point-set-topology

[tender halo]

meh, thats the case with some separation axioms, and perhaps stuff like idk Frechet-Uryhson spaces/some connectedness stuff/etc

Lindelofness js a little strange

but compactness and paracompactness i think are fairly natural

[crisp lintel]

compactness for sure

and yeah I do agree that once you are familiar with the contexts where it is useful, then it seems quite natural

which happens a lot

I didn't know why locally compact hausdorff spaces were so significant until I had worked with them a lot and had seen all of the things that they allow you to do

[queen prism]

my only intuition for paracompactness is in terms of partitions of unity

[crisp lintel]

I believe its equivalent to partitions of unity (maybe only for hausdorff spaces or smth) so thats a good intuition to have

but of course its typically an easier condition to check on a space by space basis

[tender halo]

those are actually i think are the examples of just being nice enough technical conditions

[crisp lintel]

which is probably why it exists rather than defining a "partition of unity space"

yeah fair, though hausdorff and locally compact interact in just the right way to ensure you have enough functions to do a lot of useful things which is really when they clicked for me

[unreal stratus]

LCH = I can think of this as a manifold

[crisp lintel]

my favorite manifold, the cantor set

[tender halo]

local compactness is not very natural imo, and hausdofrness is only there because it implies tychonofness

[crisp lintel]

[tender halo]

and tychonofness is the natural setting

[unreal stratus]

True

[unreal stratus]

For what?

[tender halo]

for uhh

topology

[unreal stratus]

Ig depends what one is doing lol

[tender halo]

/hj

it is the minimal sane separation axiom

equivalent to being embeddable to a compact (hausdorff space)/into a product of segments

[crisp lintel]

for harmonical analysis at least is the main setting im used to

since you want enough functions

(into R or C)

if you aren't as concerned with functions and function spaces, I suppose less would be enough

[tender halo]

and (the big part of why its natural to me) equivalent to its topology being completely described by functions into R

and its the natural setting for doing stuff with ultrafilters (you get the nice duality between ultrafilters and zero sets)

which lets you do stone cech, etc etc

[tender halo]

well its equivalent modulo uhh T0 i think

but everything is T0

except pseudometrics i guess

[gritty widget]

The REAL setting for topology is continua

[upbeat kiln]

No, the ACTUAL REAL setting for topology is clearly condensed sets! /s

[ashen rivet]

hi everyone~
i'm not very used to topology problems. i'm trying to show that if f : omega1 -> R is continuous (for the order topology), it has an upper bound

suppose it doesn't have an upper bound. u_n = f⁻¹( (-inf, n) ) is a sequence of open sets. All the u_n are countable, otherwise the image of an adherent point of u_n would be an upper bound for f (that's the point i have some intuition for, but not sure i know how to justify)

the union of the u_n is a countable union of countable sets, so it is countable and therefore in omega1. But f doesn't have an upper bound, so the union of the u_n is omega1. contradiction

can anyone help me to justify that the u_n are indeed all countable? (if that's the correct path to solve this)

[tender halo]

well the easy way to prove it is to just say that omega_1 is countably compact and therefore its image is countably compact and therefore compact and therefore (closed and) bounded

but for your way hmm

its kinda invovled i guess

if some u_i is uncountable, it means it is cofinal but not the whole space

ok what i can do is say lets take v_n to be preimages of (-inf, n], and w_n to be preimages of [n, +inf) and say
if f has no upper bound, then all w_n's are cofinal. if some v_i is uncountable, then both v_i and w_{i+1} are cofinal

which is bad because then we can take a common adherent point of v_i and w_{i+1} whose image is now both leq i and geq i+1

i think that works

i guess you can do the same with preimages of open rays instead of closed rays, the argument is the same

i like closed because then its like club sets

[ashen rivet]

oh i like this argument!

thank you, got it :)

[fringe oasis]

If I’ve got three topological spaces X, Y and Z and a continious map $f : X \times Y \rightarrow Z$, I define $\hat f : X \rightarrow Z^Y$ where $Z^Y$ is the set $C(Y,Z)$ with the compact-open topology and and where $\hat f(x)(y) = f(x,y)$.

How can I show that $\hat f$ is continious ?

[gentle osprey]

aaaa

[opaque scroll]

So it's enough to verify continuity for the subbasis of the compact open topology. I.e. if K is compact in Y, U is open in Z and V(K, U) is the set of continuous functions with g(K) < U, then you want it's preimage to be open in X.

In other words the set of x with
f(x, K) < U
should be open.

Use continuity of f, the basis for the product topology and, and compactness of K.

[lucid ocean]

does the converse hold, and if not, what must be imposed for it to hold

[opaque scroll]

Something like X being locally compact I think

No, Y being locally compact I mean

[lucid ocean]

that makes sense

I need to get back to topology and finish that project once and for all, after I'm done being sidetracked by QFT and national exams

[fringe oasis]

Thanks, but that doesn’t help me a lot 😅, that what I’ve been trying

[fringe oasis]

Thanks, but that doesn’t help me a lot 😅, that what I’ve been trying

Idk where to use the compacity of K

[opaque scroll]

Okay, so pick one x such that
f(x, K) < U.

Then what you want is a neighborhood of x where the same is true.

Now for any k in K, f(x, k) is in U. Since f is continuous that means there exists open sets
V and W in X and Y such that
f(V, W) < U and V contains x and W contains k.

Now the problem is that W might not contain all of K. Here you need to use compactness somehow

Remember that compactness has something to do with covers, so you want some kind of cover of K

[fringe oasis]

Okk so I do that for all k, and I take a finite number of Wk

[opaque scroll]

Indeed, and corresponding Vk

[fringe oasis]

And intersection of Vk ?

[opaque scroll]

Well, does that work?

[fringe oasis]

Yes

Can I ask where do u work ? (I assume ur a mathématician)

[opaque scroll]

Right now I'm in the process of graduating, so currently working nowhere I guess

[fringe oasis]

But in wich country do u study ? And wich uni?

[opaque scroll]

In Norway, NTNU

Why?

[fringe oasis]

Just like that

So now, I learn I can go to Norway for math

[opaque scroll]

I'm sure they have math in most countries in the world

[fringe oasis]

Yeah, but the education isn’t the same in all country, and also the reasarch system

[tender halo]

a topology on Y^X is called proper if for every space Z and a continuous map f: X \cross Z -> Y the corresponding f bar: Z -> Y^X is also continuous

this happens if Y^X is not too fine

a topology of Y^X is called admissible if for every space Z the reverse is true - for a continuous map g: Z -> Y^X the map g bar: X \cross Z -> Y is also continuous

this happens if Y^X is not too coarse

if \Tau_1 is a proper topology on Y^X, and \Tau_2 is an admissible topology on Y_X, then:
any topology coarser than \Tau_1 is proper
any topology finer than \Tau_2 is admissible
\Tau_1 is coarser than \Tau_2

therefore, there is at most one topology on Y^X that is both proper and admissible

the compact-open topology is always proper

and if X is locally compact, then it is also admissible

[viscid blade]

Why did bro choose to live in the arctic circle

Also first time i saw jagr outside groups rings fields

[opaque scroll]

Trondheim is over 300km south of the Arctic circle

[viscid blade]

Trondheim is same altitude as Umea

[opaque scroll]

Pretty close at least yeah

[viscid blade]

Are you doing postdoc as well at ntnu ?

After doctorate

[prime elbow]

How's the ntnu ?

[desert vortex]

you can come to my house and I will give you cookies and tea and we can chat about math

when you come to sweden hit me up

[viscid blade]

where in sweden are you ?

[desert vortex]

göteborg

[viscid blade]

ohhh nice

[desert vortex]

chalmers university

[viscid blade]

i will come one day

and you will give me cookies and tea

Fika

[desert vortex]

yes welcome

[viscid blade]

you're going to actually have to do it tho

cuz i'm a few hours from goteborg

[desert vortex]

of course

🙂

where do you live?

[viscid blade]

I dmed you

[stone goblet]

Can someone please explain this topic to me

So, I understand what an open ball intuitive is just set of all points are closer than the distance r

[austere flare]

This definition says that, in a metric space, open sets contain a ball around every point

[warped helm]

closer than a distance r from the center

otherwise the set you described looks very different

[stone goblet]

okay I see thanks for that I learnt a little more it now

[warped helm]

why did munkres write this last part about "we assert that the sets U_i X V_i..."

i don't understand what this adds to the proof that isn't apparent from just elementary set theory

i feel like im missing something

[rancid umbra]

he is saying that in the following paragraph, we will show that the sets U_i x V_i ...

he does this because you need to show that the tube W x Y is contained in N

[warped helm]

yes, but i guess my confusion is as to why this needed as much deliberation as it got

[rancid umbra]

it sounds like you are upset about the opening sentence lol

[warped helm]

like munkres has jumped over larger details than this, so im thinking there was some fine detail that im not seeing, but if it's really just set theory then

[rancid umbra]

i don't think that he is skipping over any details here?

[warped helm]

i'm not saying he is

i'm saying that other proofs omit larger details than this

[rancid umbra]

oh. sure, could the proof have been shortened, yea. but it was just a paragraph i suppose

finishes off the proof nicely

and completely

[viscid blade]

Why do math textbooks have this font

Is there like a website which can safely change font to something more legible like nice lateX

[umbral hamlet]

think of open sets as the sets than can be expressed by open balls in standard set operations

so like open sets can be unions of open balls and also FINITE intersections of open balls

finite is important bc u can get some stuff u dont want happening otherwise

[warped helm]

This is just a scan of a print of munkres

as far as i can tell it's just times

i don't know of any such website

[umbral hamlet]

the tube lemma is burnt into my head bc i read this chapter of munkres while waiting in line for the velocicoaster at universal

oh btw @warped helm how are u doing with topology now

my confidence in it has increased significantly bc i got a 92 on my topology midterm

[warped helm]

that's like me with the contraction mapping theorem in R; i had it as a problem on a hw and while in line i was doing it in my head

eh idk, i've been slacking lowkey

i have to lock in for my midterm on friday

i've been doing better though because i've been sleeping more consistently

[umbral hamlet]

yeah

i learned i gotta eat better and sleep better bc i fumbled my algebra midterm really bad

bc i had the worst test anxiety ive ever had out of nowhere

[warped helm]

my topology class is not so bad, just very boring. the professor doesn't add much apart from giving us a grade

everything written on the board is directly from munkres

[umbral hamlet]

my prof is not good at teaching but shes a really good person and very good at topology

just too good for our sakes

[warped helm]

i think i'll have to wait to take alg top until next year unfortunately

[umbral hamlet]

that sucks

[warped helm]

lowkey im fine with it since i dont think it's immediately relevant for much of the analysis im interested in

[umbral hamlet]

algtop is cool as hell

[umbral hamlet]

fair enough

what analysis do u do

[radiant stone]

algtop

[warped helm]

well, not much since i'm only in an intro real analysis course rn, but after this semester i think i can finally move onto more serious differential equations stuff. up to this point most of my experience with it has been computational

eventually i hope to do Serious Probability and maybe some variational calculus stuff :3

oh and i guess complex analysis too, that seems like great fun

and fourier...

[ruby delta]

that is surprising, i thought you would have taken a RA course already

[umbral hamlet]

im in the boat for complex like u are with algtop

was gonna take it next semester but probably not

[warped helm]

i think a lot of the stuff i talk about here on this discord i've absorbed through osmosis

hm well not a lot but some of it, i tend to not help people in topics that i haven't had a course in

[prime elbow]

If X is topological space and if I take Y = X u { ∞}, and define topology on Y such that all open sets of X union open set containing ∞ is of the form Y\K, where K is a compact set of X.

So Y is compact topological space, right?

It is similar as one point compactification, just i just removed Hausdorff condition and locally compact

[umbral hamlet]

what is an open set containing infinity

theres no notion of that yet

[young stone]

they are defining them to be complements of compact subsets of X in Y

[umbral hamlet]

oh, thats not how i read this

[gaunt linden]

You're almost describing https://en.wikipedia.org/wiki/Alexandroff_extension, except in the general (non-Hausdorff) case you want K to be not only compact but compact and closed; otherwise your new topology may not be closed under binary intersections.

[prime elbow]

I see

Can I get T1 space on countable set such that it is not second countable?

[empty grove]

Yeah, I think the cofinite topology works

[prime elbow]

How do I show it is not the first countable ?

I think if X is countable and T1 space with co finite topology it is first countable, and since X is countable implies topology is second countable

[empty grove]

Right it is second countable

You can take N x N with a wacky topology

Arens space works

[prime elbow]

So Arens space is T1 space which is not first countable

[empty grove]

yep

[prime elbow]

Oh

Thanks

[gaunt linden]

We can simplify the Arens example slightly for this purpose: Take N² and declare a subset to be open iff it is empty or it omits only finitely many elements in each column.
This is not second (nor first) countable, since for every proposed countable base we can diagonalize to find a nonempty open set that contains none of the proposed base sets.

[prime elbow]

Actually I don't understand how Arens topology is defined, can you tell me what is meant by omits only finitely many elements in each column ?

[opaque scroll]

For each x there are only finitely many y such that (x, y) is not in the set

[prime elbow]

Also, can someone give me a hint to find topology which is ordered topology but not first countable?

[prime elbow]

I see

So it is clearly T1 space

[gaunt linden]

The extended long line.

[opaque scroll]

You just need a very "big" order. Do you know about ordinals?

[prime elbow]

And how is Aren's topology defined?

[prime elbow]

Yes a little bit

[gaunt linden]

There seem to be two different Arens spaces; Wikipedia defines one of them at https://en.wikipedia.org/wiki/Arens–Fort_space

[opaque scroll]

Well then you might consider some very big ordinal. This is essentially what tropos already suggested

[prime elbow]

Oh

[gaunt linden]

They both work for this purpose; but the description in Wikipedia looks more accessible than the one they link to for the other (non-Fort) Arens space.

[prime elbow]

Okay if I take w be the minimal uncountable well-ordered set, so I have to show it is not first countable

Ordinals are the equivalence classes of well ordered sets?

[opaque scroll]

Yes, though you probably want to consider w+1

[prime elbow]

You mean R u {∞}?

[gaunt linden]

No, I was betting on you knowing about the long line already. But if you don't, just take Jagr's suggestion of the order topology on omega_1+1 instead.

[prime elbow]

Oh

[prime elbow]

Okay consider w+1.

I can say there is no countable decreasing sequence such that (a_1, w] \subset (a_2, w] \subset (a_3, w] ...

Because I can take A = { x | x < a_i for some i }, this set is countable. w is uncountable implies there exists x in w, then consider (x, w], then there is no a_i such that (x, w] \subset (a_i, w].

How do I proceed now?

[opaque scroll]

So then you're done.

You assumed you had a countable basis, then found a neighborhood not covered by that basis, badabim badaboom, not a basis

[gaunt linden]

there is no countable decreasing sequence such that (a_1, w] \subset (a_2, w] \subset (a_3, w] ...
well, there is, it just has a nontrivial limit.

Fortunately that doesn't harm the actual argument that follows.

[prime elbow]

I think I have to write my argument again, no countable decreasing sequence such that (a_1, w]\subset (a_2, w ] ...., such that any open set U containing w there exists a_i such that U \subset (a_i, w]

[gaunt linden]

You can just omit speaking about sequences in the argument at all. They're useful for intuition here, but it turns out the argument that results doesn't need them.

As Jagr stated, just start by assuming you have some countable neighborhood basis for the top element. It doesnt really need to be a decreasing sequence for your argument about the union of complements being countable to work.

[prime elbow]

So that means, every proper open set of w is countable?

[gaunt linden]

Uh no.

[prime elbow]

Sorry I am not following the argument

Let's say U_i's are the countable basis of w

Then what should be my next step?

[gaunt linden]

For each i pick an a_i such that (a_i,w] is contained in U_i.

[prime elbow]

Got it

[gaunt linden]

Then let x be the supremum of all the a_i's.

[prime elbow]

Okay

[gaunt linden]

By your argument above, x is countable and is therefore in particular not w.

[prime elbow]

But what if x is one of the a_i's?

[opaque scroll]

ai is not contained in (ai, w]

so just do (ai+1, w]

[prime elbow]

x is the supremum of a_i's, right? If I consider a_i's as set then supremum means \bigcup a_i's?

[opaque scroll]

Which is again countable, and same for x+1

[prime elbow]

I know w is well-ordered set which is uncountable and for any x in w, the set { y in w | y < x } is countable.

And i think you are working with sets, so w is a well-ordered set which is uncountable and contains sets which are countable, right?

Or can you please define ordinals for me?

[prime elbow]

So (x+1, w] is an open set containing w, but no U_i's contained in (x+1, w]

I got it, thank you @opaque scroll @gaunt linden

Just, I am not very confident about ordinals but I understand

[opaque scroll]

So an ordinal is a well ordered set such that every element is also a subset.

The first uncountable ordinal w is a set containing all the countable ordinals, then w+1 contains all the countable ordinals and w

[gaunt linden]

In fact, this could be replaced by: for each Ui pick some ai element of Ui and not equal to w, full stop.

Or, without even picking, let ai be the minimal element of Ui.

[prime elbow]

So here I am confused, every element is also a subset, why I can't say every element is set, or do we recursively define x +1 = x union {x}

[opaque scroll]

Yes, you define
0 to be the empty set
x + 1 = x union {x}
and the supremum of a set of ordinals to be their union. This then recursively defines all ordinals.

[prime elbow]

So is it the same as Von neumann construction of natural numbers?

[gaunt linden]

Yes, except for the natural number you don't take the third rule -- there are some sets of naturals that don't have a (natural) supremum.

[prime elbow]

[gaunt linden]

(As far as the topological question is concerned, we don't really need the full power and glory of ordinals here -- they're just a convenient way to make a totally ordered set such that the whole set is uncountable, yet { x | x < c } is countable for every c in the set.)

[prime elbow]

Yes

[prime elbow]

I have to show I^2, I = [0,1] is not metrizable with dictionary order, so I am thinking about to show it is not the first countable

[rancid umbra]

pretty sure it’s first countable.
on any slice, it looks like I, so you can transport over a local nbhd basis

the endpoints require some care

unless i’m missing some small detail

[prime elbow]

So how do I show it is not metrizable? Is it separable?

[gaunt linden]

How about d((x,y),(z,w)) = 2 if x=z, otherwise |y-w|?

[prime elbow]

I think it is not separable

[gaunt linden]

Does "metrizable" require separability for you?

[opaque scroll]

Compact + metrizable implies seperable is the idea I think

[prime elbow]

No

It is question in Munkres

[rancid umbra]

it does have a countable dense set though, the rationals I^2

[gaunt linden]

But [0,1]² with the dictionary order and order topology is not compact as far as I can se.

[opaque scroll]

That's not dense

[prime elbow]

But i think I can make uncountable disjoint open sets, what if I take {x} ×(0,1) for each x in [0,1]

[gaunt linden]

Yeah.

[rancid umbra]

oh yes, the slice coordinate doesn’t have to be rational

[prime elbow]

It is compact

Because it has the least upper bound and it is closed interval

[opaque scroll]

You get a lot of bandwidth from having to cover (x, 0) and (x, 1). Looks pretty compact to me

[prime elbow]

I^2 is well-ordered set, right?

[opaque scroll]

It's not well ordered no

[gaunt linden]

Even a single slice is not well-ordered.

[prime elbow]

Yes it is not

[gaunt linden]

Aahh, good point.

[prime elbow]

Can I say it has least upper bound, i think it has

[rancid umbra]

every subset has a lub in the square, yea

[prime elbow]

So then I^2 is compact

[prime elbow]

Nice idea

[quartz horizon]

What intuition should I have for G_delta and F_sigma sets

[rancid umbra]

this kind of reasoning works to show that it is not separable.

a countable subset projects down to a countable subset of I x 0 = I. this can’t be all of I, so it misses some points. but then for a point that it misses, say t, t x (0,1) is an open set in I x I not containing any of the points from your countable subset of I x I.

[prime elbow]

Yes

[crisp lintel]

Tbh I just think of the whole borel hierarchy as just slowly creating more and more weird sets

idk if there's like a better way to think about them

[prime elbow]

So here I am thinking two cases,

One is A is open iff (x,y) in A then (x, z) in A for all but finitely many z.

Second is A is open iff for all x in N, (x,z) in A for all but finitely many z.

[quartz horizon]

There’s some result that

A subset of a complete space that’s completely metrizable is G_delta

[crisp lintel]

Yeah

[quartz horizon]

why

[crisp lintel]

it's an if and only if actually for completely metrizable

Id have to go look at the proof again

[alpine nest]

Yeah, same; I think the countable operations make it fairly easy to lose the intuition.

[crisp lintel]

ok so the proof first does the result for open sets then for a countable intersection it embeds within some countable product

as a closed subspace

so the key result I guess is that a countable product of completely metrizable spaces is completely metrizable

for open sets you basically take the usual metric but augment it by the distance to the complement to get rid of the only way completeness can fail

basically you just grab the edges of the open set and pull them towards infinity I guess is the way to think about it

[quartz horizon]

@.@

[tender halo]

the surrounding space need not be complete

[alpine nest]

Yeah, "Open subset of complete metric space is completely metrizable" is a neat exercise in itself.

(including the observation that it might not be complete in the original metric)

[quartz horizon]

I don’t really understand any of these results…

[tender halo]

completely metrizable spaces are characterized as those that are only embeddable in a metrizable space as a G_\delta

[crisp lintel]

another thing that seems relevant for G_delta is that a continuous map A -> Y on a subset A of a metric space where Y is complete can be extended to a G_delta B contained in the closure of A

[tender halo]

yea

this is where the G_\delta comes from

[crisp lintel]

I think a fairly natural case where they come up is when you can construct for each natural number some open neighborhood of each point in a set

then take the intersection of that union

which is a kinda common thing to do at least in metric spaces

this already gives you a pretty good way of approximating any set as a G_delta set

much better than an open approximation via the interior or closed approximation via the closure

[crisp lintel]

(namely for each n, take a ball of radius 1/n about each point and take the union. then take the intersection of these open sets over all n)

idrk if that's a standard thing to do but maybe that gives some better idea about what G_delta sets look like or how they can arise in natural ways

[alpine nest]

For Q in R that would just give you all of R though, if I understand correctly

[crisp lintel]

no it would give you Q

or wait

nvm

[tender halo]

it would give you R, Q is not a G_\delta

[quartz horizon]

Maybe this stems from me not really understanding the baire category theorem

[crisp lintel]

yeah

[alpine nest]

But there does exist a G_delta set of Lebesgue measure 0 which contains Q

[crisp lintel]

honestly I feel like so many ways of presernting the BCT are really unintuitive

what finally made it click was the terry tao blog post on it that presents it as a sort of version of the pidgeonhole principle

[quartz horizon]

It was a big result in my functional analysis course

?

[alpine nest]

Oh yes, Baire is huge in FA

[crisp lintel]

yah although moreso its consequences

[alpine nest]

That's why Banach spaces (or Hilbert spaces) are the primary object of study rather than general normed vector spaces/inner product spaces

Well yes

[crisp lintel]

basically as an analogue of th is

he says that null sets are sort of "measure theoretic nowhere dense sets"

and then this is the version of baire

[quartz horizon]

um…

[crisp lintel]

that I find is by far the most intuitive

[quartz horizon]

I don’t see the analogy 😭

[alpine nest]

"measure zero sets" are analogous to "nowhere dense sets"

[crisp lintel]

well null sets are measure theoretically sets that have very little significance measure theoretically

and nowhere dense sets are kinda similar for topology

[alpine nest]

"The union of countably many measure zero sets can't have positive measure" translates to "the union of countably many nowhere dense sets can't have nonempty interior"

(in a complete metric space, so unlike the measure theory statement; the topological statement doesn't hold universally)

[quartz horizon]

Honestly the only time I’ve ever thought about nowhere dense sets is for Baire

I’ve never used them outside that

[alpine nest]

My preferred way to work with Baire has been via contrapositive, with dense open sets (or dense G_deltas)

[crisp lintel]

though I will say even terrys framing I prefer to just always frame it in terms of closed sets so I can actually remember it

aka if a countable union of closed sets contains a ball, then one of the individual closed sets contains a ball

which is p much directly a pidgeonhole

[quartz horizon]

Is this even true

[crisp lintel]

honestly I don't like dense open sets because I have a very hard time kinda imaging whats going on

[alpine nest]

It is in a complete metric space

[crisp lintel]

in a complete metric space yes

that's exactly the baire category theorem

[quartz horizon]

I thought it was a density thing

Having nontrivial intersection with an open ball isn’t the same as containing an open ball right

[crisp lintel]

sure but you don't need that if you restrict to closed sets

[quartz horizon]

Uh..

[crisp lintel]

condition 6 here

is the only one that makes any intuitive sense to me really

yeah wikipedia comments on this similarity as well

[crisp lintel]

also its the version that shows up in all the functional analysis proofs

[tender halo]

thinking about BCT in terms of cech-complete spaces is helpful to me

[crisp lintel]

in my class on metric spaces I think we introduced it in terms of the intersections of dense open sets and idk why because its super hard to think about imo

although maybe that's the setting where you prove it that could be why

partly I just think the term nowhere dense is kinda confusing and it always takes me a second to parse what it means maybe thats a skill issue

[alpine nest]

The most useful way probably depends on what you're doing with Baire, I've often used it to establish existence-type results; where to prove that an object satisfying <condition> exists, you phrase <condition> as a conjunction of countably many simpler condition each of which can be shown to hold on a dense open set.

[crisp lintel]

but closed sets are nicer

[alpine nest]

Therefore the set where all conditions are satisfied is dense, and in particular it's nonempty

[crisp lintel]

ah that's a nice way to think about it

[alpine nest]

It feels like a weirdly overkill way to show existence, since in fact you show that "most" objects satisfy your condition, even if you only need one.

[crisp lintel]

I think goign through the proofs of the uniform boundedness and open mapping theorems gives you a good sense of whats going on for the closed set variation

[tender halo]

my definition of nowhere dense is every open set has an open subset that doesnt intersect with our set

thats kinda how i think about them

[crisp lintel]

that's a good way to think about it actually

[tender halo]

or better yet, an open basic subset that doesnt intersect with out set

[viscid blade]

this is just all the set {(a,b) I a in [0,1] and b in [0,1]}?

[gaunt linden]

Yes, but with a particular topology applied. (And I was later convinced that it is indeed compact).

[viscid blade]

i guess i will learn wht a topology is next sem

[gaunt linden]

Uh ... it is generally assumed that people who come here do.

[alpine nest]

It was probably mentioned, but a closed interval is always compact in the order topology (and [0,1]^2 with the order topology has a least and greatest element, so it can be written as a closed interval)

[gaunt linden]

Though metric spaces are also on topic here, so perhaps not.

[viscid blade]

I come here because i will learn topology

[viscid blade]

i know what those are at least

they motivated me to study topology

[gaunt linden]

Hmm, I didn't know that, actually.

[opaque scroll]

Given a complete order you mean?

[alpine nest]

Yeah, that's what I was about to type, I missed an important assumption (the least upper bound property/order completeness)

But the lexicographic ordering on the square does have that.

[gaunt linden]

Okay, crisis averted :-)

[alpine nest]

The set of rationals between 0 and 1 isn't compact in the order topology

[prime elbow]

Jagr any comment?

[gaunt linden]

I'm not following you there. Two cases for what?

The topology you describe in the first of your cases just gives you the disjoint union of countably many copies of (N, cofinite topology), which is second countable.

[prime elbow]

I am trying to understand this one

Here how the open set looks?

[gaunt linden]

An open set is either empty or all of N² except for finitely many points in each column.

[prime elbow]

So I can say,
A is open iff for all x in N, (x,z) in A for all but finitely many z.

[gaunt linden]

That's what I said.
(Except you also need to count the empty set as open in order to satisfy the definition of a topology).

[prime elbow]

Yes

[prime elbow]

How do we diagonalize here?

I am thinking if U_1,... are countable basis elements, then for each U_i associate A^(i)_{n} = { (n,z) | (n,z) not in U_i }.

Then thinking to define a new open set with associate A_n such that A_n = A^{n}_{n}.

But it didn't work

[gaunt linden]

The idea is to make an open set A where we have chosen each column such that it prevents A from containing one of the U_i's.

One way to proceed would be, for each i pick an x_i such that (i,x_i) is in U_i.
Then let A = N² \ { (1,x1), (2,x2), (3,x3), (4,x4) ... }.

[prime elbow]

Yes I am thinking exactly same

Make A such that in each column n, we delete one point which is in n column of U_n

Thank you, I got it

I have one question, how does someone think about such construction of these types of topology?

[gaunt linden]

I don't know how to answer that.
The kind of topological intuition one gets from R^n or manifolds is pretty much useless for this kind of counterexamples. I tend to think of them more as exposing loopholes in the definition by constructing deliberately unnatural examples.

[tacit drift]

is there a notion of collinearity in arbitrary metric spaces where x,y,z are collinear if d(x,y)+d(y,z)=d(x,z)

obviously it depends on the orientation of your triplet but

[tacit basin]

If $U\subseteq \R^n$ is homeomorphic to $V\subseteq \R^m$, do we have $n=m$?

[gentle osprey]

Sara

[warped helm]

No

[tacit basin]

Ty

o wait im stupid i just thought of a counterexample lol

[warped helm]

Take n = 1, U = R and V = some line through the origin in the plane

[tacit basin]

Right ofc, thank you

[quartz horizon]

i think if U and V are open subsets this is true?

it's called the invariance of domain theorem

proving it is a little tricky afaik

[tacit basin]

that's R^n -> R^n i think

[quartz horizon]

actually you only need U to be open

wait hang on

i may be tripping

you're right

hm then what was i thinking of

[queen prism]

take U = V = {}...

[opaque scroll]

A open nonempty subset of R^n contains an open subset homeomorphic to R^n. So therefore open subset of R^n and R^m can't be homeomorphic (unless they're empty)

(or n=m)

[opaque scroll]

See "consequences" on the same wiki page

[tacit basin]

right okay, thank you :)

[prime elbow]

Any hint for the converse part?

[quick delta]

Mimic the proof of Urysohn’s lemma, being specific in which opens you choose to be the opens containing A

[prime elbow]

Say A = countable intersection of U_i's

Urysohn's lemma give me that there is continuous function f_i on X -> [0,1], on A it gives 0 and on U_i^c it gives 1.

[quick delta]

You don’t want to use Urysohn directly, you want to mimic its proof

[prime elbow]

I see

[quick delta]

Or rather, do it’s proof, but be more careful in how you choose your open sets during the proof

[prime elbow]

But that's a very long proof

[alpine nest]

I think you could prove it by invoking the Urysohn lemma in a resonable clever way, don't have to mimick that proof.

[alpine nest]

You should be able to use those functions to get the function that the exercise is asking you for

[prime elbow]

I am thinking of taking sup of f_i's

But sup of bounded continuous function, is it continuous?

[alpine nest]

The supremum of countably many continuous functions needn't be continuous

As an aside; it's going to be lower semicontinuous but that doesn't help you here.

Also the supremum would be the indicator of A^c (and the indicator of an open set is indeed lower semicontinuous)

[prime elbow]

I see

So any hint? How do I use those f_i's?

[alpine nest]

Well, you want a function that will be 0 on A, and that will be nonzero where at least one of the f_i's is nonzero.

Supremum is indeed one way of obtaining that, but it will ruin continuity.

[prime elbow]

Yes

[alpine nest]

What other ways can you think of of making a new function out of a sequence of functions?

[prime elbow]

Lim sup?

[alpine nest]

Again, continuity will be a struggle.

What operations on functions do preserve continuity?

I'm trying to come up with a hint that won't just be the answer, and it's tricky

[prime elbow]

Finite addition?

[alpine nest]

Yep, addition is a word that should be on your mind.

[prime elbow]

But countably addition don't give continuity, right? So we have to modify something here

[alpine nest]

Yes and yes

[opaque scroll]

If only there were some countable sums that converged

[alpine nest]

Countable addition gives continuity sometimes, if you play your cards right

[prime elbow]

How?

[opaque scroll]

So your functions are all bounded by 1. But unfortunately
Sum 1 doesn't converge

How can you make it into a sum that converges?

[prime elbow]

I don't get your question, but I am thinking of replacing 1 by 1/n^2

[opaque scroll]

Alright, so you have a function bounded by 1. Can you make it into a function bounded by 1/n^2 ?

[prime elbow]

Yes divide it by n^2

[opaque scroll]

So then you got it?

[prime elbow]

And how do I show f is continuous?

[alpine nest]

If X is a topological space, Y is a metric space, and (g_n) is an a sequence functions from X into Y, uniformly converging to a function g : X-> Y, then g is also continuous.

(i.e. uniform convergence preserves continuity)

[bitter shell]

Check the very last sentence of the proof where Friedl says 2b' follows in same manner as 1b by reducing it to proof of 2b

But this assumes that a collection ${ U_i }_{i \in I}$ being localy finite means {
${ {U^c}i }{i \in I}$ is locally finite as well

How else can it work

I read alternate proofs on stackexchange, but what is the author trying to say here

[gentle osprey]

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

[amber musk]

he is just re-using the same form of argument, not the same hypothesis

when he says same argument as (1b)

i think

[prime elbow]

I thought the same

[prime elbow]

So here \sum f_i(x) converges to f(x).

So if I take g_n(x) = \sum_{i=1}^{i=n} f_i(x), then g_n(x) is continuous and g_n(x) converges to f(x).

Since |f_i(x) | ≤ 1/i^2 for all x, therefore g_n is uniformly converges to f.

And clearly all they are uniformly continuous.

Hence, g is continuous.

Is it correct?

[prime elbow]

Are you using Friedl, on a side note there are some mistakes in exercise so be careful, I wasted my time on one question which I found wrong later

[prime elbow]

7, I have to show the converse.

By Lindeolf I can write X as countable union of metrizable space.

But i don't know that metrizable space is Lindeolf or not.

But i know every Lindeolf metrizable is second countable

[bitter shell]

I dont see how this follows since it would require that if Ui is a locally finite collection then so is the complements of Ui

Check the very last sentence of the proof where Friedl says 2b' follows in same manner as 1b by reducing it to proof of 2b

But this assumes that a collection ${ U_i }_{i \in I}$ being localy finite means {
${ {U^c}i }{i \in I}$ is locally finite as well

How else can it work

I read alternate proofs on stackexchange, but what is the author trying to say here

I REALLY don't see how this follows as this would require the ${ {U_i}^c }$ to be a locally finite collection whenever ${U_i }$ is a locally finite collection

[gentle osprey]

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

[prime elbow]

Friedl said you have to use 2 a'

[bitter shell]

That clearly doesn't work because then you need to apply 2a' for ${{U_i}^c }_{i \in I}$ which does not need to have local finite property
That is what I am confused about
He is telling to mimicm 1b for 1a in 2b' for 2a'

[gentle osprey]

Leno

[prime elbow]

Can you please see this carefully, because he proved 2 a' and in that proof he used locally finite property of U_i

[bitter shell]

Yes, but when you are taking complements in 2b', to invoke 2a', in the same manner he used it in 1b using 1a, you will have to need the locally finiteness of U_i complement not U_i

[bitter shell]

Do you understand what i am saying
The proof of 2b' using 2a' needs an extra condition to invoke 2a'

[prime elbow]

Yes I see the problem now

[prime elbow]

In order topology if I take an arbitrary intersection of closed interval subset, will the resulting set have to be closed interval?

Yes it is

[kind marlin]

can you use urysohn

[prime elbow]

Yes but how?

I have to show it is second countable

[kind marlin]

find a metrizable neighborhood Ux of each point x in X.

from regularity, find an open set Vx inside Ux whose closure is also contained in Ux. closed subspaces of lindelof spaces are lindelof, and subspaces of metrizable spaces are metrizable, so each Vx closure is necessarily second countable (lindelof + metrizable -> second countable), and any subspace of a second countable space is second countable, so each Vx is second countable

the Vx neighborhoods cover X, so we can find a countable subcover of Vx sets that covers X. each Vx has a countable basis as well. the union of these bases is a basis for the entire space, which you can verify, and it's a countable union of countable sets, so this is a countable basis for X

[prime elbow]

I got it, thank you ❤️

[pearl valve]

If you take [0,1]x[0,1] minus the point (0,0) (so the square minus bottom left corner) and then quotient where you identify the bottom and left edges, is this homeomorphic to a disk without an interior point?

Or homeomorphic to a disk without a boundary point

I feel like it’s the first case, but I don’t know how to check

Im just imagining taking a paper square and gluing the bottom and left edges together, u get a cone( which is homeomorphic to a disk). and then the bottom corner moves to the center not the boundary

But idk if this is like an artifact of poor intuition regarding quotients and homeomorphisms

[plush folio]

By "identify the bottom and left edges" you mean (x, 0) ~ (0, x) for x in (0, 1]? And the other edges are not identified in the same way?

[pearl valve]

Yessir

[plush folio]

Yeah seems like it would be a punctured disk. I can't think of an easy proof right away tho

[pearl valve]

Yeah that’s what the folks over on MSE said I just don’t really know why. Can’t find any similar problems or explanations online

[plush folio]

I think you can first prove that this quotient, without removing a point, is homeomorphic to a disk, then it's sufficient to show that the point you're removing is not on the boundary

[pearl valve]

Yeah and idk if that’s true cause like

[gaunt linden]

In polar coordinates, map theta,r to 4theta, r/min(sec(theta),cosec(theta)).

[pearl valve]

It’s definitely a boundary in the original space before

[plush folio]

What they're saying here is not the same thing tho, S^1 x [0, 1) is not homeomorphic to a punctured disk

[pearl valve]

Really?

It’s just a cylinder without boundary right

On one end

[plush folio]

Maybe I'm misunderstanding how you're gluing the edges, or the people on MSE are answering a different question

[pearl valve]

This was the question I asked

[gaunt linden]

It is, but [0,1) goes in reverse -- 0 is the boundary, 1 is the center.

[plush folio]

Oh damn, I'm stupid

Nvm

[pearl valve]

It doesn’t matter if you say (0,1] or [0,1) right?

They are the same?

[gaunt linden]

They're homeomorphic, yes.

(But of course not the same set).

[pearl valve]

Were u saying u can use this to show its homeomorphic to punctured disk?

[gaunt linden]

Yes.

[pearl valve]

Okay I’ll give that a whirl

Thanks

[gaunt linden]

It's perhaps simpler to say: First scale each ray out from (0,0) such that the edge of the square is moved in to a distance of 1 from the origin. Then multiply the polar angle of each point by 4.

The points that are glued together in the original picture end up mapping to the same point on the punctured disk, and the tranformation is injective otherwise.

[dawn frigate]

Here’s my professors proof for theorem 23.6 how does that prove X cross Y is connected

[warped helm]

the union of connected spaces that all share a point is connected

that's what the theorem cited at the end says

[dawn frigate]

So it shares point (x,b)

[prime elbow]

Limit point compactness implies sequentially compactness if X is metrizable space.

I think it is true when X is first countable and Hausdorff space( T1 also).

Now can I get a topology such that it is limit point compact but not sequentially compact?

[gritty widget]

Do you want a counter example or a hint for constructing one? Don’t want to spoil it if it’s the latter.

[prime elbow]

Hint

[kind marlin]

Maybe try making a topology where a point is in the closure of every other point, and there’s an “unbounded sequence” in the space

[prime elbow]

Unbounded sequence, what is the notion of boundness in topology?

[kind marlin]

yeah it’s definitely not a formal notion, I guess I’m speaking somewhat geometrically

basically, for any point, and any neighborhood of that point, im saying that we should have our sequence eventually leave and stay outside of that neighborhood. If you think of neighborhoods as a description of what’s “nearby”, then we’re kind of saying that our sequence eventually gets “far away” from every point in the space

that’s just my internal intuition for thinking about neighborhoods but that’s not important, the point is that this “guaranteed exit” property would definitely ensure that we don’t have a convergent subsequence

[prime elbow]

I have to construct a topology such that there is a point p which is in closure of every {q}, where q ≠ p.

[kind marlin]

I don’t think q not equalling p is an important distinction, and I also don’t know if this is the only way of doing it, but it was the most natural method to me

[prime elbow]

Oh p needs to be in closure not necessarily limit point

[kind marlin]

ahh got it

oh I guess a property I was loosely using is that the closure of a set contains the closure of each point in the set

Which is usually kind of obvious when singletons are closed LOL

but more useful here maybe

[prime elbow]

Oh deleted point topology works for that

So take R with deleted 0, so 0 is in closure of {x}, so take sequence 1,2,3,4,5,.... it has no convergent subsequence

Is it correct?

[kind marlin]

umm what’s the deleted point topology

[prime elbow]

X with deleted point q is the topology where open sets are those which doesn't contain q, and X.

[kind marlin]

hm but then 0 isn’t in the closure of any singleton?

like the set {1, 2, 3…} doesn’t have a limit point

oh wait I misread

X is just X

Not the old topology on X

okay lemme think

[prime elbow]

But there is a problem {1,2,3,4,...} is convergent to 0.

[kind marlin]

yeah that’s true we don’t get that “eventually exit” property

every point is nearby to 0 here

[prime elbow]

Any idea?

[kind marlin]

yeah I already have one but do you want me to give it

I think you can land on it

like the issue with your example is that 0 was too “close” to everything, its neighborhoods don’t distinguish between any of the other points

[kind marlin]

aaa sorry I misspoke

The closure of one point should be the entire space, aka every point belongs to its closure, not the other way around

[prime elbow]

Okay

[kind marlin]

It’s a really simple topology, you can define it on the naturals

|| {x : x < n} for each n in N ||

[prime elbow]

I see, in this topology every infinite set has a limit point because if S is an infinite set and by well ordering we get s in S which is the smallest element of S, then s+1 is the limit point of S.

And the sequence 1,2,3,4,...has no convergent subsequence

[queen prism]

why is it always the elon one

[rancid umbra]

<@&268886789983436800>

[gaunt linden]

Especially in the case of spambots, it would be useful to write something to that effect in your modping, or make it a reply to the spam.
When a spambot is banned, all its recent posts are automatically deleted so we don't need to hunt them down one by one -- but that means that there will be a lot of orphaned modpings in channels other than the one we first found it in, and the quicker we can ascertain "oh, this was just for the spambot I just banned" rather than some interpersonal problem that needs further action, the easier for us.

[orchid perch]

how do you show that open and closed parts of R^n are Lebesgue measurable with respect to the standard norm?

[rancid umbra]

do you know that open rectangles (products of open intervals) are lebesgue measurable?

[orchid perch]

Yes

[rancid umbra]

try to use the fact that any open set can be written as a union of open rectangles

oh

hmm

this may not work

[quartz horizon]

(countable union, right?)

[rancid umbra]

yea

but i wonder if that is true

[quartz horizon]

it is

[rancid umbra]

sweet

[quartz horizon]

given any point in an open set, you can find an open rectangle with rational coordinates containing it

and the number of such rectangles is countable

[rancid umbra]

ye, second countable

nice

[quartz horizon]

mhm, but being explicit about what the countable basis actually is

[orchid perch]

And is there anything special I need to do w it saying the standard norm

[vital shale]

Form the standard norm you get the standard metric so you get open balls which contains open rectangles

U can use the fact that max metric is equivalent to standard so they have the same open sets

[pearl valve]

How would you show the half open cylinder and open cylinder are not homeomorphic

S1 x (a,b) vs S1 x [a,b)

Is there anything straightforward via connectedness or comapctnesss?

[gaunt linden]

S1 × [a,b) contains points that don't have a neighborhood homeomorphic to R^2.

(For example, R^2 is simply connected but removing any point makes it stop being simply connected; a neighborhood of a point on the closed edge of the cylinder can't have that property).

[pearl valve]

Is there a reason that can be generalized to any kind of top space? What I mean is I read that number boundary components is invariant under homeomorphism, and I’m trying to arrive at this result from what I already know

[gaunt linden]

Hmm, that sounds like overkill -- what I'm saying is essentially that S¹ × [a,b) has boundary points at all (no matter how many or few components they split into), whereas S¹ × (a,b) doesn't.

But if you want to do it, then it should be as simple as observing that a homeomorphism takes boundary components to boundary components (and so does its inverse) because the property of a subset being a boundary components is defined only in terms of data that are preserved by a homeomorphism.

[pearl valve]

Do you know of any spaces where you would need to make such an argument? Or should you always be able to argue as such as the R^2 neighborhoods

[gaunt linden]

I can't immediately rattle of an example where it's the only way through.

I also think I'd prefer to decompose the "same number of boundary components" into two facts:

1. If two manifolds-with-boundary are homeomorphic, then so are their boundaries (each with the subspace topology).
2. If two topological spaces are homeomorphic, then they have the same number of connected components.
   of which the first can be useful without combining it with the second.

[pearl valve]

Yeah my professor mentioned the first point

I wonder why it’s sufficient to consider only manifolds though

I haven’t studied manifolds but from what I understand they are subset of topological spaces

Is it possible to discuss boundaries without discussing manifolds?

[gaunt linden]

I said manifolds because I'm not sure "boundary" makes sense for something that is not a manifold.

Well, "boundary" makes a different sense for a subset of an ambient topological space, but with that meaning, property (1) above doesn't hold.
The left half-plane and the open unit disk are homeomorphic as subspaces of R^2, but their boundaries as subsets of R^2 are not homeomorphic.

[pearl valve]

I think maybe I should set aside this problem until I learn more about manifolds

/boundaries

[unreal stratus]

I think you can consider the number of "ends" of these spaces. Basically the idea is that there are arbitrarily large compact subsets K of S^1 x (a,b) whose complements have two connected components- indeed consider the guys of the form S^1 x [a + 1/n, b - 1/n].

But this is false for S^1 x [a,b) because if you write it as a union of compact subsets then at some point you have to include a

[unreal stratus]

Ig this is not necessarily a particularly standard way but it is kinda cute and imo somewhat visual

[pearl valve]

Interesting I will take a look

thanks

[rancid umbra]

i mean, this just amounts to showing that (a,b) and [a,b) are not homeomorphic. if they were, then the two product spaces would be homeomorphic.

then you can just use the standard cut-point argument

right? am i messing something up?

[pearl valve]

If B is not homeomorphic to C, then AxB is not homeomorphic to AxC

I don’t know if this statement is true

[tender halo]

R^\omega, R, R^2

[pearl valve]

But yeah this was my first instinct

[pearl valve]

Is this a counter example

[tender halo]

yea

[rancid umbra]

ah yea

i was hoping you could just do something with the projections and relate each factor

but R^\omega is absorbative

[pearl valve]

Is R omega the same as R^natural

[crisp lintel]

yeah

[pearl valve]

Oh yeah I see

[crisp lintel]

So it isn't enough to show the open and half open interval are not homeomorphic unfortunately

My inclination would be to say something like in the half-closed curve if you remove the image of any closed curve the result will have a compact component

But formalizing that seems hard

cuz you kinda need the jordan curve theorem

or something analogous

[rancid umbra]

i liked potato’s solution with ends

[unreal stratus]

Thabk u

[rancid umbra]

you're welbcome

[rancid umbra]

why is the wiki page so cryptic here

it doesn't specify which definition \star is referring to and doesn't give the \hat{\star} construction

i can't find any other sources discussing this issue with local compactness; the wiki page doesn't list any sources either the source was listed on the wiki page, it was the springer book below. smh
i might add to this page. not having the other construction present is a bit annoying. there is also little discussion about the colimit and limit topologies on the join

https://en.wikipedia.org/wiki/Join_(topology)

apparently the source is this book
https://link.springer.com/book/10.1007/978-3-319-23488-5

in the section on joins

but this doesn't list any counter-examples either, it just claims this

[rancid umbra]

ofc an explanation is in topology and groupoids

https://mathoverflow.net/a/111139/167473

also here

[rancid umbra]

i hate that the final topology is used for colimits, initial top spaces

and that the initial topology is used for limits, terminal top spaces

so is the nlab wrong here? it claims that the join defined as a colimit (with the final, or colimit topology) in Top is associative

[gritty widget]

Because it's the best topology textbook

[rancid umbra]

i would honestly use it more if the font didn’t give me such a headache

[gritty widget]

I have it physically but I for whatever reason usually end up reading it on my computer online

It is in the desk my computer is on

[frigid sedge]

Why the definition of compact is so so so strangeeeee in metric spaces .. like whyyyyy mathematicians defined this way .. what's the motivation

[umbral hamlet]

wdym definition in metric spaces?

do you mean the "complete and totally bounded" corollary?

[frigid sedge]

Like I just started rudin .. so here the compactness is defined in metric spaces .. I don't know about other topological space

[hidden ravine]

Like the cover definition?

[umbral hamlet]

i see

so just the one that says

[frigid sedge]

Yesh😭

[umbral hamlet]

for any open cover of a space, we can get a finite subcover

[frigid sedge]

Yeshhh why 😭.. like I don't see anyyyy motivation like why someone wants to define something they call compactness in this way

[umbral hamlet]

compactness essentially acts like finiteness but for continuous functions

for example if you have a finite set A and a function f: A to R, where R is the real numbers, then f has a max and a min

thats pretty intuitive i hope

or clear to see at least

[frigid sedge]

Yesh and then in continuity this happens if domain is a closed bounded interval

[umbral hamlet]

yes, but we can go further

[hidden ravine]

https://math.stackexchange.com/questions/371928/what-should-be-the-intuition-when-working-with-compactness

[umbral hamlet]

if you have a compact space A and a continuous function f:A to R, then f has a max and a min

[hidden ravine]

Look at the answer by Qiaochu

[umbral hamlet]

turns out this result will be incredibly useful for some theorems in real analysis that youll see soon

[queen prism]

it's one of those things that makes sense only after you've been studying it for a while (on the order of a class or two. or ten.)

imo

[frigid sedge]

Umm hmm 🙂.. hope so

[umbral hamlet]

itll show up i promise

i was also confused as hell when i first saw it

speaking of confused as hell, im not doing great with the homotopy questions on my hw

its late tho so ill use that as an excuse

[queen prism]

how dare you turn in your hw late

[umbral hamlet]

oh no my hw isnt late

its due in like 22 hours

but im doing it late at night

[quartz horizon]

so there are 2 ways i like to think about compactness

[frigid sedge]

What what?

[quartz horizon]

in metric spaces, it's equivalent to sequential compactness, which you can view as a kind of "topological pigeonhole principle"

you may have heard that compactness generalises finiteness in some ways

if you have an infinite sequence in a finite set, then it has to hit some value infinitely many times

similarly, if you have an infinite sequence in a compact space, it has to "cluster" around some point (which you can view as having a subsequence that converges to the point)

this is the sense in which compact spaces are "inescapable"

more formally, you say that a sequence $(x_n)$ in a space $X$ has a cluster point $c$ if, for every neighbourhood $V$ of $c$, $x_n \in V$ infinitely often

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

the statement is that, in compact spaces, every sequence has a cluster point

[unreal stratus]

This is cool – not heard it called a topological pigeonhole principle before, but it feels very apt

:)

[quartz horizon]

it generalises nicely to compactness for arbitrary topological spaces

you just use nets instead of sequences

compactness is equivalent to every net having a cluster point

[frigid sedge]

😭i just started with compact sets 🥺🙂

[quartz horizon]

that's alright#

the other way i like to think of compactness is as an induction principle

[unreal stratus]

Compact sets are sets which are less than about a metre long

[quartz horizon]

https://www.drmaciver.com/2015/03/topological-compactness-is-an-induction-principle/

[frigid sedge]

Da best!

[quartz horizon]

you can do a kind of "induction" for compact spaces in the following way

suppose you want to define something globally over your whole space $X$, or maybe prove something is true over your whole space $X$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

if $X$ is compact, you have access to the following strategy:

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

1. define/prove your thing locally - meaning that, for every $x \in X$, find some open set $U_x \ni x$ where you can define/prove your thing

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

2. show that you can extend your definition/proof across finite unions - that if it works for open sets $U$ and $V$, it will also work for $U \cup V$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

step 1 is the "base case", and step 2 is the "inductive step"

compactness then guarantees you can extend your thing to all of X

so it works as a kind of "local-to-global" machine

whenever i've used compactness in practice, it's been one of these two viewpoints

either as a pigeonhole principle, or as an induction principle

[fickle tendon]

the nice thing about compactness is that it generalises nicely to posets

the proof that e.g. Zariski topology is compact follows from the fact that the top ideal of a ring is a compact element

[quartz horizon]

Yeah this is the notion of compact object in a category

One where Hom(X, -) preserves filtered colimits

This means a bit more towards the pigeonhole view I think

[novel stump]

is this the right approach for proving that the lower limit topology is a topology?

bc if thats the case then i believe i can do the intersection part, however the union part is still a bit confusing just bc of how many unions there are

[hidden abyss]

a_i and b_i should use the same index but otherwise yeah that's what the open sets are

[novel stump]

ahh okay, thank you!

[hidden abyss]

Unions might look daunting but it should actually be the easier part: think through what an union of unions is

Another way to do this exercise is to prove that half open intervals define a basis for a topology on ℝ, if you know what is. That way you don't need to keep track of the indices and how unions and intersections interact with each other

[novel stump]

my professor said we probably wont go over what a basis is in this class 😭

which i find very strange

[warped helm]

that seems awfully inconvenient

[novel stump]

shrug, maybe we'll learn about it next semester?

[warped helm]

idk it seems weird because then how do you talk about products

or metric topology

or a number of basic results

[rancid umbra]

it’s going to be
U_{i in I} U_{j in J} ([a_i, b_i) n [a_j, b_j))

[a_i, b_i) n [a_j, b_j) is either empty or [max(a_i,a_j), min(b_i,b_j))

so the double union is just a union of intervals of the form [a,b)

[novel stump]

shrug

[novel stump]

yeye

thats what i got :]

[rancid umbra]

i can sort of see this without need for a basis

[rancid umbra]

for finite products you can use the box topology and be fine by just fully specifying what the open sets are

[warped helm]

atp arent you just talking about bases without calling it by name

[rancid umbra]

just skipping over the steps that involve the basis, but yea it has that feel

my questions would be, how are you going to talk about first and second countability?

how are you going to talk about the different definitions of continuity, especially on the real line (and more generally in metric spaces)

how are you going to speak of generating new topological spaces?

[novel stump]

am i allowed to ask people to check my work/logic here?

[warped helm]

yep

[novel stump]

sick!

i apologize in advance for the crappy cropping, i am currently charging my computer and i had to crop out my irl name, but does the logic here/work make sense?

[warped helm]

the only expectation is that it isn't something like an exam/quiz/whatever that your professor implicitly or explicitly asks that you don't consult external sources on

[novel stump]

oh yeah this is just hw

and we're allowed to get external help

[warped helm]

this part right here is not true as written

the only observation needed is that is U and V are open in the particular point topology and nonempty, then (0 \in U \cap V)

[gentle osprey]

josemom2

[novel stump]

aw rats

[warped helm]

but for example [-1,1] and [0,1/2] are open in that topology but their intersection is [0,1/2]

[novel stump]

oh well clearly 0\in U \cap V if 0 in both

[warped helm]

do you want feedback on wording?

[novel stump]

yes!

i would love to improve my proof writing lol

[warped helm]

ok, this part right here is true, but there is a less wordy way of stating it:

if 0 \in U_i for some i, then 0 \in \bigcup U_i since U_i \subseteq \bigcup U_i

i only say this because there's some ambiguity with the wording "any union of the rest of the open sets"

you could say with, but overall this isn't a huge point

[novel stump]

ooh okay

[warped helm]

everything looks pretty good!

however

i would show this equality

if for nothing else than to get some slight bit more practice with inf

[novel stump]

hmm, okay, idk if i can show it but i can justify it

[warped helm]

well, good thing i caught it: you need to be able to do routine verifications of those sorts of things

you can proceed straight forwardly with a double inclusion argument and the definition of the infimum of a set

[novel stump]

double inclusion argument?

[warped helm]

A subset B and B subset A implies A = B

[novel stump]

ahhhhh

[warped helm]

for this part, i think you are doing more work than is necessary

i admire the indexing efforts

but i think you can just say that the union of the unions of half open intervals is still a union of half open intervals

[novel stump]

that was the only way i could think of but i'm also not surprised its too much work LOL

[warped helm]

it is correct nonetheless

[novel stump]

yay!

[novel stump]

i would imagine we can just say, take an x \in big cup blah blah blah, since every (a_i,infty) is getting bigger, but still bounded below, we have that x \in (inf a_i, \infty) right?

[gritty widget]

Someone here once posted it, don’t recall their username, but you can think of compactness sort of like an induction principle

If X is compact, P is a property of subsets of X, then all you need to prove P(X) is the “base case”, that every point x has an open neighborhood U s.t. P(U), and the “inductive step”, that P is closed under finite unions. Then, P(X).

In fact, this completely characterizes compactness. Compact spaces are precisely the spaces where, for any property P of sets, the above “base case” and “inductive case” imply P(X).

[warped helm]

this is the correct intuition, but i think personally some rigor would help

[gritty widget]

Oh they already said it here lol

[novel stump]

ah :/

[warped helm]

what can we say about (s = \inf { a_k : k \in K}) in relation to all the (a_i)s?

[novel stump]

rigor is not my forte lol

[gentle osprey]

josemom2

[novel stump]

that s is the greatest lower bound, so there is no lower bound greater than it

[warped helm]

sure, sure, but how about (s \leq a_i) for all (i \in I) ?

[gentle osprey]

josemom2

[novel stump]

(also warning i still struggle with suprema and infema even though i learned about it a year ago)

[quartz horizon]

Oh is this universal property of inf

[novel stump]

yes we can say that

[warped helm]

so does the inequality [
a_i < x < \infty
] look suspicious?

[gentle osprey]

josemom2

[novel stump]

maybe?

[warped helm]

how can we add in s there?

[novel stump]

s\leq a_i < x < \infty

[warped helm]

right, so the chain gives you s < x < \infty

therefore x in (s, \infty)

now for the other inclusion, it will be helpful to recall that if s is a greatest lower bound for A, then for all r > 0, there is some a in A such that a < s + r

really what this says is that, if s the greatest lower bound, then anything bigger than s is not a lower bound for A

[novel stump]

mhm

[warped helm]

now see if you can apply this to x in (s, \infty)

[novel stump]

i'll try 😭

[rancid umbra]

bump?

[rancid umbra]

oh, btw, for this min-max thing, you may want to specify what you mean by min(b,d) when one or both of b and d are infinite.

[hidden abyss]

The illustrations in that book are really cool

Oh one of the authors (Fomenko) is the artist

[rancid umbra]

oh wow

thats incredibly cool

[tiny obsidian]

What

[rancid umbra]

think they put some of his works at the start or end of each chapter

[young stone]

woahhh

[fickle tendon]

this is amazing art

[hidden abyss]

Unfortunately he's also a pseudohistory crank

[young stone]

that explains the super weird website I found when looking him up

[hidden abyss]

Anyway, I tried proving this (via A★B := pushout of M(p_A) <- A × B × 1 -> M(p_B) which should be equivalent I think?) and ran into the issue that pushouts dont preserve - × C in general

[fickle tendon]

A⭐B

[rancid umbra]

yea, i’m pretty sure it’s just false if you use the colimit topology

[radiant stone]

Hodge star operator 🤑

(\star)

[fickle tendon]

hodge ✨ star ✨ operator

[twin swift]

If I have an irreducible topological space X and U an open subset, how can I describe the connected components of X using the connected components of U in the subset topology?

Mainly I want to know what connected components intersect U, if there are any

[warped helm]

is this what people who are good at topology see

[unreal stratus]

Is there a reason why this is struckthrough

[twin swift]

solved

[unreal stratus]

But yes irreducible implies connected and irreducible goes to open subsets so the answer is simple ig

[twin swift]

exactly

[unreal stratus]

Noice

[twin swift]

I thought it was the other way around

this is why I asked

[fickle tendon]

is there a special name for the intersection of all neighborhoods of a point?

i know in like Hausdorff spaces these are not interesting but in coarser spaces they are

[warped helm]

@umbral hamlet wow i had a

moment today in topology mid term

not grade ruining but so silly

for some reason i said that [0,1] was connected in the lower limit topology

brain was not braining

[umbral hamlet]

Rip

It happens lowkey

I say dumb shit in classes all the time just comes with the trade

[warped helm]

my brain rushed to that one theorem in munkres that says if its a linear continuum with the order topology then its connected

but the ll and order genuinely differ

everything else was fine

[rancid umbra]

wouldn’t know lmao

[umbral hamlet]

Rip

I mean ig its understandable to make a quick booboo

Btw have u seen the topological proof of the infinitude of primes

Im giving a presentation on it next week and its pretty cool

[warped helm]

i have not

what does it involve?

next semester i need to lock in and start doing talks and stuff O_o

[umbral hamlet]

you take $\mathbb{Z}$ and induce what is called the evenly spaced integer topology on it, where the basic open sets are given by $S(a,b) = {an + b : n \in \mathbb{Z}}$ with $a \neq 0$

[gentle osprey]

hiidostuff

[umbral hamlet]

the open sets can be thought of as arithmetic progressions or the set of integers that are b mod a

either way its routine to show that this induces a topology on Z

the basic open sets have some interesting properties though

firstly theyre clopen

since the integers that are b mod a can be thought of as all the integers that are not x mod a for some x \neq b

i.e. $S(a,b) = \mathbb{Z} \setminus \bigcup_{k = 1}^{a-1} S(a, b + k)$

oh i messed smth up

ah

[gentle osprey]

hiidostuff

[umbral hamlet]

ok @warped helm this is the right notation for this

pretty much just subtracting all the things that arent b mod a which can be given as basic open sets themselves

also note that nonempty open sets have to be infinite

since each basic open set is infinite

now we notice that $\mathbb{Z} \setminus {-1, 1} = \bigcup_{p , \text{prime}} S(p,0)$

[gentle osprey]

hiidostuff

[umbral hamlet]

on the right we just have all the integers that are divisible by some prime

now by the fundamental theorem of arithmetic thats every single integer except for -1,1

now, the set on the left cant be closed as its complement is finite and thus not open

however, on the right we have a union of closed sets

if theres finitely many such closed sets then we get a closed set, a contradiction

thus there must be infinitely many primes

another cool fact about this topology on Z is that its homeomorphic to Q with the standard topology

theres a theorem that says that countable, metrizable topologies with no isolated points are homeomorphic to Q

[gritty widget]

Q^n=Q is an interesting corollary

Never really thought about that

[umbral hamlet]

clearly Z with this topology is countable and has no isolated points since every open set is infinite

[umbral hamlet]

yeah this is a really strange result

goes to show that Q is super weird

[umbral hamlet]

Z w/ this topology is metrizable by urysohn's metrization theorem

since its regular hausdorff and is second countable by definition

regular by the clopen-ness of the basic open sets

hausdorff by looking at the basic open sets as arithmetic progressions

[kind marlin]

it would be funny to call Q an idempotent topology

are there other interesting examples of that

I think this is true for sequence topologies

[umbral hamlet]

We just gotta find topologies that are unique in some sense up to homeo

Well also with properties entirely preserved by finite product

Q works bc countability metrizability and lack of iso points is preserved

[gritty widget]

Edrős space is 1 dimensional separable metrizable totally disconnected iirc, and it’s square is itself

I think a lot of infinite products will be examples too right

Like 2^omega(homeomorphic to cantor set), omega^omega(irrationals), etc.

That’s quite neat actually

There’s a partition of R into two spaces which are homeomorphic to their own square

[gritty widget]

Given an infinite discrete space X, it is also the case that the product of X with itself is homeomorphic to X, and the disjoint union of X with itself is X.

I wonder if there are any non-empty non-discrete spaces satisfying both of these simultaneously?

[umbral hamlet]

heres a problem from munkres

r* is the homomorphism induced by r

so that r*([f]) = [r of f]

it feels like this is pretty straightforward so i wanna check that im not crazy

for some [f] in pi1(A, a0), the representative f is a loop of a0 in A and thus is a loop of a0 in X

but importantly the range of f is contained in A

thus r of f = f as r restricted to A is the identity of A

and so the preimage of [f] is [f] in pi1(X, a0)

is that right?

[gritty widget]

the preimage contains [f], but need not be only [f]. But yeah that's the proof

[umbral hamlet]

right my bad

thank you though

[gritty widget]

np

By virtually the same proof we get that the homomorphism from the fundamental groupoid of X on A basepoints to the fundamental groupoid on A is full

Also bijective on objects, so that the fundamental groupoid on A is a quotient of the fundamental groupoid of X on A basepoints by the kernel

[umbral hamlet]

ok heres another one

i think i just dont know what it means to extend a function

[crisp lintel]

it just means there's a function with the larger domain that restricts to the same function on the smaller domain

[umbral hamlet]

im predicting that the end will be a result of the fact that pi1(Rn, a0) is trivial

[umbral hamlet]

ah thats simple enough

[kind marlin]

ooh thats cool

[kind marlin]

right, i think for any topology S and any infinite set K, S^K is homeomorphic to its square

what's kind of cool is Q is not homeomorphic to any such form, bc S^K is always either finite or uncountable

this is kind of a cheat but erdos space can be written as this if you just let S = E and K be any countable set :P

if that's allowed as part of the S^K family (which is maybe not a good idea lol), then Q already feels kind of special

any countable "idempotent" topology would similarly be ruled out, but I'm not sure if there are uncountable "idempotent" topologies that don't have an S^K representation (implying at least that the homeomorphism fails at infinite powers)

i also dont know what other countable idempotent topologies are

[gritty widget]

Erdos space is homeomorphic to a countably infinite product of itself? neat

[gritty widget]

What about an uncountable disjoint union of Q's

Would the square be homeomorphic to itself?

[kind marlin]

i think so

i havent really seen coproduct topologies before, is this basically like the box topology?

[gritty widget]

coproduct topology is just the union of the topologies(e.g. a basic open set of X union Y is an open set of X or Y)

[kind marlin]

hm, that also includes all "combinations" of an open set in X and an open set in Y right

just by closure under unions?

or okay what i mean is im confused why that isnt the same thing as X x Y, bc it seems like you can uniquely identify the X set and the Y set that make up an open set of X sqcup Y

okay i think im just not thinking correctly, theyre obviously different bc u have |X| |Y| points in one and |X| + |Y| points in the other 😭

oh wait okay i stopped brainfarting, the entire coproduct topology is basically the "basis" of the product topology, you can never have a coproduct topology open set that isn't decomposable into factors from each individual topology

okay that was silly moving on

i wanna say the product of coproduct topologies = the coproduct of pairwise product topologies

[gritty widget]

Yeah top seems like a distributive category but I'm too busy to prove it rn fr I'll try later

[kind marlin]

in that case, you get a square uncountable (so uncountable) disjoint union of square Qs, and each QxQ is homeomorphic to Q, so yes? this feels incredibly handwavy

okay i can visualize why the coproduct product thing is true so i can buy into this

okay dope and that cant be homeomorphic to S^K for infinite K bc we have countable open sets here and every open set in S^K is uncountable

[gritty widget]

Yeah so Q times any discrete set K should satisfy this

Generally, if X is a space s.t. X^2=X, X times k for discrete k should satisfy this for the same reason right?

So if X~=S^K, X times k isn't either

[kind marlin]

wait wow that feels weird in my brain

oh wait yeah of course K x K is homeomorphic to K when K is infinite

thats so much more obvious written like that lol

[queen prism]

= for homeomorphisms ;-;

[kind marlin]

hm i forgot the symbol for homeomorphic 😭

is it just the ~ over =

[kind marlin]

this is specifically true for K infinite right?

[gritty widget]

two ~'s

[gritty widget]

Uhhh, yeah mb

[kind marlin]

omg but isnt that like approximately equal 😭 that feels cursed

[gritty widget]

Qx2 isn't homeomorphic to it's own square

Idt

[kind marlin]

wait is Q homeomorphic to Q x k generally

where k is finite

this is countable and metrizable and doesn't have isolated points

so.... yes?

so this does hold for arbitrary finite k i think 😭 but specifically for X = Q 😵‍💫

[gritty widget]

Yeah

Wait

oh it is

huh neat

[kind marlin]

why is Q amazing

Q is my favorite topology ever now

irrational numbers were a mistake

[gritty widget]

But yeah in general if X is idempotent, Xxk is too for infinite discrete k, and X^k for infinite k is just always idempotent(regardless of X)

I wonder if there's an example of a space such that the series (X, X^2, ...) is eventually constant but X isn't idempotent, e.g. the least n s.t. X^n is idempotent is > 1?

In general thinking about "square roots" of spaces and whatnot is quite neat

[kind marlin]

ooh that is interesting

[gritty widget]

I recall seeing a proof there is no space X s.t. X^2 is homeomorphic to R but I completely forgot how it goes

I don't even know how you'd go about reasoning about that haha

[kind marlin]

yeah this is already breaking my brain a bit </3

i wonder if you lose an ordering property

[kind marlin]

oh i randomly saw this online, something about covering dimension?

i think the covering dimension of (X x X) = 2 * dimension of X

in that case, the covering dimension of R is 1 so it cant be the square of something

the covering dimension of Q is 0 :0 so that would be another necessary condition for idempotency

https://math.stackexchange.com/questions/57375/mathbb-r-x2-as-a-cartesian-product

whoa this gives a clever argument too

[gritty widget]

I don't think that's true in general? Erdos space has dim 1(in the brouwer dim sense), is homeomorphic to it's own square, and is separable metrizable(so all 3 dimensions are equivalent for it)

[kind marlin]

hmm im seeing online that covering dim (X x Y) is <= sum of dims generally so that makes sense

apparently equality is achieved for Euclidean spaces but I don’t know why that’s true

oh and that anyway doesn’t help bc the square root doesn’t have to be Euclidean

[gritty widget]

Iirc it's relatively complicated to prove dim(R^n)=n, but this result just follows from R^nXR^m=R^(n+m) and dim(R^n)=n

[kind marlin]

right that makes sense

[gritty widget]

Is it easier to characterize when a space is homeomorphic to the disjoint union with itself? I feel like that's simpler