract alg and linalg, and they all seem to explain the things in the same way, and then looking at topology books.. first, there aren't as many of them, second, they're all really different in how they teach the subject. maybe any recommendations?

[quartz horizon]

If you’re into cat theory you could check out Tai Danae-Bradley’s “a categorical approach to topology”

[tender halo]

i mean, munkres?

is peoples go to really

i would argue it doesnt teach most interesting parts of pointset, but an average person is not interested in pointset

if you want to actually know what the subject is about, i would read Kuratowski

Engelking is a good source but very dense

[waxen urchin]

unfortunately im not yet good at cat theory, and they are serious about it. the books heavy on colimits, actively uses cat theory theorems and stuff

[tender halo]

i think the best way, if you have enough mathematical maturity, is to pick up engelking and not slack on the exercises

and kuratowski to kind of understand the why of it all

[waxen urchin]

by mathematical maturity what do you mean exactly?

btw i read some bourbaki on topology and they seemed nice

i know its an encyclopedia but still

[tender halo]

the ability to reconstruct the point of definitions and theorems from their statements and proofs

[radiant stone]

I like bredon topology

[tender halo]

its geometry not topology

[radiant stone]

Lmao

[waxen urchin]

like.. to understand what's going on basically?

[tender halo]

more or less yea

understanding the mathematical substance of a given construction, proof, or method

[hoary breach]

theres a wiki for it https://en.wikipedia.org/wiki/Mathematical_maturity

[waxen urchin]

lol

[waxen urchin]

they define it as "fearlessness in the face of symbols". people are so scared of symbols lolol

[wide kayak]

I think Kelley is a widely used reference

I only skimmed part of it back in the day so can’t really speak to how nice it is

[waxen urchin]

I've found General Topology by Stephen Willard and it seems nice. I'll look Kelley's up

[alpine nest]

Yep, Willard is solid

[tender halo]

if you will happen to want more advanced stuff but are not ready to read the handbook, nagata's modern general topology is a good choice

alternatively, everyone should read gillman

[waxen urchin]

eh, i dunno. from what I've read on topology, it seems that among mathematicians the opinion on topology is rather the same and is that topology is more or less just a language for formal description of the notion of continuity and as a science is already long ago “done” (meaning that there are no active researchers and discovery in the field)

so having this impression i think I'll just have to master the “language”

[tender halo]

i mean of course there are reasearchers and discovery, but yeah its no particularly alive

[warped helm]

this doesn’t sound accurate for the entire field of topology

it sounds more like of a description of point-set topology

[tender halo]

we are talking about pointset yea

[viscid blade]

what they rsrching then ?

[tender halo]

stuff? idk

pointset is a very haphazard field, because "topological space" is a very weak condition

so there is a lot of stuff to research, but relatively few big research questions

C_p theory is somewhat active

[kind marlin]

https://www.tiktok.com/t/ZTrY9EfQ5/

idk if this is novel but this seemed cool

[pearl valve]

so i understand this proof, but dont undertsnad this picture circled in red

could anyone help me understand whats happening here

is it just showing X and the homotopy H between h and the constant point? what are the lines extending out from the point

[opaque scroll]

Yes, and the lines are the same lines drawn in the other two pictures.

[waxen urchin]

what's the book?

[vagrant stirrup]

check out Viro et al. "Elementary Topology. Textbook in Problems" -- it's very nice, basically they give you a path by smartly scheduling lots of problems for you to prove, and then eventually you will be able to prove some non-trivial theorems, all by yourself: https://www.pdmi.ras.ru/~olegviro/topoman/index.html

it's also free, and even Allen Hatcher likes and recommends it

I've solved many problems from that book and I can attest that it was a fun way to study general topology (unfortunately I got distracted and stopped at some point)

[waxen urchin]

I've already been using it for some time and i like it too. But i saw the homotophy stuff on the screen above and wanted to ask specifically for a book that contains homotopy theory. And Viro isn't usually recommended for algtop afaik

[vagrant stirrup]

mm, but your question was explicitly about general topology

not about algebraic topology

[waxen urchin]

i was inquiring about homotopy theory

[vagrant stirrup]

not sure I can parse that from that message above: "btw are there any nice books on general topology?"

[waxen urchin]

oh man, sorry, i thought you were responding to the message above

[vagrant stirrup]

but anyway 🙂

[waxen urchin]

this one

[vagrant stirrup]

no, that's for Jack to answer

[waxen urchin]

i should've paid more attention lolol

thanks for Viro though, it really is a good source

[tender halo]

i like rotman for algtop

[vagrant stirrup]

I've bought myself Armstrong "Basic Topology", it looks like it's an undergrad book that focuses not only on General Topology, but also on Algebraic Topology, and does it quite early, so I am looking forward to that, but haven't started it yet

[pearl valve]

Munkres

[queen prism]

oh boy

I remember someone ranting hard about this one

[vagrant stirrup]

what did they say? 🙂

[queen prism]

I think it had to do with unconventional notation/terminology

I haven't read it though so idk

[vagrant stirrup]

ah, fine, terminology is not so important to me 🙂

I am reading books that define f . g = g(f(x)) and still survive

[queen prism]

superhuman

[tender halo]

thats normal

functions should be applied on the right

[viscid blade]

right so shouldn't it be f(g(x)) ?

[tender halo]

it should be ((x)f)g

[queen prism]

but morally correct

in a different timeline that could've been

[vagrant stirrup]

yeah, it makes sense: f goes first in that "f . g" combo, so should be applied first

and you are right @tender halo , that book also uses "xf" to denote "f(x)", so it's consistent

[tender halo]

life if functions composed from left to right

[vagrant stirrup]

so f . g = xfg

[cursive stirrup]

are you a native speaker of English? Some of mathematicians prefer to write topology via fancy descriptions and intuitive methods like they’re writing novels . they prefer to let u imagine how an object is folded/bended/torn up/ twisted…instead of mathematically claiming it, maybe hard for non native or those do not like intuitively thinking.

[tender halo]

you can find this rationalization and many more in my new book "why hatcher is not an abysmal F tier textbook"

[vagrant stirrup]

I am not, but I think I am quite proficient 🙂 at least I know and can imagine what "folded", "bended" or "twisted" means

in general I am not as good with geometry as with algebra/analysis though, but that has nothing to do with my command of English language...

[cursive stirrup]

that’s fine , maybe i should rephrase,I just wonder if you accept this style .For me i like set-theoretically discussion like we treating a topological space purely as some sets meet some conditions instead there are really some vivid examples in geometry.

it is up to you though, for those who care about knots in our real world this is really important

[vagrant stirrup]

I don't know, depends on a book and concrete examples. In general, illustrations of some kind often provide additional insights (Pugh wrote in his real analysis book: beware of books without pictures)

[cursive stirrup]

i d like to read it

[young stone]

"properly discontinuous action" is the worst name ever

[iron bolt]

there's also two inequivalent definitions of it and I never know which one people mean

one requires every point to have a neighbourhood U such that U is disjoint from gU for all g ≠ 1, the other just requires it to be disjoint for all but finitely many g

I would call the latter "properly discontinuous" and the former "properly discontinuous and free", but some people don't

[gritty widget]

topology & groupoids covers some basic homotopy stuff, but idk if that's what you want

[cloud kindle]

Do you have a hotkey bound for topology and groupoids

[gritty widget]

what a great idea

[iron bolt]

I have nlab added as a search engine into my browser

yet don't use it because it's so slow that using a standard search engine and clicking on the first nlab link is faster anyway

[queen prism]

https://ncatlab.org/Please_dont_download_the_nLab.html I was gonna say

[queen prism]

tbh I should start doing this tho

[rancid umbra]

why is the nlab set up so poorly 😭

[cloud kindle]

I was reading the compact Hausdorff page earlier and there’s a bit that just says something like “you already know what a compact Hausdorff space is so let’s have some fun” and further “this is probably also equivalent but I haven’t checked”

Unserious website

[rancid umbra]

i tried contributing and this is the way that you have to get access to the pages

[queen prism]

yea the authors can be funny

[queen prism]

oh that's a different tier of funny

[rancid umbra]

either i kept getting the wrong answer or they set it up wrong

i never ended up contributing

[queen prism]

did you try emailing them?

idk

[rancid umbra]

yea, but this was months ago. i might try doing it again, but it was such a pain to just get to this point

the nlab won’t have a native dark mode for the foreseeable future

[queen prism]

join the light side

[rancid umbra]

never

it hurts my eyes 😭

[viscid blade]

bro needs to activate light mode

[rancid umbra]

aghhh the contrast

[viscid blade]

https://tenor.com/view/ron-swanson-parks-and-rec-its-so-beautiful-gif-15644547

[queen prism]

I was considering doing that

[iron bolt]

do you mean contributing to the backend? because to edit articles you don't even need an account, let alone a hard to obtain password

[iron bolt]

ah, I see.

[iron bolt]

regarding this though, the nlab can actually be incredibly helpful at getting ideas across and finding you references

you just need to keep some healthy skepticism about what you read there - if it seems like it's too strong of a claim and might be missing some assumption, there's at least some chance it is

it's more like getting a peek at the lab notes someone took just for themselves than a carefulles curated encyclopedia - there might be some missing pages, a lot of stubs and some mistakes, but it's also a really condensed way of seeing how others (at least a certain kind of people) think about a topic

[strong lantern]

It feels like breaking the fourth wall

[novel acorn]

you see shit like this and realize people use that website to learn math

nlab has actually ruined a generation

[strong lantern]

For me it has been excelent for studying infinite categories. but, it is clear that it is not meant to be use for learning stuff for the first time.

[cloud kindle]

I actually tend to find the more niche the idea the better nlab is

I don’t know if that’s because the people writing take it more seriously or if the niche shit I need to find on nlab is already so fucked that I don’t notice, but still

[strong lantern]

Probably the second option

[unreal stratus]

I have been misinformed by nlab a few times I think lol

[cloud kindle]

To be fair, given that it’s like open source, pretty advanced pretty niche stuff, the fact there’s so much coverage of maths, and that it’s generally at least ok (if brain rot) is quite impressive

Like beyond stuff that’s straight up just recent research ideas, I don’t think there’s anything I’ve not been able to find on nlab

[strong lantern]

There's a reason why people quote the stacks project but not nlab

[cloud kindle]

Stacks is even scarier

[iron bolt]

just interrupting here to quickly post my favourite nlab screenshots I've collected over the years

and the differential equation article of course, I don't have a screenshot at hand

[strong lantern]

science of logic is a funny one

[cloud kindle]

I should really start a folder of nlab moments

[novel acorn]

I'm like 90% sure lawvere and morava had the same coke dealer

[unreal stratus]

Lol

[pearl valve]

This comment here is referring to an unpointed homotopy right?

for in this specific case, if we had a pointed (not necessarily strong) deformation retract, it would automatically be a strong deformation retract, which is impossible?

[sly geyser]

so just a question

disconnected means, vaguely, that you can split the set into two disjoint open sets

but, of course, in actuality, it's the open sets as interpreted by the subset S

so S ∩ A and S ∩ B

now is what is needed the disjointness of S ∩ A and S ∩ B or the disjointness of A and B, and is there a practical difference?

[tender halo]

wdym vaguely

that is what it means

a disconnected space is the one that is a union of two disjoint open sets

[sly geyser]

ok so it's the former

[tender halo]

disconnected subset is a subset that is disconnected as a subspace

[sly geyser]

because S is treated as the space, ok got it

is it equivalent to the latter?

[tender halo]

the two are not equivalent no

take uhh a normal space that has a non normal subspace

or something

that will break it

basically if the closures of S \cap A and S \cap B are not disjoint it can be bad

[sly geyser]

?

[tender halo]

what

hmm can that happen actually

i feel like it can and im too lazy to think about it

[sly geyser]

what I was thinking about

was if there was a case such that

1. A and B were closed
2. A and B had no non-intersecting open supersets (so A and B being an example of normality failing basically)
3. A and B had a pair of open supersets such that the intersection of those supersets was disjoint from the union of A and B

[tender halo]

basically the question is if there is a disconnected subspace all of which supspaces are connected

well its not the question but uhh if what i say is true

then its an example of non equivalence

and it sounds true

[opaque scroll]

That's totally possible yeah. Consider a space X = {a, b, c} with open sets
{a, c}, {b, c}, {c}, Ø and X.

Then A = {a} and B = {b} are closed, and there are no pairs open neighbourhoods of them that don't intersect. But AuB is discrete as a subspace

[warped helm]

hilarious

[unreal stratus]

Probably correct though lol

The start of pointset top is half definitions and getting used to them

Which can be hard lol

[gloomy pivot]

I love that little blurb. He doubles down and says "Yeah, the proofs in this book are probably unnecessary before this point, you could do them yourself."

[pallid lake]

Which book is that from?

[naive fossil]

Is the complement of the unit ball in an infinite-dimensional normed vector space path-connected?

[rancid umbra]

the unit sphere S should be path connected. your subspace is the image of the continuous map (1,oo) x S —> V given by (t,v) |-> tv

[pearl valve]

Are there ever any cases where E is simply connected, so the lifting correspondence is a bijection, but it is not a homomorphism? If not, why? and if not, then why not state the theorem that if E is simply connected then phi is an isomorphism?

My guess is that, you can always make it a group homomorphism by defining a group operation on pinv(b_0) based on the bijection with pi1 since pi1 is always a group. Is this correct?

[strong lantern]

There's no reason to expect a natural group structure on the fibers. Like if you give it one, there's no reason for it to mean anything in relation to the space E.
The case R->S^1 is highly specific if you think about.

[lament steppe]

Been thinking about this problem from John Lee's book and Im not sure I even understand the question. I think he's asking for me to show that the union of any collection of bases from across all open sets of the open cover can be an element to a new basis that lives on X...

But wouldnt a counter example be that if we have 2 open sets in the cover that overlap we could potentially have an intersection of the elements of this new basis that isnt another basis element (or contain an element of the basis)?

[alpine nest]

If your 2 open sets of the cover overlap, their intersection is an open subset of both, so it will be attainable using their bases

[cosmic mirage]

im having some trouble understanding what your question is, but i can tell you how i am interpreting lee's question. you want to show that you can construct any open set in X by taking unions of any of the open sets coming from the basis of any of the U's

[alpine nest]

One key thing to keep in mind would be that since you're considering open sets, their subsets are open in the relative topology if and only if they're open in the original topology

[strong lantern]

This is a fancy way of saying that a set is open in X if it its intersection with each element in the cover is open.

[pearl valve]

Are there any spaces where the identity map is not a covering map?

[strong lantern]

No, use the definition of covering map to prove it.

[lament steppe]

Thanks all, I think you've cleared up all the things!

[opaque zodiac]

nLab is so fucking annoying jfc

By volume, it's like 50% 'did you know William Lawvere was the greatest fucking thinker of the 20th century and he totally didn't lose his first academic position because he punched his department chair or anything like that'

[quartz horizon]

Seems like it’d be good to have a change of guard, so to speak

[night orchid]

Hello!
I am working on this exercise but I'm not sure how to proceed. May I get a hint, please?

[tender halo]

for any net N and point x you can make a net N' that is convergent iff N is convergent to x

[crisp lintel]

use closed sets

[tender halo]

yea and then you can use closed sets, the main difficulty of the exercise is that the nets can happen converge to different points

[night orchid]

I mean I had in mind showing that nets converge to the same point

[tender halo]

well, if you can do it that will also solve it sure

[opaque scroll]

The problem is that the limit of a net isn't unique. So you have to be a little clever to ensure that your net can only converge to one specific point

And I guess a hint could be that ||a space is T1 iff a constant net only ever converges to one value||

[crisp lintel]

Ah yeah it doesn't specify that they converge to the same point that's a bit trickier

[strong lantern]

Lawvere fixed point theorem was the steam engine of the 20 th century

[umbral hamlet]

in munkres i have to prove that the "figure eight space" X and the "theta space" Y have maps f: X to Y and g: Y to X that are homotopy inverses of each other

isnt this just by virtue of the two spaces being homotopy equivalent since theyre deformation retracts of the doubly punctured plane

actually wait this might go in alg top

whatever

i do also have to describe f and g

[rancid umbra]

i would just describe them in words

[umbral hamlet]

so a map from Y to X that im thinking of is just mapping the points on the y-axis line to the single point connecting the two copies of S1 in the figure eight space

[umbral hamlet]

fair enough

hopefully the one i just gave is correct

not fully sure about a map from the figure eight space to the theta space

i guess i could just morph the connecting point plus some segments of the S1s to the vertical line

[rancid umbra]

for 8 to 𝛳, to start, the point connecting the two circles should go to the center of 𝛳

[umbral hamlet]

yeah

maybe i should draw a picture of what i was thinking

[rancid umbra]

yea

colors r helpful too

[umbral hamlet]

red stuff gets sent to the vertical line

in the most obvious way ig

[rancid umbra]

yea

[umbral hamlet]

i guess in words i can just claim that theyre constructed in such a way that its very clear that their compositions are homotopic to the respective identity maps of each space

at least to me its obvious that thats possible

[rancid umbra]

your picture was very clear tbh. maybe add a color-coded picture of theta as well

to show where the red and blue end up

[umbral hamlet]

sounds good

first time im adding pictures onto my math homework

thatll be fun

[rancid umbra]

post-rigour

[umbral hamlet]

hell yeah

seems thats how topology works

at least from my profs perspective

she does some significant topology-related research but her way of understanding the subject was so foreign at first

but then i read hatcher and it is pretty much just that

[rancid umbra]

lmao yea. just freestylin fr

[umbral hamlet]

yeah

i mean its in a good way though because its not too handwavy to the point where its unclear

finally get to let loose with some visual proofs

bc they actually kinda work to some degree

[umbral hamlet]

im confused by this tip, its still true that the S2 wedge S2 has trivial fundamental group

maybe they are just pointing out that this result isnt trivial or something idk

it follows pretty immediately from van kampen tho

[unreal stratus]

Yeah

I think the point was like they want to be careful to prove that the "obvious" answer is correct lol

[unreal stratus]

But agreed

[umbral hamlet]

alright cool

just needed a sanity check

[strong lantern]

There's probably a mess up top spaces which doesn't satifices the van kampen theorem, which at this point, I wouldn't worry about it.

[umbral hamlet]

yeah i mean im in the very very beginning of algebraic topology

pretty much every space weve seen is very easy to break apart into overlapping open sets and thus can just be van kampened to death

[strong lantern]

I bet it is two topologist comb glue together in some weird way

[strong lantern]

this is the case for almost all the objects of interested in algebraic topology. Or at the very least, this is kind of true for manifolds and finite CW-complexes

[umbral hamlet]

how can i know exactly which properties homeomorphisms preserve?

i suppose itd just be the ones that can be expressed in purely topological terms

[kind marlin]

thats a good way of thinking abt it

for instance homeomorphisms dont preserve completeness of metrics bc that isn't purely topological

actually maybe thats a bad example bc you do get complete metrizability...

[iron bolt]

I think it's a good example

boundedness is another classical one - whether a metric space is bounded depends very much on its metric, and can't really be recovered from its topology

I mean, under some circumstances you can guarantee that any metric you could possibly put on a space has to be bounded, such as when the space is compact. but on the other hand, any metric space can also be given a bounded metric that induces the same topology

in other words, every unbounded metric space is homeomorphic to a bounded one

[queen prism]

these are usually good candidates, but there are some surprising topological invariants outside of these, too

like de rham cohomology groups, though surely there's a simpler example

[iron bolt]

oh, good point

hyperbolic volume in knot theory is another example - but not a simpler one

[umbral hamlet]

ok am i crazy

i have to find the fundamental group of S1 \times B2

where B2 is the unit ball in R3 including the boundary

so even though this is a solid torus, i know that the fund. group of a product is the product of fund. groups

S1's fund group is Z

B2's is trivial right

[unreal stratus]

indeed the space is contractible

may as well say S1 \times B2 \simeq S^1 gg lol

[umbral hamlet]

i see

i was worried since its closed but doesnt matter i guess

idek why i was worried about that

[unreal stratus]

Ye

[lament steppe]

You're saying that you essentially have to assume that the basis elements of the individual U's need to be open in X, right? ( Like it's a necessary assumption because you define open sets in a given topology and it's not something you prove in this case, correct?)

[cosmic mirage]

that is fine though, an open set in U with the subspace topology is the intersection of some open set V in X with U, but U was already open so that this intersection is open in X

[umbral hamlet]

okay idk how im struggling with this so much

i have to determine the fundamental group of the subset of R2 whose points have l2 norms greater than 1

im tempted to say its the same as S1 because the points that arent there just form a big point

but idk

[tender halo]

well its like homeomorphic to S1 x R

[umbral hamlet]

why

i guess i can kinda see why

[tender halo]

polar coordinates

[umbral hamlet]

i see

so S1 x R gets mapped to by plotting the angle and magnitude of each point i suppose

[tender halo]

its obviously a continuous bijection and its easy to see that its, say, open

[gaunt linden]

Alternatively, R^2 minus the disc deformation retracts to a slightly larger circle.

[prisma garnet]

wait, what's the example mentioned?

[umbral hamlet]

Wdym

Its S2 wedge S2 I think

[gaunt linden]

I think Darq's question is, what's an example of X and Y that are each simply connected, but where X wedge Y isn't.
(I'm curious about that too).

[gaunt linden]

Hmm, perhaps it could just be weasel-wording that "the union of two spaces having a point in common" is not necessarily a wedge sum. For example we could consider S^1 to the the union of (-pi,0] and [0,pi] parts, which are each simply connected and have one point in common.

[opaque scroll]

There's this funny cone of the topological earrings
https://math.stackexchange.com/a/1790520/306319

[gaunt linden]

Ah, devious! Thanks.

[umbral hamlet]

Couldnt the lack of an example be shown by van kampen

Oh nvm jagr showed a counterexample

[opaque scroll]

There is an opennes condition in van kampen that can fail

If you're not locally contractible around the base point

[umbral hamlet]

Ah I see

Interesting

[quasi drift]

Yo

Whats a filter intuitevly

Im having a hard time understanding it

Im.having a hard time with it and i cnat comtinue without it

[gaunt linden]

There are several possible intuitions. One I like is that it's a generalization of "neighborhood" in a metric space:
If p is some point in the metric space X, then { A subseteq X | A is a neighborhood of p } is a filter, and "filter" in general means a collection of subsets that behaves "more or less" like the collection of neighborhoods of a point.

Some people like to say that each filter specifies a sense in which a subset can have "enough points" for some particular purpose.

[prisma garnet]

dude that's trippyyy

I love it

[quartz horizon]

i like to think of filters as generalising "eventual truth"

or axiomatising the statement that "p(x) is true for sufficiently [BLANK] x"

[tender halo]

filter is a map that helps you find stuff

map as like a literal map

[quartz horizon]

specifically, if you have a filter on a set X

you can view it as a map $\forall^F : (X \to {0, 1}) \to {0, 1}$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

which sends a predicate $p$ on $X$ to the truth value "$p^{-1}(1) \in F$"

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

the reason why you can denote it as $\forall^F$ is that it behaves a little like a "generalised forall" quantifier

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

specifically, the filter rules correspond to the following:

If $\forall^F p$ is true and $p \implies q$, then $\forall^F q$ is true

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

and if $\forall^F p$ and $\forall^F q$ are individually true, then $\forall^F (p \land q)$ is also true

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

for example, you can consider the cofinite filter on $\mathbb{N}$ consisting of all subsets whose complement is finite

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

this models predicates that are "eventually true", or "true for sufficiently large n"

[quartz horizon]

so if p is true for sufficiently large n, and p implies q, then q is also true for sufficiently large n

[tender halo]

A \in F means that what you are searching for is at A, which is why F is upward closed and finite-intersection closed: if the "object" is in A it is also in any supset of A, and if it is both in A and B then it is also in the intersection of A and B

[quartz horizon]

and if p is true for sufficiently large n, as well as q, then at some point, both p and q are simultaneously true for sufficiently large n

[quartz horizon]

i guess this sounds a little like principal filters?

one way to create a filter on a set $X$ is to select some $x_0 \in X$, and define $\forall^F : (X \to {0, 1}) \to {0, 1}$ by $p \mapsto p(x_0)$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[tender halo]

the object can be virtual

[quartz horizon]

mhm, true

in topological spaces, filters often take the form of "p is true sufficiently close to x"

and this x could either be an actual point of your space (in which case it's a neighbourhood filter)

or maybe a "virtual point" - for the cofinite filter on N, you can think of it as modelling "p is true sufficiently close to infinity"

[quartz horizon]

this kind of filter arises if you're working with discrete topological spaces

in which case, being true sufficiently close to x is the same as being true at x

indeed one cool thing that filters let you do is axiomatise a qualitative notion of "closeness", as opposed to the quantitative notion offered by a metric

in general, because of the sorites paradox, it doesn't really make sense to ask "is 3 close to 3.01" as a yes/no question

but you can make it quantitative, by saying they're within 0.2 of each other, for example

however, filters let you rigorously define what it means for a predicate on $\mathbb{R}$ to be true "sufficiently close to 3"

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

even though it doesn't make sense 'pointwise' to ask whether some $x \in \mathbb{R}$ is close to $3$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[tender halo]

here its more that the filter is a notion of directions, abstracted from the object(s) it is the directions towards

filters goes... somewhere

[quartz horizon]

that sounds more like a directed set

though i suppose every filter is a directed subset of P(X)

[gaunt linden]

Filters are directed sets under the superset-of relation.

[tender halo]

yea

[alpine nest]

Well, I'd say it does make sense to say that 3 is close to 3

[quartz horizon]

yes but reflexivity is essentially all you can get

and my statement was really about generic real numbers

[alpine nest]

Yes, of course, the strikethrough was intended to convey that I was being facetious

I do agree with your points (no pun intended) entirely

[quartz horizon]

ah ok, i couldn't quite tell whether you were just being facetious or intending to mock all the points i made

[alpine nest]

Definitely the former. Also I'd add that even the ostensibly "quantitative" metric only really tends to give you the "closeness" qualitatively, since virtually all definitions/theorems say something about points whose distance to x is "sufficiently small"

But whether "sufficiently small" means d < 1, d < 0.1, or d < 1e-100, you rarely know

[quartz horizon]

yes that's essentially the point of the neighbourhood filter of 3

you say $p(x)$ is true "sufficiently close to $3$" iff $\exists \delta > 0$ such that $\forall x \in (3 - \delta, 3 + \delta)$, $p(x)$ is true

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

but because of the existential you don't care which delta it is, which is what makes it qualitative

this ties a little into soft vs hard analysis

all the qualitative stuff is "soft", but sometimes it's useful to have an effective version of a theorem where you actually get bounds for how large you can make delta

like the effective version of the inverse function theorem

that'd be more "hard" analysis

[alpine nest]

Well yeah, and also the fact that a metric is a real-valued function, does let you bring the "quantitative" aspect into it in some way, I've lately been thinking of that in connection with the Urysohn lemma.

[quartz horizon]

mhm?

[alpine nest]

In a metric space you construct the Urysohn function directly and trivially, by giving a formula in terms of distance of point to set.

[quartz horizon]

yes yes ofc

this is part of the proof that metric spaces are T_6

[alpine nest]

In a general normal space you can still get the Urysohn function, but it's considerably more involved to do so

[quartz horizon]

mhm, it's a very cool construction using the sandwich formulation of normality and dyadic rationals

[alpine nest]

Yeah, so fundamentally you don't need the metric, but having immediate access to something real-valued helps you a lot

[quartz horizon]

i think that's why it's good to know both sides of analysis and not get stuck in either the soft or hard style

the soft "infinitary" style can make some statements significantly easier to state and prove

but it's also easy for things to feel a lot more "wishy-washy" without hard numbers

[dawn frigate]

If we define an equivalnce relation on X = R with the standard tooology by setting x sim u if x-y in Q give the familiar space to which X minus sim is homeomorphic

Idk how to prove this

Equivalence class should be [x]=x plus Q

Quotient map is R to R/sim

[opaque scroll]

Do you know how to describe the topology on a quotient space?

[dawn frigate]

Hmm 🤔 meaning saying its discrete, indiscrete etc?

[kind marlin]

what does X - sim mean?

[opaque scroll]

I mean given a space X and an equivalence relation ~ do you know what the open sets in the topology on X/~ is?

[opaque scroll]

They mean X/~

[kind marlin]

oh i see

[gaunt linden]

I think the question was, can you complete the definition: "In a quotient space X/~, the open sets are the sets such that ..."

[opaque scroll]

And spoiler warning, the next question I will ask you is what do you get if you apply the definition to your example

[dawn frigate]

Pi inverse of U is open in X

Okie

[opaque scroll]

Indeed. And it might even be easier to think about closed sets instead of open sets here. Because ||what happens if a closed set contains one of these equivalence classes x+Q||

[dawn frigate]

so indiscrete4 topology

[dawn frigate]

Is this right method to finding out what this is homeomorphic to

[proven kernel]

Yes it's the right approach

[novel stump]

i'm very confused on how to not be able to extract a finite subcover from this question, i understand that we have an open cover {U_i | i in I} in the lower limit topology, meaning theyre all unions of infinitely many intervals [a,b), but after that i'm so lost

i think we have to find an open cover such that we cant obtain a finite subcover, but i have no idea how to go about doing that

[queen prism]

start from the right end of the set [0, 1] and work your way toward the left end

it'll probably be easier if you restrict to working with disjoint open sets like [0, 0.1) and [0.1, 0.2), which is possible in this topology

[novel stump]

do we want like a sequence?

[queen prism]

probably works

[novel stump]

i dont see how we can get an open cover though bc it has to be infinite right?

maybe [n, n + 0.1) for n >= 0?

[warped helm]

open covers can be infinite

[warped helm]

this won't work

[novel stump]

not surprised

[warped helm]

the first set is [0, 0.1) and the second is [1, 1.1). you immediately miss most of [0,1]

[novel stump]

oh youre right

rats

[queen prism]

well start with a finite open cover and look to see if you can turn that into an infinite open cover, maybe by cutting up your open sets somehow

[novel stump]

i can think of an immediate finite open cover to maybe be [0,2)

[warped helm]

you don't need both end points to be different in your open cover

[novel stump]

?

wdym

[warped helm]

you can fix the left end point and build up your cover by going towards the right towards 1

so something consisting of things that look like [0,a)

[novel stump]

hm

[queen prism]

true that works

[novel stump]

well [0, 1/n) obviously doesnt work

[warped helm]

well because 1/n -> 0

[novel stump]

oh yeah

duh

omg

[warped helm]

but i'm sure you can think of a sequence that that lives in (0,1) that converges to 1

[novel stump]

[0, 1 - 1/n)?

[warped helm]

that's promising

verify the details

[novel stump]

it converges to [0,1] but it doesnt ever equals 1

but its still above it

so it contains [0,1]

[warped helm]

it does not converge to [0,1]

it is exactly [0,1)

1 isn't in any of the sets [0, 1 - 1/n) and so can't be in (\bigcup_n [0, 1 - 1/n) )

[novel stump]

rats

[gentle osprey]

josemom2

[novel stump]

so nvm

[warped helm]

well it's not a huge issue you can fix this immediately

(how?)

[novel stump]

i guess then we could try 1.1 - 1/n?

[warped helm]

how about adding [1, a) to your open cover

[novel stump]

oh

so we have [0, 1 - 1/n) u [1,a)?

[warped helm]

recall that covers just have to satisfy [
A \subset \bigcup_\alpha U_\alpha
]

[gentle osprey]

josemom2

[novel stump]

yeye

[warped helm]

yes

now you should check whether:

• it's a cover
• it contains a finite subcover

[novel stump]

well its clearly a cover since we have unioned a closed 1 on it

i would think it contains a subcover bc we can consider n = 10, then we just have [0, 1.1) which contains it and is finite

[warped helm]

uh no

[novel stump]

:')

[warped helm]

try showing that the collection { [0, 1 - 1/n) } actually covers [0,1) explicitly

the fact that 1 is on our cover is resolved by the fact that we threw in [1,a)

[novel stump]

i dont know how to do it explicitly other than it converges to [0, 1+epsilon) maybe? which contains [0,1)

[warped helm]

you can do it in cases, but the starting step is always to let x in [0,1] and show that it actually lives inside one of the sets of your cover.

• consider x in [0,1/2]
• consider x in (1/2, 1)
• consider x = 1

[novel stump]

i mean

• x in [0, 1/2] is clear
• x in (1/2, 1) seems pretty clear as well since 1 - 1/n < 1 for all n > 1

[warped helm]

< *

[novel stump]

but then that makes me think x = 1 is in there

wait omg

i'm thinking about this wrong

[warped helm]

sorry if this seems annoying, but i do recommend being able to explicitly produce a set in the collection that something in (1/2, 1) lives in

[novel stump]

i thought we were going from tight to left instead of left to right

[warped helm]

it's going to be important for when you check whether this collection has a finite cover

[novel stump]

its not annoying i just dont know what im doing

so my answers will be extremely ignorant

[warped helm]

it's ok dw, it's not as bad as you think. i always tell myself that it's just thinking really hard about numbers and quantities

let's say x = 0.75, which set(s) in our collection does this live in?

[novel stump]

(1/2,1)

[warped helm]

the thought process is something like:
we need some set [0, 1 - 1/n) where the right end point is bigger than 0.75, so 1 - 1/n > 0.75
we can certainly find a natural number n such that 1/n < 0.25, say n = 10 works. this gives 1 - 0.1 = 0.9 so we may say that 0.75 lives in [0,0.9) which is one of the sets in our collection.

[warped helm]

ah, they need to be sets in our collection that we're trying to check is a cover

so we need to be able to find some [0, 1 - 1/n) that works

[novel stump]

hm

[novel stump]

i don't understand this problem at all then LOL

[warped helm]

the reason i suggested checking the second case is because 1 - 1/n \geq 1/2 for all n \geq 1/2

[warped helm]

so, we're trying to show [0,1] is not compact in the lower limit topology. this means that we need to be able to find an infinite cover of [0,1] that doesn't contain any finite subcover, i.e if we take any finite collection from said cover, it's no longer a cover for the entire set

[novel stump]

ohh

i didn't know that last part

[warped helm]

something that should clue you into what the open cover should look like is the fact that we're in R_l and sets like [0,a) are open. it's true that regular open intervals (a,b) are open in R_l as well, but [0,1] is compact in the regular topology and so we might not be successful if our cover just consists of regular old open intervals

[novel stump]

R_l?

the lower limit topology?

[warped helm]

yes

[novel stump]

ohhhhh

[novel stump]

yeye i understand that

[warped helm]

sorry, it's commonly denoted (\mathbb{R}l, \mathbb{R}\ell, \mathbb{R}_L)

[gentle osprey]

josemom2

[novel stump]

ohhhhhh

[warped helm]

anyways, if 1/2 < x < 1, can we pick N such that 1 - 1/N > x ?

[novel stump]

yes

wait

no yea i think we can

[warped helm]

if you rearrange it, you get that 1 - x > 1/N

[novel stump]

yep

[warped helm]

now, is 1 - x positive?

[novel stump]

yes for all 0 \geq x < 1

[warped helm]

right, and 1/2 < x < 1 so that checks out

[novel stump]

yep

[warped helm]

the archimedean property says that some natural number N exists such that 1/N < 1 - x

so x lives in [0, 1 - 1/N)

[warped helm]

beyond just knowing the name of the property, it should be intuitive to expect that you can take some really large natural number N and have 1/N < e for any positive number e

[novel stump]

yeye i understand the Archimedean property

[night orchid]

Hello everyone! The other day I was asking about this problem, but I honestly don't know how to start.

[tender halo]

for a net N think about what happens if you insert an immediate successor to each element of the directed set and send it to a fixed point of the space

[opaque scroll]

So two things to note.

If T and T' are different then there is a set that is closed in one topology, but not the other.

A point is in the closure of a set if and only if there is a net in that set that converges to the point.

[night orchid]

So this would be the better approach to follow? T = T' iff cl_T(S) = cl_T'(S) for any subset S of X?

[opaque scroll]

Well you can do that if you want. But what I'm suggesting is assume T and T' are different and pick S such S is a closed set in T, but the closure is bigger in T' (or vice versa)

[quasi drift]

@gaunt linden @prisma garnet @quartz horizon @tender halo thank you guys for helping me understand the concept of a filter, btw im not even studying topology yet but my prof in real analysis 2 really wants us to understand functions as from a filter to a metric space so i was having a hard time intuitevly understanding filters

i just read what you guys said so im sorry for not responding yesterday

i was bussy studying and going to lectures

[queen prism]

wtf is your prof doing

[tender halo]

||bussy studying? tell me more||

[gaunt linden]

A point is in the closure of a set if and only if there is a net in that set that converges to the point.
Hmm, but it doesn't seem to be assumed in the problem that a net converges to the same point according to T and T', so some extra footwork will be needed.

[quasi drift]

i just want to start of with the fact that when we get into first year of uni we from the beggining start with realanalysis/calc in one and he starts of the first lecture with introduction to what is a filter but doesnt explain it but just putting it out there to confuse people, but i got him in 2nd year so now he thinks we already understand the core concepts

sorry for talkin to much out of context of the cnannel

[opaque scroll]

Indeed, and some tips have been given for that extra footwork.

But I think it's probably a good idea to take this problem one step at a time

I.e. first step, can we come up with any sort of net at all relevant to the problem?

[opaque scroll]

As bussy beaver has basically said already: ||If you take a net N and introduce an immediate successor to every element with value x, then the new net converges iff N converges to x||

[tender halo]

yea also we need to like prove that T' is also T1 at some point... i think

im unsure

[night orchid]

I'm not so sure I can show T' is also T1. It would be helpful, but not sure it can be done

[tender halo]

oh i guess we can have an alternating sequence of x1 x2

[opaque scroll]

It's not so hard, so could be a warm up I guess

[night orchid]

If a net N converges to x, and D is the directed set, then x_d, x, x_d, x, .. must converge to x

[tender halo]

that wont converge in a T1 space

[night orchid]

Kind of thinking as a sequence, but I think that is what bussy mentioned too

[gaunt linden]

The fact that we only have a T1 assumption on one of the sides leads me to expect that Jagr's approach will lead to a somewhat awkward case analysis.
I think I would aim for proving that if some net converges to p according to T' then it also converges to the same p according to T, and then use the T1 assumption to show that the two topologies agree on the "net converges to such and such point" relation, which then fixes what the topologies are.

[opaque scroll]

Yeah, so more formally you can do Dx{0<1}

[opaque scroll]

I mean, I don't know if the two cases T' is T1 and T' is not T1 is that akward, but you might be able to do it without

[gaunt linden]

Hmm, perhaps it's the same amount of work in the end.

[tender halo]

just make an alternating sequence of x1 x2 if x1 is in x2's closure in T', it doesnt converge in T but converges to x1 in T'

[ripe creek]

I'm going to homeomorph you.

[rancid umbra]

ur about to get homotoped making threats like that

[ripe creek]

I'm going to turn you into a disconnected topological space.

[prime elbow]

I don't get the hint, can anyone help me here?

[opaque scroll]

Like you don't understand what the hint means or you don't understand how it's helpful?

[prime elbow]

I don't get what it means

[tender halo]

they are asking you to consider functions whose graph is a broken line (polygonal chain, polyline, idk what name you use for those)

[opaque scroll]

So the graph is a subset of [0, 1]xR and then you should consider when the graph is just made up of straight lines between rational points

Such a graph will be determined by just which rational points

[uneven bronze]

I'm not super sure where to ask this. Suppose I have a family of (probability) measures $\mu_{\omega}$ for $\omega\in [0,1]$. These are measures on the Borel $\sigma$-algebra of $[0,1]$. Suppose further that \begin{equation}\mu_\omega\left(\bigcup_{n=1}^\infty A_n\right)=\sum_{n=1}^\infty \mu_\omega(A_n)\text{ for a.e. }\omega,\end{equation} and for all collection of disjoint sets $\mathcal{A}={A_1,A_2,\ldots}$. Then, for some $\omega=x_{\mathcal{A}}\in[0,1]$, we are allowed to redefine $\mu_\omega$ on $\bigcup_{n=1}^\infty A_n$ so it doesn't equal the RHS of (1), while still (1) being true a.e.\

My question is; is $\mathcal{A}\mapsto x_{\mathcal{A}}$ surjective (or even bijective)? In other words, if for each $\mathcal{A}\subset\mathcal{B}([0,1])$ we redefine $\mu_\omega$ at $\omega=x_{\mathcal{A}}$, we will eventually have redefined on each point in $[0,1]$?

[gentle osprey]

psie

[novel acorn]

Send this in #advanced-analysis probably

[tacit drift]

wikipedia says that a space is sequential iff it is a quotient of a metric space