point-set-topology

[zealous berry]

I have some lecture notes on hand that do stuff with universal properties and apply it to topologies

[cosmic mirage]

we dont need to do anything special to account for the empty set and X in a subbasis/basis right? like the empty intersections are X and empty unions are empty (intersection = product and union = coproduct in the poset category, and empty lims/colims are final/initial objects)

like we dont actually need the subbasis to be a cover for this reason

[rancid umbra]

i think there are two definitions of subbase and one requires that they cover

[zealous berry]

Yeah the def I worn with just says a subbasis covers X

[rancid umbra]

but you’re right that they don’t need to cover. you freely generate the basis from the subbasis and this contains the empty intersection

[zealous berry]

Emptyset doesn't need to be in the subbasis

[cosmic mirage]

yeah sorry this is just 0 \not\in \mathbb{N} propaganda from the 0 \not\in \mathbb{N} lobby

[zealous berry]

When doing finite intersections and arbitrary unions, emptyset should appear in the topology

And X ofc

Bases don't need emptyset or X yeah?

[cosmic mirage]

nah

[rancid umbra]

nope. they get freely generated for the same reason

[cosmic mirage]

but the empty set will be an empty union and we should require the basis to be a cover

by should require i mean, it follows from things we require

[zealous berry]

Now what are the implications of this for the simple harmonic oscillator modeled by a = -w^2 x

[warped helm]

lots eventually

check strogatz

[zealous berry]

[rancid umbra]

continuity?

[zealous berry]

No I was kidding

I have an ap wxam tomorrow

[rancid umbra]

good luck altanis

[zealous berry]

Oh yeah I forgot about this

Or wait no I didn't

That's how {} is forced into a topology

Or I guess you can view it as emptyset satisfging "U being open implies there is a basis element B such that x in B subseteq U bla bla bla" vacuously

[rancid umbra]

the conditions of a topology are usually stated as:

1. it contains the whole space and the empty space
2. closed under arbitrary unions
3. closed under finite intersections

[zealous berry]

Yep

[cosmic mirage]

hmm i have a technical question about this adjunction, lemme ponder the phrasing a bit more and then ask in the category theory channel

[zealous berry]

And a basis generates a topology because emptyset is forced in

[rancid umbra]

the 1st point is redundant by the 2nd and 3rd points

[zealous berry]

Cuz the emptynunion is defined to be {}

[cosmic mirage]

given the question might include the word "grothendieck fibration" it might not be good for this channel lolol

[zealous berry]

Umm is empty finite intersection left undefined

I think munkres addressed this somewhere in ch1

[cosmic mirage]

i implore you to take the empty finite intersection of subsets of X to be X

[rancid umbra]

might not be good for me either lmao. idk about grothendieck fibrations too well

[zealous berry]

Wtf?

Is this a joke

[rancid umbra]

no

like

[zealous berry]

Why 💀

[cosmic mirage]

all good, sorry my phrasing was just i have a question, not i expect you specifically to answer this

its this thing i said earlier. P(X) is a poset so you can view it as a category, and non-empty intersections and unions clearly are products and coproducts in that category

[zealous berry]

[rancid umbra]

the intersection of sets is like a filter (in a non-technical sense). it is increasingly difficult to be in the intersection of a family of sets as the size of the family grows. as the size of the family shrinks, it becomes easier. so easy that when it is empty, everything should be in it

[cosmic mirage]

so the most obvious way to extend to empty intersections/unions is the empty product and coproduct, which are the final and initial object of the category respectively

[zealous berry]

I see

Yes okay this makes sense

[cosmic mirage]

yeah also like @zealous berry what does it mean for x to NOT be in the empty intersection

[zealous berry]

Smaller set less restriction

[cosmic mirage]

it means there is some subset you are intersecting over such that x is not in that set

[zealous berry]

Ooh

[cosmic mirage]

which there just. isnt

[zealous berry]

True.....

[cosmic mirage]

yeah this sort of thing is called a vacously true statement

[zealous berry]

It was to my understanding this was left to the textbook author to decide

And not simply vacuous

[cosmic mirage]

its like saying like

[zealous berry]

Like saying all elements of {} are positive and negative and zero simultaneously is vacuously true

[cosmic mirage]

every solution of e^x = 0 provides a counterexample to the riemann hypothesis

[zealous berry]

Riemann hypothesis solved,,,

Yeah I understand, but I don't think this is vacuous perse?

Because you can probably find a different way to think about empty intersection

[cosmic mirage]

i mean in a sense unions and intersections of sets dont exist

like shouldnt you only take these inside of ambient grothendieck universe

this is really only a problem for unions

[cosmic mirage]

okay this is a bad phrasing but the point is that it is natural to only think of intersections inside of some ambient bigger set

[zealous berry]

This reminds me of something

[cosmic mirage]

and my argument for this is that empty intersections deserve rights

[rancid umbra]

yea, is your point that like, the empty intersection really should be a proper class?

[cosmic mirage]

actually i dont want to make any points, ignore all i said; what is true is that basis things are always relative to an ambient set so everything works

[rancid umbra]

you can make everything relative to U {your family of sets}

[cosmic mirage]

okay i asked a precursor to my question about fibrations in the cat theory channel to make sure i phrase the actual question correctly

[cosmic mirage]

ok grothendieck fibration questioned asked lol

[cosmic mirage]

is the msg from 24 days ago just like

markov babbler vibes lol

[zealous berry]

Yes

[robust drum]

@cosmic mirage you probably got answered this earlier but no every universal property is just initial/terminal object in some category. Since a limit is a terminal object in category of cones over the diagram, and a colimit is an initial object in category of cocones

[alpine nest]

Is "cocones" really a term?

[tender halo]

yea its a type of pastry

[opaque scroll]

I guess we should be calling then nes instead

[zealous berry]

[tribal palm]

this footnote goes crazy

[quick delta]

Subgroups of F_2 correspond to 4 valent graphs
My favourite fact in maths :3

[tribal palm]

F_2 being… the free group on two elements?

[quick delta]

Yes

[tribal palm]

lovely

[zealous berry]

didnt know math had chemistry in it

[midnight umbra]

wait until bro learns about the look and say sequence

[zealous berry]

💀

mfw Spec Z isn't the bright light spectrum of Z

[cosmic mirage]

when i first heard the term Balmer spectrum, I thought the person was talking about the Balmer series in the light spectrum for hydrogen lol

[zealous berry]

[empty crypt]

does the pasting lemma also work for open subsets?

[polar storm]

yes

iirc, this was one of the exercises in Munkres' book.

[empty crypt]

I have this in my book, it seems like a same version

[unreal stratus]

In fact it is much simpler for open sets

[zealous berry]

beautiful

[wide kayak]

sheaf property detected!

[hidden spade]

for this, could we use the local characterisation of continuity using neighbourhood bases?

[polar storm]

u mean for the proof of pasting lemma for open sets?

[hidden spade]

yes

if we define continuity as a local property

using neighbourhood bases

and show thats equivalent to the standard definition

it should just fall out instantly

[polar storm]

yeah. Actually, the proof follows immediately from the definition of continuity (preimage of open sets)

[hidden spade]

oh ofc, that would be way easier

[hidden spade]

if you havent done this already

[prime elbow]

I want a hint, so I know that in metric space if f: X -> Y is a function such that if x_n converges to x then f(x_n) converges to f(x), then f is continuous.

Now in topological space, it is not true, I want an example, hint?

[prime elbow]

I got it

[rancid umbra]

go through the proof of why, in a metric space, sequential continuity implies continuity, and see what assumptions you are using

[prime elbow]

So here we need first countability right?

What's the relation between nets and continuity?

[opaque scroll]

The statement becomes true in general if you replace sequences with nets

I.e. f is continuous iff whenever you have a net xn converging to x, then f(xn) converges to f(x)

[rancid umbra]

yes

[prime elbow]

So what's the statement if for each nets (x_\alpha) converges to x implies f(x_\alpha) converges to f(x) then f is continuous

Let me prove this

[rancid umbra]

R with the ||cocountable|| topology is not first countable. the ||identity function (R, cocountable) —> (R,euclidean)|| is not continuous, but is sequentially continuous

[prime elbow]

Yes

Btw how do we define subnet?

[gaunt linden]

IIRC there are several different definitions, not all equivalent. (!)

https://en.wikipedia.org/wiki/Subnet_(mathematics) has some discussion.

[empty crypt]

Is it true that every function with the indiscrete topology on its codomain is continuous

I feel like it is

[gaunt linden]

Yes, that follows immediately from the definition of "continuous" in general topological spaces.

[midnight umbra]

yes

the only open sets in indiscrete are empty and whole space

preimage of empty is empty is open

preimage of codomain is the entire domain which is open

[prime elbow]

Okay thank you

[gaunt linden]

Weirdly enough that page doesn't even state they Kelley definition, but it can be found at https://en.wikipedia.org/wiki/Filters_in_topology#Non-equivalence_of_subnets_and_subordinate_filters

[prime elbow]

Yes

[empty crypt]

is the intersection of two connected subspaces connected?

[gaunt linden]

Not necesarily.

[cosmic mirage]

for example consider a line cutting through a circle in R^2

[empty crypt]

but for example intervals are connected in R

[gaunt linden]

Yeah, that's a pretty specific property of R.

[empty crypt]

is every interval in [0,1] connected too?

alright but why?

well it's a way i am proving [0,1] to be locally connected

[gaunt linden]

An interval in [0,1] is also an interval in R, which you might know already is connected.

[empty crypt]

yes

but does it mean that all interval in [0,1] will be connected as well

?

[gaunt linden]

How couldn't it?

[empty crypt]

Wouldnt that mean that intersections of intervals in R are connected

[gaunt linden]

It's true that intersections of intervals in R are either empty or connected.

That's just not the case for all topological spaces.

[empty crypt]

Alright but how would I justify it to prove that [0,1] is locally connected, is the intersection of an interval in R with [0,1] connected in the subspace [0,1] then?

[warped helm]

i mean where are you having difficulties proving that?

pick an x in [0,1], produce a connected neighborhood

[gaunt linden]

"Connected" is independent of any larger space. It's just a matter of how the set itself looks like with the subspace topology -- and that is the same whether you view the interval as a subspace of R or as a subspace of [0,1].

[gaunt linden]

(Beware that for "locally connected" it's not enough to produce one connected neighborhood for each point -- you need a whole neighborhood basis of them).

[warped helm]

ah true

but thats also not hard

[empty crypt]

i want a neighbourhood U around an x in [0,1] to contain an element of the base {(a,b) \cap [0,1] : a,b in Q}. And since all intervals are connected in R i hoped that element in the base would be connected

with what i am hearing here it is the case but i have no real result in my textbook to justify the last conclusion

[warped helm]

did your class not prove intervals are connected?

[empty crypt]

yes we did

[gaunt linden]

The justfification is that the intersection of two intervals is either an interval or empty.

[warped helm]

then what is the issue

you can describe exactly what the intersection of two intervals in R is

[empty crypt]

Yeah but wouldnt that mean that intersection is connected in the space itself and not the subspace (i know this will be the case)

[gaunt linden]

As I already wrote, it doesn't make sense to talk about "connected in the space itself".

[empty crypt]

i am getting what you're saying

[gaunt linden]

A set with a given topology -- such as your interval -- is either connected or not. it doens't matter what you consider it to be a subset of.

[gaunt linden]

Then what is your problem? As long as (a,b) cap [0,1] is not empty, it will be an interval and intervals of R are connected.

[empty crypt]

Okay right then its trivial really

Since not being connected in X will continu to be in subspaces if that subspace contains that set

and since the intersection of intervals in R is always connected that intersection being in [0,1] will keep that property

[warped helm]

whats the reason for this definition?

[unreal stratus]

This is a bit of an annoying issue but usually when you say something is locally X you should be able to check this after shrinking to smaller neighbourhoods / on arbtirarily small neighbourhoods

[warped helm]

i encountered it in my point set class but forgot it required a whole neighborhood basis, but i cant see why other than potentially that its the right definition for doing things down the line

[unreal stratus]

(though there will be uses of locally that also conflict with this probably lol)

[gaunt linden]

It it just meant that every point has a connected neighborhood, then every connected space would be locally connected -- but the concept is intended to capture pathological behavior that can happen even within a connected set.
The standard example would be something like Q×[0,1] cup [0,1]×{0}.
The whole space is connected, but all sufficiently small neighborhoods of (0,1) are disconnected.

[unreal stratus]

Good example

[opaque scroll]

If X is locally connected at x, but a neighborhood U of x is not, then it wouldn't be very local, but instead pretty global

[prime elbow]

So which one do I have to prefer?

[gaunt linden]

My personal conclusion would be to use filters whenever sequences are not enough.

[robust drum]

connectedness and compactness are intrinsic properties. Meaning: A subseteq B subseteq C then A is connected relative to B iff it is relative to C

[limber wyvern]

Hi, can anyone give me a hint about how to proceed with this proof?

“ Prove that a space (X,T) is Hausdorff if and only if the diagonal

Δ={(x,x)∈X×X}

is a closed subset of X×X endowed with the product topology."

[stuck geyser]

Open sets in the product topology of two spaces are generated by open boxes U x V.

||What does that mean if an open box is disjoint from the diagonal?||

[limber wyvern]

Well, geometrically speaking , it means it's separated like two triangles

[limber wyvern]

right?

[stuck geyser]

That’s basically the proof

Is the spoilered part

Brb

[limber wyvern]

I just looked at your name xD, nice metric

[stuck geyser]

||the box U x V around (x,y) disjoint from the diagonal is the box formed by the open sets separating x and y||

Another problem using the “box” topology idea is to show that if A x B is a box formed by compact A and B then it is enclosed in an open box U x V

Hint: ||start with covering singleton {a} x B with open boxes||

[limber wyvern]

I haven't reached the "compact" definition

[stuck geyser]

You can Google it and get the idea

[limber wyvern]

Well, I knew this was like a triangle, but I don't know how to proof it with words xD

I drawed this

[stuck geyser]

Draw a box around a point away from the diagonal, then draw it’s “shadow” on the axes

Those are your open sets separating the points

In a very drafting-centric pov

[limber wyvern]

mmm

I don't see it

[limber wyvern]

why the ghosting😔

[polar storm]

which direction r u struggling with?

[limber wyvern]

both xD

[polar storm]

so for the forward direction, try to show that the complement is open instead. That would be easier

[polar storm]

^

[limber wyvern]

Yeah

Im trying to do that

but um

Im currently in

[polar storm]

What can u say about the pair (x,y) in X^2 \ Δ

[limber wyvern]

"Suppose A is a closed subet in XxX, so XxX\A is open"

I'm stuck here xD

[urban zinc]

Do you mean XxX \ A is open

[polar storm]

Is it possible to have x = y?

[limber wyvern]

yes

[limber wyvern]

Nope

[polar storm]

so for every (x,y) in X^2 \ Δ, x and y must be distinct. Now apply Hausdorffness.

The goal is that we want to find a neighborhood of (x,y) contained in X^2 \ Δ. This is essentially the same as finding a basis element (which has the form UxV) that cannot intersect the diagonal. Here under which condition do we need for U and V such that UxV cannot intersect the diagonal? Think about this.

[polar storm]

What does Hausdorffness say?

[limber wyvern]

Im thinking about what you said

but

given x,y in U,V respectively, the intersection U, V is empty

[polar storm]

yes. Then is it possible for UxV to have a point of the form (z,z)?

[limber wyvern]

nope

[polar storm]

Done

That proves the forward implication.

[limber wyvern]

mmm

give me a sec I write it down formally

btw, how do you see this proofs so quickly

like

in 10 seconds you know how to do it

[polar storm]

mmm it's like a standard trick u see for Hausdorffness.

[limber wyvern]

what trick?

like

[polar storm]

like u can see the immediate connection b/w distinct pair (x,y) not contained in the diagonal and the hypothesis "Haudorffness"

[limber wyvern]

yes

I was missing the fact that if it is open then x != y

even though is obviously

[polar storm]

Also, I've done this exercise from Munkres book Chp 2 tho it has been like a yr ago.

[limber wyvern]

Im writing it down

and I have a question

like

even though the points are like x!=y

Why cannot there be a intersection that is non empty

[stuck geyser]

You can also prove this idea visually

[limber wyvern]

Like, couldn't be two opens like.

U = {(1,2),(2,3)}
V = {(1,2)}

?

[polar storm]

they r closed.

finite sets are always closed in Haudorff spaces

[limber wyvern]

oih

I didn't know that nor it come in my uni book

[polar storm]

they are not open in R^2 with the standard topology

[limber wyvern]

why not?

[gaunt linden]

It's a good exercise to show that every singleton set is closed in a Hausdorff (or even just T1) space.

[polar storm]

Can u form a neighborhood of (1,2) (such as an open ball) contained in {(1,2),(2,3)}?

[limber wyvern]

oh

Okay

im totally misunderstanding

because this is R^2 the set {(x,y)} is considered to be a singleton?

[gaunt linden]

Yes, "singleton" means a set with one element.

[limber wyvern]

okay

yeah

It's late and I was thinking about {n}

or something like that

yeah, I have that in my book and I have the proof😅

[gaunt linden]

Note that if we're in R², then the notation "(1,2)" must mean the point with coordinates 1,2 rather than the interval of R that has endpoints 1 and 2.

[limber wyvern]

yes

that was my problem

I was thinking about the interval

[gaunt linden]

It's rather pesky that the two notations look exactly alike, but we seem to be stuck with them.

In French notation, the interval would be ]1,2[, which cannot be confused with a point.

[limber wyvern]

what is that notation lol

[gaunt linden]

French interval notation uses inwards-pointing square brackets for endpoints that are included and outwards-pointing ones for endpoints that are excluded from the interval.

[limber wyvern]

I see your point, in order to be open it has to be infinite, because if not you cannot find a neighborhoud, right?

[gaunt linden]

I like it (it's what I was taught in school), but many people here absolutely loathe it.

[limber wyvern]

yes

It seems a bit massive tbh

xDDDD

I mean

You find it useful because you have been seeing it all your life

but first time seeing it it's a bit strange

[gaunt linden]

Its main benefit is exactly that open intervals cannot be mistaken for points/pairs.

Some authors use <x,y> for pairs instead of (x,y), which also solves the problem -- except that <x,y> clashes with other notations, so there's no easy win.

[limber wyvern]

there exists so many things that finding a notation for it is hard

[limber wyvern]

is my point valid?

like the concept of being open or closed, that in a first glance it may seem like a "door" in the way that a set can only be closed or open, but no, it can be both

so it's strange because we usually think about things being closed or open not both

[polar storm]

yeah lmao

we call such sets “clopen”

open and closed

[stuck geyser]

I’m at the beach lmfao

But here

,r

The disjointness can be seen by reflecting it about the diagonal and projecting

[limber wyvern]

HAHAHAHAHA

LOL

[limber wyvern]

Is my answer before valid?

like

this one

[limber wyvern]

.

[limber wyvern]

The married 80 y/o couple seeing a grown man drawing triangles and lines in the sand:

[gaunt linden]

Yes, at least if we're speaking about R^n with the usual topology.

[limber wyvern]

I see

Wow

There are so many aspects to consider before applying a concept😅

[gaunt linden]

(Except the empty set is pretty finite and always open).

[limber wyvern]

so

{}

is open?

[gaunt linden]

Yes.

In a general topological space, that is explicitly required by the definition of "topology".

[kind marlin]

maybe one interesting thing to think about is that a space is connected iff there is no nonempty proper subset that is both open and closed

so if you find a nontrivial subset that is open and closed then your space is not connected

[limber wyvern]

I think I finished the exercise

[limber wyvern]

I don't know yet what connectedness is😅

[kind marlin]

ooh okay nvm then

[limber wyvern]

but thanks for the information 🙂

It's always good

$Let \space (G,V \in \mathcal{T}) be such that (G \cap V = \varnothing). The open sets in (X^2) are of the form ((x,y)), but since the open sets are disjoint, we must have (x \neq y). Hence, the open sets in (X^2) with the product topology (\mathcal{T}_2) are of the form
[
X^2 \setminus {(x,x)},
]
therefore the set
[
\Delta={(x,x)\in X\times X}
]
is closed, besides being a subset since it only contains the cases ((x,x)).$

[gentle osprey]

S0S4 - Feel free to ping
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[limber wyvern]

what do you think

[gaunt linden]

You seem to be confusing "the open sets in X×X are" ... and "these are a basis for the open sets in X×X".

And even so, the basic open sets in X×X are not of the form {(x,y)} -- that would be singletons, which are not open! -- but of the form A×B where each of A and B is open in X.

[limber wyvern]

https://klipy.com/gifs/jgmm-monkey-think--k01KRM6BN2TT4ZZV66RNW964RJE

[gaunt linden]

X²\{(x,x)} is definitely open, but those are by far not the only open sets in X×X.

[limber wyvern]

yeah

I see

I am always confusing the notation

I'm so used to think about (x,y) as interval

and working in the usual topology

😅

[gaunt linden]

I'm not sure if this has been covered in the conversation already, but you should really be clearer about whether you're proving "Hausdorff => diagonal is closed" or "diagonal is closed => Hausdorff".
I can't immediately tell whether the above attempt was supposed to be one of the other.

[limber wyvern]

What's the difference between being open in X and being open in the topology?

[limber wyvern]

I'm going for <= because I finded =>

[gaunt linden]

None, that's just wording variants.

[gaunt linden]

So "diagonal is closed in X×X => X is Hausdorff"?

[limber wyvern]

"X is hausdorff => diagonal is closed in XxX"

[gaunt linden]

I see.

So in other words you need to prove that { (p,q) in X×X | p != q } is open in X×X.

Okay, I can sort of see which way you're going with the first sentence.
You're looking at an arbitrary (x,y) in X×X in X×X-D and want to show that (x,y) has a neighborhood in X×X-D.
So you're using the Hausdorff property of X to select disjoint open G and V in X, such that x in G and y in V, right?

If that's the case then you need to explcitly state that

• you're looking at arbitrary x != y,
• you're using the Hausdorff property of X,
• you get x in G and y in V.

[limber wyvern]

https://klipy.com/gifs/head-explode-mind-blown-1

give me one sec

because I saw my => proof is not well written

[gaunt linden]

Right, I'm trying to help you clean it up.

[limber wyvern]

yes

I appreciate it

I'm back

I'm so bad proving

and it's my first class in where I really have to proof things

[gaunt linden]

You didn't get any proof experience in real analysis? That's too bad.

[limber wyvern]

I didn't have real analysis

xD

I'm in my first uni year

[limber wyvern]

more or less yeah

[gaunt linden]

Point-set in first year is a bit unconventional, but alright.

[limber wyvern]

Like, my idea was saying that, in a product topology for X^2, the open sets are of the form: AxB where A and B are open in X. But, because of the space being hausdorff, the intersection of open sets is empty, Therefore the elements of the open sets cannot be equal (i.e x != y)

Because they cannot be equal, then, the open sets in T_2 are of the form X^2 - {(x,x)}, so, the set A = {(x,x) \in XxX} is closed

And, A is a subset of X^2, because it only covers one specific situation where x = y

[limber wyvern]

Yeah, it's f*cking my life xD

And I think is not the best approach, because I miss so much base concepts that I'm missing lot of things

And I'm loving it, but It's being so hard to follow

[gaunt linden]

Uh, let me pick that apart piece for piece.

in a product topology for X^2, the open sets are of the form: AxB where A and B are open in X.
All of the sets of that form are open, but they are not the only open sets. An open set in X×X is generally a union of (possibly infinitely many) sets of the form A×B.

because of the space being hausdorff, the intersection of open sets is empty
That's not what Hausdorff says.
The Hausdorff property is that given two different points a and b, there exist some open sets A and B such that A contains a, and B contains b, and A cap B = Ø.
But that doesn't mean that any two random open sets have empty intersection.

[limber wyvern]

So, it can be like AxBxAxB?

[gaunt linden]

What wouldn't be a set of the right kind of things, but it can be A×B \cup C×D \cup E×F, for example.

[limber wyvern]

but that would not be X^2

oh

I see

[gaunt linden]

It's usually convenient to think of X×X as if X is R, in which case X×X is R^2 which we hopefully have some intuition for already.

[limber wyvern]

Little intuition for R^2 but some more than for the general case xD

[gaunt linden]

(And in that case it turns out that the product topology on R×R is just the "ordinary" topology on R² which we know fr- you would have seen in real analysis).

[limber wyvern]

it's crazy for a noob like me to be having topology xD

and in september I have algebraic topology

https://tenor.com/view/corner-sit-deer-gif-10844181262590199499

my uni plan is crazy

[limber wyvern]

This changed all my proof

because now i cannot say that the open sets are of the form X^2-{(x,x)}

mmm

I think I managed to do it

$Let ( U, V \in \tau ), with ( U \cap V = \varnothing ). The open sets in ( X^2 ) are of the form ( A \times B ), where ( A ) and ( B ) are open in ( X ).

Since we are in a ( T_2 ) space, given two points ( (x_0,y_0) ) and ( (x_1,y_1) ), there exist two open sets ( A,B ) such that
[
(x_0,y_0) \in A, \qquad (x_1,y_1) \in B,
]
and
[
A \cap B = \varnothing.
]

For this to hold, the open sets must be of the form
[
X^2 \setminus {(x,x)}.
]

Let
[
\Delta = {(x,x)\in X\times X}.
]$

Then $\Delta$ is closed, since its complement is open.

Moreover, it is a subset containing only the cases ((x,x)).

[gaunt linden]

The bot is complaining because you seem to be enclosing the post both in dollar signs and \(...\)/\[...\].

[limber wyvern]

It exploted so whatever xD

[gentle osprey]

S0S4 - Feel free to ping
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[gaunt linden]

The open sets in ( X^2 ) are of the form ( A \times B ), where ( A ) and ( B ) are open in ( X ).
This is still only the basic open sets, not all the open sets.
Since we are in a ( T_2 ) space, given two points ( (x_0,y_0) ) and ( (x_1,y_1) ),
Um, we haven't been promised that X×X is Hausdorff, only that X is. But something of the form (x0,y0) is a point in X×X, not a point in X.
For this to hold, the open sets must be of the form X^2 \setminus {(x,x)}.
And that's not true either -- there are many open sets that don't have that form, including most of the A×B ones.

[limber wyvern]

[gaunt linden]

(Most recently the bot is complaining that you said \Delta outside math mode -- you can see in the output that isclosed,sinceits... gets typeset with math lettershapes, because TeX thinks you had just forgotten the opening math marker).

[limber wyvern]

but the math mode is there

like

at the beggining and at the end

[limber wyvern]

I'll finish this tomorrow because I need to sleep because tomorrow I work

[gaunt linden]

But you shouldn't put all the non-math text (that is non-formula text) in math mode ...

[limber wyvern]

And how can I do so the text is shown as text and the math as math xD

but it doesn't explote?

[gaunt linden]

Do not put dollar signs around everything. Only either dollar signs or \(...\) around each piece that should be typeset as math.

[limber wyvern]

Okay

But isn't there like a generic way so I do not have to put $$ around each math piece?

[wide kayak]

if there's going to be a mix of math mode and non-math mode, all the math mode has to be specified

[stuck geyser]

That would go against the point if delineating text and math mode lowk

[limber wyvern]

Okay

Thank you very much for the help, time and patience 🙂

See you later! 🙂

[stuck geyser]

@limber wyvern an easy follow up exercise is to show that for a continuous function $f\colon X \rightarrow Y$ where $Y$ is Hausdorff, then the graph,
[\Gamma (f) = {(x, f(x)) \in X \times Y}]
is closed.

[gentle osprey]

Tits metric

[stuck geyser]

hint:

|| The preimage of a closed map is closed.||

[polar storm]

Which book are you studying from?

[hidden spade]

what does $$f(\alpha_j)$$ mean here? and shouldn't it be "neighbourhood in $$X_{\alpha_j}$$"?

[gentle osprey]

CrazyCuber217

[unreal stratus]

X^A can be identified (or really, is defined as) the set of functions A -> X. and so given f in X^A you can evaluate it as alpha_j to get some point in X

Idk what you mean by X_{alpha_j} other than as notation for a copy of X

[hidden spade]

oh yeah, myb, X is used as notation for each copy, they dont need to be reindexed

[hidden spade]

so just the projection map, right?

[unreal stratus]

Ye

[hidden spade]

why do we require continuity to declare that these spaces are vector spaces? Is this solely to have a "nicer" structure on them?

[gaunt linden]

We don't need continuity to get vector spaces, no. But vector spaces of arbitrary functions tend to have too little structure to be very interesting -- for one thing, the topology of X would become irrelevant.

[covert nexus]

moreover we are very interested in continuous, differentiable, smooth, lipschitz etc. functions and the fact that they form nice vector spaces is like the foundational fact of functional analysis

[hidden spade]

ofc, that makes sense, ty

[gaunt linden]

(The topology of X is already irrelevant for B(X), but at least B(X) can become interesting because it has a norm, namely the sup-norm, no matter what X is).

[limber wyvern]

Its a uni custom one, but I'm also following munkres, and topology manifolds

[limber wyvern]

Ill try to do It today

[silk light]

i like squares

[foggy quartz]

Let $(X,T)$ be a topological space that is first countable with the T1 property. The only compact subsets of $X$ are the finite subsets. Prove that $T$ has to be the discrete topology on $X$.

[gentle osprey]

you_are_me

[foggy quartz]

this is the T1 property idk if the name is used everywhere

so anyway i have been staring at a blank page for a while now

i think i should do a proof by contradiction where i say that it isnt the discrete topology and then iuse that it has the T1 property and is first countable and then conclude that there is a non finite compact subset

but i dont rlly see a way to get there

[opaque scroll]

Wouldn't the rational numbers be a counterexample to this?

No, I guess it has other compact subsets, nvm

[unreal stratus]

dw, this is quite standard terminology

[opaque scroll]

Alright, pick a point such that {x} is not open.

Find a sequence converging to x (using first countable assumption) observe this to be compact

[foggy quartz]

alright will try

[limber wyvern]

Hi, I'm working with this proof again, and I see that second comment:

"Um, we haven't been promised that X×X is Hausdorff, only that X is. But something of the form (x0,y0) is a point in X×X, not a point in X."

But, because T2 is a multiplicative property, XxX is Hausdorff is X is Hausdorff, which is the case, right?

[gaunt linden]

Well, it is true that X×X is Hausdorff iff X is but (a) you then need to refer to that result explicitly, and (b) needing to rely on it is definitely not the straightest way to what you need to prove.

[limber wyvern]

So, proving that XxX is T2 if X is T2 it's going to help with this proof

[gaunt linden]

Not really.

[limber wyvern]

And okay, I'll recall that is because it's a multiplicative property

it's given as a proposition in my book

like

We consider a product space non empty $X_1 \times X_2$. Then $X_1 \times X_2$ is a Hausdorff space iff the spaces of both factors are Hausdorff spaces

[gentle osprey]

S0S4 - Feel free to ping

[gaunt linden]

Yes, as I said it's true. But it's not something you ought to need here.

[limber wyvern]

oh

Okay, sorry, I wasn't understanding you right

mmm

You recommend me start again

and erase what I've got?

Because I see a mess if I don't have to apply the fact that XxX is T2

[gaunt linden]

In general the last attempt you posted was something of a jumble, so I'd recommend starting over and perhaps deliberately make it informal and not attempt to "make it sound like a proof" before you have a clear idea of how to get through.

(Most importantly, wait until we have the structure before you do battle with LaTeX).

So if I'm recalling correctly, we're assuming that X is Hausdorff, and we want to prove that Delta is closed.

[limber wyvern]

Yeah, my problem Is that I write it down in my native language, then ask AI to translate it using latex 😅

and it all explotes

[gaunt linden]

It may even help with focusing to write it directly in English, if that means you have to break things down into smaller parts mentally.

[limber wyvern]

Proof that a space $(X,\mathcal{T})$ is Hausdorff if and only if, the diagonal $\Delta={(x,x) \in X\times X }$ is a closed subset of $X \times X$ with the product topology

[gentle osprey]

S0S4 - Feel free to ping

[limber wyvern]

I'll do it in english for you

Okay

So

My idea is

This is Hausdorff, because, making the diagonal closed, will mean that the open sets surrounding the points at both sides of the diagonal have empty intersection

Something like this

So, by making the diagonal close, it forces it to be T2

[gaunt linden]

Okay, lets stop here.

[limber wyvern]

https://tenor.com/view/corner-sit-deer-gif-10844181262590199499

[gaunt linden]

First you really should start by making it explicitly clear which direction of the "if and only if" it is you're proving.

[limber wyvern]

okay

so

The right implication (=>) is proved

So I'm going with the same direction as yesterday (<=)

[gaunt linden]

When I said "explicitly clear" I mean you ought to write it down explicitly. "We assume [such and such], and we then want to prove [this and that]".

[limber wyvern]

Okay, but, before that, Is my mental image going in the right direction

or is that also wrong

[gaunt linden]

Next, when you say

the open sets surrounding the points at both sides of the diagonal have empty intersection
that sounds very weird. There's no "the" open sets surrounding points around the diagonal -- there are many possible open sets; some of them have empty intersections, others don't. It won't be true to say that they have empty intersections until you have selected some particular open sets to speak about.

[gaunt linden]

Yes, good.

[limber wyvern]

By the way, how do you so that the text is like this?

[gaunt linden]

Start the line with >.

Ah, but perhaps I'm misunderstanding what you say about those open sets.

[limber wyvern]

What I wanted to express is that, for the nearest points to the diagonal in left and right. There exists two open sets containing that points whose intersection is empty

[gaunt linden]

Yeah, okay, I can see that knowing X×X is Hausdorff makes you think in those terms -- but it doesn't actually seem to help you here.
In particular, knowing that G and U are disjoint from each other is not what you need to know. You need to find open sets that are each disjoint from Delta.

[limber wyvern]

mmm

[gaunt linden]

And for that what you need is not that X×X is Hausdorff, but just that X is.

[limber wyvern]

because knowing they are disjoint to Delta,makes G and U disjoint?

[gaunt linden]

No. It's that we don't care whether G and U are disjoint. (At least I don't see why we should).

Your goal is to prove that X×X\Delta is open, right?

[limber wyvern]

yeah

that's true

[gaunt linden]

And it most be open in the product topology, and the product topology is defined by something like

• A subset of X×X is open iff it is a union of sets of the form A×B where A and B are open in X.
  So we want to prove that X×X\Delta is such a union.

[limber wyvern]

https://klipy.com/gifs/jgmm-monkey-think--k01KRP8TRB03DGA3Z3015PN4VAG

[gaunt linden]

So the natural strategy would be to say, okay let p be an arbitrary point of X×X\Delta, and then find an A×B, such that (a) A×B contains x, and (b) A×B is a subset of X×X\Delta.
If we can find such an A×B for every point in X×X\Delta, the union of all of them will be X×X\Delta, and then we have proved that X×X\Delta is open.

[limber wyvern]

I think I'm cooking

[gaunt linden]

What does that mean?

[limber wyvern]

like

I think I'm getting somewhere

[gaunt linden]

Ah, good.

[gaunt linden]

Let's hear where you're getting.

[limber wyvern]

Im losing my mind xDDD

nah

I literally have nothing

[gaunt linden]

Do you follow as far as this?

[limber wyvern]

I do not find any AxB that satisfies:

A×B for every point in X×X\Delta, the union of all of them will be X×X\Delta, and then we have proved that X×X\Delta is open

[gaunt linden]

Well, of course you'll need a different A×B for each point.

[limber wyvern]

I see

[gaunt linden]

Recall that a point p in X×X\Delta is really a pair of point (x,y) with x and y both in X, and x!=y because the point is not on the diagonal.

[limber wyvern]

because every open only has one point in it?

This is where i am xD

pretty sad tbh

[gaunt linden]

No, I'd say just because we can. We're allowed to use as many A×B as we want to cover X×X\Delta, and it doesn't matter if some of the points are covered by more than one of them. So it simplifies the task if we just pick a new A×B for each p, so that p is the only one we need to think about when we pick A and B.

[limber wyvern]

I need to study more on product spaces😅

[gaunt linden]

Now you have x and y in X and x != y, and we have also (back at the beginning) assumed that X is Hausdorff. Does this remind you of anything?

[gaunt linden]

That's what you're doing -- the exercise is in part to get some experience with working with the product topology.

[limber wyvern]

Well, if the space is T2 and x !=y then, there exists atleast one open set containing x and one containg y that they must be disjoint

[gaunt linden]

Yes.

(When you said you were translating, is your native language one that doesn't have a word like "the"?)

[limber wyvern]

no

Im just stupid

and my brain goes faster than my fingers

So I may think about a full phrase, but then write it missing parts, and I do think that I write them down, but most of the time I just imagined that I writed them down xD

https://tenor.com/view/caucaflocka-squidward-im-insane-guys-its-ok-squidwardgif4ava-gif-18811288

[gaunt linden]

So this is the right direction. Can you see where it's going from there?

[limber wyvern]

well

We can construct the subset of AxB by taking the sets AxB that are disjoint?

[gaunt linden]

Yes.

[foggy quartz]

Prove that every path connected hausdorff space is also arc connected. You may use the following theorem without proof:
Every compact, connected, locally connected, regular, C2 space is arc connected

well so the obvious way to start here is to prove all the requirements of the theorem but it appears to be quite hard lol like i get so little information about the space that idk what to do

[gaunt linden]

I think suspect the strategy here is to pick a path, and then look at the image of that path with the subspace topology.

[foggy quartz]

alright thanks, I will give it a shot

[limber wyvern]

What do you think?

[gaunt linden]

Okay, so I think you've gotten the right idea, but we need to work some on the presentation.

[limber wyvern]

https://tenor.com/view/corner-sit-deer-gif-10844181262590199499

[gaunt linden]

What I had in mind was something like for each $p = (x,y)$ in $X\times X\setminus\Delta$, choose open disjoint $A_p$ and $B_p$ with $x\in A$ and $y\in B$, and then see that $X\times X\setminus\Delta$ is $\bigcup_p A_p\times B_p$.
The $p$ subscript shows the reader that there are different $A$ and $B$ for each $p$.

However, the way you wrote it makes me think we can also write it first saying, look at the set
$$\bigcup_{\substack{A\in\tau\B\in\tau\A\cap B=\varnothing}} A\times B$$
(which is obviously open in $X\times X$) and then show this set equals $X\times X\setminus\Delta$.
That might be neater, but I think we had probably better continue the way we started, to avoid further confusion.

[gentle osprey]

Troposphere

[limber wyvern]

I prefer the first way

not like 3 subindexes to the union

[gaunt linden]

Okay, we'll continue the first way.

You're writing

Let A,B be open sets in ...
but that is not the right phrasing, because "let ... be" is used when we want to prove something for every possible choice of A and B, and that's not what's going on here.
Instead once we have picked p (and for that "let ... be" is the right wording), we want to use the Hausdorff property to pick just one set of A and B that will work for that p.

So the way to phrase that would be

Because X is Hausdorff (or T2), we can pick A_p and B_p such that x is in A and y is in B and A cap B = Ø.

[gaunt linden]

Yes, good.

You'll probably want to add some more lines to justify that X×X\Delta is that union, but each of the points you need to make are straightforward.

Then explicitly state that X×X\Delta is open because we have written it in the form open sets have, and therefore Delta is closed.

[gaunt linden]

Ah, you lost the detail that p is (x,y).

[crisp lintel]

I don't see why you need a big union at all

a set is open if and only if each point in the set has an open neighborhood contained in the set

I guess you basically proved that inside the proof

but if you know that fact already, its cleaner and faster to just focus on a single point and end the proof there

[gaunt linden]

That's true (and I agree it would be valid to phrase it that way), but I wanted to tie more directly up to how the product topology is defined.

[gaunt linden]

I would want to say explicitly somewhere that because Ap and Bp are disjoint, Ap×Bp doesn't contain any points on the diagonal.

[gaunt linden]

Yeah, then just remove "by definition" since you now have a better argument.

[gaunt linden]

And yeah, that's what I would do myself if I was not focused on not confusing SOS4 after we had already discussed what the open sets of X×X are in those terms.

[limber wyvern]

okay

Thank you very much for the help and time 🙂

[foggy quartz]

ok so i defined a path between x and y and i have proven that the image of the path is compact, connected and regular but i am stuck on the locally connected and the second countable proof once i got that i can say that the image is arc connected and then that means that the two randomly chosen points are arc connected thus the entire space is arc connected

cuz can be done with any random points

anyway so the second countable and the locally connected proof is what i am stuck on

any hint?

for the second countable im think that maybe i can construct a basis for the unit interval and then transfer that over in some way to the image

[gaunt linden]

It seems to be true that images of paths in a Hausdorff space are locally path connected, but I haven't found a proof yet.

[foggy quartz]

i got the C2 property proof i had to use a result from 2 questions ago

so only locally connected remains

ok its time for a break
imma be back at it in like 90 min

[foggy quartz]

alrighty tieme to roll again

[silent garnet]

trivia: what is this surface homeomorphic to

[novel acorn]

RP^2

Reminds me of Boy's surface

[novel acorn]

Oh it isn't

Interesting

[midnight umbra]

is it the sphere

(I'm guessing)

[zealous berry]

fidget spinner

[covert nexus]

if my brain is working right, this is a torus?

[zealous berry]

new nickname :3

[gaunt linden]

Looks like it's a Klein bottle.

(Three Klein bottles joined end-to-end, but two of the orientation flips cancel out, so up to homeomorphism it's just Klein).

[zealous berry]

is this too handwavy

[polar storm]

looks good.

[zealous berry]

awesome

[polar storm]

This is how I wrote in my tex notes for point-set topology

[zealous berry]

[quick delta]

||This is a triple cover of a klein bottle, so it’s just a Klein bottle, no?||

[robust drum]

Yes I agree

[quick delta]

Ok I haven’t proved this rigorously
But I can definitely see that this triple cover is a Klein bottle

[gaunt linden]

As far as I can see it's even smoothly homotopic to the usual immersion of the Klein bottle.

[quick delta]

Diagramattic proof

[robust drum]

Yep

I think homotopic through immersions too

[gaunt linden]

That's what I was trying to say.

[robust drum]

https://youtu.be/kQcy5DvpvlM?is=9gZyjwZimTcn2af2

Just do this shit to it

[gaunt linden]

Which shows that two Klein bottles end-to-end is just a torus.

[robust drum]

Yeah, but I mean rather play it backwards from slightly after the beginning

Since EG the situation right after this frame

Is basically analogous to what we have

[gaunt linden]

Oh, I got the point.

[robust drum]

With the gluing out of frame different but that doesn’t matter

[cosmic mirage]

they're calling it the most enlightening proof ever discovered

[zealous berry]

nah it's second to this

[cosmic mirage]

honestly why isn't that how every analysis proof is presented

jk epsilon delta chasing can be a bit fun

[zealous berry]

I see.

[zealous berry]

i wrote something nearly identical

except i didnt do trivial

[rancid umbra]

https://math.stackexchange.com/a/5137002/803927

this was fun/annoying to think about…

i learned about a curious definition of closed maps today from this

[warped helm]

but one of the key points is cauchys mean value theorem

[zealous berry]

cauchys mvt is also used for lhopital 0/0

[silent garnet]

yes!!

the answer is klein bottle

[lavish mountain]

(i) duh and (ii) obviously qed

[lavish mountain]

you can still see the light in this user's eyes

[limber wyvern]

Hi! I'm trying to imagine geometrically the following proposition:

In (X,T), A c X is closed iff A = \overline{A}

If I think of A as an interval in R is easy to see that, if it's closed (e.g [0,1]) it does not have any point infinitely close to it such that a neighborhood of the point intersects with the set, but the point is not in the set, therefore the adherence is equal to A.

But trying to see this as a closed set in more general topologies, and thinking of closed set as the complementary of an open, I'm finding it hard to visualize it.

[polar storm]

I usually like to think of this as saying that A must contain all of its boundary points. Here boundary points of A are points that can't be "separated" from A via neighborhoods. You can clearly picture this by imagining a bunch of open balls in R^2, which is what I usually do to build some intuition for arbitrary topological spaces.

[broken crag]

Think of closure this way:

[
\overline A={x\in X:\text{ every open neighborhood of }x\text{ intersects }A}
]

So (x\in \overline A) means: the topology cannot separate (x) from (A) by an open neighborhood

Now use the complement definition of closed

[broken crag]

,tex
[
\overline A

{x\in X:\text{ every open neighbourhood of }x\text{ meets }A}.
]

[
A\text{ is closed}
\iff
X\setminus A\text{ is open}.
]

If (A) is closed and (x\notin A), then (x\in X\setminus A).
Since (X\setminus A) is open, it is an open neighbourhood of (x) which does not meet (A). Hence

[
x\notin\overline A.
]

So no outside point is in (\overline A), and therefore

[
\overline A=A.
]

Conversely, suppose

[
\overline A=A.
]

If (x\notin A), then (x\notin\overline A).
So there exists an open neighbourhood (U) of (x) such that

[
U\cap A=\varnothing.
]

Thus (U\subseteq X\setminus A).
Every point of (X\setminus A) has an open neighbourhood contained in (X\setminus A), so

[
X\setminus A\text{ is open}.
]

Hence

[
A\text{ is closed}.
]

[
\boxed{A\text{ is closed}\iff A=\overline A}
]

[gentle osprey]

Dex

[broken crag]

If (A) is closed, then (X\setminus A) is open. So if (x\notin A), then (x\in X\setminus A), and this open set is a neighborhood of (x) that completely avoids (A).

Therefore (x\notin\overline A). So no outside point belongs to the closure, hence (\overline A=A)

If (\overline A=A), then every point outside (A) is outside (\overline A)

Every outside point has an open neighborhood contained in (X\setminus A). That exactly says (X\setminus A) is open. Therefore (A) is closed

A point is close to (A) if every open set around it is forced to touch (A)

[limber wyvern]

Yeah, I imagine it as R2 balls, but, what relation does it has so that if its closed, is the same as his adherence?

[polar storm]

@limber wyvern I think Dex's explanation answers it quite well. Like if A is closed and a point x is not in A, then u can make sufficiently small room around x such that it is completely isolated from A. This is basically the opposite of the notion of adherence.

[limber wyvern]

let me write down what I think of

This is a vague drawing

but you can get the idea

like

The green drawing are the closed

how tf that is adherent to a point

[polar storm]

can u clarify a bit more?

[limber wyvern]

So

let A = X\T (The green lines), how is any point on the topology adherent to A?

I may be misunderstanding everything😅

[polar storm]

So A is the green region with the black-line boundary in ur drawing?

[limber wyvern]

affirmative

[polar storm]

What is T?

[limber wyvern]

I may draw it better xD

T should be the topology containing the open sets

[polar storm]

so X\T means the X setminus T? or u meant X \setminus (union of all the open sets in X)

T is a collection of subsets of X, so it doesn't make sense to take a complement with X. So I guess u meant X \ (union of all open sets in X), which is just empty.

[polar storm]

You can also think of closure in terms of sequences. That is, if there exists a sequence in A that converges to x, then x must be in the closure of A. I think this also captures the notion of closeness/adherence well and very intuitive cuz the notion of convergence matches well with the notion of closeness/nearby.

Unfortunately, the converse does not hold in general (counterexample: cocountable topology on R). But the converse holds in metric spaces (which we usually intuit on). More generally, the converse holds in first-countable spaces.

[limber wyvern]

Sorry I didn't read your messages, pls ping me in every message xD

So

Forget about the drawing

But

Sometime the closed sets can be sets that are not in the topology, and so, they are in the space

like, freely

why are those sets adherent to some point in the open sets?

[polar storm]

@limber wyvern oh so u mean u want to know the proof why the following is true?

Let $X$ be a topological space, and let $A \subseteq X$. Then $x \in \overline{A}$ if and only if for every neighborhood $U$ of $x$ in $X$, we have $U \cap A \neq \varnothing$.

[gentle osprey]

Euiseok (Class of 1929 + 200)

[limber wyvern]

I mean, I read the proof like 9 times

and yeah

I understand it

but i cannot picture it

[warped helm]

what points are added in the closure of (0,1) as a subset of R for example

[limber wyvern]

[0,1]

[warped helm]

thats the closure, which points did you add in

[rancid umbra]

the geometric intuition i have for closures is that they are encasings, or shells, around your set.

[limber wyvern]

0 and 1

[warped helm]

why did you add them in?

according to this theorem

[limber wyvern]

So you cannot find any neighborhood around that points and so, the set closed

[warped helm]

huh?

hm

[limber wyvern]

What do you mean

[warped helm]

do you agree that every neighborhood (1-e, 1+e) intersects (0,1)?

for any e > 0

[lone vessel]

hey guyss you mind helping me with topology

[limber wyvern]

yeah

Thats what i was referring, you cannot introduce any neighborhooud in [0,1] because every neighborhood you add, stays out of the set, therefore it's closed

[warped helm]

then yes, 1 is in the closure of (0,1) according to this result

[limber wyvern]

First of all, amazing drawing.

So, I can see that because it's closed it has all the points in the frontier and in the interior and so thats why his adherence is all the set, because all the points intersect the open sets in A

I think my problem is that I was thinking about random open sets inside a topology and the rest of the space

something like

[warped helm]

this isnt really a relevant condition

you dont talk about adding in neighborhoods when talking about closures

and its not true that every point of closure has every neighborhood intersect the complement

[lone vessel]

Exercise 18Given 10 points generated from two concentric (nested) circles:Inner circle: radius 2, consisting of 5 equally spaced points.Outer circle: radius 5, consisting of 5 equally spaced points.The distance between an inner point and its nearest outer point is 3.a) Describe the critical epsilon ($\varepsilon$) thresholds.b) At $\varepsilon = 3.0$, has each circle closed into a loop yet? What is the value of $\beta_1$?c) At $\varepsilon = 4.0$, have the two circles connected with each other? What is the value of $\beta_1$ at this point?d) At $\varepsilon = 6.0$, what happens to the holes? Explain.
can yu helpp me with this problem

[gentle osprey]

minnhtri45

Exercise 18Given 10 points generated from two concentric (nested) circles:Inner circle: radius 2, consisting of 5 equally spaced points.Outer circle: radius 5, consisting of 5 equally spaced points.The distance between an inner point and its nearest outer point is 3.a) Describe the critical epsilon ($\varepsilon$) thresholds.b) At $\varepsilon = 3.0$, has each circle closed into a loop yet? What is the value of $\beta_1$?c) At $\varepsilon = 4.0$, have the two circles connected with each other? What is the value of $\beta_1$ at this point?d) At $\varepsilon = 6.0$, what happens to the holes? Explain.
 can yu helpp me with this problem

[limber wyvern]

Thats why its closed, right?

[warped helm]

no

[limber wyvern]

So, that point in red has no neighborhood that can be fully inside of the set

therefore that point has to be adherent and closed

[limber wyvern]

Where the red are closed sets

[warped helm]

you dont say a single point is closed

a singleton set {a}, sure

but it makes no sense to say that elements are closed

the set is closed if it contains all its limit points

[warped helm]

that would mean its a point belonging to the boundary

[limber wyvern]

Okay, I meaned a singleton, in my language there is no distinction

[warped helm]

well no i dont think you meant that

because we werent talking about the singleton set being closed

[limber wyvern]

Like, the singleton red has no neighborhood around it such that the whole neighborhood is inside the topology

[limber wyvern]

well, but in (R,T_u) singletons are closed

[warped helm]

we arent talking about whether the singleton is closed or not

a singleton set being closed has no bearing on whether the point it contains belongs to the closure of another set

every neighborhood intersecting the exterior and interior is not the condition for belonging to the closure

its sufficient but not necessary

[limber wyvern]

so

lets start from the beggining

If a set has no neighborhood that is fully contained in an open set

it's closed

?

[warped helm]

no

this is not even a relevant question

[limber wyvern]

so what is your definition for closed set

[warped helm]

it contains all its limit points

[limber wyvern]

Okay, and could you do one that involves neighborhoods?

[warped helm]

every neighborhood of every point intersects the set

[limber wyvern]

mmm

[warped helm]

is {1/n : n = 1,2,…} closed?

[warped helm]

actually this isnt precise enough

[warped helm]

this is still what you want

[limber wyvern]

well

I normally don't work with that definition

but it's okay

[limber wyvern]

nope

[warped helm]

why not?

(this isnt a rhetorical question, just asking for justification)

[zealous berry]

[limber wyvern]

well

not all the limit points are contained (0)

[warped helm]

yep

and why is 0 a limit point?

[limber wyvern]

well

because is not in the set, but the intersection of the neighborhood around it with the space is not empty

[warped helm]

limit points can be in the set

the condition youre alluding to is that every neighborhood of 0 intersects the set

its not enough that only a single, specific neighborhood intersects the set

[limber wyvern]

I see

[zealous berry]

Yeah x is a limit point of a set X if every open neighborhood about x intersects X at a point other than x

Then X is closed if it contains all its limit points

[sly geyser]

epsilon neighbourhood?

[zealous berry]

what's up

oh my bad

open neighborhood

[deep surge]

\textbf{Theorem.} Suppose $\mathcal{A}$ is the basis for some topology on $X$. Show that the topology $\mathcal{A}$ generates is the intersection of all topologies on $X$ that contain $\mathcal{A}$.

\begin{proof}
  Suppose $\mathcal{A}$ generates a topology $T$. Let $\{\mathcal{T}_\alpha\}_{\alpha \in A}$ be the family of topologies on $X$ such that $\mathcal{A} \subseteq \mathcal{T}_\alpha$ for each $\alpha \in A$. Let $T' = \bigcap_{\alpha \in A} \mathcal{T}_\alpha$---we show $T = T'$. Of course, $T \in \{\mathcal{T}_\alpha\}_{\alpha \in A}$, and so $T' \subseteq T$. Now we seek to show $T \subseteq T'$. Equivalently, we show that for any open set in $T$, it follows that aforementioned open set is also in $\mathcal{T}_\alpha$ for every $\alpha \in A$. Fix $\alpha \in A$ and $O \subseteq T$. Note $\mathcal{A}$ is a basis for $T$, and so for each $x \in O$, there is some $B_x \in \mathcal{A}$ such that $x \in B_x \subseteq O$. Thus $O = \bigcup_{x \in O} B_x$. But note $B_x \in \mathcal{A} \subseteq \mathcal{T}_\alpha$ for every $x \in O$, and so the arbitrary union of each $B_x$ is in $\mathcal{T}_\alpha$. Thus $O \in \mathcal{T}_\alpha$, which completes the proof.
\end{proof}

is this proof OK?

[gentle osprey]

0DIMENSIONAL LOVE FEELINGS

[rancid umbra]

you need O \in T, not O \subset T

[deep surge]

right my bad

that was just a typo

[rancid umbra]

looks good then

note that a very similar proof works if you replace basis with subbasis

[deep surge]

indeed, that's the next exercise

that i'm not doing because the proof is analogous

[rancid umbra]

there is some intersection business that goes on, but otherwise, yes, it is analogous

you can even break it into two steps

subbasis —> basis —> topology

[deep surge]

yes i've realized this already

[unreal stratus]

(This is amusing because I would define the topology a basis generates as being the intersection of all topologies containing it)

[deep surge]

same tbh

[rancid umbra]

top-down vs bottom-up approaches

[rancid umbra]

guessing you define the subgroup generated by a subset to be the intersection of all subgroups containing it?

[unreal stratus]

i think this is a good definition yes

though for subgroups it is sort of trivial to show that this coincides with the other more concrete one

[rancid umbra]

yea. my thoughts are that it depends on what u r using it for

[midnight umbra]

well, this should surely be the definition of "the coarsest topology making some collection of sets open," as it doesnt use anything of the structure of a basis

[deep surge]

i am so gay right now

[midnight umbra]

same

[rancid umbra]

wdym 💀

[queen prism]

point-set just does that to you

[alpine nest]

It encourages openness

[alpine nest]

Hell yeah, same

[deep surge]

topology is the study of holes 🤤

[unreal stratus]

Sure, my perspective is to just say that any collection subsets generates a topology (the coarsest one containing it) and then show that if your original thing is a basis then you have an easier description of this generated ting

[kind marlin]

is the easier description you're referring to the intersection of all topologies containing it? isn't that equivalent to the coarsest topology containing it (for any collection, including non-bases)

or are you referring to something else (i.e. the topology generated by a basis is the collection of all possible unions of sets in the basis)

[midnight umbra]

the latter I would think

[kind marlin]

right yeah that makes sense

[autumn inlet]

Later in Lemma 2, Tverberg restricts closeness in param space S^1 to max √3

Why that? Anything less than 2 should work right for ensuring uniqueness (?)

Is it just because √3 is nice

[autumn inlet]

Why not √2 or 1.99?

[digital sun]

Hello. I can't figure out why $\mathbb{R}^{\omega}$ with the uniform topology is not separable.

[gentle osprey]

Gol D Roger

[digital sun]

Can anyone give me an intuitive approach?

[gaunt linden]

A dense subset will need to contain something close to every element of {0,1}^omega in particular.

[crisp lintel]

Yeah try to find an uncountable collection of pairwise disjoint open sets

[spare igloo]

🎉

[digital sun]

So I know ${0,1}^{\omega}$ is a discrete uncountable subset of $\mathbb{R}^{\omega}$, so I can take the uncountable set of balls of radius $\frac{1}{2}$ over all points in A, all of them which are disjoint .

[gentle osprey]

Gol D Roger

[digital sun]

I don't know what density plays there

[crisp lintel]

Yeah then you're done

remember that if a subset is dense, it must intersect every open set

[digital sun]

Ohh I see, I just realized that

Thanks!

[lone heath]

Hello guys

I believe that i can ask this here

How much problematic is the axiom of choice?

Like is fundamentall for maths or not much?

[polar storm]

axiom of choice is TOO GOOD.

[lone heath]

Yeah i saw that is very important but. Why are people who are so scandalized with it?

[unreal stratus]

I don't think many people particularly are but it is largely hangovers from like 50+ years ago

[polar storm]

mmm I don't really know about set theory, but I guess it's bcuz there are some results that are equivalent to AC and sounds very counterintuitive such as well-ordering thm.

[wide kayak]

it leads to some highly counterintuitive results

[lone heath]

Yep

[wide kayak]

and it's somewhat not obvious what it means to make uncountably many choices or if that's in fact possible

[lone heath]

I didn't understood the order thing. But i know has a relation with ac

[polar storm]

well-ordeing means that every nonempty subset has a minimal element. Now well-ordeing thm states that every set has a well-order. Now think about this for R.

[lone heath]

And what we have to do? The critics i saw were about a function that cannot be constructed or something like that. Am i wrong?

[wide kayak]

perhaps you saw something about the existence of non-measurable functions relying on the axiom of choice? And the banach-tarski paradox?

[lone heath]

Yes but that talks about nor usual orders right? What is a not usual order?

[polar storm]

not the usual order

[lone heath]

Yes but the thing wasn't about consequences but about the statement itself. A rule i believe.

[polar storm]

also, AC implies the existence of basis for every vecto space. Now this means the vector space R over Q has a basis. Now can u imagine it?

[wide kayak]

to get a well-ordering of R using the axiom of choice, you basically go "Choose a real number r1 to be the smallest real number. Now choose any r2 to be the next real number after r1. Then choose an r3, and r4, and so on, until you have chosen all real numbers"

[lone heath]

Well,idk what "imagine" means but i think it has sense,and is importante for linear algebra. Right?

[gaunt linden]

For the most part, not problematic at all, and in most of everyday mathematics argument that require it are mainstream and accepted without even noting explicitly that it's being used.

There were some doubt about it in the earlier part of the 1900s, but after Gödel proved that it doesn't introduce any new inconsistencies, and doesn't allow one to prove false things about finite computations that can actually be carried out, this has mostly faded away.

"Constructive" mathematics, which is a bit of a niche topic, avoids appeals to choice, but there's much more that it also avoids.

[unreal stratus]

(I thought this was more Cohen than Gödel?)

Oh sure different things were proven by each towards this lol, and you are right in referring to Gödel here

[gaunt linden]

Gödel proved that the axiom of choice and the continuum hypothesis are both consistent with ZF. Cohen later proved that their negations are, too.

[lone heath]

There is something that constructive mathematicians don't accept that mainstream yes? Like without the axiom of choice what things you don't have?

Sorry for the questions i'm not an expert in this jsjaja

[gaunt linden]

Constructive mathematics uses intuitionistic logic, which rejects proof by case analysis, that is, the schema

Consider the claim P which I'm hereby pulling out of a hat.
If P is true then (bla bla bla) and therefore Q.
If P is false, then (bla bla bla) and therefore also Q.
Therefore Q always holds.
unless the proof also shows a procedure for determining whether P is true or false.
In order to reject this, intuitionsistic logic also has to deny that "P or not P" is provable in general, or that "not not P" is the same as P, and a number of similar reasoning principles that are otherwise mainstream.

It also rejects some but not all applications of indirect proof. It is okay to prove ¬P by showing that assuming P leads to a contradiction, but it is not intuitionistically valid to prove P by showing that ¬P leads to a contradiction.

[lone heath]

Ok i didn't understood but is for me you know jsjsja

Intuitionistic logic has relation with type thory or not necessary? Because is a different way to found matjamtics without set theory right?

[gaunt linden]

Yes, type theory is also usually done with intuitionistic logic, and is the most common formalization framework for constructive mathematics.

[grave solstice]

what kind of existence proofs do constructivists/intuitionists accept? Or something exists only when an explicit example is given and proven?

[lone heath]

Is true that type theory doesn't have axioms? How start if don't?

[gaunt linden]

Yes, essentially.

[gaunt linden]

Not quite -- but in some formulations of type theory there are only very few and simple axioms (at the level of "you can conclude anything you have assumed" or "the input to the function you're currently defining exists") and all the interesting stuff happens in the rules of inference.
(This style of presentation discards the otherwise common distinction between "logic for pure reasoning" and "axioms to define what it is you're reasoning about").

[grave solstice]

But sometimes it can be proven that a set of objects exist but for which no particular concrete example can be proven to be in. Eg. #foundations message what would intuitionists/constructivists think in this example?

I think btw Brouwer rejected formalization (via axioms I mean), but not all intuitionists/constructivists

[gaunt linden]

If no example can be given (and one cannot even describe a procedure that would produce one, assuming unbounded resources and patience), then the existence would not be considered constructively proved.
For the particular post you point to, I think I have too little context to say anything intelligent (and it's only showing a claim not a proof anyway).

[gaunt linden]

Yeah, that matches what I've learned. I think most who do constructive mathematics nowadays don't do it because they're followers of Brouwer. My sense is that for the most part they don't seriously doubt the mainstream results are true, just that having a constructive proof of something is "extra good" and in itself an interesting property that is worth investigating, for example, for the computational connections, or because such proofs can often be translated to "internal" statements in categories that don't accept full classical logic.

[grave solstice]

The proof is: construct an algorithm that enumerates all theorems and proofs of that given axiomatic system (which we should assume it's finite or recursive or something like that). It's in this wikipedia article https://en.wikipedia.org/wiki/Kolmogorov_complexity#

[gaunt linden]

What I meant is, I'm not even sure what all the notation in the claim is.

[cosmic mirage]

kind of a brainrot question but is there an easier way to fully convince someone of what the product/quotient/etc topologies ought to be than

1. the forgetful functor Top --> Set has a left and right adjoint (discrete and codiscrete topology)
2. that means you can compute the underlying set of a lim/colim of spaces using the same lim/colim in sets
3. stare at the poset of topologies on a set and work out which one is the finest/coarsest so that the appropriate maps are continuous, to get the correct universal property

when i was learning pointset stuff a lot of these constructions seemed a bit arbitrary and i didn't understand why they had to be that way (stuff like why the box topology is the "correct" one on products) until embarassingly much later

[stuck geyser]

What

[gaunt linden]

(But the box topology isn't the correct one on (infinitary) products).

[cosmic mirage]

ah sorry, i meant to say why it isnt the correct one 💀

[stuck geyser]

I mean first you need to understand how different families of set are courser or finer (i.e contain more or less data about how points are related by neighborhoods) then the idea of the “least amount of data” or the “most amount of data” arise naturally

The motivation of this for me was how you can pullback a topology via a preimage since the preimage is super well behaved in respect to intersections and unions, and how that is the “least amount of data” needed for the map to be continuous

Same shit for sigma algebras too or really set algebras generally

Limits and colimits follow that same principle especially with the cone intuition as almost an inf or sup of a diagram

Quotients are just removing data by lumping it into clusters specified by a relation and only dealing with those

Or just disregarding it

[cosmic mirage]

yeah i think this is a good way to think about it, I'm just not sure how to motivate that this is what we should want to have in the first place (at least, without the motivation of it working in other contexts, like sigma algebras like you said / what is forced by category theory if you accept the definitions from there are correct)

[stuck geyser]

Well it’s what a topology is.

Relating points by nicely compatible families of neighborhoods

Like how we relate people by their street and towns and such

How much data or neighborhoods we consider is more data and gives a finer topology

Like viewing townships instead of counties

[cosmic mirage]

hmm i see, yeah i think that's a good way to think about it

[stuck geyser]

That’s how I try to intuit topologies as a point set idea to people who aren’t too abstract math brained

Same with quotients as getting rid of “redundancies” by grouping things together and only dealing with those collections, like how we can gather all cars into “make and model” and only compare that without caring about the details of what day it was made and where

Unfortunately topologies are not as well behaved as algebraic things. E.g. equalizer is not necessarily closed unless Hausdorff

The box topology on like a countable product is a good example of the fact that you don’t have enough data

[cosmic mirage]

makes sense, yeah. when i was learning these definitions it was sort of just arbitrarily given to us and then as homework we proved they were the finest/coarsest topologies such that the relevant functions were continuous. these days i have no problem accepting that its kind of backwards (we want the most efficient topology such that ..., and those are the topologies which realize that) but still unclear about how i would explain that philosophy to someone still learning the subject

[stuck geyser]

Like 90% of my math background is self study so I have spent a lot of time trying to build intuition instead of just trying to remember rote definitions lol

[cosmic mirage]

good strategy lol

[stuck geyser]

Are you teaching a course or tutoring or smth

Also your approach is super categorical lol

[cosmic mirage]

im mentoring an undergrad in a directed reading program about pointset topology yeah

[cosmic mirage]

yeah I mean the first two steps arent really doing anything relevant to the question i'm asking, I now realize. I don't think anyone is complaining that products and quotients of spaces are just the products and quotients of the underlying sets, with some topology

[stuck geyser]

I’m doing this next sem for classical mechanics :3

[stuck geyser]

This is a great introduction to category theory ideas though

[cosmic mirage]

but its worth thinking about at some point in one's life why that is true in topology. coproducts and cokernels in groups are much more interesting than the respective colimits in sets, for example. limits are well-behaved which is clear from the free group adjunction

[cosmic mirage]

nice! they are great programs

[stuck geyser]

The idea of data and limits and colimits easily lead to the categorical ideas

[cosmic mirage]

yeah agreed. i mean category theory was basically developed for algtop lol

[stuck geyser]

The most or least amount of data needed to “cap” the top and bottom of the diagram

[cosmic mirage]

anyways yeah i should've just excluded the first 2 bullet points lmao, like i said no one is confused about why the product of spaces is a product. just want to make sure i have better answers than some unclear philosophy about efficiency / the definitions in topology have proven to be useful / the definitions of category theory have proven to be useful and they force the definitions in topology to be what they are

but yeah i like the way you explained it, thanks!

it is definitely important to have intuition that extends beyond "it is forced by category theory"... i once saw someone define homotopy equivalence of spaces by defining a category, defining isomorphisms, defining the homotopy category of spaces, and then just saying those are the isomorphisms in that category. its efficient but completely unhelpful

[stuck geyser]

I mean sure, and it provides a framework to describe it in but it just adds more overhead to communicate

[stuck geyser]

<@&268886789983436800> kill

Good job

[robust drum]

What’s cokernel in grp is it the same as in Ab

[stuck geyser]

Probably take normal closure

[cosmic mirage]

i agree with @stuck geyser. this was my bad but this specific type of coequalizer should probably not be called coker unless you work in Ab

well

idk i wouldnt call something coker unless your category is preadditive, which Grp is not

as a sanity check: abelianization will preserve colimits, being left adjoint to the inclusion Ab --> Grp (Ab is a localization of Grp). so abelianization of coker is coker of abelianization. if everything is already abelian then you just get G/im(f) which will be abelian, so everything agres

this is kind of weird, coproducts of abelian groups computed in Grp will still be annoying free-group-type things but i guess coequalizers are well behaved. a priori you only know limits (in Grp) of abelian groups will be abelian, again by being a localization

[robust drum]

They’re more well behaved but still different

[cosmic mirage]

wait are they not the same

its kind of late here so i'd definitely believe i said something false lol

also @robust drum lets move to #groups-rings-fields probably

[deep surge]

R's topology is the order topology, and its basis is just open intervals in R
then Z's subspace topology has a basis of all open intervals of R, intersected with Z
so
isnt the subspace topology of Z the discrete topology (as a subspace of R)?
and thus every f: Z --> R is continuous
i havent gotten to a def of subspace topology or continuity but i would imagine:

1. subspace topology is just the set of all open sets intersected with the subset
2. continuity is every inverse image of an open set is open

[ruby delta]

yep, that's all correct

if you haven't seen subspace topology before you might want to quickly check that that definition does give you a well-defined topology

[deep surge]

awesome thx

[deep surge]

its in the section of my book haha

i verified it in my head though

emptyset and original set are obviously in the subspace topology

then finite intersection and arbitrary union you just use set rules

wait a minute
if f: Z-->R is always continuous, then wouldn't a sequence a: Z+ -> R+ always be continuous too

[ruby delta]

depends on what you mean by a sequence being continuous

[deep surge]

like the function generating the sequence

a function mapping Z+ to R+ (in this case i am specifying entirely nonnegative sequences)

[ruby delta]

well then sure

by definition

[deep surge]

any implications from this

[ruby delta]

we don't typically talk about sequences being continuous

because as the way you've defined it there's no reason why I wouldn't just look at the original function instead

[rancid umbra]

if you have a sequence x : N U { ∞ } —> X where X is a top space and N U { ∞ } has the order topology, then x is continuous if and only if x converges to x_{∞}

[deep surge]

🤔

ok

is there a formal name for the theorem saying that if a cover locally refines every point in a topological space, then it is a basis for that topology

[deep surge]

the content of this lemma is that subspace topologies are subsets of parent topology yeah?

im not reading the proof cuz i like to prove theorems without seeing proof occasionally

[rancid umbra]

only if Y is open. otherwise, this may not be true

[deep surge]

but, suppose O in topology on Y. then O is the arbitrary union of basis elements of form Y cup U lambda. rewriting yields Y cap (arbitrary union of sets open in X), which is a finite intersection of open sets in X. then O is in the topololgy of X

yes

that part is used in my last sentence

ah wait so this is um

it means

tau_X is finer than tau_Y right

for an open subspace Y \subseteq X

[rancid umbra]

well

you really only compare topologies on the same set, to my knowledge

[deep surge]

right i see

well whatever

nvm

[warped helm]

ok wait scratch this im just tired LEL

[silent garnet]

Is there a self-homeomorphism of closed disc minus 2 disjoint closed discs that has no fixed points?

the answer is pretty easy for n≠2 holes and if the holes are open. but i have no idea how could this be proved for n=2 and closed holes.

[gaunt linden]

Wiseguy answer: let one of the disks you subtract touch the boundary of the outer disk.

[deep surge]

ping?

🤔

ok

[gaunt linden]

I suppose you mean a closed disk minus two disjoint open disks, so the resulting space is closed in R².

[silent garnet]

Nope, that one is easy

[gaunt linden]

Ah right, I just found a solution for that too.

[fierce mesa]

Hi. For Proposition 8.14, can anyone explain why A being contained in B being contained in closure(A) implies f^-1[{k}] = B?

[polar storm]

it's bcuz the closure of A in the subspace topology on B is B.

[fierce mesa]

Can't the closure of A be bigger than B though?

That would mean closure(A) wouldn't be in the subspace topology of B (on that note, is the subspace topology the same as the relative topology of B?)

[polar storm]

oh I'm talking about the closure of A with respect to the subspace topology on B.

[lavish mountain]

we take the closure in B, not in X

[fierce mesa]

Yeah wait that's what I'm saying

How can the closure of A be part of the subspace topology of B

[polar storm]

so it can't be bigger

[fierce mesa]

The largest element of the subspace topology of B is B

[polar storm]

since the ambient space is B, not cl(A) or X.

[fierce mesa]

I'm still confused sorry. Taking what you said to be true, how does that imply that f^-1[{k}] = B?

[lavish mountain]

f^-1[k] is a closed (in B) set containing A and so is cl(A), the smallest closed (in X) set containing A (when restricted to B, it's still closed)

this is another theorem saying the closure of a set A can't be that much bigger than A, in this nice case that A is connected

[fierce mesa]

Ohh wait I think I get it

So f^-1[k] is a closed set in B that contains A, closure(A) is a closed set that contains A.
If you restrict closure(A) to be in B, that restriction should still be closed (by definition of closure(A)), but more so we have that B = restriction of closure(A) under the topology of B [since closure(A) is bigger than or equal to B in X]

Which means that B has to be closed in B

Thank you both!

[solar turtle]

I'm wondering about the differences between compactness and sequential compactness. I have not formally learned topology, so I'm not going to delve very deep and I'm just asking out of curiosity. I read https://en.wikipedia.org/wiki/Sequentially_compact_space which says the two notions are equivalent for metric spaces, which I can understand, but apparently

there exist sequentially compact topological spaces that are not compact, and compact topological spaces that are not sequentially compact.
There are examples of both, but I fail to understand them.
First, for sequentially compact but not compact, the example is:
The first uncountable ordinal with the order topology
Now, I think I understand what the first uncountable ordinal is, and I can kind of process what an order topology is, but I don't understand what the result of putting the two together is.
Second, for compact but not sequentially compact, the example is:
The topological product of 2^ℵ_0 copies of the closed unit interval
Here I think I just don't understand why this is not sequentially compact.
Can someone give me some intuition?

[polar storm]

The usual compactness and sequential compactness are non comparable notions in general topological spaces, but they coincide in metrizable spaces

[solar turtle]

Yes I got that, I'm specifically asking about the examples

[opaque scroll]

So in the order topology you can check a sequence for convergences by doing the liminf and limsup, just like you would with a sequence of real numbers for example.

In in the first uncountable ordinal every countable set has a supremum

[polar storm]

For counterexample, {0,1}^P(N) is compact by Tychonoff thm, but u can easily show that it is not sequentially compact

[opaque scroll]

Consider e_i to be the element that is 1 in degree i and 0 otherwise

Then look at the sequence {e_i}

[polar storm]

Conversely, the minimal uncountable well-ordered set is sequentially compact, but not compact.

[solar turtle]

Wait is that for the first or second example?

[opaque scroll]

For 2^N

Actually, there might be a mistake in that example. Hold on

[solar turtle]

Oh ok so that's compact because it's a product of compact spaces but not sequentially compact because of that sequence

[opaque scroll]

This sequence converges to 0

[solar turtle]

Ok that left me more confused

[opaque scroll]

Sorry, I just gave a bad example

[solar turtle]

2^ℵ_0, where ℵ_0 is the cardinality of the natural numbers, is definitely not finite, right?

[opaque scroll]

No, it has the same cardinality as the real numbers

So you might as well use the real numbers as indexing

[solar turtle]

Right yes it's the mathfrak c thing

[polar storm]

oh nvm, u already gave the specifc counterexample. sorry, didn't ready the question properly

[solar turtle]

So with that many copies of the closed unit interval, how would your sequence ever converge?

[opaque scroll]

Convergence of sequence in the product topology is the same as pointwise convergence.

So the idea here is that since you have 2^N factors you have enough room to go crazy in each factor that you can't tame all of them at once by going to a subsequence

Alright, here's what you can do:

We think of 2^N as subsets of N (the natural numbers).

Define a sequence xn to be 1 at those indecies where n belongs to the set and 0 otherwise.

Now for any subsequence we pick, we can make a subset of N containing about half of those elements and there the sequence will fail to converge pointwise as it will flip flop between 1 and 0

[solar turtle]

Ok I think I understand that but I'm missing why the simpler example fails

[opaque scroll]

The other example was actually about 2^N and not
[0, 1]^(2^N)

[solar turtle]

Not sure what you mean... 2^N is just a number, what space are we talking about?

[rugged escarp]

No its set of maps from natural numbers to 0,1

X raised Y where X and Y are sets means maps from Y to X

2 denotez the two point set

[solar turtle]

Oh okay that makes more sense

And so the sequence {e_i} converges to 0 because in every degree the limit is 0?

If that's right then I think I got it for this part