point-set-topology

[paper wedge]

ioirc

or even latere

and it had like fundamental groups, homology

euler char

[cloud kindle]

I think ive seen someone talk about this, im not sure if here or in a book, as a a pedagogical issue. We teach maths in a very bottom up way, staring from the axioms and building the theory, but this is the exact opposite of how these things are done in practice.

[paper wedge]

yeah textbooks suck

like actually suck

the only good textbooks are the textbooks taht teach u through problems

[cloud kindle]

People had some problems, some properties they wanted and they worked out what they needed for it, over time what they needed got refined into the axioms we have no, see dropping Haussdorfness, so it can be hard to see why this is the case when you dontknow the motivating problems that lead to those definitions

[paper wedge]

yeah

thats why vladimir arnold is goat

and his views are the right one

😄

[paper wedge]

s

[cloud kindle]

Im not sure I woud agree with this, I actually think that the bottom up approach is fine, its more efficent for learning, but it can certainly lead to a lack of motivation for stuff

[paper wedge]

i think for me

its more efficient for learning but like not more efficient for being able to do harder stuff imo

like u get to see the subject polished correctly but it doesn't train you much to be able to like go to the known from the unknown

which is kinda what a mathematician does

imo

[cloud kindle]

@rapid olive As for the specific problems that lead to the definitions we have now, im not sure, history of maths is not my strong suit. Its an interesting question though so let me know if you find anything!

[queen prism]

I think a lot of learning incl. math is best done in several passes

each pass more thorough than the last

[paper wedge]

yeah def with anything not just math

imo

[fringe thorn]

somewhat related: Arki had a spiel about this many months ago
#advanced-lounge message

[rapid olive]

ok that's helpful

[rapid olive]

thanks

I'll try and prove the typical axioms follow from this

[queen prism]

if I were to add something
a major turning point in the history of math was when mathematicians realized how much information you could glean from studying the functions on a space, particularly their invariants
topology, in particular, centers around those functions that are continuous and those properties preserved under continuous maps, such as compactness (another mathematical idea that took ages to perfect) and connectedness
so whatever definition of a topological space you have should respect this

[paper wedge]

true

[rotund venture]

does anyone know a good example of a regular space which isn't normal? i'm having some trouble wrapping my head around the difference

[dire dove]

Sorgenfrey plane

(it's the usual example for "product of normal space is not necessarily normal" so i guess it's a "good" example)

[red yoke]

Every LCH space is T3.5

But not necessarily T4

[white oxide]

This insight is I think a bit more modern. People like poincare did their stuff before most of the modern language was established. The viewpoint that the most important aspect to look after is the functions between objects is sometimes coined the relative viewpoint that Grothendieck pushed a lot.

That is to say, there was no real turning point. I mean, in AG it definitely was but outside of that it was a more subtle change over time. The prototype of this transition would be group theory and their actions imo

[paper wedge]

truee

[meager yoke]

hi

i believe this channel may be more appropriate for my question

I've been trying this for a while now

[paper wedge]

I BELIEVE YOU OWE DETECTIVE BATISTUA AN APOLOGY

[meager yoke]

what

[paper wedge]

https://www.youtube.com/watch?v=D5HkiYxPcC0&ab_channel=DexterRecap 0:35

[meager yoke]

my bad I thought this channel was for questions on point set topology

I'll go away

[paper wedge]

no

what

im just joking

what do u need help with

[meager yoke]

I was just joking back

but yeah that's the question

I've tried a bunch of variations of multiplying the euclidian metric by a term that gives 1 in the circle, goes to infinity near 0 and goes to 0 in far away points, but it always breaks the triangular inequality

tried some other things too but they don't come close to working

[paper wedge]

whats the question

[meager yoke]

the question is there before your joke

[meager yoke]

.

[paper wedge]

maybe normalize the norm?

[meager yoke]

then how will the ball be unbounded

[paper wedge]

maybe like

the maximum of the usual metric and 1 or something

[meager yoke]

no lol

then the ball is bounded

there's a bunch of stuff it needs to satisfy

• equivalent to euclidian metric
• coincides with it in the circle of radius 1 centered in 0
• any open ball in the euclidian metric is unbounded in the delta one
• the set of x² + y² > r is bounded in the delta metric

I'd have a suspicion that this is impossible, but every time I thought the question was wrong turns out I was wrong

[paper wedge]

yeah i honestly dk

[meager yoke]

yeah that's why I'm asking the question here

I've thought for a lot of time about it

if the question is in fact wrong it probably comes down to showing that delta can't be finer than the euclidian

[paper wedge]

can u just argue that R^2-{0} is homemoprhic to S^1

and then equip S^1 with the standard metric

these balls >r must be bounded

[meager yoke]

what is the homemorphism?

[paper wedge]

wtf am i saying my bad

S^1 x R not just S^1

[meager yoke]

yeah

[paper wedge]

meant homotopy equivalent ig

lmao

[meager yoke]

I'm doing metric spaces so anything beyond the basics is probably not needed to solve the question

[meager yoke]

I was thinking about this

because since x² + y² > r is bounded in delta, then there must be a ball in delta that contains it, which makes it very hard for me to imagine that a ball in delta could ever be contained inside a ball in euclidian metric

[paper wedge]

ig yeah it has to do with lilke the topology and heine borel

[meager yoke]

unless there are "2 types" of balls in delta, the ones that contain those x² + y² > r and the ones that don't, which are smaller balls that can be contained inside euclidian balls

sounds like it shouldn't happen

[paper wedge]

well what ur saying is that

the metric shouldl be equivalent

to the euclidean

[meager yoke]

I'm arguing on why the question may be impossible because of that requirement

but I wouldn't hold my breath for that

[paper wedge]

i meant like since they are equivalent they ahve the same open sets

just replying to ur "2 types" thing

[meager yoke]

oh ok

didn't know about that

[paper wedge]

yeah thats what i was thinking of like idk hows that impossible they should have the same bounded sets too

how s that possible*

[meager yoke]

if that is true then the question is impossible

[paper wedge]

yeah it is

d_1(x,y) <= Ld_2(x,y)

[meager yoke]

wait but that's strong equivalence no?

[paper wedge]

idk whats that

[meager yoke]

well the way I learnt it is that there is:

[paper wedge]

two metrics are equivalent iff the identity function is a homeomorphism

so like

it should be impossible

[meager yoke]

oh ok that is what I learnt as weak equivalence

[paper wedge]

oh your talkigng about like

unifomrly euqivalent

[meager yoke]

and strong would be that there is constants A and B such that Ad1 <= d2 <= Bd1

[paper wedge]

nah forget abuot htat

yeah yeah

but in general equivalent metrics do induce the same topology

[meager yoke]

hmm

ok how would I go about proving that they have the same bounded sets

if they don't then there is a set in one which cannot be contained by a ball in that metric which then cannot be contained by a ball in the other metric ig

ok probably smth like that

[paper wedge]

nah thats not true

[meager yoke]

huh?

[paper wedge]

the standard metric and the standard bounded metric are equivalent

yet the latter has every set bounded

i meant like bounded sets go to bounded sets

[meager yoke]

standard bounded metric being d/1+d?

[paper wedge]

yeah

or min(1,d(x,y))

those induce the same topology but one is bounded

[meager yoke]

wait so what exactly do you mean

[paper wedge]

thats not true

[meager yoke]

.

[paper wedge]

.

[meager yoke]

.

[paper wedge]

what 😄

[meager yoke]

you're saying what you said is true or not true?

[paper wedge]

the statement that if two metrics are equivalent then every bounded set in one is bounded in another is false

[meager yoke]

ok

then what did you mean by "they have the same bounded sets"

[paper wedge]

thas what i thought

then i thought about it

[meager yoke]

oh ok

[paper wedge]

and its wrong.

i think ur idea should be the right one

or wait ig

[meager yoke]

ok thats only strong equivalence

still

it does sound weird for delta to be finer than euclidean

[paper wedge]

yeah it does for strong equivalence

by just basic algebraic inequality playing

where did u find this problem

[meager yoke]

list of exercises from a book

hmm

I can think of one thing that wouldn't be wholly unnatural

maybe balls in delta centered in 0 are just the x² + y² > r, and anywhere else it becomes a ring

maybe like 1/|x| > x² + y² > r

brb

back

[paper wedge]

i mean yeah you can define it piece-wise

on like the open ball and the complement

but it would be anightmare to prove that this is a metric

[meager yoke]

yeah

[obsidian vault]

There is a very easy way to do this.

Try to find a bijection f to some suitable metric space (X, d) and then consider the metric (x, y) -> d(f(x), f(y)) on the space you started with

[meager yoke]

oh ok I imagine it must be the stereographic projection perhaps

I'll think about it

[obsidian vault]

That’s a good idea

[meager yoke]

hmm ok maybe I just didn't think about it well enough but

I tried doing the projection back from R² - {0} to S²

that makes x² + y² > r bounded, but x² + y² < r is also bounded

unless I use some other metric that makes the bottom of the sphere unbounded, but then it feels like I'm just going back to the plane again

actually nvm I'm being dumb, will think of another thing

[obsidian vault]

Oh wait, my bad, forget this comment

Maybe it helps if you think about the complex plane

[meager yoke]

the idea crossed my mind but never in a very meaningful way

right now considering d(x/|x|², y/|y|²)

[obsidian vault]

That’s almost z -> 1/z if you think about it

[meager yoke]

yeah

I feel like this problem I'm damned if do, damned if I don't

I need to "invert" the scales to make the bounded ball unbounded and vice versa

but if do that then how are the metrics gonna be equivalent

[obsidian vault]

Oh I totally overlooked that. Well that’s impossible. The property of being bounded is not changed by passing to an equivalent metric, so it’s impossible to find such a metric

[meager yoke]

that's what we were talking about before

the other guy said this might not be true

is there a proof for that somewhere?

[obsidian vault]

It’s really just spelling out the definitions, start with the definition of boundedness and use the inequalities from the equivalence to prove that a set is bounded wrt to the Euclidean metric if and only if it is bounded wrt delta

[meager yoke]

but isn't that strong equivalence?

the question is referring to weak equivalence, the identity from (M, d1) to (M, d2) is continuous and its inverse is too

idk if that's the terminology in english

[fierce lily]

It asks me to use the above theorem to show that product of Xi must be compact, but I think I can apply tychonoff thm to argue that Xi is compact implies any product of Xi is compact directly, so I am confused whether this can be related to the above statement.

[obsidian vault]

Oh that’s super confusing wording

Usually "equivalence" means strong equivalence

[meager yoke]

lol

yeah here its the other thing

that's pretty much the discussion we were having before

[obsidian vault]

In that case, stick with the homeomorphism idea

[meager yoke]

strong equivalence its obvious but I'm not sure about the weak one

[obsidian vault]

So this, but use a homeomorphism

[meager yoke]

eeeh

I've been at the question for 2 days

I'll try it some other time, can't really think of anything new

[obsidian vault]

Weak means homeomorphic as topological spaces?

[meager yoke]

no idea I didn't see topological spaces much yet

[obsidian vault]

It’s the only thing that would make sense to me

[meager yoke]

weak equivalence here just means: $\forall \varepsilon > 0, \exists \delta > 0: B_{d_1}(p, \delta) \subset B_{d_2}(p, \varepsilon)$ and vice versa with $d_1$ and $d_2$ switched

[gentle osprey]

Sires

[obsidian vault]

You could use Tychonoff, but that is not what the exercise had in mind.

[fierce lily]

how you will answer this question using the preceding thm in that exercise

[obsidian vault]

Yes, this means that the metric spaces are homeomorphic, i.e. they have the same topology

Try to prove this with z -> 1/z, it should just work (TM)

[meager yoke]

you mean the thing I was doing?

[meager yoke]

.

[obsidian vault]

That’s not exactly the same, it should be
$(x/\sqrt(x^2 + y^2), y/\sqrt(x^2 + y^2))$

[gentle osprey]

wolftoeter

[meager yoke]

isn't that just normalizing the vectors?

[obsidian vault]

Oops yes, leave out the roots!

[meager yoke]

which is what I was doing

[obsidian vault]

Almost

[meager yoke]

?

[obsidian vault]

I want to divide by (x^2 + y^2) in both components, you want to divide by x^2 in the first and y^2 in the second

[meager yoke]

oh those are not components

x and y are vectors

x/|x|² meaning $(x_1, x_2) / x_1^2 + x_2^2$

[gentle osprey]

Sires

[meager yoke]

so I was struggling to find how to prove this new metric is finer than the euclidean one

for every ball in the euclidean one I have to show I can find a ball in this new one with the same center that fits inside

[obsidian vault]

Ah yes sorry, this is the same thing, good!

[meager yoke]

I may be mistaken but it looks like balls in this one are like x² + y² > r

[obsidian vault]

Not all of them

[meager yoke]

say y is the center, x are the points in the ball
$$
r > \delta(x, y) = \left| \frac x {|x|^2} - \frac y {|y|^2} \right| = \left| \frac x {|x|^2} - \frac y {|y|^2} \right| + \left| \frac y {|y|^2} \right| - \left| \frac y {|y|^2} \right| \geq \left| \frac x {|x|^2} \right| - \left| \frac y {|y|^2} \right|
$$

[gentle osprey]

Sires

[meager yoke]

so
$$
r + \frac 1 {|y|} > \frac 1 {|x|} \Rightarrow \frac 1 {r + 1/|y|} < |x|
$$

[gentle osprey]

Sires

[obsidian vault]

Use that the image of a compact set under a continuous map is compact

[meager yoke]

so yeah I don't see how its not similar to x² + y² > r

[obsidian vault]

You are essentially reproving that x -> x/|x|^2 is continuous

[meager yoke]

ok but does my point not stand?

delta(x, y) < r implies that the magnitude of x is greater than something

maybe there is an upper bound too?

that's the hope I can find

[obsidian vault]

Yes

[obsidian vault]

You might have an easier time if you use the continuity of x-> x/|x|^2 directly

[meager yoke]

you mean that if f is continuous then d(f(x), f(y)) is equivalent to d(x, y)?

[obsidian vault]

It also needs to be invertible with continuous inverse, but then yes

[meager yoke]

oh great

[obsidian vault]

That’s a general fact you can try to prove

[meager yoke]

yeah I think I can see there is an upper bound too

the bigger |x|, closer x/|x|² is to 0, so if 0 is not in the ball there must be an upper bound

thanks a lot for the help

[obsidian vault]

I hope it wasn’t more confusing than helpful, I should read more slowly

[meager yoke]

nah, all those little things I already stumbled upon during all the time I was trying to solve

[obsidian vault]

Also, look at your favourite Calc book’s proof of the continuity x -> 1/x if you want some inspiration

[meager yoke]

oh I know how to do that

[obsidian vault]

I am too old for that so I forgot 😂

[meager yoke]

I am having to do that too many times nowadays so I can't forget

the funny thing this x/|x|² was one of the first things I tried but for some reason I assumed that "inverting" things does not make it a metric

I was struggling to prove that if x != y then d(x, y) > 0

now that I tried again it wasn't nearly as difficult as I thought lol

anyways thanks a lot

[chrome plank]

I'm trying to self study general topology(started with Martin D. crossley's essential topology)

And I've got a question I'm curious about

If we consider the set of real numbers with the standard topology (R,τ), what would its boundary, ∂R, be?

[cloud kindle]

What is the definition of the boundary in your notes?

You should have a nice characterisation that makes this quite clear

[chrome plank]

∂S={x∈X/ ∀N(x), N(x)∩S=A, and N(x)∩S'=B}

N(x) any neighbourhood of x

B and A generic nonempty sets(since I can't write the not equal symbol in here cuz I do not know LaTex)

X the total set of the topological space(wrote R there before because I was writing it in terms of R)

Oh, and S' the complementary set of S, forgot to specify that one

[queen prism]

I will say
no matter what you believe about \mathbb, it will never be as bad as \mathsf{R}

[unreal stratus]

Oof

[chrome plank]

Anyone here who could help me with this?

[rancid umbra]

hint: ||R is a neighbourhood of x for any real number x||

[sterile jungle]

no, {X} is an open cover of X and X can be an element of open covers of subsets of X

[chrome plank]

Hm, then that means no real number could be within its boundary

[rancid umbra]

yup!

the boundary, you mean, right?

[chrome plank]

My idea was that ±∞ could be the frontier of R, but then again, since infinity behaves weirdly, I'm not sure I can affirm it

[chrome plank]

Yup, sometimes I get the notation messed up

[rancid umbra]

i wouldn’t think of it as an “infinity being weird” thing

it’s the fact that R is connected

you can play the same game with R^2

what is its boundary?

[chrome plank]

A connected set can't have a boundary? Don't see how the definition of connectedness gives way to that

[rancid umbra]

in a connected space, the only non-empty subset with empty boundary is the space itself

[cerulean oriole]

Visually, the point is that ±∞ are not in the ambient space, so they can't be in the boundary of any subset.

[chrome plank]

OH

[queen prism]

well
the boundary of any topological space with respect to its own topology is always going to be empty...

[chrome plank]

So it's essentially because the boundary is contained within the space, duh

And cause R is connected, so there exist no “holes" in it that could constitute part of its boundary

[paper wedge]

prove that if a set has empty boundary then it must be closed and open at the same time

[chrome plank]

Isn't that vacuously true?

[cerulean oriole]

This is true, but it doesn't explain why ∂ℝ is empty IMO. The other half, that in any space the empty set and the whole space have empty boundary, does.

[paper wedge]

idk how ur saying that

forget about intuition for a second and drawings and shit

boundary(A) = cl(A) intersc cl(X-A)

[cerulean oriole]

Tautologically, but not vacuously.

[paper wedge]

now write out elements and prove stuff

then think about it

[chrome plank]

A set is open if it contains its boundary, and closed if it doesn't, if the boundary is empty, the set both contains and doesn't contain its boundary, right?

[paper wedge]

what

[rancid umbra]

yes, but the key is that it’s the only one

[cerulean oriole]

(Assuming ∂A := closure(A) \ interior(A).)

[chrome plank]

That's what comes to me inmediately, of course, I still gotta think about it

[paper wedge]

yeah too much weird things for me here haha im not following anyone

[rancid umbra]

no, this is just wrong

[paper wedge]

what happened to definitions 😄

good luck everyone

[chrome plank]

Alright

[paper wedge]

one last thing

open up munkres

and go to the section that has closed set and limit points

theres an exercise that outlines everything here

with definition

all u need to know is what a closed and a limit point is (by definition)

[chrome plank]

Guess I'll still have to think about this, but dunno if I'll be able to get it

[paper wedge]

it's not that deep

[rancid umbra]

that’s why i said it’s the only non-empty one with empty boundary

[paper wedge]

ur overcomplicating it by avoiding definitions

[queen prism]

why does this matter if we’re trying to find bd(R) in R

[chrome plank]

I assume what you mean by “definition“ is that a set is closed if its complementary is open?

[paper wedge]

yes

and a set is open if its in the topology

a limit point of a set is a point such that any opepn set containing this point intersects the set in a point other than the original point

[chrome plank]

This confuses me cuz I've seen multiple statements regarded as definitions but that seems to be the most prominent one

[paper wedge]

the closure of A is the intersection of all closed sets containg A

[chrome plank]

Alright

[paper wedge]

which is = A U set of limit points of A

prove this

and then define the boundary to be CL(A) intersec CL(X-A)

[rancid umbra]

my bad. ur right.

[paper wedge]

you can prove (easily) that this is = to cl(A) - int(A)

[sharp aurora]

why is there no precise definition for an open set in a topological space?

[paper wedge]

and by easily i mean just unfolding the definitions

[paper wedge]

an open set is a set in the topology

[sharp aurora]

is that really all it is?

[paper wedge]

once u do that it becomes clear that if a set has no boundary then it must be both open and closed

a seperation for a space X is a pair of disjoint nonempty open sets that have union X

if a set is both closed and open , call it A

consider X-A

then A and X-A is a seperation

thats it

[chrome plank]

Isn't a limit point a point such that any neighbourhood of it has a non null intersection with both the set and its complementary

Or am I again complicating it?

[paper wedge]

no that would imply that every element of A is a limit point

u must intersect not just non emtpy but in an element that is not the original point

the intuitino behind this is just to define sequences

and limits of sqeuences

in metric spaces the limit points are exactly the limits of sequences u can form from elements in this set

only the converse true in gen top spaces

[chrome plank]

For any non limit interior point, there would exist a neighbourhood that doesn't intersect the complementary though, no?

[paper wedge]

so for example {1/n |n in N} has limit point 0

[paper wedge]

an interior point is a point in the set such that there exists a nbd contained in the set

so what ur saying doesn't really follow

or ig yehaa

what ur saying is a limit point is either a point in the boundary or a point in the interior ?

[chrome plank]

Hm, well, I don't really know how to explain it all that well, since I'm still a beginner in this field

[chrome plank]

Yeah

[paper wedge]

then ur right

[paper wedge]

i am too

and i think u should stick to the definitions

[chrome plank]

And that for any point in the interior, there exists a neighbourhood that doesn't intersect the complementary

[paper wedge]

if a point is in the interior then yeah theres a nbd that can't intersect the complement

[chrome plank]

That's what I was trying to get at

[paper wedge]

yes this is just set theory

[chrome plank]

But regardless, I'll take that into account

[paper wedge]

A subset of B iff A intersec B^c = empty

[cerulean oriole]

That's the boundary (not necessarily by definition, depending on your definition).

[paper wedge]

yes

you can think about the boundary as points that intersec both the set and its complement

like think of the circle in R^2

the boundary of this circle (in R^2) is the circle

[chrome plank]

See, something that really confuses me is that I've seen and heard of different statements being regarded as definitions

[paper wedge]

every point indeed in the circle has points touching outside and inside

[chrome plank]

And isn't a limit point just a point in the boundary?

[paper wedge]

a point in the boundary is a limit point of both the set and the complement

[chrome plank]

Alright

[paper wedge]

bd(A) = cl(A) intersec cl(X-A)

[chrome plank]

Hm, alright

[cerulean oriole]

That's because they're equivalent and all useful ways to think about the concepts. It helps to pretend one of the definitions is the "definition" and the others are "properties" when learning it, of course, but different textbooks would have different preferences on which one to give you as the definition.

[paper wedge]

you should be able to prove everything

like you should be able to

[paper wedge]

start with one, say this ,

and be able to prove all the equivalent definitions

[chrome plank]

Most likely, but I've still got a long ways to go before I can tackle any and all proofs on topology

[paper wedge]

why

[cerulean oriole]

I would suggest you pick one set of definitions and make that the "real definition" in your head, and prove that the others are equivalent to it as you come across them, until you feel comfortable switching between definitions.

[chrome plank]

So I'm first trying to build intuition and follow the book I've chosen

[paper wedge]

point-set topology i think has easier proofs (in the begining level i mean)

and by easier i mean like usually the proofs and exercises are like

almost straightforward

alot of the proofs and good properties boil down to basic set theory

(not always tho, uryshons lemma for example lmao )

[chrome plank]

Cuz I started with a book less than a week ago, and I'm a first year undergrad that started about 1/2 months ago, so I've still got lots to learn

[paper wedge]

yeah i think u can check out munkres and begin with chapter 1

[chrome plank]

I mean, most of the ones I've seen so far aren't all that hard, but again, I'm (naturally) still lacking in rigor, and not yet accustomed to formal proofs

[chrome plank]

Well, right now I'm trying to make use of Martin D. crossley's Essential topology

So far, so good, lemme check the book you mentioned real quick

[paper wedge]

i havent heard about that book tbh

[paper wedge]

yeah munkres is pretty known

i hope topology doesn't get dry for you tho

it can get dry very quickl

point-set atleast

[cerulean oriole]

Does that book use one set of definitions consistently?

[chrome plank]

I haven't gotten all that far yet, so I wouldn't be able to tell you, but I'm fairly sure it does

So far, it's stuck to R, it generalizes the notions by chapters 3 and 4

[cerulean oriole]

Ah.

[chrome plank]

And the only actual important general definition it's given me, the closed set one, it gave me in terms of the complementary being an open set

[cerulean oriole]

So I suppose open, closed, etc. would all be defined with epsilons?

[chrome plank]

For now, yes

Oh, and it defined an open set over R as a set such that for any x in the set, there exists an open interval that contains it

Which, I assume, is generalized by replacing the notion of interval by the notion of neighborhood

[white oxide]

You mean there exists an open interval sandwiched between x and the open set

[cerulean oriole]

That's one way. Another is to take open (instead of neighbourhood) itself as basic and define everything else in terms of thate

[chrome plank]

In the words of the book: that there exists some “breathing space“, that is, an interval (x-ε,x+ε)⊆A ∀x∈A

[white oxide]

Yeah, just saying since your previous message was a bit unclear

[white oxide]

The "algebraic" and "geometric" ways to conceive of it lol

[sharp panther]

Is wedge sum of two compact spaces again compact?

[red yoke]

A finite union of compact spaces is compact

[mighty hull]

Yes. The disjoint union of two compact spaces is compact, and every quotient of a compact space is compact.

jinx

[sharp panther]

oh yes

thanks

[prime elbow]

i think this question is not correct if Y is not Hausdroff space, because take X = {1,2} = Y = Z, X with discrete topology, Y with indiscrete topology and Z with discrete topology.

then f and gof is continuous but we can construct identity mapping g such that it is not continuous, correct?

i mean Hausdroff is sufficient not necessary

[trim gate]

The phrasing of this question is a huge red flag anyway.

But your counterexample looks correct to me.

[prime elbow]

why?

[trim gate]

Wrong arrows for functions, bad English grammar, the assumption about g is phrased as a statement.

[prime elbow]

i see

[trim gate]

If my students wrote down sentences like these, I would definitely subtract points.

[low flame]

max{p(x1,x2), p(y1,y2)} = 0 implies that p(x1,x2) = p(y1,y2) right?

then p(x1, x2) = p(x2,x1) so then max{p(x1,x2), p(y1,y2)} = max{p(x2,x1), p(y2,y1)}

Then, i can take (x1,y1) (x2, y2) (x3, y3)

max{p(x1,y1), p(x1,y1)} <= max{p(x1, y1 ), p(x3,y3)} + max{(p(x3, y3), p(x2, y2)}?

is that how i can show this?

because i can write these balls in a metrix as B( (x0,y0) < r) in {X \times Y | p((x,y), (x0,y0)) < r}

[low flame]

so does that mean its better to not answer?

[prime elbow]

But how will you show that this metric induces the same topology as X × Y ?

[low flame]

This is the hint

so i think im kinda doing it?

idk haha sorry

[prime elbow]

You showed that p is a metric function but you have to show they generate the same product topology

[low flame]

is this like saying p(xo,x) < r, and p(yo, y) <r

then p(xo,x) \times p(yo,y) < R?

so then idk X \times Y < R?

[prime elbow]

You know the basis element of X × Y product topology?

[low flame]

yeah its the intersection of the preimages

[prime elbow]

Oh

Can you show { U × V | U is open in X and V is open in Y } is a basis for product topology on X × Y ?

[low flame]

i was thinking of sub-base sorry

[prime elbow]

No problem

[low flame]

and its gonna be take the product of U intersected with the product of V is open

[prime elbow]

I think you can show by definition of product topology

Now show that if you pick arbitrary open set U in X and arbitrary open set V in Y then U × V is open in metric space X × Y with metric p

Maybe this is not the correct way but if I want to do this question then I would do in this way

[low flame]

oh i dont know man im just a hobbiest

not a professional. so i appreciate someone talking to me about it

[paper wedge]

R has the order topology (standard)

but also R is a metrizable with the standard metric

RxR has the product topology which is also metrizable

how did u go from the metric on R to the metric on RxR?

generalize

@low flame

[low flame]

I think of it as (a,b) X (c, d)

then you just use the metric on it. which would be the line between two points i believe in R times R.

Thank you.

Also its like gonna be B(x,e) X B(y,e)

[sharp panther]

is every locally euclidean and compact space second countable?

Let X be such a space. Then for every point x \in X, there exists a neighborhood U_x such that it is homeomorphic to R^n. Consider {U_x}_{x \in X}. Then it is an open covering of X. Since X is comapct, there exists a finite subcover {U_x1,...,U_xn}. Let V be an open set in X. Then V = U (V \cap U_x) which is a finite sum. The problem here is that I have to add these intersections to my collection of sets which is supposed to be the basis. However, this set may be uncountable if there are countably many open sets V. How do I resolve this issue? Or what am I missing?

[near idol]

how does munkres use the intermediate value theorem here? ivt needs a function to go from a connected space to a space with the order topology. but S^1 does not have an order topology, right?

[knotty vine]

But U does

[chrome plank]

I was wondering since it showed up, is the empty set both closed and open under any given topological space?

or does there exist some topological space such that the empty set is just open, just closed, or neither?

[cedar jungle]

Every topological space contains an empty set and the entire space. Closed sets are complements of the open sets. Open sets are all the sets that are in the topology. So empty set and the full space are always both open and closed at the same time.

[chrome plank]

alright

[cedar jungle]

How can I show that every isometry is an embedding? I can show that isometry is injective and continuous. Do I need to prove anything else?

[tender halo]

well you need to mention why images of open sets are open in the subspace topology of the image

[cedar jungle]

Maybe I can look at images of open balls?

[tender halo]

sure, that works

[worldly magnet]

On its face, it is debatable - seeing as though undefined algebraic objects can be embedded over the complex plane.

I should say the embedding itself uses modus ponens, but the implication of complex solutions assures its reality

[meager yoke]

hi

once again i was doing a metric space question and I feel like this should be impossible but I'm probably wrong

just want to discuss it

so I had to prove that given M, N two metric spaces and $A \subset M, B \subset N$ then $\partial (A \times B) = (\partial A \times B) \cup (A \times \partial B)$

[gentle osprey]

Sires

[meager yoke]

ok so I was thinking

what if we just take the reals for the spaces and $A = (0, 1)$ and $B = (0, 1)$

[gentle osprey]

Sires

[meager yoke]

then 0 is frontier of both A and B

and the point $(0, 0)$ seems to me pretty obviously frontier of $A \times B$

[gentle osprey]

Sires

[meager yoke]

but $0 \not \in A$ and $0 \not \in B$, which contradicts the assertion

[gentle osprey]

Sires

[meager yoke]

any ideas on where I'm going wrong?

[meager yoke]

I'll try and prove this just in case:
Let $r>0$, let's use the maximum metric, take $p = (r/2, r/2)$, then it follows that $d(0, p) = \max{|0 - r/2|, |0 - r/2|} = r/2 < r$, so that's a point inside $A \times B$, now, for a point outside take $-p$, the distance is the same due to the absolute value and $A \times B$ does not contain points with negative coordinates, thus, $(0, 0) \in \partial(A \times B)$

[gentle osprey]

Sires

[meager yoke]

ok at this point I'm pretty convinced the question is wrong, but I'll leave this here just in case

[light herald]

yea i think it's wrong

yea, A and B should be closed on the right side

[gentle osprey]

ТТерра

[light herald]

does it also work if you add $ \partial A \times \partial B $ to the original one

[meager yoke]

Yes, I proved that much but then was trying to show it cant be in that set

And couldnt find any way to do it

Thanks for checking if im sane

[light herald]

i also want a sanity check
every metric space is sequential right? and by extension unitary and normed spaces too with induced topology

[meager yoke]

I believe there is some profound conexion here, I heard we use partial to indicate both frontiers and derivatives for a good reason, didn't bother to look into it though. Cool observation, didnt notice

Ill look into it, eating rn, thx

[heady skiff]

is (iv) asking me to give a counterexample or prove it for all nonconstant f

[paper wedge]

no

prove it by contradiction

[heady skiff]

so the former right

[paper wedge]

yeah

latter

or whatever

[heady skiff]

wait wdym 😂

they're different

the first one or the second one

[queen prism]

give a counterexample to what statement exactly

you use a counterexample to prove an implication false, not true

[paper wedge]

prove it for all nonconstant f

[paper wedge]

sec

[heady skiff]

no i was confused as to if it was asking fo ra counterexample or not lol

obviously counterexample does not mean everything is false

[heady skiff]

okay 👍

also I have no clue why this is true

[queen prism]

you have a statement of the form “if P then Q”
you would use a counterexample to prove “there are examples of P yet not Q”

[heady skiff]

i can show that it's equal to p for 0 <= t <= 3/4

oh wait

then I can use that to show it's equal to p on 3/4 <= t <= 1

I hope

let's see

yeah never mind I have no clue

Suppose that $(i_p * f)(t) = f(t)$ for all $t \in \mathbb{I}$. Then if $0 \leq s \leq \frac{1}{2}$,
[(i_p * f)(s) = (i_p)(2s) = p = f(s)] so $f(s) = p$ for all $0 \leq s \leq \frac{1}{2}$. Suppose that $\frac{1}{2} \leq s \leq \frac{3}{4}$. Then $0 \leq 2s - 1 \leq \frac{1}{2}$, so
[(i_p * f)(s) = f(2s - 1) = p = f(s)] by above.

Like there's no way to proceed from here for s >= 3/4, can't use other cases

because then 1/2 <= 2s - 1 <= 1

so something tells me this result is false

since we could have f(s) not equal to p for t >= 3/4 but still i_p * f = f

[gentle osprey]

okeyokay

[heady skiff]

damn the proof requires some finicky induction stuff

and picking k such that t < 1 - 2^-k if f is nonconstant

wild

[wise mist]

$$A=\bigg{\sum_{i=1}^{\infty} \frac{a_i}{5^{i}}\ :\ a_i\in{0,1,2,3,4} \bigg} \subset \mathbb{R}.$$

$A$ contains a open interval

[gentle osprey]

nastasya

[wise mist]

but how ?

dont those 5 individual series give me individual points, and it can only get to rational points no ?

[unreal stratus]

No because you have an infinite sum

[wise mist]

its GP

so can i not say its a_i5/4

let $(X,d)$ be a metric space then $X$ has exactly how many dense subsets from following
$$ 3,4,5,6 $$

[gentle osprey]

nastasya

[wise mist]

for this i have thought like this

X \ {a}, X \ {b}, X \ {a,b}

only 3 can be constructed like this, is this a good argument ?

[wise mist]

i have issue with the statement (3)

isnt [a,b) satisfying the condition as we get a as a cluster point ?

still there is a sequence converging to b but b isnt in the set

idk where im stuck !

[rancid sand]

The subset {b-1/n |n>N_0} for some N_0 is infinite (countably) and yet doesn't have a cluster point (in the set).

[wise mist]

for it to be coompact u need the converging point to be included to the set no ?

[queen prism]

hence [a, b) is not compact

[a, b) itself has a cluster point, but not every infinite subset (in the metric topology on [a, b)) does

(in fact every point in [a, b) is a cluster point of [a, b))

[wise mist]

so its "every subset of X....." ?

[prime elbow]

Every infinite subset of X

[wise mist]

i totally missed the subset

and i took it as every infinite X

[strong briar]

Hello you all, I'm new here
I wanna know is this enough to answer
and excuse my writing it's all over, I will rewrite it later (not in english)

[cerulean oriole]

It looks correct to me.

[strong briar]

Thanks

[bronze wadi]

are graph isomorphisms stronger that graphs homeomorphisms? i.e. if 2 graphs are isomorphic, are they automatically homeomorphic?

thx

[wispy dirge]

isnt this wrong

$E^o$ here is the set of all interior points of $E$ so $E=E^o$ means that every interior point of $E$ is a point of $E$ and every point of $E$ is an interior point of $E$

[queen prism]

what should the correct statement be

[wispy dirge]

each of its points is interior doesnt guarantee that there is no interior point of E which isnt an element of it

[queen prism]

does such a point exist?

what does it mean to be an interior point?

[wispy dirge]

ah wait nvm now i get it

tysm

[tawny tundra]

Does anybody know of any resources that explore convergence in ℓ-sequences where ℓ ∈ ORD? (for example, a special case is ω-sequences where this gives us the notion of a sequence as a function f : ℕ → _)

[chrome plank]

May be too basic of a question but was wondering and wanted to be sure

Is the definition of a closed set a double implication? Like, we know that a set is closed if its complement is open

But, can we assure a set is open if its complement is closed?

[queen prism]

wait

ok

let A be a subset of X and suppose X \ A is closed in X

what does it mean for X \ A to be closed?

[plush folio]

In general all definitions are bidirectional, even though they may not be written like that. Say if I define X to be a foobar if it satisfies some predicate P(X). That really means X is a foobar if and only if P(X)

(assuming of course this is how you defined a closed set. If this is a theorem then you need to prove both directions)

[queen prism]

very true although this is a different problem in this case
since it's not "my subset A is the complement of an open set, does that mean it's closed?" but "my subset A is the complement of a closed set, does that mean it's open?"

the latter is something you'd have to prove

unless you started with something like "Def. A set is open if its complement is closed."...

[plush folio]

Ah, good point

[chrome plank]

That its complement is open?

[queen prism]

ok

write down its complement

what are the elements of this set?

[civic verge]

point b is not an adhesion point.

[worldly magnet]

If there were an class of infinitesimals associated with the infinite set - the points would cluster, otherwise - we wouldn’t associate the infinite set with countability

[queen prism]

which chatbot tech is this

[plush folio]

Definitely a bot, just spouting nonsense and never replying

[worldly magnet]

Me?

Are you suggesting as a mathematician - I’m not allowed to posit the necessity of a class of infinitesimals associated with the character of the set?

Or perhaps your Booleanesque dichotomy really is insufficient to describe phenomena in a world of entangled scalar patterns

[queen prism]

hold on this is better than a chatbot

[red yoke]

Christmas hyena

[wise mist]

im incapable of understanding that language

[queen prism]

the following statements about a metric space M are equivalent:
M is compact
every infinite subsequence of M contains an accumulation point
every sequence in M possesses a convergent subsequence
M is complete and totally bounded

and I don't know this language at all, bless romance languages

[dapper kestrel]

If I prove that $\mathbb{R}$ with the cocountable topology is not locally compact because:

Given a $x \in \mathbb{R}$ and U a neighborhood of $x$, I have that U it's $\mathbb{R}$ minus some (countable) points, then the closure of U would be $\mathbb{R}$, and as $\mathbb{R}$ is not compact which means that that topological space is not locally compact (?)

That last part is what bothers me, because it's like i'm proving it's not compact by saying it's not compact, but local compactness and compactness are two different thing so it could be right?

[gentle osprey]

pablo

[wise mist]

yeah its every infinite subset of X

there is a subtle error in the text im reading and that why i was confused

wrong tag sorry, i meant to reply blade

also, kevalt, can i hope to get more analytic and organic explanation of your poems, they arent bad, im just too sort of a dumb idiot for comprehension

[queen prism]

oops this is wrong too, "has" not "contains"

[worldly magnet]

I look at math with my own linguistic lens and library

My research include matrix method analysis for string theory, moyal weyl star product, and non-commutative scalar fields, and this thing I call phenomenological velocity, which is connected to spinors

https://zenodo.org/records/13997317

https://zenodo.org/records/8433050

https://zenodo.org/records/10962859

https://zenodo.org/records/13959848

I won't send more though, I don't want to blow up this message box with my paper

I'm still learning though, and any time I can find a bridge between things I know and other parts of math language, it's helpful

I hope I don't get modded for posting this

[wise mist]

my subhumanistic brain doesnt comprehend anything of it

[worldly magnet]

yeh...

well, no, but I was hoping you could prove my bizarrely complex mathematical conjecture

oh well... I guess we'll just throw it on the pile of other bizarrely complex mathematical conjectures

[worldly magnet]

In point set topology, a Cantor set is a
perfect, totally disconnected, compact set that is uncountable, yet has measure
zero in R. Embedding such a set in R^2 suggests that the level set exhibits
extremely intricate, self-similar structures at every scale.

[dapper kestrel]

Can someone check this please :)

[languid patrol]

slay

[worldly magnet]

Yes, that’s a theoretical sketch - but the updated version is here: https://zenodo.org/records/13953707

[worldly magnet]

https://www.authorea.com/users/852917/articles/1240975-proof-of-the-riemann-hypothesis-using-sweeping-net

[worldly magnet]

since the closure of any neighborhood (U) is (\mathbb{R}), which is compact, the space satisfies the condition for being locally compact at every point.

[gentle osprey]

\blue {kevlat}

[wispy veldt]

read what they wrote again, R is not compact.

[wispy veldt]

your proof is fine, the connection between comactness and local compactness depends on your definition of local compactness and compactness lol

in certain cases compactness is stronger

with certain definitions you can have compactness without local compactness

if you define local compactness as "every point has a compact neighborhood" then ofcourse compactness is enough

https://math.stackexchange.com/questions/727635/find-an-example-of-a-compact-space-which-is-not-locally-compact?rq=1 in the other case you need hausdorffness

[worldly magnet]

I think the assumption is false

People treat R like a field, but in reality, it is a projection quotient spaces under modular reduction

“ the projective limit of a system
of quotient spaces under modular reduction”

[worldly magnet]

ok, by using the symbol, "0," you've inherently embedded a contradiction in your notational language, because the symbol itself exists. - you might not like it, but that's a fact

[worldly magnet]

true

[warped rover]

slander, #foundations is real math

[chrome plank]

Any element within the space and “outside“ A?

[mighty hull]

In general, if you have a set A ⊆ X then

X\(X\A) = A

That's true just from the basic properties of sets and the definition of set complement/difference.

[queen prism]

yea
just be careful when you’re trying to evaluate something like A \ (A \ B) (which pop ups when you’re working with subspaces) because in general it isn’t the same as B, unless B is a subset of A

[tender halo]

if f: X -> Y is a closed continuous map and U \subseteq X is an open set, is there a special name for the set f^-1(Y \ f(X \ V))?

[red yoke]

I guess you could call it $f_\forall(U)$

[gentle osprey]

Arki

[red yoke]

If you denote also denote the image of U by f∃, you have an adjunction of posets

[tender halo]

ooh i see, thats cool

[red yoke]

$f_\exists \dashv f^{-1} \dashv f_\forall$

[gentle osprey]

Arki

[red yoke]

So this translates to the fact that "not ∃x not φ(x)" is the same as "∀x φ(x)"

[low plaza]

Lets consider the topology on R τ={(a; +infty), a£R}U{Ø,R}, and let τ' be the euclidean topology.
Characterize the continuous functions from (R, τ) to (R, τ').

I tried to do it, my result is that such functions are all and only the constant functions, is it correct?

[mighty hull]

How'd you arrive at the result?

[golden bronze]

best intro to point set topology textbook? i need a suggestion

[gentle girder]

munkres.

[golden bronze]

that’s what i’ve heard as well

thank you!

[gentle girder]

what are you using it for? just to learn point-set comprehensively?

[golden bronze]

what would you say the prereqs are?

[gentle girder]

depends

[golden bronze]

my goal is to move to differential geo

a lot of textbooks suggest point set as a prereq

[gentle girder]

you should know calculus then

[golden bronze]

i know up to multi variable

[gentle girder]

hhh it’s hard because like, you don’t really need to use munkres if you only want to learn about manifolds

because munkres has a lot of silly spaces too

[golden bronze]

i’m interested in algebraic topology as well, I just figured point set it the place to start

[gentle girder]

but i think it’s interesting in its own right.

[golden bronze]

i’m also not saying i’m only learning it to move past it, i’m interested in the content as well

[gentle girder]

check out munkres, also try checking out Bredon’s topology and geometry

and also Lee’s introduction to manifolds

and also Tu’s introduction to manifolds

[golden bronze]

yes I wanted to read lees, although i scanned the appendices and my knowledge of topology and algebra seemed too little

[gentle girder]

okay, maybe you might like Loring Tu’s book, which other people might not like as much but i really like it

to start

[golden bronze]

i’ll check it out, thanks for all the info!

[deep lotus]

were you reading introduction to topological manifolds by lee or introduction to smooth manifolds?

the former includes an introduction to point set topology, with an emphasis on what will be important in differential geometry

the latter does asume you’ve already learned some topology, so it sounds like you may have been looking at the wrong one

[white oxide]

Potentially consider just using this
https://pi.math.cornell.edu/~hatcher/Top/Topdownloads.html

It's 50 pages of the topology thats sufficient for doing alg top, and only really lacks stuff on metric spaces of diff geo possibly. If you enjoy the subject you can do the other chapters in the above books.

If you don't like Hatchers writing style just use the above as a high priority to-do list ig

[versed pivot]

Topology can be axiomatized via sets of limit points A' (and many other notions like closure, boundary, etc, see here: https://en.wikipedia.org/wiki/Axiomatic_foundations_of_topological_spaces)

Is the set of isolated points A^i = A \ A' also sufficient to determine the space?

I can't exactly solve for A' from that definition so it's not quite obvious

An easy answer would be to find two distinct topologies on the same set for which every subset has the same isolated points in both cases...nothing comes to mind for me though lol

[tribal palm]

ahaha i love this exam problem

don't state the urysohn lemma, but carefully state the urysohn lemma

[granite slate]

doesnt this retraction violate the "no ripping" intuition of continuity?

[mighty hull]

I'd get the idea of "no ripping" out of your head when thinking about continuity. You're probably thinking of progressively shrinking the torus to a circle, which is called a deformation retract. Or maybe just homotopy equivalent, a weaker condition than being a deformation retract.

After all, any constant function from the torus is continuous.

[versed pivot]

"No ripping" could be interpreted as preserving connectedness which is true and not violated by that example

[mighty hull]

Maybe, but was trying to interpret it in a way that made his reaction sensible!

[versed pivot]

Yes your explanation is very helpful I was going to follow up with something similar but you said it more elegantly than I would have 😂

[sonic crane]

math profs love doing that lol

[chrome plank]

So the book was trying to show that the subspace topology on the rationals is not equivalent to the discrete topology, but it made the assumption that within any given interval on the reals, there exists more than one rational number

why is that? Because any given interval is connex?

[gritty widget]

because rationals are dense in real numbers set

[chrome plank]

that means that between any two given rationals there exist infinitely many

[gritty widget]

yes

[chrome plank]

but to generalise that we would need to show that within any given real interval at least 2 rationals exist

[rapid olive]

*between any two given reals

[gritty widget]

between any two real numbers there exists rational number

[chrome plank]

hm, alright that makes more sense

thanks

[rapid olive]

e.g. in the interval (a,b) there is a rational q
then in (a,q/2) there is another distinct rational

and you can keep going to get infinitely many

[gritty widget]

now assume there is two real number a < b, then there exist rational r such that a<r<b, now apply again on a and r we get rational number r_1 such that a< r_1 < r < b, continuously you will get infinitely many rational numbers between a and b

[cerulean oriole]

What topology should be put on the direct sum of a family of topological (resp., locally compact Hausdorff) abelian groups to make it the coproduct in the category of topological (resp., locally compact Hausdorff) abelian groups?

[white oxide]

The topology should be related to its universal property. You probably already fixed the inclusions using the group structure so what's left is determining how to make
A. Them continuous and B. The induced map continuous

[cerulean oriole]

Well yes. I was hoping for a more explicit description of the topology. It should be at least as fine as the product topology (because the inclusion is characterised categorically and should be continuous). Is it the product topology? How is it related to the final topology with respect to the inclusions?

[cerulean oriole]

Alright, the join of group topologies is a group topology, so for any topology on a group, there is a finest group topology coarser than it. It follows that the answer (in the category of topological abelian groups) is the finest group topology coarser than the final topology wrt the inclusions, which is somewhat sort of slightly explicit.

And infinite products of locally compact groups aren't locally compact unless all (but finitely many) are compact, so dually I "shouldn't" expect the direct sum to work normally unless all (but finitely many) are discrete.

[knotty vine]

Why do you expect the categorical (co)product of these things to be the same as that of their underlying topological spaces?

[cerulean oriole]

The argument shows that it won't (although note that it's also no longer clear that it exists), and I'm slightly less interested in knowing the answer if not.

Mostly a matter of priorities.

I "expect" the restricted direct product to be more "nice" for generally locally compact groups.

[knotty vine]

Im not sure about locally compact Hausdorff abelian groups, but the category of compact Hausdorff spaces has (infinite) products and coproducts given by the Stone--Cech compactification of the underlying spaces

[cerulean oriole]

Right, that makes sense.

[balmy nexus]

the theorem: if K is a family of compact sets in $\Omega = \mathbb{R}^n$, s.t. for any finite subset $K'\subsetK$, $\cap K' \neq \emptyset$, then $\cap K \neq \emptyset$

my proof:

assume for contradiction that the intersection of K is empty. fix $k \in K$. Then

$(\bigcap K)^c = \Omega$

$(\bigcap_{k'\inK} (k\cap k'))^c = \bigcup (k\cap k')^c = \Omega$

so as for all k', $(k\cap k')^c$ is open, this is an open cover. k is compact, so there is a finite subcover. Thus:

$k \subset \bigcup_1^n (k \cap k_i)^c$

$k \subset \bigcup k_i^c$

so

$k\cap \bigcap k_i = \emptyset$

reaching a contradiction

can you check my proof? is there an intuitive explanation? Im not sure "why" this works

[gentle osprey]

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

[balmy nexus]

i also tried with converging sequences but i couldnt prove for uncountable Ks with that method

[prime elbow]

Why do you take k'?

You can use your idea that all elements in K are closed set

So intersection empty means the complement of elements of family F makes open cover for R^n

No wait

See it's called FIP which is equivalent to compact space in metric space

But I am missing something here

I think your underlying must be compact

Underlying set

[balmy nexus]

it was a question on R^n

but i didnt refer to that in the proof so where am i missing it?

[prime elbow]

Can you explain your proof?

[balmy nexus]

not much other than "it works"
we choose a specific k in K which we will use to reach contradiction. We then rephrase the intersection as intersections of k with everyone else so we can use it later on
we assume that the intersection is empty, which means the complement of the intersection is all of the space.
using de morgan we rephrase the intersection to an union of open sets (as compact sets are closed in R^n)
then since the union is all of R^n it is also a cover of k. k is compact, so there is a finite sub-cover, and then we de-morgan again to reach the contradiction

I dont know about compact sets on more general spaces so maybe they arent generally closed?

my current explanation to why this works is that the finite subcover property lets us move from an uncountable property to a finite one, and that's, so far, the only way I can see of doing that.

[prime elbow]

Yes they are not closed in arbitrary topological space, they are closed in Hausdorff space

[balmy nexus]

it seems to me that an intersection proof like this should go through lemma of zorn style arguments, but I couldn't see how

I even looked up the proof we used back in logic for compactness of logical sentences which was much more of a zorn style but i didn't see how to translate it

[prime elbow]

It looks correct but I am not sure

Which book is that?

[balmy nexus]

herbert enderton an introduction to logic

he expands the logical formulas he has to a maximized size set

[prime elbow]

But this topology question from which book?

[balmy nexus]

my question? it's a geometric infitisemals class

[prime elbow]

Okay

[balmy nexus]

if you're interested but it's nothing special

[prime elbow]

Thank you

[balmy nexus]

thank you! 🙂

[prime elbow]

Have you taken a metric space before?

[balmy nexus]

you mean a class?

[prime elbow]

Yes

[balmy nexus]

prof. vinikov does all the point-set concepts you can see in the sheet in R^n context

[rich cliff]

you look like alex kruckman @balmy nexus

[balmy nexus]

i suspect the class also functions as a introduction to intrdocution to topology class

[balmy nexus]

why thank you, i suppouse

[prime elbow]

I mean you do your some assignment questions in general metric space setting

[balmy nexus]

well we can also go through coordinate convergence if we want to in that sheet so it makes it easier

[odd gulch]

Is this proof OK?

this is C btw

[sharp aurora]

how are holes defined

[paper wedge]

rank of hom

[balmy nexus]

i thought about it a bit and you dont need to take k' it just needlessly complicated the proof

[prime elbow]

Oh

[balmy nexus]

if you have the time problem 12 in that page was really fun. Nice buildup of concepts

[prime elbow]

Yes I already did this one and I did in both settings, in general setting

You can try in general metric space

You can refer rudin

[balmy nexus]

its nice how in 3) you use the metric from 2) to show there is a minimum distance between the compact set and the complement of the open set

you can probably prove this by getting your hands dirty, but instead it was very elegant. I felt my brain expanding a bit

[prime elbow]

Yes

[exotic hemlock]

Just started on topology, one of the property that doesn't generally hold is that any infinite intersection of open sets is open. Is there an example of a topology where this holds? Excluding the discrete/indiscrete topologies

[queen prism]

take a look at Euclidean space

say, the real line

[exotic hemlock]

Wouldn't the intersection of the nested open intervals (-1/n,1.n) which is 0 not be open

[queen prism]

oh where it holds

the topologies you are looking for are called https://en.wikipedia.org/wiki/Alexandrov_topology

[exotic hemlock]

ah yes this was what i was looking for, thanks for the link

[odd gulch]

how can I see that O(x) is the smallest open nh of x

[mighty hull]

O(x) is the intersection of all open sets containing x. This always exists and is open in a finite topological space (why?).

cl({y}) is the intersection of all closed sets containing y.

[low flame]

is this a good?

[paper wedge]

its pretty unclear honestly i cant follow

a basis for XxY is {U x V| U and V are open in X and Y respec}

and hence an open set would be a union of sets of this form

now apply ur projection map

[low flame]

yeah i thought of this. Also, ive shown that x is an interior point, and x is arbitary, thus all points are interior points, so the set is open

[iron bolt]

yeah, the proof does work like that

I just think it's not written down very clearly

but the idea is indeed correct

[warm kettle]

What properties of a topological space significantly affect the structure of its bases? The "best" topological spaces are those that admit a partition being their base. If a space is separable, it admits a countable and dense base (note: if a space admits a countable base, it's also separable). But in the case of an arbitrary topological space, we cannot say anything about its bases; they can be arbitrarily complex. What can I assume about a space to make it easier to analyse its bases (assuming I can choose any of them)? Or what else can I assume about a base itself?

[iron bolt]

idk what you mean by dense base - if a basis of a topology is a set of sets, while density is a properties single subsets of a space can have

that aside, spaces with countable bases are also called second-countable - so the note you already mentioned is that second-countable spaces are separable

aside from that though, I don't know any topological properties that concern bases of the whole topology. properties about bases at a point are a bit more common

for example, a space is called first-countable if each point has a countable neighbourhood basis

and you can also more generally say some nice things about spaces where every point has a linearly ordered neighbourhood basis, though I forgot which

[unreal stratus]

I don't get what you mean with the best spaces having a base partitioning the space. That'd make the space disconnected or indiscrete in general

[warm kettle]

Sorry. I went a bit ahead of myself. But if a space is metric and it’s separable, it admits a countable base (which may be a useful property)

[unreal stratus]

Well not just disconnected but also such that each component is indiscrete

[warm kettle]

If a space has a partition that also happens to be its base, then we’re dealing with a discrete or an “almost discrete” space. Analysing it is extremely easy in most cases (you can, for example, easily classify all of homeomorphic spaces with a given one that satisfies that property. If a base is more complex, it’s far more difficult)

[unreal stratus]

Sure but that doesn't mean it's "the best", but rather a somewhat degenerate space

[warm kettle]

Fair enough

[median sand]

Let E be an equivalence relation on X. I know that if X->X/E is open and E is closed in X^2, then X/E is Hausdorff. Does the same work if X->X/E is closed?

[tender halo]

wdym by "E is closed in X^2"?

[median sand]

E is a relation on X, by definition it's a subset of the product X^2=XxX.

[tender halo]

ohhh

[sharp panther]

does product of mobius band and unit interval have Klein bottle as a boundary?

I am trying to show that Klein bottle is a boundary of some 3-manifold

[low flame]

Rational numbers are dense in R right. For any open set U of (a,b) in the standard topolgy we can find rationals s,t where a < s < x <t<b for each x in U?

[odd palm]

how are you in topology not knowing that

[odd palm]

I think circle instead of mobius band

could be wrong though

actually you have to glue an additional side from the mobius x circle product for a boundary to be the Klein bottle

[warm kettle]

Mate, they’re trying to learn it, so they ask questions

[thorny agate]

yes this follows from the definition of density

[low flame]

thank you

[paper wedge]

yes

[sonic crane]

Fr!

[dusky wasp]

Is there any prerequisite for learning topology?

Btw , what the heck is topology really?

I am not getting much feel of what it is ?

[warm kettle]

The basics of foundations of mathematics. A bit of real analysis may also be helpful

But even if you haven't done anything, you can give it a go

[random sinew]

Prove that second-countability means any open cover of the space has a countable subcover
does the following proof work
every set in open cover is a union of basis elements
let S be the set of basis elements that are in one of these unions for one of these sets
S is a subset of a countable set so must be countable and it obviously covers so we're done

[warm kettle]

Almost. Keep in mind you need to take the "whole" open sets from your original cover (unless I misunderstood your description. I hope I didn't)

[random sinew]

what exactly do you mean?

[warm kettle]

You showed that you can use a countable number of basis elements that your open sets are made from to cover the whole space. However, your task is to prove that there exists a countable subcover of the family of those open sets

[random sinew]

ohhh i forgot what a subcover is lmfao

this is why i shouldnt do maths at 10pm

[warm kettle]

sometimes it do be like that haha

[iron bolt]

in other words, second-countable implies Lindelöf

compactness if countable is good enough to you

[random sinew]

is this a thing we care about

“Lindelof”

obviously having a finite subcover is nice, is having a countable one also nice?

[mighty hull]

LOST was a great TV show

[random sinew]

whar

[mighty hull]

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

[random sinew]

ah

[tiny obsidian]

I mean Lindelöf is nicer than not being able to say anything about sizes of open covers

[warm kettle]

Let $\mathcal{T}$ be the family of all sets $U \subseteq \mathbb{R}^2$ such that the intersection $U \cap L$ with every line parallel to one of the axes is open in the Euclidean subspace of that line.
\begin{enumerate}
\item Verify that $\mathcal{T}$ is a topology on $\mathbb{R}^2$ and the space $(\mathbb{R}^2, \mathcal{T})$ is T2.
\item Show that if $C \subseteq (0, \infty)^2$ is an infinite set, then there exists a $U \in \mathcal{T}$ such that $\vb 0 \in U$ and the set $C \setminus U$ is infinite.
\end{enumerate}

[gentle osprey]

Thingoln

[warm kettle]

I did 1., but I don't know what to do in 2. It's easy to see that any open set in the Euclidean topology is also open in T, but I can't think of any example of a set that wouldn't be open in R^2 but would be in T

[frigid rose]

Is a closed euclidean ball in R² without a boundary point homeomorphic to a closed euclidean ball in R² without two boundary points? I think not, but whats the proper justification?

[warm kettle]

Maybe the connectivity of the border of the set would be a problem here?

A closed ball without one boundary point would have a connected border, but if two boundary points are missing, the border is can no longer connected

(Just an idea. Idk if that works)

[frigid rose]

Idk, how would one go about defining the border of a topological space?

[warm kettle]

The closure minus the interior

Oh, wait. I see what you mean

[frigid rose]

Yeah but thats for subsets of topological spaces

When looking at a ball in R² without a point by itself it should be a topological space on its own and i dunno if looking at its border makes sense

I dunno maybe its definable somehow

[warm kettle]

Another idea I have in mind is that you could assume a homeomorphism exists, connect the two missing points and consider the subspace of that segment. If those spaces are homeomorphic, then so are the segment and homeo(the segment)

[frigid rose]

I could do this and they would be homeomorphic i think

[warm kettle]

Fair enough

Pretty solution ✨

[warm kettle]

bump

[frigid rose]

If by "one of the axes" means you just fix, say, the x axis and check only for lines parallel to it then ]0,1[x[0,1] would be open in T but not in the euclidean topology

[warm kettle]

The line can be parallel to either the X, or the Y axis

Sorry for the ambiguous wording

[frigid rose]

If not then i agree that they would probably be the same topology

Yeah no worries

[warm kettle]

That's the issue because then the problem doesn't make sense. If there existed an open set U containing zero and satisfying the other condition, then I could say that C = {(1/n, 1/n) : n natural}. Then |C \ U| would be finite because almost every element of C would lie in a ball B(0, r) no matter how small r I would take

[frigid rose]

Oh i see what you mean

Huh yeah it would need to have a boundary point at 0

Or something like that

[warm kettle]

Exactly

[frigid rose]

Wouldnt R²\C work in that case?

[warm kettle]

It might not be open

[frigid rose]

Why not? The lines passing through 0 parallel to the x and y axis are both subsets of R²\C so the intersection must be open

The rest are isolated points so no issues there

[frigid rose]

This holds no matter how C converges to 0 since it cant touch the axes themselves

[warm kettle]

Not necessarily. Consider C = (0, infty)^2

Then we could take e.g. the line x = 1, whose intersection with R^2 \ C would be closed in the Euclidean topology of that line

[unreal stratus]

What do you mean by "closed euclidean ball"

[warm kettle]

I was thinking that maybe the fact the space is T2 could help construct an open set, but I still can't even imagine an open set in that topology that wouldn't be open in the Euclidean R^2. On the other hand, that propery alone is most likely insufficient because otherwise the Euclidean R^2 would also yield an open set satisfying those conditions, but it doesn't

[frigid rose]

${(x, y)\in\mathbb{R}^2:x^2+y^2\leq1}$

[unreal stratus]

Oh lol

[gentle osprey]

TheItalianGame

[unreal stratus]

Sure so when you said "without a boundary point" you mean like closed ball minus boundary point

sure

[frigid rose]

Yes

[frigid rose]

Say, that minus (0,1) and that minus (0,1) and (0, -1)

[unreal stratus]

Well one thing is that their one point compactifications aren't homeomorphic

so they can't be homeomorphic

[frigid rose]

Their what?

[unreal stratus]

Given a locally compact Hausdorff space you can consider its "one point compactification" which is a compact Hausdorff space with one extra point

Homeomorphic spaces have homeomorphic one point compactifications

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

Here the one point compactification of D^2 minus a point is just D^2

But for D^2 minus two points you have to "join together" the two gaps and you get something more complicated

Like you take a disk and glue two points to one another

[versed pivot]

I think you can consider a set shaped like this https://en.wikipedia.org/wiki/Maltese_cross

[warm kettle]

So like the quotient space of a disc where you choose two points at a boundary and divide by it?

[warm kettle]

Oh, I wouldn't have thought about something like that, thanks! But wouldn't that shape also be open in the Euclidean R^2?

[frigid rose]

Ohh i see

[versed pivot]

no it's not a neighborhood of the center point

[frigid rose]

Interesting

[warm kettle]

Riiight. That makes SO much sense now

[frigid rose]

And homeomorphisms preserve these compactifications?

[unreal stratus]

Sure, D^2/(a ~ b)

[unreal stratus]

Yes

[warm kettle]

Btw. Potato, I said this based on my geometric intuition, but does it make sense to embed topological spaces in a bigger one (e.g. in R^2 like here) and consider objects like the borders of the original spaces in it when trying to determine if they're isomorphic?

[runic moss]

any good textbooks/websites/videos on topology?

[runic moss]

thanks, gonna download this

[unreal stratus]

Not unless you show that smth is independent of choice of embedding or smth

[sonic crane]

The Goat

[warm kettle]

Does there exist a simple/known example where two spaces are isomorphic, but they're not in an embedding? Or that aren't isomorphic on their own, but they are in an embedding

[sonic crane]

Third edition cover better though

[warm kettle]

I did some googling, but I didn't find anything (but chances are I used the wrong keywords)

[tender halo]

can you rephrase?

wdym by "not in an embedding"

[warm kettle]

Let's say you have two spaces: X and Y. Assume they can be seen as subspaces of a bigger space, say R^n, so you can embed X and Y in R^n and try to analyse them there. My question was: are X and Y isomorphic if and only if they're isomorphic when embedded in the bigger space?

For example, you can see two segments as topological spaces on their own or as subspaces of R

[tender halo]

spaces are either homeomorphic or not homeomorphic, being embedded into some larger space is irrelevant to the question

[warm kettle]

Yeah, it would also probably be difficult/unnatural to rephrase the definition of a homeomorphism to match embeddings

The motivation behind my question was studying properties like the border or the interior, or properties like closeness/openesss

[ember hazel]

Would anyone be able to help me with a question on the metric space C([0,1], R)?

I proved that there is a subset of this space with sup norm less than 1 that is closed and bounded

Now I need to show that it is not sequentially compact.

My prof said I explicitly used this fact in my argument and needdd to state it. But I can’t find this theorem anywhere and can’t understand what he means.

[rancid umbra]

they are saying that you need to prove that a sequence of functions in that space is pointwise converges if and only if it is uniformly convergent.

[ember hazel]

This is the proof that he said is supposedly incomplete. He says I should explicitly state the assumption. Does this mean that pointwise convergence implies uniform convergence?

The assumption being the previous image I sent.

[small gyro]

it is because of the norm no ?

i mean under sup norm

im unable to get where you are stuck at tho i would suggest your to go through a book called topology of metric spaces

[small gyro]

take sequence take converge but not u.c. ?

[hexed steppe]

the uniform limit, if it existed, would coincide with the pointwise one

[random sinew]

general advice for this qn? im finding it hard to come up with examples that fit one and not the other two, or two and not the other one, is there any sort of way i can work backwards from properties i want -> sets and a function satisfying them

an open map is one that sends open sets to open sets, a closed map sends closed to closed

[mighty hull]

What have you looked at?

[cedar jungle]

Let $f: X \cup X' \rightarrow Y$ be a function. Let $f| _{X}$ and $f| _{X'}$ be continuous in $x \in X \cap X'$. Show that $f$ is continuous in $x$.

[gentle osprey]

Gapi

[paper wedge]

hint: f^-1(A) = f_X^-1(A) U f_X'^-1(A)

@cedar jungle

[cedar jungle]

So if I take an open set that contains f(x), let's call it A, then f^-1(A) is also open?

@paper wedge

[paper wedge]

i mean this is the problem we are doing

like this is what it is asking u to do no?

[iron bolt]

as you are stating it, this is wrong for technical reasons

I mean, to start with, I think the terminology you are looking for is "continuous on X ∩ X'", not "continuous in x ∈ X ∩ X'"

I mean, I think it's somewhat understandable either way, but the former is a bit more standard

[paper wedge]

i dont think u need continuity u just need that the restricitions agree

[iron bolt]

you do need continuity if you want f to be continuous

the thing is, you don't just need continuity on X ∩ X'

you need continuity on X and continuity on X'

[paper wedge]

i mean yeah for it to make sense i meant

like for it to be a function

[paper wedge]

yes def

[iron bolt]

and you also need X and X' to be both open or both closed. otherwise there's also easy counterexamples

[paper wedge]

it just follows ffrom here ig

[paper wedge]

yes

the closed form is more known as the pasting lemma

and a nice(?) fact is that you can do this for infinitely many X_is

open

and only locally finite X_is

closed

[iron bolt]

I think pasting and gluing lemma are both used for both the open and closed version

but yes, they're definitely the right key word if one wants to look this up

[paper wedge]

yeah ig whatever the names ig i remember in munkres the open set form wasnt called anything

while the closed set form was called pasting lemma

butw hatevers ig

[hollow geyser]

Every limit between topological spaces X->Y is unique if and only if Y is Hausdorff, correct?

[paper wedge]

what does limit between top spaces mean

[hollow geyser]

Is this a question where you do not know, or where the explanation will make the answer to my question very obvious?

[novel acorn]

do you mean the limit of a net?

[hollow geyser]

to be honest, my book never really mentioned limits in topological spaces, just points of continuity. I tried to transform the metric limit definition into a topological variant. Here's the definition I came up with:

Let $f :X\to Y$ be a function between topological spaces. For some $L\in Y$, we say $\lim_{x\to a}f(x)=L$ if, for any open $V\subseteq Y$, if $L\in V$, then there exists some open $U\subseteq X$ such that $a\in U$ and $f(U\setminus{x})\subseteq V$.

This may not even be a valid definition. I just came up with last night in a thought train

[hoary breach]

i was about to ask if you meant your question to be this https://math.stackexchange.com/questions/1369459/if-every-convergent-sequence-has-a-unique-limit-point-in-space-x-then-is-x before this last post

[gentle osprey]

SWR

[hollow geyser]

Minor correction. I meant to say a in U, not x in U

[visual rock]

presumably you mean to remove a. but yes this works

[hollow geyser]

oh right yeah I did.

But okay cool. And I'm assuming that every limit that exists is unique if and only if Y is hausdorff

[visual rock]

the issue people took with your original statement is that it is phrased as "limit of f", not "limit of f at a"

[hollow geyser]

ah

The common error of getting too deep into your own work and forgetting the lack of perspective others will have when you inevitably reach out

[visual rock]

and the statement is not true. Y being hausdorff implies the limit is unique but the converse is false. even if the limit at every point is unique, that does not mean Y is hausdorff

[hollow geyser]

Even for every existing limit at every function between the two spaces?

I suppose then it has to be hausdorff since I can just do limits of constant functions

and they'd only be unique if distinct points had disjoint open sets

[visual rock]

this argument should just show that Y is T1