point-set-topology

[forest pasture]

But if I ask you to prove something, it would be better to reformalize it

[woven trail]

since it kinda guides you into what the solution should look like by knowing what it shouldn't look like

[woven trail]

oh yeah

[forest pasture]

So you'd see where you can apply the rules you know

I mean like

All of first year abstract algebra for me will be to reword the questions with sets

[woven trail]

oh that makes sense

I mean, a big part in understanding a subject is understanding how it teaches you to formalize problems

like, a lot of problems could just be stated with too much words but what the tools give you is a way of formalizing it in an approachable way

[forest pasture]

Hmmmm

I don't know if it's appropriate to ask in topology, but

Wouldn't computer generated proofs be all about reformalization?

[woven trail]

why

[forest pasture]

Idk

[woven trail]

I think it's ok
once in a while some channel just starts a random ass discussion and it just stays there
ig we should only care if someone appears actually wanting to talk about topology

happened in #real-complex-analysis one time

[forest pasture]

Ok so to formalize it, I think any proof takes a logical state and known lemmas and transforms it using logical rules into another logical state

[woven trail]

oh right

[forest pasture]

The question is, in a computer proof, would a logical state that is dual to another logical state and they lead to dual proofs (I don't know if I'm using dual right, I mean the category theory thing)

Would they be represented in the same way?

If not, we'd clearly need to reformalize

[woven trail]

ym like, computer generated proofs supposedly would or do work with us reformalizing a problem in a way that the computer can apply stuff it "knows" to come to a conclusion?

[forest pasture]

I'm very sorry if I sound dumb I didn't start uni yet so I didn't study most of this

[woven trail]

I don't think so

[woven trail]

me neither B)

[forest pasture]

Highschool grads ftw

My uni starts tomorrow

[woven trail]

I didn't even finish highschool lmao

I'm thinking this has a strong metaphor potential with (non-modern) algebra

[forest pasture]

Gigachadette

I did watch some lectures on how Isabelle and similar programs work

[woven trail]

like, computers can only simplify algebra so easily because we have an amazing formalism for it to work with

(god thank viéte)

[forest pasture]

I have no clue what you're referring to

[woven trail]

ok I need to dinner rn I'll explain it later

[forest pasture]

I haven't added numbers by hand since 2023

[woven trail]

this isn't what I'm talking about

[forest pasture]

Wanna dm about it? I don't think it's appropriate to converse on it in topology

Alternatively there's chill

Or discussion

[civic verge]

Has anyone read Grothendieck's work on Topological vector spaces?

[next tinsel]

Given $G$ a topological group with identity $e$. Let $a \neq e \in G$, I want to show there exists a neighbourhood $V$ of $e$ such that $(V \cdot a) \cap V = \emptyset$. What I tried is that I firstly suppose for the contrary that for all neighbourhood V of e, we have $(V \cdot a) \cap V \neq \emptyset$, then $\exists x,y \in V, a = x^{-1}y$, that means for all neighbourhood $V$ of $e$, we have $a \in V^{-1}V$. In short, $\forall V$ containing e, we have $a \in V^{-1}V$. I want to somehow say, $\bigcap_{e \in V} V^{-1} V$ = ${e}$ probably using the T1 axiom. But I don't know how.

[gentle osprey]

lmwang4321

[red yoke]

When working with topological groups it's nice to keep the continuous map G×G → G in mind

Not just the translation maps

[tribal palm]

from Viro, that second paragraph is an interesting way to start off the book

Do not hurry to fall in love with it, do not let an imprinting happen.

[sonic crane]

Man like how do i show this map between quotient space and S1 x S1 is continuous

The open sets are just hard to write out

Its like R2 \ ~ -> S1 x S1

Equivalence relation is when p-q in Z x Z

[real star]

i think you can use a covering like the exponential map from R^2 -> S1 x S1, (x,y) -> e^(2pi * i * x),e^(2pi* i * y) and then that this is also well defined on the quotient

then something something universal property of the quotient

[unreal stratus]

You basically always want to use the "universal property" of quotients

Oh lol said above

But yeah like you write down the map from R^2 -> S^1 x S^1 which should be easy to check is continuous, and then check it respects the relation so that it automaticlaly factors through

[sonic crane]

Man like my prof barely covered these properties tho is the thing

Oh ok fine fine

[sonic crane]

Yes i think this one was covered

[sonic crane]

How can I show inverse map is continuous? S1 x S1 -> R2 \ ~

I guess its the same thing similar

[tribal palm]

my prof talked would always mention these universal properties whenever introducing a new construction, and many of his arguments would be in vaguely categorical language, but he never bothered to define a category

kiand what book are you following btw

[unreal stratus]

So your original map from R^2/~ -> S^1 x S^1 is a continuous bijection and you can show it is open by showing it is a local homeomorphism

[sonic crane]

my prof is using janich but i dont like it

i like munkres but actually rn we are doing a lot of top group stuff and munkres doesnt really cover that

[iron bolt]

no point in defining categories just to talk about specific universal constructions like that tbf

I think it makes sense to talk about universal properties early to get familiar with them, but only introduce categories later when they're actually useful

[sonic crane]

is this what ur referring to??

bro i cannot read my profs notes

like what is this saying

[tribal palm]

yeah i loved his lectures

[novel plank]

I need help with something. For context, let $(R, <)$ be a dense linear order and consider the order topology on $R$. Let $n$ be a natural number and $i_1,...,i_n \in {0,1}$. A subset of $R^n$ is an $(i_1,...,i_n)$-box if it's of the form $A_1 \times ... \times A_n$ where $A_k$ is a singleton if $i_k = 0$ or $A_k$ is an open interval if $i_k = 1$.

Given $A\subseteq R^n$ an $(i_1,...,i_n)$-box, we define it's dimention to be $i_1 + ... + i_n$.

Also, a box in $R^n$ is simply a $(i_1,...,i_n)$-box for some $i_1,...,i_n \in {0,1}$

[gentle osprey]

Eduude

[novel plank]

I'm trying to prove the following:

If $A\subseteq R^n$ is a box, and we can write $A = C_1 \cup ... \cup C_k$ as the disjoint union of boxes, then $$\dim(A) = \max {\dim(C_1), ..., \dim(C_k)}$$

[gentle osprey]

Eduude

[novel plank]

This makes sense and it's not so difficult to see why it is true, but for some reason I'm having a lot of trouble formalizing this into a proof

how would you go about this?

idk why but this is giving me algebraic topology vibes. However there is probably some simple argument I'm not seeing

[sonic crane]

What does that mean

Show its a local homeomorphism?

[iron bolt]

it means that any point in the domain has a neighbourhood on which the map restricts to a homeomorphism between open subsets of the domain and codomain

you can prove that any such map is open

[forest pasture]

Hello

I have a proof that every compact subset of a metric space is closed idk if it's right

The proof is as follows

Take the K ⊂ X a metric space, K is open. Take a point in K call it p. Take a disk and circle around it (closed set) and say it has a radius of r and is entirely within K call it C

Denote by C_i a circle like C but with a radius of r/i

An infinite cover of K is the union of all K-C_i∪p (an open set)

It has no finite subcover

An immediate issue I noticed when I started writing this is that in topology closed and open aren't antonyms, but I think that generally proves at least every compact subspace isn't open

I would like it if someone would scrutinize my proof

[iron bolt]

is it?

[forest pasture]

Btw I used the words circle and disk fast and loose a little, but I mean the points q where d(p, q) ≤ r/i

[forest pasture]

Yeah I think

no wait

Union p

[iron bolt]

(closed disk and open disks are the words I'd use - but it's clear what you meant)

[iron bolt]

you mean adding the set {p} to the cover? or adding p to every set in the cover?

[forest pasture]

Doesn't matter

[iron bolt]

either way - it is indeed a cover then. but is it open?

[forest pasture]

I don't think it changes

[forest pasture]

Oh yeah

P makes it not open

Ok so

Wait a sec

[iron bolt]

yup. {p} could be open depending on the space, but it's usually not

[forest pasture]

I got it

Just a sec

So the radius is a more complicated function

[opaque scroll]

Compact sets can be open though, so proving that is not a good thing

[forest pasture]

Yeah yeah

I just noticed it won't work

[iron bolt]

you are close to proving that compact sets in metric spaces are closed though

[forest pasture]

It was born out of this dumbness

[forest pasture]

I know a proof they are closed

I wanted to construct my own proof

I will no cri

But like

Can you gimme an example of an open compact set?

I think the open disk is one

No wait it isn't

If it has radius r

[iron bolt]

it's not compact

[forest pasture]

Take the union of all open disks on the same centre or radius r*(n-1)/n

[opaque scroll]

Being open is relative to the ambient space, while being compact is not.

For example the whole space is always an open subset of itself

[forest pasture]

Oh yeah

Lel

Thanks all you are all so nice

[iron bolt]

that's the open disk of radius r again, lol

[forest pasture]

Yes

But contains no finite subcover

Proving it isn't compact

[iron bolt]

ah, yup. I see

I thought you mean that as another example of a compact open subspace

[forest pasture]

Back in my age galiois started developing Galois theory

While I can't understand what compactness is...

[forest pasture]

I did

I literally thought of this proof 5 minutes before as to why it isn't compact

It is why I asked the question

That was a generalization of that proof

Is every compact set only open wrt itself?

As far as I understand, which is near, the ambient space is the object itself?

[forest pasture]

I have good news

Galois was one year older when he submitted his paper

I have one year to become galois

Galois speedrun

[clever dagger]

We still have some time to develop an entirely new way of understanding polynomials

[iron bolt]

nope. but since compact sets are closed (in metric spaces at least), every open compact set must be both open and closed, which means that it is disconnected from the rest of the space

so in connected spaces like R^n, every compact set is non-open or empty

[forest pasture]

Question

Can we have like

Say

This

[sonic crane]

That doesnt imply continuity, just open map?

[forest pasture]

The metric space is the open disk at the origin, and everything where x and y > 2r

And the open disk is a compact subspace

[iron bolt]

should also imply continuity I think. though usually continuity is take to be part of the definition

as in, "a continuous map f : X → Y is called a local homeomorphism if..."

[iron bolt]

if you make that a closed disk yes. other wise it's still not compact

[forest pasture]

Is this a legal metric space in topology or will the topology police get me for it?

[forest pasture]

Yeah true

[iron bolt]

it is for a very simple reason

every subset of a metric spaces is a metric space

[forest pasture]

I just finished linear algebra

So like

Having holes feels weird lol

[iron bolt]

welcome to topology, I guess 😆

later in #alg-top-geo-top you'll learn to count and classify them

[forest pasture]

Thanks a lot

I want to study it so bad

[paper wedge]

AT is cool

but many flavors

[forest pasture]

Is there bread flavour?

[paper wedge]

idk i only know the algebraic and geometric ones

[forest pasture]

I heard everything in maths is bread flavoured

[paper wedge]

ie may and hatcher

[forest pasture]

I see I see I quite understand...

[round oyster]

how can a set be open and closed at the same time, excluding the set it self and the empty

like intiution for this set, how can u even draw it / feel it

[forest pasture]

Ok so

Closed isn't related to open

Closed means contains all the limit points

A limit point is a point that's right next to a point in the set simply

More formally, a limit point is a point that if you drew a circle/sphere around that, no matter how small you make that shape (so long as its radius isn't 0) there will be points from the shape in it

This zooming in circle thing applies to 2 kinds of points, the border of the shape, in which some points in the shape are in the set we want and some are out

And what mathematicians call interior points, points where all the points in some small enough circle are in the set

Is that clear?

An open set is a set which only contains interior points

A point contained within a border, no matter how close it is to the border, if it doesn't touch it, there is a small enough circle that is all in the border, centred at the point and encircling it

Is that clear?

Please note, limit points are the border and interior points, not just the border

[alpine nest]

Interior points

[forest pasture]

@round oyster , if you want, I can draw it for you

[forest pasture]

Exactly, I always make that error

@alpine nest is my explanation clear? I hope I didn't fuck it up lol

[alpine nest]

You've explained the definition, but I'm not sure that's what MadAbdo was confused about, more about the intuition of what a clopen set is.

But I agree it's important to be clear on definitions, and in particular that "open vs closed" is not a dichotomy.

Generally clopen sets show up when your space is disconnected, for example if its the set {0,1}

{0} is a closed and open subset of it, because it has no limit points outside of it (it has no limit points at all), and also it's open, beacuse its complement is closed.

And more directly, it's open because 0 is its interior point, because all elements sufficiently close to 0 will be in {0}

(because the only such element would be 0 itself)

So the intuition about clopen sets is that they're "pieces" of your space that are "cut off" from the rest

[tiny obsidian]

As a slightly more complicated example, consider the space X = (0,1) u [2,3]. Then both sets (0,1) and [2,3] are open and closed in X. In R, which is probably what you're thinking of as intuition, (0,1) isn't closed because it doesn't contain limit points 0 and 1; these aren't even in X, so (0,1) doesn't not contain them in order to be not closed. Similarly in R, [2,3] isn't open because 2 and 3 aren't interior points, as 1.9, 1.99, 1.999 or 3.1, 3.01, 3.001 etc. are arbitrarily close to them; in X, again these points don't exist to make [2,3] not open.

[alpine nest]

Yep, and again, note that the space in this example is in two separate "pieces"

And each of these pieces is a closed-and-open (clopen) set

[mighty hull]

Yeah, it's not right to say closed and open are "unrelated."

They are related, just not in the naïve way their names suggest, that "closed" means "not open".

[tiny obsidian]

Yeah there being closed and open sets (other than empty and whole space) is in fact equivalent to being in multiple pieces; I hoped to give an example where it's clear why this break would lead to sets being closed and open, and one where its clear exactly how both conditions are satisfied and where the usual intuition (viewing sets as lying in connected R^n, specifically n=1 for the example) breaks down

[wanton heart]

Is it necessary to take delta=min(delta1, delta2, ...) here?

I am a little confused on exactly what I'm supposed to do at that stage

I want to show that for any t0 in h^(-1)(U) that there exists an open set in the standard topology on R containing it which is a subset of h^(-1)(U)

[tiny obsidian]

The issue with this is that you don't necessarily have delta>0

(and technically you mean inf in this case)

what's the codomain of h? you didn't say

[wanton heart]

oh right since there are infinitely many delta

[wanton heart]

R -> R^(omega)

Do you have an idea for how it is meant to be done then?

[tiny obsidian]

the fact sup|...| < ε is a stronger condition than |...| < ε for all n

more specifically, there is ε' (e.g. the supremum itself) with |...| <= ε' < ε; in other words |...| doesn't get arbitrarily close to ε

I'm not yet sure how precisely to use this fact, but I suspect this allows you to either pick the δn a smarter way, or even δ directly

probably triangle inequality...

[round oyster]

yes please

i would really appreciate

[forest pasture]

Hey

A sec

So, C is a circle without border, or an open disk

C' is a circle with a border

The dashed line is just to signify we have to border

In sending a drawing

I could do these in desmos but I'm an arteest

So

[round oyster]

the picture still loading for me sec :)

[forest pasture]

C, in IR ², is open

[round oyster]

ok i see now

[forest pasture]

It doesn't contain the border although the border touched some points of C

C' is closed

But C, with respect to C, is closed

It is still open, every point is surrounded by points

There isn't an edge, only points arbitrarily close to it

Cut since the edge doesn't exist in C, then C is "closed" because it contains all the limit points of C that are in C

It contains all the borders of C that are in it, that is, none of them

[round oyster]

so there is no edge, and it contains the border?

[forest pasture]

No

[round oyster]

like, i'm trying to imagine it not in like R^2, in general toplogy would it make any sense

[forest pasture]

Just a sec

[round oyster]

ok

[forest pasture]

Slow down

Firstly, I'm using edge and border interchangeably

Secondly, begin with this example then I'll generalize it

[round oyster]

k

[forest pasture]

As I was saying, if you live in R², you'd be walking and suddenly you'd be like "holy shit I fell into a circle! Someone should put an edge around this!"

But if the circle was everything, you can't get outside the circle

You can't be like "oh look, an edge"

You can't touch it

[round oyster]

got u

its basically an open set

[forest pasture]

You can get arbitrarily close

[round oyster]

yes

[forest pasture]

It is open because you are always in the circle

In both cases

But because you can't get outside, you can't touch the edge

[round oyster]

i imagine it you see a prism, whenever you get close to it, it feels a little bit more far, yet its still fixed in its place

[forest pasture]

The edge doesn't exist in that frame of reference

So it is closed with respect to itself

[round oyster]

got it

[forest pasture]

Kinda

Now, on the number line, the open interval (0, 1) is an open set

[round oyster]

ye

[forest pasture]

There are no points that touch a number between 0 and 1 and touch the outside world

They are all contained within

But

With respect to R², it is neither open nor closed

The points are all edges

Right?

The points all touch other points from the set, while also touching points from outside

So it isn't open

But also

It doesn't contain 1 and 0, which do touch a point from the set, but also the outside world

[round oyster]

so you saying points from outside the set are touching points from inside the set if its not closed nor open?

[forest pasture]

No

[round oyster]

wait, i'll tell you how i think about it, i want to restore my thinking

i had problem understanding your language

[forest pasture]

Closed means contains all points that touch points from it

[round oyster]

yes agree with this

[forest pasture]

Now that means at most adding the border

Ok,

There isn't a point outside the set with the border that touches the border

It will always be some distance away

Even if that distance is super close

[round oyster]

can you formualte this in math operators, sorry but i'm struggling to understand your language

[forest pasture]

Oh gladly

So, for now, P ⊂ X
p is a point in P
X is a metric space

Neighborhood of p, denoted N_r(p) is the set of all q such that
d(p, q) < r
r is a real positive number
d is distance

P is an open set if for all p∊P, the exists some r >0, for all q∊N_r(p) => q∊P

that is, there is a small enough circle such that all points in that circle are in the set

Or p is surrounded by points in the set

Something obvious worth noting is that, if that works for r, then it works for all reals in (0, r)

Now

[round oyster]

yes sure this makes sense, so basically for any point that is inside the circle, there is a small enough circle within the set of neighborhood of any point, that is included in the circle

[forest pasture]

If for all q∊X, if for all r>0, N_r(q) contains some point in P, then q∊P, then P is closed

[forest pasture]

Not necessarily a circle in the first sentence

A set can be dinosaur shaped

But the distance is a circle

[round oyster]

ye

any neighborhood

[forest pasture]

No,

Some neighborhood

[round oyster]

i mean exist one

yes

[forest pasture]

Is the definition of closed clear?

[round oyster]

yes

so far so good

[forest pasture]

That's kinda it

[round oyster]

for any neighborhood of the points, they share a point with the set

[forest pasture]

As you can see

The set of q we can pick from depends on X

If X = P, then any open set is closed

[round oyster]

first i asked you this question, because i was trynna to shape how the clopen sets are in connex spaces

[forest pasture]

I have 0 clue what this is

I didn't study this far

Lol

[round oyster]

i mean, for the set it self X, and phi are both clopens but i'm looking to draw it

[forest pasture]

If you know what it is, please tell me

I don't know what clopen is

When you know the answer, please dm me

[round oyster]

let X be some topological space, we say that it's connex if it can't be written as a unionn of two opens / two closed

[forest pasture]

Makes sense

[round oyster]

there are some weird propertys with clopens

connex = connected, i forgot to translate that to english

[forest pasture]

What's a clopen

[round oyster]

closed and open set at the same time

oh wait, in connected spaces, only open / closed are X and the empty set

but in other toplogical spaces, i have no idea how they look like, if i find an answer i will tell you

right

[forest pasture]

That makes so much sense

I was struggling to visualize it otherwise

[round oyster]

R is connected, since you can't do with with two disjoint

[forest pasture]

Connected I'd say means no missing points

No wait

That isn't right

[round oyster]

there are ones in the middle

you can't put in either side

[forest pasture]

(-∞; 0) U [0, ∞)

I don't know what that is

[round oyster]

your set will never be split up, thats why it's called connected

[forest pasture]

Please note I am dumb

[round oyster]

two opens / two closed

u have

[forest pasture]

Makes sense

[round oyster]

(-oo, 0) U (0,oo)

[forest pasture]

So

Lemme draw something

[round oyster]

so there is always a link between any two opens i think

[forest pasture]

So

Pardon my shitty drawing it's 1:23 am

So

[round oyster]

not necessarly, but generally there is path between some opens, if you think of opens as mountains there is canyon in between. i think

[forest pasture]

In this example

A dashed line means some areas are closed and some are open

B U C =A

So clearly

[round oyster]

i dont get what u mean, but i suppose your thinking of it as one is completing the other, but there is emptiness those gears cannot be filled

[forest pasture]

I think any set can be then a combination of 2 sets of different types

[forest pasture]

What

I mean

For any set, an open and a closed can make it

[round oyster]

oh, of course that will always work

any toplogy

consider any toplogy thats not trivial, take any element as an open set, the complement is closed and their union is X it self

[round oyster]

hmm, wait lemme think about it

[sonic crane]

I dont actually understand this

[forest pasture]

Hey

I must sleep

I'd really love to help but I can't

[round oyster]

that could be true

i'll think about it tmr

good night

i have to sleep too, bye

[forest pasture]

When you reach anything

[round oyster]

i'll tell yo

[forest pasture]

Ping me or dm me

Ok?

[forest pasture]

Byeee

[sonic crane]

Idk how 😢

[real gazelle]

I'm thinking about a dumb generalization of metric spaces where instead of real numbers, the set of distances is a set D, equipped with an arbitrary relation ("less than or equal to"), a bottom element ("zero") that relates to every other element on the left, and a binary operation (for defining the triangle inequality). I can define convergence, show constant sequences converge, show a sequence converges iff. every subsequence converges, define the discrete metric when D={0,1}, and turn it into a topological space by taking the generating set to be the "open balls", but that's about it.

[sonic crane]

If you have a quotient map X -> Y, is it like X / ~ with ~ being x = y if phi(x)=phi(y), and give quotient topology to it makes X / ~ homeomorphic to Y?

[real star]

if it's surjective i think yes

[sonic crane]

Yeah quotient map defined to be surjective in munkres

[real star]

yea i guess it's also where the word quotient map comes from

[unreal stratus]

Yup

[rancid umbra]

less than or equal to has to be total order. otherwise triangle inequality may not hold in all cases

[sonic crane]

Im pretty stuck on showing the maps are continuous still

For R2 \ ~ -> s1xs1 I showed R2 -> s1 x s1 is continuous by showing that the map to the individual factors are continuous, and then use universal property to see the quotient map version is cts

But showing S1 x S1 to R2 / ~ is cts i dont know how

Or showing the original thing is an open map

[wise mist]

incase of finite fields vector subspaces of an NLS are bounded subsets

or does that even make sense ?

[plush folio]

One idea could be to show that R^2/~ is compact, for example by showing that it is homeomorphic to [0, 1]^2/~

dunno how easy that is though

[sonic crane]

Yeah, ive seen arguments like this online but it seems suspect to throw that in

He did no such thing in lectures

[plush folio]

it's just the closed map lemma, so it shouldn't be too far-fetched

it's IMO the nicest lemma in basic topology

[sonic crane]

True but I wouldnt have come up with that myself yk

wait, what theorem are u even referring to? continuous bijection X compact Y Hausdorff implies map is homeomorphism?

is what i wrote even correct lol, been a while

maps into products can be checked to be continuous by checking each factor, what about maps from out of products?

no such simplification ?

is the map ((cos(2pi t), sin(2pi t)) -> (cos(pi t), sin(pi t)) From S1 to S1 easily seen to be continuous and open?

Like i just said it is in my homework cause like damn idk going through the nitty details of what open sets look like here just seems annoying

[plush folio]

yep, a continuous map f from X compact to Y hausdorff is closed, and in particular if f is bijective then it is a homeomorphism

[sonic crane]

oh, closed sets in compact spaces are compact, and compact sets in hausdorff spaces are closed

i would like to not use this though

[plush folio]

I'm not the best person to ask, but I think in general there's no easy way to check. I don't think (cos(2pi t), sin(2pi t) -> (cos(pi t), sin(pi t) is open, as it maps S^1 to a half-open circle in the upper plane

[sonic crane]

for 0 <= t < 1 right

poo

[plush folio]

yeah, you would need the restriction 0 <= t < 1, otherwise the map is not well-defined

btw, maybe it is possible conclude R^2/Z^2 =~ S1 x S1 by using R/Z =~ S1 somehow? For example, is R^2/Z^2 isomorphic to R/Z x R/Z?

[sonic crane]

But it sends open proper subsets of S1 to opens in S1 right

[low flame]

would maximality be something like this?

X = (-1,1) cup (5, 15)

subset of R

would (-1,1) and (5,15) be maximal sets

since theyre the largest contained sets in R

and each point in each of these sets will be connected through their unions?

[mighty hull]

Yes. When talking about sets, a maximal set M with property BLAH is a set such that if there's another set with property BLAH and M ⊆ S then S ⊆ M.

[low flame]

is the blah a fill in the blank for me?

[mighty hull]

Yeah, whatever you want.

The idea of "maximal" is something you can frame generically in any partial order.

Let (P, ≤) be a partial order. Let C ⊆ P some subset of elements in C (these are the ones with the property we're interested in).

An element m ∈ C is maximal if s ∈ C and m ≤ s implies s ≤ m.

Here the partial order is (P(X), ⊆), where P(X) is the power set of X, and C ⊆ P(X) is the collection of all connected subsets of X.

[low flame]

ah okay this makes sense, it hits my intution, but i just want to verify it 🙂

[mighty hull]

A maximal element might not exist.

For example, look at ℝ and let C be the set of all compact subsets of ℝ. ℝ itself isn't compact, but given any compact subset X ⊆ ℝ I can always find a strictly larger compact subset containing X.

Hence the need for 11.G, to show that every point is inside at least one maximal connected subset.

[low flame]

thats very interesting actually

[low flame]

Thank you very much cufflink 🙂

thats a beautifiul example btw

[mighty hull]

Also, don't think the connected components have to be open intervals.

Modifying your example, this space has two closed connected components:

X = [-1, 1] ∪ [5,15]

[mighty hull]

Last thing. One relevant theorem is that a space X is connected if and only if it has exactly two subsets that are both open and closed (aka clopen): the empty set ∅ and the whole space X, which are both clopen in any space.

If a subset S ⊆ X is both closed and open then X\S is also closed and open. They're always disjoint, simply as sets, which means

X = S ∪ (X\S)

is a decomposition of X into two disjoint open subsets.

So if you're looking for connected components it can help to look for non-trivial clopen subsets.

[low flame]

so if i have [-1,1] as X and I set S equal to {1} then i got a clopen set, which is then also connected?

[mighty hull]

No, {1} is not open in [-1,1].

[-1,1] is connected, so that theorem says the only clopen subsets are all of [-1,1] and the empty set.

But look at something like X = [-1,1] ∪ [10, 15].

[-1,1] is both closed and open in X.

Or if you want a single point to be open (in the subspace topology), look at a space like

X = {-2} ∪ [10, 15]

[low flame]

ah i see then okay!

that is a clopen space

[mighty hull]

A space is always clopen in itself. Subsets of a space may be closed, open, neither, or both.

[low flame]

ah!! i understand i think now

X is open based on its topology. and its closed if its compliment is the empty set which is closed

[mighty hull]

A topology is defined in terms of open sets and one condition is that a topology on X includes both X and ∅ as open sets.

If ∅ is open then its complement in X is closed, but that's all of X.

If X is open then its complement in X is closed, but that's the empty set ∅.

So that part just follows from how we define topologies. Every topological space X has at least two clopen subsets: X and ∅.

It has exactly two if and only if its connected.

[low flame]

beautiful

[chilly topaz]

Are all interior points limit points?

[sonic crane]

Well maybe If X = {1} U [2,3] and consider A = {1} then A is open so 1 is an interior point but its not a limit point of A?

Well a better example is probably X = {0,1} U [2,3] then A = {0,1} is open but we can see 0 and 1 are not limit points

[chilly topaz]

I think 1 is still a limit point of A, because in this example the only open set that contains 1 is {1} and clearly {1} intersects with A,

[sonic crane]

Limit point x in A is if every open set containing x intersects A-{x}

[chilly topaz]

And here, 0 and 1 are interior points?

[sonic crane]

Thats also not true. There are other open sets that contain 1 that are not {1}

[sonic crane]

Yes because {0,1} is open, so everything in that set is interior points

[chilly topaz]

Ohhhhh

So then . . . can I say no interior point of A is a limit point because there is always an open set u that is a subset of A (aka it does not intersect X - A) ?

[chilly topaz]

Example?

Wait,

The whole set itself,

Is open and contains 1, yeah makes sense,

[sonic crane]

Yeah but also {1} U [2,2.5)

Remember subspace topology

[chilly topaz]

Right yes yes,

[sonic crane]

In that example before, 1 is not a limit point because open sets containing 1 does not intersect A - {1}

[sonic crane]

But i think this example is better to look at

[chilly topaz]

Are we constricting ourselves to the subspace A or do we look at the whole X? . . . If we only look at A, the ofcourse and open sets of A does not intersect X-A, where would make no point a limit point? . . . But if we look at all X, then the open set (1/2, 3/2) contains 1 and intersects X - A, so how can 1 not be limit point?

[sonic crane]

For a subset A < X, x in A is a limit point of A if every open set containing x intersects A-{x}

In that example, there are open sets contaning 1 that do not intersect A-{1}, like that case is even worse because every open set containing 1 does not intersect A-{1}... because A-{1} is just empty

[sonic crane]

You dont look if it intersects the complement of A

So why are u looking at X-A

U dont look at X-A

[chilly topaz]

Ohhhh okay okay,

A - {x}

Not X - A,

[sonic crane]

Yea

A take away the limit point ur considering

[chilly topaz]

And because litterally any other open set that is not {1} doesn’t contain 1, 1 is not a limit point,

?

[sonic crane]

A is just {1} so A-{1} is empty

so of course any open set containing 1 isnt gonna intersect A-{1}

so 1 is an interior point of A (bc {1} is an open set) but 1 is not a limit point of A

[sonic crane]

i like this example better

[chilly topaz]

And here, 0 and 1 are not limit points because there is an open set that contains 1 but does not intersect A - {1} (same with 0) . . . Is that set {1}? Why is {1} open?

[sonic crane]

sure, that set could be {1} but also {1} U [2,2.5)

u should review subspace topology

{1} is open because {1} = (-0.5,0.5) intersect X

this is all subspace topology stuff

definition of subspace topology

[mighty hull]

Share the definitions of "interior point" and "limit point" that you're working from.

[chilly topaz]

Why is the intersection not (0, 0.5) ?

[chilly topaz]

Yes yes I should,

[sonic crane]

it could be different things that was just an example showing {1} is open

oh

well that intersection is just {1} cause X = {0,1} U [2,3]

[sonic crane]

none of that is in X

besides 1

[chilly topaz]

Ahh wait, {0, 1} is just point 0 and 1,

[sonic crane]

ye

thats prob why u confused

[chilly topaz]

I read that as the interval of 0 to 1 my bad,

Yes yes,

Okay got it got it makes sense,

So intuitively it just point that are “alone” (do not belong to interval) that can be interior but not limit points?

[chilly topaz]

https://m.youtube.com/watch?v=Z0e4gZZCxBM&list=PLAvgI3H-gclZa-DVTMyUIAxM-X8NSikwu&index=35&pp=iAQB

^ from here ^

^ where I (wrongly) generalized that a limit point p of an arbitrary topology is when there is an open set that that contains p that intersects with A (when it should intersect with A - {p}, otherwise the weird lonely points like {1} would be limit points).

[mighty hull]

That first definition is not the usual definition of limit point. It implies every element of M is a limit point.

It's what is sometimes called an "adherent point".

A limit point has the additional condition that every neighborhood of x doesn't just contain some element of M, but that it contains an element other than x.

Because if x is in M then every open ball around x contains an element of M, namely x itself.

So the definition is that x is a limit point of M if every open set containing x also contains an element m of M, with m != x

[alpine nest]

As a side note, I think when working with closures/closed sets, thinking in terms of adherent points is often preferable to thinking in terms of bona fide limit points, since the specific distinction involved in being a limit point is often completely irrelevant to the proof.

[chilly topaz]

Okay okay, so my open ball must contain an element of M that is not x, otherwise every point in M would be a limit point, including the single points like {1},

But in practice I can just use adherent points (where x can be said element of M),

Got it got it, thank you all so much 🧡

[wise mist]

are they trying to make sense for isolated points ?

[mighty hull]

Annoying terminology. Its normally the other way around! Haha.

[mighty hull]

But yes, it's trying to account for isolated points because every point in a set is a limit point under this definition.

[wise mist]

alr

[alpine nest]

As Cufflink points out, most authors define limit point as one that satisfies definition 2.2.23, and points satisfying 2.2.1 are called adherent points. I wouldn't call the definition you posted wrong, because it does show up in literature, but it's worth keeping in mind that in other texts "limit point" is very likely to be one satisfying 2.2.23 (and also that the distinction between 2.2.1 and 2.2.23 only occasionally matters)

[wise mist]

noted

[alpine nest]

In particular the statement "a set is closed if and only if it contains all its limit points" works for both defintions of "limit point", as does "the closure of a set is the union of itself and all its limit points"

[wise mist]

what about the closeness of G ?

can we create something like a subgroup in the rationals, then take a sequence converging outside of it
(namely the important angles in unit circles are irrational)

but i would love to get an argument via preimage of cont maps, if thats desireable

[low flame]

Prove that if A is a proper nonempty subset of a connected space, then FrA!=∅.

so i was thinking something like this here. If As boundry is empty, then well A is open, but its also closed since it contains all of its boundry points right? This then would be a contradiction to the fact that X is connected since X and the empty set are the only ones allowed to be clopen?

[low flame]

Suppose the closure isnt connected.

then U cup V is the closure of A

then that means the the boundry of fr A = (Fr A cap U) cup (FrA cap V)

now since the boundry is connected, as per the question, this implies that one of these intersections is empty.

without loss of generality let (FrA cap U) be empty. Thus we have a set U that is open, since it contains no boundry points, and closed since it contains all its boundry points. This contradicts that its part of an larger connected space, and thus is the boundry is connected, then also is the closure.

[fierce lily]

I suppose X is hausdorff here, then for every point x,y in X, they are separated by open sets U and V. Since X has cofinite topology, then X-U, X-V is finite as well, but it contradicts that X is infinite. Am I right?

[gritty widget]

How does it contradict that X is infinite?

[fierce lily]

I mean if U is an open set that contains x, then X-U is finite. and since U is a finite set( in hausdorff, every pair of points can be seprated by open sets, so I just let U only contains x), but X-U is not finite because X is infinite, hence it contradicts.

[gritty widget]

But how can you choose U open set such that U contains only x ?

[fierce lily]

How you will argue this question?

The only thing I can come up with is to point out X hausdorff here will make X finite as my purpose

[gritty widget]

Say there exists non-empty set U and V such that U cap V is empty, then I can write X = X - U cup X-V.

But X-U and X-V is finite implies X is finite.

[mighty hull]

Yes, but you have to say why. Let's prove the contrapositive: every Hausdorff space with the cofinite topology is finite.

\\

Let $X$ have the cofinite topology and assume it is Hausdorff. Pick $u,v \in X$ and separate them by disjoint open sets $U$ and $V$. In general, if $U \cap V = \emptyset$ we can write $X$ as \begin{align*}X &= X \setminus \emptyset \ &= X \setminus \left(U \cap V\right) \&= (X \setminus U) \cup (X \setminus V)\end{align*}

But if $X$ has the cofinite topology then $X \setminus U$ and $X \setminus V$ are both finite, which means $X$ is also finite.

[gentle osprey]

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

[unreal stratus]

Yes

[gritty widget]

Just take the closed unbounded sets in R

[cedar pier]

for the proof?

how do i even start this

[gritty widget]

So are you working in metric space?

If yes then you can use sequential compactness

[cedar pier]

yeah

[gritty widget]

You know the sequential compactness?

[cedar pier]

maybe?

can you elaborate more

[gritty widget]

Every sequence has a convergent subsequence which has a limit in that subset.

If K is a compact subset in metric space X then every sequence in K has subsequence which is convergent in K.

[cedar pier]

do we use this theorem?

[gritty widget]

Yes

Take x_n in K_n

x_n in A_n

Now use that

To show the intersection is non-empty

Can you guess which point must be in intersection?

[cedar pier]

uh

the 0 point?

[gritty widget]

0?

What is 0?

[cedar pier]

null set?

[gritty widget]

No you have to show intersection is non-empty

Let x_n in A_n, so your sequence is {x_1, x_2,...,x_n,...} where x_1 is from A_1, x_2 is from A_2,....

Now (x_n) is a sequence in A_1, correct?

So it has convergent subsequence which is convergent in A_1, say limit is x

[cedar pier]

we need to show that x is in that set?

[gritty widget]

Which set? Intersection

[cedar pier]

yeah

[gritty widget]

I used A_1 is compact so x in A_1

[gritty widget]

Yes we need to show x is in the intersection

Now that subsequence, if x_1 occur in that subsequence remove it, remaining subsequence is convergent to x and sequence in A_2, so x in A_2

[gritty widget]

Here I used that compact space is closed

[cedar pier]

so can we keep doing this so that x is in all of them?

[gritty widget]

Yes

If you are not comfortable in this method, there are other methods too

[cedar pier]

i get it

[gritty widget]

Finite Intersection Property

This one is generalization and equivalent to compact definition

https://en.m.wikipedia.org/wiki/Finite_intersection_property#:~:text=Applications-,Compactness,property has non-empty intersection.

[cedar pier]

ill check that out, thanks

[gritty widget]

So decreasing sequence follows FIP

[cedar pier]

how about c and d?

[heady skiff]

if f and g are continuous and H(x) = G(f(x), g(x)) where G is continuous is it the case that H is continuous

[paper wedge]

yeah in product top

[heady skiff]

ok cool thanks

[gritty widget]

c is closed, take the convergent sequence in that set and shows that it is convergent in that set

[gritty widget]

d is open, think about interior points, for each point you can choose the radius around it such that the entire open ball contained in given set

[wise mist]

Z[sqrt2] dense in R

well should i treat this as a ring

still im only converging to zero

i dont like the epsilson arguemnt here

the next generalisation is density of Z[a] in R

$\mathbb Z$, $\sqrt 2 \mathbb Z$ are closed\
but $\mathbb Z[\sqrt2]$
is dense in $\mathbb R$

[gentle osprey]

yeshua

[iron bolt]

if it was closed and dense it'd have to be the space itself, which it is clearly not

oh, my bad. I misread and thought you said Z[√2] was closed

[pallid comet]

hi point-set-topology! today i just read something that blew my mind from wikipedia

apparently Z with the Furstenberg topology is homeomorphic to the rationals as a subset of the reals with the usual topology

the Furstenberg topology being the topology on Z generated by the basis of arithmetic progressions {an+b | n in Z} for integers a, b (a nonzero)

does anyone know how to construct the homeomorphism? (googlr isnt being helpful)

[red yoke]

The more I stare at this the more my head hurts

[opaque scroll]

The proofs I could find for them being homeomorphic is quite far from being constructive.

So I'm assuming any explicit construction must be absolutely horrible

[fierce lily]

Can someone help me with part(b)? and what does the set of all sequences that are eventually 0 mean here?

[paper wedge]

what is the complement of the solid torus in S^3

[mighty hull]

Another solid torus

[paper wedge]

xd

[red yoke]

What if it's knotted

[fringe thorn]

what are some examples of interesting (whatever that means to the reader) adjunction spaces?

I'm just looking for some examples to keep in mind as I go forwards

[paper wedge]

how do u picture this

[bronze wadi]

I have a disk, 2 open sets U, V on its boundary, 2 points x, y on the disk. Can I always find 2 open sets in the disk, one containing x and having exactly boundary U, the other containing y and having exactly boundary V?

It sounds true but i have actually 0 idea how to show that rigorously

[unreal stratus]

Wdym by "on its boundary'

[bronze wadi]

well ig id have to consider the disk as a manifold

[unreal stratus]

This sounds impossible unless U = V = disk as V would be open and closed

[bronze wadi]

and thats the boundary

as a manifold

[unreal stratus]

Well

Sure, manifold w boundary

But uh yeah

Do you mean U and V are open subsets of the boundary

[bronze wadi]

yes

[unreal stratus]

Okay that was unclear lol

[bronze wadi]

sry about that

[unreal stratus]

But yeah you'll have U and V closed and open

If they are boundaries of smth else then they are closed

[bronze wadi]

they dont need to partition the boundary

ok ill draw for image

[unreal stratus]

And I didn't use that

[bronze wadi]

oh right

[unreal stratus]

It's just if V is the boundary of smth else then V is closed

[bronze wadi]

thats just the topology

woops

Isn’t this red set open

im not sure i understand why its not true

[unreal stratus]

The like big thing or the bit on the boundary

Oh

You still mean boundary in the sense of manifolds for that open subset

OK then yes sure

[bronze wadi]

Again sry for the lack of clarity lol

[unreal stratus]

You can just like radially scale or whatever

Like e.g. consider the segment you get from that open arc

(And remove the centre)

[bronze wadi]

this type of property is not true for all manifolds right

[bronze wadi]

and right that does make sense

thx a lot

[unreal stratus]

Hm not sure

Though note that when you say manifold that means no boundary by default

[bronze wadi]

do we say like manifold with boundary?

[unreal stratus]

Ye

[bronze wadi]

ah fair enough

do we require compactness on manifolds

or can they extend infinitely

i think compactness is only assumed if we say its closed, but i forgot exact nomenclature

[cerulean oriole]

Closed manifold = compact manifold-without-boundary

Pretty sure

[fringe thorn]

bumping this

[red yoke]

It just seems to be a very common construction in general

it's how you glue things

[bronze wadi]

I’m a bit lost at how to understand the nhood of the vertex here. I’m fairly confident this isn’t a surface but I have trouble showing the vertex cannot have a nhood in R^2 or R_+^2

I tried to glue and going from sides to sides, but I don’t really understand what’s going on honestly

[red yoke]

What if you pick a point on the edge instead?

(That's actually the same as picking a vertex)

[bronze wadi]

do i get like

3 cones around a point

[red yoke]

3 cones?

[bronze wadi]

well idk like i dont get where it should get glued first

[red yoke]

I mean just a small nbhd of the point

[bronze wadi]

ok ill draw that sure gimme a sec

Is this not right?

[red yoke]

Yea

So it's 3 semidisks glued at an edge

[bronze wadi]

basically yeah

well i could have them meet in the center

doesnt rly matter ig

[red yoke]

Vertices are the same

[bronze wadi]

So is this the picture then

[red yoke]

No

You only glued one point

[bronze wadi]

Yes but aren’t I looking at it locally

Oh right these edges are the same damn

[red yoke]

You can apply a homeomorphism to this diagram

[bronze wadi]

right its the same thing

sry im not sure i understand what youre getting me to do

[red yoke]

Homeomorphism

Like, you push the green inwards and pull the blue outwards

[bronze wadi]

right i get its a homeomorphism

[red yoke]

Gluing the blue arrows is the same as gluing the red arrows

So edge point and vertex are literally the same

[bronze wadi]

well they have similar nhood yeah

[red yoke]

So nbhd of vertex is this too

[bronze wadi]

well i feel like 3 semidisks at an edge cant be similar to a nhood of a plane or half plane if thats what youre getting at

[red yoke]

Yea

[bronze wadi]

hm yeah, its just hard to tell becauses surfaces can bend weirdly

but it sounds right

thx

[red yoke]

I guess one way to prove this would be to apply invariance of domain on two disks glued at an edge

[wise mist]

continous maps in from MnR to MnR

how do i think about the open ball?

do i have to rely upon a metric ?

also is there any expmple of discontinous map in this space

[opaque scroll]

As a space MnR is just R^n^2, so just think about it in the same way you would R^n

[wise mist]

oh right, sequence and open ball both

[knotty vine]

Two lines attached along their endpoints are a circle, two disks attached along their boundaries are a sphere (this generalizes)

Not sure if that answers your question... Attaching doesnt really do anything pathological unless you put weird spaces in.

[iron bolt]

if only

glues two real lines together by their rational points

I guess you could say that the rational numbers are weird tbf

not being locally compact and all that

still, I think having the spaces in the adjunction be nice still gives your gluing maps plenty of room to be pathological

[knotty vine]

Fair enough

[low flame]

The integral of 1/x is Ln|x| + c

The domain of 1/x is not connected, since it cannot take a value at 0. IE, (-inf, 0) union (0, inf).

Thus, by the def of a map being locally constant, this means that we can have 4 + ln(x) for x in (0, inf). Then we could have -14 + ln(x) for x in (-inf, 0) and the derivative of this still gets you 1/x.

Even more interesting is that this can be generalized further: Locally constant functions are a vector space of dimension c, where c is the number of connected componets. The basis of this vector space is the characteristic function which is 1 on given component and 0 on another.

so my question is does this mean that the + c in integration is realted to localy constant functions?

sorry i know its a calculus question more maybe?

but i came across the idea doing topology tbh

[red yoke]

Yea this is related to de Rham cohomology

The differential map (f → df) is a linear map from the R-vector space of 0-forms (smooth functions) on the smooth manifold (-∞,0) ∪ (0,∞) to the R-vector space of 1-forms (smooth function * dx)

The kernel of this map is the vector space of locally constant functions

It's called the 0th de Rham cohomology group

[low flame]

okay thats a lot for me to read, ive never seen this before 😛

[red yoke]

One can prove that de Rham cohomology groups agree with singular cohomology groups, so this is indeed related to a topological property of the domain

[low flame]

does this kinda explain why you add c? because you can get a lot of different splits of domains on R?

[red yoke]

If the domain is connected, that means locally constant functions are just constant functions C

But if the domain is split in 2, locally constant functions can be + C for < 0 and + D for > 0, for instance

[red yoke]

So you can see the dimension of this kernel reflects the number of topological components

[low flame]

okay, i think i understand. yeah i just made a simple example when i was reading this because 1/x has a simple split domain and this whole thing came up. so when they teach you that its just ln|x| + c, its in a way wrong since you could have two different constants

and then i went well if you have more connected sets, then youre gonna have more of this

its good to hear that my intuion was correct

and its related to something ive never seen before haha

[red yoke]

Algebraic topology is fun

[low flame]

lemme get through regular topology first

[red yoke]

We eagerly await the day of your enlightenment

[low flame]

Thanks! have a good day 🙂

[low flame]

I am wondering. Could i just apply the intermediate value theorm and say that on [0,1] there is some f(x), then its connected. and then being a path is a continous map, then if the domain is continous, so is the image.

[low flame]

yes sorry, i meant connected

thank you

[cedar pier]

how do i show it's open, closed, or neither?

[novel acorn]

Write it as a countable union of open sets

[cedar pier]

take x to be a point in S where S is the set from that question, take the union of another point?

[novel acorn]

Idk what you mean

[cedar pier]

how do i write it as a countable union of open sets?

[novel acorn]

Hint: the set for R^1 is U_i (i,i+1)

[cedar pier]

😭

[novel acorn]

Basically you're taking the union of everything in between the integers

[cedar pier]

i dont think my prof covered that 😭

[novel acorn]

Didn't cover what lol

Do you know what the open sets in R look like?

[cedar pier]

all i know is that the defn of open sets is that there exists an open ball that is completely inside the open set

[novel acorn]

Ah okay

Okay so

The idea is that like

I'm trying to describe what this space looks like as best I can

[cedar pier]

i dont understand what the space with no integers looks like, is it just filled with gaps like swiss cheese?

[novel acorn]

You take R^n but you throw out all the hyperplanes parallel to the coordinate hyperplanes that pass through a point with at least one integer entry

[cedar pier]

if so, we just have to take an open ball placed at the edge of an integer, if there exists an episilon that is greater than 0, the ball wont be contained inside

[novel acorn]

Think of this space like a bunch of little n-dimensional boxes with edges of length 1
If you pick a point in this space it'll be in one of the little boxes
So you just have to see which edge it's closest to and pick a ball of size less than that distance

[cedar pier]

how do i show that mathematically?

[novel acorn]

take a point in your set x = (x1, ..., xn)
Take the distance of each coordinate to the nearest integer
Take the minima of all these distances
This is the distance from your point to the nearest point containing an integer (show this)
Take a ball less than this distance
Show my minimality there are no points with an integer coordinate inside this ball

[cedar pier]

ok i need to think on this a little more

thank you though

[heady skiff]

what is f^-1 here? it isn't the inverse mapping, since the inverse may not be defined; is it a function from Y to the quotient set Y / ker f (with fibers identified)?

We first note that $f^{-1}$ has domain $Y$ and codomain $Y / \text{ker}(f)$. Hence we actually have the composition
[Y \xrightarrow{f^{-1}} Y / \text{ker}(f) \xrightarrow{h} Z]

so is this true?

[gentle osprey]

okeyokay

[heady skiff]

otherwise this doesn't make sense

never mind

[paper wedge]

whats ker

f is a surjection

[heady skiff]

yeah i was able to figure it out

can somebody help me see this? here X / A is the quotient space obtained by identifying all the points of A (and everything else is a singleton)

this amounts to showing that h(x) = h(A) for every x not in A right

[iron bolt]

nope

it's probably simpler than you think - what are the fibers of that map X → X/A? ||A itself and a bunch of singletons||, right?

it's easy to see that h is constant on each of those if it is constant on A

[heady skiff]

yea my brain was fried at this point

thank you lol

[wanton heart]

What is the difference between R^inf and R^omega typically?

is it that R^inf has only finitely many non zero entries

[iron bolt]

I wouldn't call either of those well-defined without further context

[wanton heart]

Munkres

[mighty hull]

He'll define them specifically somewhere.

[heady skiff]

can somebody explain to me how corollary 1.9 shows that varphi is continuous

is it because $\varphi = h\upsilon^{-1}$? then $\upsilon$ is an identification if we give $X / \text{ker}(h)$ the quotient topology, but this also requires that $h$ is constant on each fiber of $\upsilon$

[gentle osprey]

okeyokay

[heady skiff]

and i know this looks like the previous remark, but ker h isn't a subset of X

[heady skiff]

by constant on each fiber you mean h(v^{-1}({[x]})) = z' for all [x] in X / A right

in other words the image of each of the fibers is the same point in z

[iron bolt]

it's not the same point for each fiber - but each fiber gets mapped to a single point, yes

or in other words, whenever you restrict v to a fiber, that restriction is constant

[heady skiff]

oh okay, i was mixing up my definition then

[heady skiff]

also do you know why this holds?

nvm

i think i got it

[iron bolt]

I don't know what they mean by identification - presumably what others call a quotient map?

ah, good

[heady skiff]

To see that $\varphi$ is continuous, note that $\varphi = h\upsilon^{-1}$ where $\upsilon: X \to X / \text{ker}(h)$ is the natural projection. In other words, the following diagram commutes:
[\begin{tikzcd}
X && Z \
\
& {X / \text{ker}(h)}
\arrow["h"', from=1-1, to=1-3]
\arrow["\upsilon", from=1-1, to=3-2]
\arrow["\varphi", from=3-2, to=1-3]
\end{tikzcd}]
Now $\upsilon$ is an identification since $X / \text{ker}(h)$ (presumably) has the quotient topology. Also, for each $[x] \in X / \text{ker}(h)$,
[h\bigl(\upsilon^{-1}\bigl({[x]}\bigl)\bigl) = h\bigl(h^{-1}({x})\bigl) = x] by the definition of $\text{ker}(h)$ (and using the fact that $h$ is surjective). Thus Corollary 1.9 applies and $\varphi$ is continuous.

[gentle osprey]

okeyokay

[heady skiff]

What a headache

[heady skiff]

yes

had to break out the commutative diagrams for this one

[rain nova]

Sorry guys but can an open set has a closure? I get confuse by the definition of closure of any subset$A$ of a topological space $X$ is the least closed set contained $A$. Thanks for the response!

[gentle osprey]

Tanxs

[plush folio]

Every set has a closure, including open sets. For example, the closure of (0, 1) is [0, 1]

[iron bolt]

yep. the closure is the smallest closed set containing A, and you can always construct it by taking the intersection of all closed sets that contain A

since arbitrary intersections of closed sets are closed

[opaque scroll]

Why couldn't O have boundary points in X besides itself?

Is there an assumption you haven't stated?

Like if X is the space of real numbers and O is (0, 1) then it's closure is [0, 1], i.e. O plus it's boundary {0, 1}

[rain nova]

My fault I thought I delet already but it did not! The question solved once I realize this will be a issue when the space $X$ is connected.

[gentle osprey]

Tanxs

[rain nova]

Nope, I think I get confused by the definition of closure, it did not require set $O$ to has an boundary point as it element (which caused the question asked because open set contains none of it boundary point).

[gentle osprey]

Tanxs

[tidal lynx]

If S is closed in X and int(T) is empty then why must int(S u T) = int(S)

**int is the interior and u is union

If U is an open subset of S u T, I need to show U is contained in S / does not intersect T

Now I’m stuck

[opaque scroll]

Consider U \ S

[tidal lynx]

it’s open

and a subset of T

Ok thank you

[fluid herald]

Does anyone here know anything about the Ellentuck topology? I just learned about it and it seems very interesting. The space is the set of countable infinite sets of natural numbers and the open sets are those that can be defined as follows. Take a finite set $s$ of natural numbers and an infinite set $x$ of whole numbers such that $max(s) < min(x)$. The open set corresponding to this would be the set of infinite sets of whole numbers $z$ such that $s \subseteq z \subseteq s \cup x$ It has the neat property that a set is nowhere dense iff it is completely Ramsey null, and has the Baire property iff it is completely Ramsey. I wonder what other interesting or cursed properties it has. Not sure where to put this.

[gentle osprey]

θωθ^-1

[low flame]

would [0,1] -> S1 : t -> cos(2pit) + sin(2pit) make a path connected curve for the circle?

no its (cos(2pit), sin(2pit))

[paper wedge]

this is a parametrization of the circle

[paper wedge]

the continuous image of anything path connected is path connected

[low flame]

well yes, i agree its the parameterization of the circle, but its important to define this i believe right if i wanted ot actually prove it.

[low flame]

and i can use this to say that S^n is path connected whenever n > 0 right

[paper wedge]

no ur using that [0,1] is path connected

[low flame]

yes, but you have to have a function that maps [0,1] to S^1

[paper wedge]

yeah which is what u have no?

[low flame]

yes i do? haha im sorry

because like if this was [0,1] to R^n i would have t: (1-t)x_i + ty_i

ah no shit i see what you mean, since [0,1] is path connected, then the image will be as well

[low flame]

Thank you for that bit of knoweldge 😛 its like regular connected

[paper wedge]

yw

[inland thistle]

I just want to check the quick fact: Let A in X be a subspace. Then if x in cl(A) (closure of A), then any neighborhood of x contains some point a in A? is this correct? what other facts do we also have

[paper wedge]

yeah with a not being = x

every nbd meets A in a point other than x

that's what it means for x to be a limit point

the clossure is the union of the set with its limit points

[inland thistle]

thx for ur clarification

[paper wedge]

yw lancer evo!

[cedar jungle]

Let $f: X \rightarrow Y$ be a continunous function. Let X be separable. How can I show that f(X) is also separable?

[gentle osprey]

Gapi

[pallid comet]

try considering f(A), where A is a countable dense subset of X. Let U be open in f(X). Then...?

[cedar jungle]

then $f^{-1}\left( U \right)$ is open in X.

[gentle osprey]

Gapi

[pallid comet]

...then?

[tender halo]

there is only O(1) things to try

[pallid comet]

also (sorry for interrupting Gapi - dm me if your question is swept away) i have a question of my own:

Are there any compact topological spaces with an uncountable collection of disjoint open sets? (i.e. does compactness imply the Souslin property?)

[cedar jungle]

Then $A \cap f^{-1}\left( U \right)$ is not empty and it is countable

[gentle osprey]

Gapi

[cedar jungle]

Then $f \left( A \cap f^{-1}\left( U \right) \right)$ is countable

[gentle osprey]

Gapi

[cedar jungle]

but what now?

[pallid comet]

f(A) intersect U is ...

so f(A) is ...

[tender halo]

sure

none of them are metrizable of course because souslin implies separable there

[red yoke]

First uncountable ordinal + 1

[tender halo]

or take [0; 1] in some high degree, that should also work

[cedar jungle]

countable

[tender halo]

high meaning at least bigger than 2^omega

so 2^2^omega works I think

wait actually no, Hewitt Marcewski Pondiczery implies that it doesnt work

[cedar jungle]

so f(A) is countable

but is it dense too?

[tender halo]

does beta (discrete space of cardinality 2^omega) work i wonder

yeah it works unless im missing something

[vague hinge]

Hey, in the standard topology on R how would you prove that the closure of [a,b) = [a,b]? Ive solved problems similar to this one in different ways but none of them are formal and precise enough for me and I can’t figure it out. Appreciate all the help

I mostly used the fact that closure of A is the smallest closed set containing A

[pallid comet]

hm. you have shown that for any open set U of f(X), U intersect f(A) is nonempty.

[vague hinge]

But Id like to do it using the formal definition

[pallid comet]

thank you!

[pallid comet]

wait actually im not sure how this would work? can't you take the open cover consisting of all the singletons?

[pallid comet]

is b a limit point of [a,b)?

let p<a. is p a limit point of [a,b)?
let q>b. is q a limit point of [a,b)?

[tender halo]

and do what with it

[pallid comet]

the cover has no finite subcover

and so it's not compact

[tender halo]

i am talking about Stone Cech compactification of 2^omega

not the space itself

[pallid comet]

...oh

[vague hinge]

So basically I say that [a,b] is the closure then show that any closed set that contains [a,b) must contain b aswell?

[pallid comet]

if you want to be formal here i think "the closure of A is A union all of A's limit points" is better / easier to work with than "the closure of A is the smallest closed set containing A", so maybe try to think in that direction

[vague hinge]

I dont think we had limit points introduced or Im just not familiar with the name in english. Our definition of a closure was the union of all closed sets containing A

[pallid comet]

wait so how did you define closed sets

[vague hinge]

a set A is closed if A^c is open

by A^c i mean X\A

[cedar jungle]

so f(A) is therefore dense in f(X). Thank you

[sly geyser]

I've just realised that the triangle inequality is actually more important for metrics than I thought
it means that for a point P in an r-ball around O, the (r-d(O,P))-ball around P is a subset of it, because
d(X,P) < r-d(O,P) -> d(X, O) <= d(X,P) + d(P, O) < r - d(O,P) + d(O,P) = r

[queen prism]

yea triangle inequality is indispensable

[low flame]

im unsure of how to even do this proof honestly? if its path connected, then for x1, x2 in A s(0) = x1 and s(1) =x2 thus obviuosly s(I) is a subset of A and are connected path

[queen prism]

so A being path-connected means for any two points x, y in A there’s a continuous map (in the subspace topology) s : [0, 1] -> A joining s(0) = x to s(1) = y

does the latter statement imply this?

and does this imply the latter as well?

[low flame]

yes, i mean obvioulsy so to me?

we pretty much wrote the same thing above. i can see though how this is in the subspace topology though

[queen prism]

yea it’s not supposed to be hard, just being careful about what is taken with respect to what

[low flame]

okay i skipped ahead and did the proof that showed that is X is path-connected then is Y, so thats prob why im overthinking this.

[heady skiff]

here, would the fibers of just be S^n x {1}, and every singleton in S^n x I \ S^n x {1}?

[mighty hull]

What is CX? The cone over X?

[heady skiff]

yea, i was able to show that it's a homeomorphism

[worn fable]

How is point set topology? I am considering taking a course for it, but I am uncertain about what it is like.

[quick crane]

I wouldn't take a course in it. I learn it as I need it for more interesting things.

[worn fable]

Interesting, why wouldn’t you take a course for it?

[sonic crane]

bruh is that Janich

[sonic crane]

what are the more interesting things that use topology?

[queen prism]

I don't think janich's examples look like that unless I'm looking at a different book

[opaque scroll]

Functional analysis, algebraic topology, algebraic geometry, differential topology, more stuff probably

[plush folio]

I would definitely take a course in it. Topology is so fundamental, so it's nice to get a thorough introduction to it

But I guess it depends on what you intend to study later, some subjects need it more than others

[quick crane]

analysis, including PDEs, differential geometry, probability theory, statistics.

[sonic crane]

how does probability theory use topology? thats interesting

[worn fable]

Thanks for your input. To be honest, I definitely don’t need it, but I am just weighing my options as to what is the most interesting to me.

[quick crane]

Metric spaces and functional analysis are generally sufficient, but you need to know some general topology for functional analysis. See https://link.springer.com/book/10.1007/978-3-031-29040-4 to see it in action.

[queen prism]

functional

[heady skiff]

nah it's rotman

[rare creek]

can I get a hint on part b. im not totally sure where to start

[steep kite]

Take a countable base for X, it will be of the form {Bn | n natural}, now for every n chose an element xn in Bn. Consider D={xn | n natural}, this will be your dense subset.

[rare creek]

holy shit

i had that written down on my pape r

and i crossed it out

💀

THANK U bruh

[steep kite]

Good luck haha

[rare creek]

hopefully there wasnt apush hw

cuz i been doing math for like 2 hours

fuck it we ball

[steep kite]

And another thing, for this to work every Bn has to be non empty, so you have to exclude the Bn’s that are empty

[pliant elk]

what's required for a real-valued continuous function to not have any "holes" in its range? (ie. if f:X to R is continuous for a topological space, are there any "nice" properties that prevent f from not attaining values in [a,b] but with values both >b and <a?)

like, if X=R, there aren't any holes. If X is the trivial topology, then any function is continuous, so you can easily make holes. If X is the discrete topology, then only constant functions are continuous, so there aren't any holes. Is there a middle ground that makes it true?

[rare creek]

i would assume connected and locally connected topologies

[red yoke]

Connected suffices

[prime elbow]

I am not sure about what the closure of A looks like.

But I think closed sets are a form of R, an empty set, (-∞, a].

And closure of A is intersection of all closed sets containing A

[red yoke]

Yes

You can pick your favorite set A and work out an example

[prime elbow]

Set of natural numbers are dense in this topology

[red yoke]

Yup

How about negative integers

[prime elbow]

Finite sets are not closed in this topology

[red yoke]

Yup

[prime elbow]

Say -N, so closure of -N is (-∞, -1]

[red yoke]

Yea

[prime elbow]

But how can I generalize this ?

[red yoke]

Do you know what a supremum is

[prime elbow]

Yes I thought about it

So if sup A exists then closure of A is (-∞, sup A), in this case sup A is not in A
If it is in A then (-∞, sup A]

And if sup A does not exist then A is dense

[red yoke]

Is (-∞, supA) closed

[prime elbow]

No😕

Say A = (0,1) then sup A = 1

Closure of A is (-∞,1], because 1 in cl A

If I take any open set containing 1 the it is form of R or (1-e, ∞) so by sup A it is intersect with A

Thank you @red yoke

[wise mist]

$\overline{A} = {x \in \mathbb{R} \mid \forall a < x, , A \cap (a, \infty) \neq \emptyset }$

[gentle osprey]

yeshua

[wise mist]

does this get a pass ?

[gentle osprey]

yeshua

yeshua

[wise mist]

so convergence of a sequence here secures UC so there is nothing to say anymore right ?

now what if the norm is L_1
is the map still continous ?

[dawn tundra]

are metric spaces compactly generated?

or do you have to assume locally compact or smth

[wise mist]

to which my answer is

[gentle osprey]

yeshua

[wise mist]

am i wrong in my arguments ?

[tender halo]

yes

you dont need metric, being first countable is enough

[jagged prairie]

I've been trying to learn about Topology, but the difference between Open and Closed sets has me stumped. They say that an open set U of superset X does not include the boundary of U, but how does U not have a boundary? Isn't a boundary inherent to an interval/area/volume? Or is it that U does have a boundary, but that boundary is not part of the set U?

[opaque scroll]

It's that U doesn't contain any part of its boundary.

For example, say X is the real line and U is the open interval (0, 1). Then the boundary of U consists of the two points 0 and 1, neither of which are in U.

Conversely the closed interval [0, 1] has the same boundary, but this time the set contains it's boundary

[jagged prairie]

Ah okay, so it's not that it doesn't have a boundary, just that the set does not contain it. Thank you

[opaque scroll]

It can happen that a set doesn't have a boundary, in which case it's both open and closed (because it both contains all of and none of its boundary)

For example the empty set is such a set

[jagged prairie]

And then (1, 3] would be neither open nor closed, because it contains half of its boundary?