point-set-topology

[dry seal]

Does anyone have any info on what this sequence could be?

[alpine nest]

You could probably take any sequence which enumerates NxN

Every neighboorhood of (0,0) will contain infinitely many elements of such a sequence

[dry seal]

Ohhhhhh

I have just misunderstood how neighborhoods of (0,0) work

We're allowed to omit infinite elements in only finitely many columns

[alpine nest]

Yeah

[dry seal]

Thanks this helped

[alpine nest]

I mean, there's no denying that this is a weird space

So it's entirely natural for intuitions to fail

[dry seal]

Yeah this is pretty freaky

But I guess all the non first countable spaces are lol

[tender halo]

there are a bunch of spaces constructed like that

really what they are are just N with a certain ultrafilter adjoined to them

that makes it much easier to reason about

you automatically get that they are not sequential (and therefore neither Frechet Uryhson or first countable or metrizable) or compact

and you also get that they are perfectly normal

[alpine nest]

I magine seeing the Arens-Fort space and going "this is a perfectly normal space"

[tiny obsidian]

(Mr. Arens and Mr. Fort, of number four Privet Drive, were proud to say that they were perfectly normal, thank you very much. They were the last people you'd expect to come up with a space X containing two disjoint closed sets of X without disjoint open neighborhoods, because they just didn't hold with such nonsense.)

[tender halo]

wait im very wrong

you just adjoin a filter...

hmm im sure you can salvage it somehow

[tender halo]

anyway that is an easy way to get a space with the above listed propeties

arens fort is actually a subspace of another space that is sequential but not frechet uryhson

there is a construction in the end of the first part of engelking

[exotic hemlock]

can someone help me check if i made a mistake for the following?
Let $A$ and $B$ be disjoint compact subspaces of the Hausdorff space $X$. Show that there exist disjoint open sets $U$ and $V$ containing $A$ and $B$, respectively.

[gentle osprey]

somethingwrong

[exotic hemlock]

Using the fact that compact subspaces of hausdorff are closed. $A^c$ and $B^c$ are disjoint open sets with $A^c$ containing $B$ and $B^c$ containing $A$, so we are done b

[gentle osprey]

somethingwrong

[exotic hemlock]

oh nvm i just realised

[ruby delta]

How are you sure A^C and B^C are disjoint? In fact, unless your entire space is just the union of A and B only the complements are never disjoint.
Take your space to be R, A = [0,1], B = [2,3]

[paper wedge]

somethingwrong

A^c and B^c need not be disjoint

[exotic hemlock]

thanks for the help, i just noticed this too

[ruby delta]

Compactness is going to be necessary here

[paper wedge]

a hint for this problem is ||use original definitions of compactness/hausdorfness, that is, obtain two open sets by seperating points manually from the hausdorff condition. how can we make them open by compactness? use it||

[shrewd mortar]

Hey guys I'm struggling in my studying of manifolds I just don't understand them it just goes through one year and out the other.I want to understand the foundational basics of them and I don't have much of a strong background.

[rancid umbra]

where specifically are your struggling?

[jolly umbra]

*ear

[crimson terrace]

This is the first chapter so no notion of closed and open sets

[unreal stratus]

Idk what you mean by this implication

Like the right hand side is just a set

[crimson terrace]

Oh yes I forgot to add a subset of A sorry

But it has been resolve thank you 👍

[unreal stratus]

Well now I'm still confused as those inclusions are obvious

[dry seal]

I've returned to this example of the open ordinal space [0,\omega_1). I can easily show it is not compact but showing it is sequentially compact has been hard. The book I'm using does it in an extremely roundabout way that doesn't really make sense for what I'm doing. (It shows it is first countable and countably compact by showing the closed ordinal space is compact and getting sequences with accumulation points in the open space implying that it is countably compact, so like 2-3 lemmas I don't use for anything else in the paper just dedicated to this step). I've tried showing it directly but I'm stuck showing that sequences of countable ordinals can converge to some countable limit ordinal. Does anyone have any tips?

[opaque scroll]

If you have a sequence of ordinals then their supremum will be a limit point

[dry seal]

The book I'm using doesn't define supremum of ordinals, I take it it's just the ordinal defined by the union of the elements of the sequence?

Or is it just the least upper bound like usual?

[opaque scroll]

In general supremum of something is just it's smallest upper bound.

That is the same as union for ordinals yes

And I guess using that it's a union makes it obvious that the supremum is again countable (being a countable union of countable sets)

[dry seal]

Ohhh alright makes sense

It's a bit hand wavey but it should work

[opaque scroll]

I mean you can absolutely make it rigorous, but it's probably fine to handwave a little

[dry seal]

tell that to my supervisor

I mean ok

There's obviously more work to be done

Because you need to be able to extract a monotonic subsequence

Not every sequence of ordinals converges by itself after all (e.g 0,1,0,1...)

I'm trying to avoid well ordering but it seems necessary here since we're taking arbitrary sets of ordinals?

Well countable sets at least

[opaque scroll]

Well you don't really need the well ordering (other than guaranteeing supremums exist). In general if you have a sequence in any ordered set you can find a subsequence that converges to the limsup

But yeah there is something to be done

[dry seal]

Alright I'm pretty sure I figured it out but I'd like to hear what you think @opaque scroll

Let the supremum of the sequence ${\alpha_n}$ be $\beta$. If the sequence is constant somewhere we are done. Otherwise, $\beta$ is a limit ordinal. We have then for any ordinal $\gamma<\beta$ a corresponding element in the sequence $\alpha_\gamma>\gamma$. So for some monotonically increasing sequence of ordinals $\gamma_1,\gamma_2,\dots<\beta$ we have corresponding elements in ${\alpha_n}$ such that each $\alpha_{\gamma_i}>\gamma_i$. So we have a monotonically increasing subsequence ${\alpha_{\gamma_i}}$ and so for any neighborhood $(a,b)$ of $\beta$ we can find some corresponding index $\gamma_a$ such that $\alpha_{\gamma_a}>a$ and so it is eventually in this (and by extension any) neighborhood.

Does this seem about right or am I missing something?

[gentle osprey]

Slashbaker

[opaque scroll]

Pretty much though there is some subtlety.

The resulting sequence won't necessarily be increasing (for example if the original sequence was decreasing), but picking alpha_gamma > gamma will still work

[dry seal]

This actually makes me think my construction is wrong, since if the original sequence is decreasing then it's constantly getting further away from the supremum and thus shouldn't be able to converge to it, no?

[opaque scroll]

Supremum should be limsup, not just the supremum as a set

[dry seal]

Even if I took some subsequence such that each alpha_gamma>gamma then I still don't have that the next element in the sequence stays in the neighborhood necessarily

[dry seal]

I can clarify the way I wrote it is just the union of the elements of the sequence

Whether or not that is sup or limsup it's the LUB as I've defined it in my paper (I want to avoid using too much terminology for what is essentially the same thing in my case)

[tender halo]

i mean... thats just kinda the proof i guess

the essence of the proof is that every sequence is contained in a sequential and compact space

which is therefore sequentially compact

[quick delta]

How do you do this?

[alpine nest]

What's your favourite topological space?

[hidden abyss]

For every x in K, take a compact neighbourhood A_x of x and use them (and compactness of K) to construct V

[cloud kindle]

The long line. Deeply humours name

This will just always amuse me

[hidden abyss]

I like it just because of my mental image of it

Its inherently silly

[quick delta]

That doesn’t help

[hidden abyss]

The interiors of A_x are open and cover K so you can use the compactness assumption

[quick delta]

Stop drip feeding it

I’ve given up on this an hour ago and I’m just trying to get it off my mind

[cloud kindle]

Being rude to the person offering to help you is a choice

[hidden abyss]

By compactness, there exist $x_1,\dots, x_n \in K$ such that $U_i := int(A_{x_i})$ cover K, so let $ V = U_1 \cup \cdots \cup U_n$. Then $\bar V \subseteq A_{x_1} \cup \cdots \cup A_{x_n}$, the right side is compact so $\bar V$ is compact. Since $\bar V$ is compact Hausdorff, its normal so you can apply Urysohn to the pair $(K, \bar V \setminus V)$ to get a continuous function which is 1 on $K$ and vanishes outside V, and then just check that this extends to a continuous function on X

[gentle osprey]

Jussari

[unreal stratus]

Lol this phrasing is legendary yes

Glad they haven't changed it

[tender halo]

either \beta N or the Niemytsky plane

first because its so esoteric and second because having discs tangent to yourself as your nbhbds is so cool

[alpine nest]

My current fave is the Arens-Fort space purely because I've only recently learned it's a perfectly normal space.

And the idea of calling that a "perfectly normal space" amuses me no end.

[rugged dagger]

Let {Ai} be a family of connected sets, let s be a path from [0,1] to A=UAi such that s([0,1]) has a non-empty intersection with for every Ai, then A is connected.

Is this true??

[tender halo]

yes

if there is a family of connected sets A_i such that there exists an A \in A_i such that A is not separated from any of the A_i, then their union is connected

for if there is a separation of the union, A lies in one of the sides of the separation and therefore all the A_i cannot lie in the other side, therefore the union lies in one of the sides

in particular, take a connected set A and the family of singletons of members of Cl(A) \ A

the condition applies, so Cl(A) is connected

[rugged dagger]

Sorry I'm not getting the relation with the path

From this to the path

You can have non intersecting sets such that the union is connected and there exists a path between

For example (0,1) U [1, 2] = (0,2] connected, with connected components and clearly a path existing but having an empty intersection

[prime elbow]

How do I think of a metric on R such that (0,1) is closed but not open?

[young stone]

Hint: || Use a bijection from R to itself that takes (0,1) to [0,1]||

[prime elbow]

I got it, thank you ❤️

So it makes a new metric on R which is homeomorphic to usual topology on R

[young stone]

yes but the points are not located where they usually are in R

[prime elbow]

Yeah

[tender halo]

the path is the A

s([0; 1]) is connected and is not separated from any A_i

[fickle tendon]

If we have a chain of refinements of partitions on a set X
... ≤ π2 ≤ π1 ≤ π0
And consider the now ascending chain of partition topologies
τ0 ⊂ τ1 ⊂ τ2 ⊂ ...
Then take the union of these topologies τ = ⋃_i∈ℕ τi, does this topology have a name? Does it come up anywhere?

This definition arose as I wanted to look at continued orbit decompositions of left quasigroups

(meaning you see the orbits of the natural action from the left multiplication group as left quasigroups themselves, and decompose that orbit into "suborbits", and so on)

[graceful solar]

I need to come up with an example of continuous bijection f: X->Y and the non-compact X and Hausdorff Y such that f-1 is not continuous, hence not homeomorphism

[rancid umbra]

there is such a map with Y = S^1

[graceful solar]

what is S1?

[rancid umbra]

the unit circle

[graceful solar]

I don't think I could use that

[rancid umbra]

why?

[graceful solar]

Because we're still in the very early stages of topology and I can only use what's on my guidebook

[unreal stratus]

i would say is one of the most basic exxamples of a Hausdorff space

[graceful solar]

Oh wait

[graceful solar]

can you elaborate a bit?

[rancid umbra]

S^1 = {(x,y) ∈ R^2 : x^2 + y^2 = 1}

with the subspace topology from R^2

[graceful solar]

is this function bijective?

[rancid umbra]

what function?

[graceful solar]

I need to explicitly say the bijective and continuous function

[rancid umbra]

think of some maps into S^1

[tender halo]

wouldnt it be the limit of the inverse sequence of t_i

[tender halo]

take any non discrete hausdorff topology and the discrete topology on the same set

[graceful solar]

this discrete topology would be the non-compact one, correct?

[tender halo]

yea

the question is equivalent to "does there exist two comparable Hausdorff topologies on the same set"

which is obviously true

for a non trivial example uhh

R and Sorgenfrey line?

[fair idol]

I found this definition of an abstract convex structure on a set in a book called theory of convex structures. I’m a bit confused by the book’s definition of a convex structure compared to a closed structured.

Aren’t closed sets stable under nested unions? It seems like any closed structure is a convexity structure in this definition. I was hoping somebody could provide a sanity check for me.

[rancid umbra]

U [0, 1 - 1/n] = [0,1) is not closed in R

[fair idol]

Thanks!

[haughty jungle]

https://mathoverflow.net/a/296605 can someone explain the part in the asterisk where it says that showing the compact sub intervals are polynomials suffices

[warped bone]

hi guys, if i want to show that two basis B and B' generates the same topology, i need to check that for every set U in B and for every x in U, i can find a V in B' such that x is in V and V is subset of U, then i do the same switching B and B', now, i want to check if B:={ [a,b) : a<b} generate the standard topology on R. Now i can see that, i can find an open interval containing x for every x in [a,b) excluding the case x=a, since i will not be able to find an open interval containing a while being subset of [a,b) i can confirm that they generate different topologies?

[burnt tendon]

Yes, the condition you give is necessary and sufficient for B and B’ to generate the same topology

[warped bone]

thanks

[haughty jungle]

You could also just show that lower limit topology is not connected

[rugged dagger]

What are the main utilities of compactifying a space?

[tender halo]

getting nice limits

[rugged dagger]

Huh

What

[tender halo]

compactification is about making all ultrafilters on your space convergent

so you need to add points to your space until all limits start converging

[rancid umbra]

i think this answer does a good job motivating it.

also the nlab gives some nice exposition on compactifications.

compactifications can be functorial the stone-cech compactification is also functorial which is convenient (it’s even a left adjoint to the inclusion functor of compact hausdorff spaces into top)

[tender halo]

conceptually, any noncompact space has "holes", the places where you can point to but there is not a point there. kind of like with Cauchy sequences in noncomplete spaces - you know something should be there, but there isnt. if you plug all the holes with points (and you can plug several holes with one point), you get a compact space - if you plug all the holes at once with a single point, thats the Alexandroff compactification. if you use a separate point for each hole, thats the Stone Cech compactification. all the others somewhere in between (that is to say, \omega X and \beta X are respectively the bottom and the top of the complete upper semilattice of compactifications, although the bottom/Alexandroff compactification exists iff the space is locally compact)

[iron bolt]

the slight asterisk there is that the Stone-Cech compactification doesn't just add points, it also identifies some points of your original space to make it Hausdorff

it's a way to turn any space into a compact Hausdorff space really, not a way to turn it into a compact space. or at least not mainly

[rugged dagger]

Thanks to everyone for the clarifications

[tender halo]

sure, i usually only think of Tychonoff spaces for compactifications

[unreal stratus]

There's also a more geometric thing which is that compact stuff often behaves well for things like cohomology lol

[warped bone]

i need to prove that if X and Y are topological spaces, and {p} is an open set of X, then every f:X->Y will be continuous in p. to do so i choose {p}=U and V an open set containing f(p) and from this i can conclude that f(U) is a subset of V and that f is continuous in p. the existence of the V open set that contains f(p) is guaranteed by the fact that no matter what topology i choose for Y, since his base will cover all Y, i know for sure that a V like this exists (eventually Y itself)?

[tender halo]

the second half ("the existence of the V open set...") is superfluous

but yeah its correct

[iron bolt]

you introduce the V because you need to prove that something holds for all V

you do that by assuming you have a V and then showing that the needed property holds for it

no need to prove that there exists a V at all, although like you noticed it indeed does

[warped bone]

thanks guys

currently spamming exercises since i have exam in 3 weeks and i'm missing lots of stuff, so i'll probably come back often to ask for help

[unreal stratus]

"Continuous in p" isn't really standard terminology - do you mean at p?

Then you can probably phrase it as follows; since {p} is open it suffices to check that the composite {p} -> X -> Y is continuous, which is immediate as any function out of a singleton is continuous

[warped bone]

sorry, bad translation by me, in italian we say 'in p' so i wrote it like that without thinking

[unreal stratus]

No worries aha just making sure

[iron bolt]

same in german, so I didn't even notice, lol

I wonder if I sometimes make that mistake too

[queen prism]

in English if you say "f is continuous in p" it sounds like you're saying the function p -> f(p) is continuous

"f is continuous at p" means the function x -> f(x) is continuous at the point x = p

[unreal stratus]

Yeah

I should've said this lol thanks for doing it for me aha

[gritty widget]

Saying “behaves well for cohomology” is a geometric thing seems like a very bizarre way to characterize it to me(at least, not really what I think of when I think “geometric thing”), but I guess that’ll change when I learn more alg top lol.

[gentle osprey]

longboard kayak

longboard kayak

[spare light]

is this alright?

and in fact it achieves a minimum and a maximum.

what to do for this extension?

[young stone]

Think a little more abstractly, do you know any theorems about maps f: X \to Y when X is compact?

and if not can you make a guess on what happens to the image f(X) ?

[spare light]

oh i can use that the image is compact too

[young stone]

yep

[spare light]

then HB

[young stone]

yeah, this also shows the first part of your question directly

[spare light]

alr

[desert lodge]

when we say C is a subbasis for the topology on X

then the topology generated by subbasis equals the topology on X?

[tender halo]

yea

[desert lodge]

To show that a set S is a subbasis for a topology on X

we just have to show the union of elements in S equals X?

[lilac musk]

To show that S is a subbasis for a given topology on X, you need to show that the given topology is equal to the collection of unions of finite intersections of elements of S.

[tender halo]

for some topology? any collection of subsets of X gives rise to a topology

the empty intersection of elements of S is X

[cyan roost]

I have a difficult time determining the closure of A in the product topology. Now I have shown that the interior of A is the empty set. Now I wanna show the closure of A. Is it the same as it is in the box topology, where it would be the Cartesian product on [0,1]? Or is it R^n? or something else entirely. A isn't open in the product topology, but that doesn't mean it is closed right? If it is closed then the closure is easy, it is just A itself.

[opaque scroll]

It might be good to think about what closed sets in the product topology should look like.

Remember that every closed set is the complement of an open set.

What is a typical open set in the product topology? What is it's complement?

[cyan roost]

an open set U in the product topology is the cartesian product on U_i, with U_i open in X_i and U_i=X_i for all but finitely many i. So its complement would be the cartesian product on X_i\U_i, which would be a much "smaller" product since U_i=X_i for all but finitely many i?

[cyan roost]

but we have a theorem that says that if the cartesian product on X_n is either the product topology or the box topology, then the closure of a set such as A in the example is the same as the product on the closures of each A_n. So is the closure of A just the cartesian product on [0,1]?

[tender halo]

yea

[opaque scroll]

The complement would not be the product of Xi\Ui no.

Think about for example
(0, 1)xR, what is it's complement? Draw it in the plane

[dry seal]

I have a bunch of nice theorems which are basically reformulating a bunch of topological notions in terms of nets. So far I've talked about continuous functions, hausdorff spaces, and compactness all in terms of nets. Does anyone know of any other nice ones? Ideally I'd like to avoid the use of filters but if there's a particularly nice one that uses them it's not off the table.

By the way it took a while to understand but the proof for compactness being equivalent to nets having convergent subnets was probably the most fun I've had in topology ever (which isn't saying much but hey it's something)

Really interesting topic that I'm surprised is omitted by most curricula

[unreal stratus]

I guess it's just most people do not do topology for the sake of pointset topology tbh

But it is fun

[unreal stratus]

Well a nice thing is the proof of Tychnoff using ultrafilters - it's a classic. Similarly Stone-Čech compactifi ation

[dry seal]

Alright I'll give them a look, thanks!

And yeah that's fair I don't really see how a lot of point set can be used outside of researching it for the sake of it

[viral snow]

am new to topology ,only done a semester of it ,is there a way to keep the various theorems and lemmas in memory, finding it realy inefficient to keep referencing stuff I should already know.

[iron bolt]

my experience is that you typically remember what you use a lot and forget what you don't

being able to rederive simple lemmas you only vaguely remember on the fly helps - so you don't need to remember everything, just the rough ideas

and when you rederive a lemma like that, you're also less likely to forget it next time

[hexed steppe]

you should try proving these lemmas for yourself when you look them up

and eventually you will build intuition for what is/isnt true

[red yoke]

Point-set is overrated

[tender halo]

its so overrated that nobody likes it

[young stone]

it's definitely important but I don't know if it's worth a whole course especially if the student has studied metric spaces already

[tender halo]

point set unfortunately suffers from having a lot of shallow material that is just... not very insightful, and therefore hard to remember

there are like a thousand theorems that go "paracompact Hausdorff spaces are normal" or "space is hereditarily normal iff all open subspaces are normal"

or "pseudocompact normal spaces are countably compact"

i mean its like fine but remembering the statements is harder than actually proving stuff, the proof is usually just "write down all the definitions and shake them"

there are like three interesting theorems in the base point set course

Tychonoff and Uryhson and uhh

[gritty widget]

If you want to go “off the beaten path” and find some extremely uninteresting pointset stuff, there is no fiber bundle p:E to B where E is T_4 and the fiber F\subsetE is not T_4

[young stone]

this also makes it kinda fun though, it's such a highly developed subject that you can just keep proving trivial things until you've suddenly proven something non trivial

[tender halo]

and tietze uryhson i guess

and ive seen people just not prove the interesting part of Tychonoff because Choice is too hard or something

[alpine nest]

Yeah, I think that's fair, any particular branch of mathematics which uses methods of pointset topology, is only going to use a very small subset of those methods, but it will be a different subset for each of them.

So a general topology course not oriented towards a particular use of it, is just going to broadly cover all the dozens of separation axioms and hundreds of variations on compactness.

[warped bone]

I have a continuous function from X to Y, and x is limit point of A, i want to check if f(x) is a limit point for f(A) or not. i consider f from X to Y as the constant function f(x)=c and choose x as limit point of A={y}, assuming that f(x) is limit point of f(A) then it means that every open V set containing f(x) intersect with f(A)-{f(x)} since this one is the empty set, i will have that their intersection is empty and that's a contradiction.

i think that's correct (?) also i need some advice on how to find an example of two non empty, closed subset A,B of R such that: dist(A,B)=0 and their intersection is empty

[alpine nest]

Yep, if f is constant on A then f(A) will consist of one point so it won't have any limit points.

[alpine nest]

The advice would be that the two sets can't be bounded.

[ruby delta]

Sniped

[alpine nest]

Great minds think alike! ||And so do ours!||

HChan had a more extended hint which explains the reason why they can't be bounded, and the keyword there is compactness.

If either of A and B was compact, then they won't work (why?), so they must both be closed but not compact (and equivalently, closed but not bounded)

I mean, if you aren't familiar with the notion of compactness then you probably won't figure out the reason, but either way the hint remains that you should look for unbounded A and B

[warped bone]

i was thinking of it but how do i make their intersection empty if they are not bounded? i'm not sure if i can use like A=[0 , 1-1/n) and B=(1+1/n , 2] with n to infinity

[ruby delta]

1 - they’re not closed, 2 - they’re not well defined sets, unless you’re talking about their infinite union (which does intersect) or their infinite intersection (which has nonzero distance)

[warped bone]

right they are not closed

[ruby delta]

If you do not know what compactness is yet then just take it to be fact that A and B both need to be unbounded

And if you’re struggling to do this in R, it’s slightly easier to do it in R^2

[alpine nest]

Here's a more constructive hint: for every n you need a pair of points a_n,b_n in A and B respectively, such that |a_n-b_n | < 1/n (and of course a_n and b_n aren't equal). If A and B were bounded, then a_n and b_n would have convergent subsequences, which is why it wouldn't work.

So you want to prevent the existence of such convergent subsequences.

[warped bone]

forgot to answer sorry , if i consider A=N-{0} and B={n + 1/(n+1) } ? they have no intersection, both closed and for n to infinity i will have that the distance will tend to 0?

[alpine nest]

Yep

You don't even have to exclude 0 from A, it doesn't change anything

A consists of naturals, and B consists of those naturals shifted slightly

[warped bone]

i removed 0 because if n=0 than one element of B is 1

[alpine nest]

Then you should remove 1 from A

Also you should define those sets more rigorously and unambiguously, which presumably you would in an actual solution

[warped bone]

yep, it's just that i avoid it just to keep it short in messages here

thanks for the help

[warped bone]

to show that they are homeomorphic i show that there's a function f such that: f(1)={2}, f(2)={1} and f=Id otherwise. f it's continuous , its bijective and f^-1 is continuous so they are homeomorphic. feels too easy if it's just that honestly (last time i ask for stuff today i swear)

[alpine nest]

Well, if you've shown it's bijective and bicontinuous, then it is indeed a homeomorphism

[tender halo]

bicontinuous means a continuous function between bitopological spaces

[gritty widget]

How can i show that in R with usual metric space, if K \subset R and {U_\alpha} are open cover of K then there exists an countable subcover of K, in general if X is separable metric space then for any Y \subset X , every open cover of Y has countable subcover ?

[alpine nest]

Start by picking your countable dense set and consider balls of radius 1/n around elements of that set

[tender halo]

||if X is separable and metric means X is second countable means Y is second countable means Y is lindelof||

[alpine nest]

Also note that any subspace of a separable metric space is also separable,so ultimately you're proving that every separable metric space has the Lindelof property (which is the name for the existence of countable subcovers)

[tender halo]

in fact, \kappa-linfeloffness is equivalent to having weight \leq \kappa for metric spaces

[gritty widget]

yeah

[gritty widget]

what is that?

[tender halo]

\kappa lindelofness is "any cover has a subcover of cardinality \leq \kappa"

weight is just the smallest cardinality for a base of a space

[gritty widget]

oh

[tender halo]

it is not true in general, a subset of a lindelof space need not be lindelof

[tender halo]

it is also equivalent to "any set of pairwise disjoint open sets has cardinality \leq \kappa"

which i think is kinda cute

[gritty widget]

so, let K is a subset and D is a dense subset of K, now for every d in D we are taking B(d, 1/n) balls. is n fixed here? I think no, are we trying to show it is second countable space?

[tender halo]

yes to your second question

[gritty widget]

okay, i know this is second countable

but how it will help me?

i see

i got it

Thank you Outsider and Bussy Beaver

[reef tulip]

Not sure if this is the right channel, but i was wondering if compactness is preserved under homotopy equivalence? I.e. if X, Y are homotopy equivalent and one is compact, is the other also compact?

[fringe thorn]

R^n is homotopy equivalent to a point

[reef tulip]

aha

thanks for the answer :)

[gritty widget]

I am thinking of example such that, f:X->X both have same metric space structure and f is bijective continuous function but f is not homeomorphism, so we need X has non compact set

any hint?

[tender halo]

uhh you can like cheat i think

lets say you have a sum of N copies of R plus N copies of a discrete space of cardinality continuum

the function is just making one of the copies of D(c) into a copy of R

by making the topology coarser

idk its a boring solution but i cant really think of a good one atm

[tender halo]

surely there has to be a connected solution

[rancid umbra]

i think you can do something with [0,1) —> S^1

[opaque scroll]

There's this pretty fun example where you take the real line and to the negative integers you glue on [0, 1) and to the nonegative integers you glue on a circle.

By shifting to the right you can map [0, 1) to the circles, but you cannot shift to the left and unwind the circles into line segments

[rancid umbra]

this is along the lines of what i had in mind

[tender halo]

oh yeah that one

[rancid umbra]

i think i recall something with a zipper too

like, zipping things up and down doesn’t work nicely in terms of being a homeomorphism

uhh

lemme see if i can draw what im talking about

[opaque scroll]

Yeah, I guess take two real lines and connect them by line segments on the positive integers.

Then add some open teeth on one of them

You can close the zipper, but not open it

[rancid umbra]

something like this

you end up tearing the space if you try to go back

i guess this isn’t a self bijection tho

[rancid umbra]

hmm. i can’t quite see this one

[rancid umbra]

oh that makes more sense

[gritty widget]

isn't [0,1) and S^1 are distinct set?

[rancid umbra]

@gritty widget yea, but you can combine them like jagr said here to get a metric space using both of them

[gritty widget]

yes i am thinking how picture looks like

[loud echo]

how can i show the one point compactification of the rationals is connected? it makes no sense in my head how compactifying makes it connected.. cant find anything on mse either

[sonic crane]

Wassup Big Bro 💪🏼

[iron bolt]

assume that you have a nontrivial clopen set and think about what that gets you

||the new point ∞ belongs to either the set or its complement - since both are clopen, you can wlog assume that it belongs to the set. what does it mean for the clopen set to contain that point?||

[rancid umbra]

this is my least favorite new fact

[loud echo]

the set is then of the form X\C for some compact C, so the clopen set contains such a nbh X\C, so all rationals not in a compact set are in the clopen set, so the complement of the clopen set is clopen => closed contained within the compact C..

ugh its starting to get messy.. trying to arrive at a contradiction

[gaunt linden]

I think you simply want that a nonempty open subset of Q cannot be compact.

(And thus C cannot exist).

[tender halo]

Q just doesnt have a lot of compact subsets

[iron bolt]

small correction: the clopen set itself is of the form X\C ∪ {∞} for a compact set C, which means that its complement (which is just that set C) is both clopen and compact. that then allows you to show that it must be empty

no need to involve neighbourhoods

[iron bolt]

it has, actually :p

[loud echo]

ah i see, ty

[iron bolt]

my most favourite (or rather only?) fun fact about Q as a topological space is that there exists a classification of its compact subsets that does not mention Q

[tender halo]

i mean only da ordinals

something something cantor derivatives

[iron bolt]

yep, lol

"all countable ordinals" is quite a lot of subsets though for a countable space

[tender halo]

i guess

[tender halo]

your second favorite topological fact about Q should be that any countable perfect metric space is homemomorphic to Q

[iron bolt]

that's neat, I did not know that

I wonder how much metrisability can be weakened here

[tender halo]

its pretty essential to the proof

countability + metrizability gives you that the space is zero dimensional

[white oak]

Hello!

fact   ℝ² \ {⟨0, 0⟩} is homeomorphic to ℝ × S₁, taking the logarithm of norm and the angle to ⟨x, y⟩ as coördinates in ℝ × S₁.
question   Is there a similar representation of ℝ² \ {⟨0, 0⟩, ⟨0, 1⟩} as a product of non-trivial topological spaces? If not, why not?

[rancid umbra]

by representation, do you mean homeomorphism?

the result is positive if you work up to homotopy equivalence. R^2 - (n points) is homotopy equivalent to a wedge of n circles, and therefore homotopy equivalent to R x (a wedge of n circles)

[white oak]

Yep, I mean homeomorphism.

Bonus points for differentiable homeomorphism.

[opaque scroll]

I would guess it's not a product of two spaces.

If it was those spaces would need to be pretty weird at least

Here there are some discussions about when the product of non-manifolds is a manifold
https://math.stackexchange.com/a/78166/306319

But I don't see any examples for R^2, so assuming no such examples exist it would need to be a product of manifolds. But it's not
RxR, RxS1 or S1 x S1

[plush folio]

Isn't it possible to use the fundamental group? The fundamental group of R² \ {(0, 0), (0, 1)} is Z * Z (https://math.stackexchange.com/questions/1802198/why-is-the-fundamental-group-of-the-plane-with-two-holes-non-abelian), and the fundamental group of a product is the product of fundamental groups, which I don't think can equal Z * Z

[white oak]

Interesting approches!

[opaque scroll]

I mean,
Z * Z = (Z * Z) x {0}

Unless there is an argument for why the factors can't be simply connected...

[plush folio]

oops, you're right 😅 but atleast it's not a product of two spaces with non-trivial fundamental group

[rancid umbra]

is there an example where attaching infinitely many spaces to some space X goes poorly? or does that colimit (i guess it would be a wide adjunction, or a wide pushout) always exist in Top?

[iron bolt]

Top is both complete and cocomplete, and the forgetful functor to Set preserves both limits and colimits

so you can take all limits and colimits, and they're just the underlying limits / colimits of sets equipped with the corresponding induced / coinduced topologies

try checking this if you want. the proof really is just "to get a limit in Top, take the corresponding limit in Set, equip it with the coarsest topology that makes all the maps belonging to the limit cone in Set continuous, and check that this gives a limit cone in Top" and the dual thing for colimits

[unreal stratus]

(hope it's ok me pointing this out, but wide adjunction isn't like rly a thing, nor is it a type of colimit)

Ah I guess you probably are using adjunction in the sense of "adjunction space"

[rancid umbra]

aren’t adjunctions colimits? they’re pushouts

[iron bolt]

adjunction spaces yes. I don't like that terminology because it conflicts with category-theoretical adjunctions

[unreal stratus]

I would like advise not using that terminology (personally I have never seen it used beyond some point set textbooks) beause of category theory

[rancid umbra]

sure. what should i call them instead?

[iron bolt]

pushouts I guess

or when talking informally, just "X glued to Y along A"

same thing

[unreal stratus]

or make sure to say "adjunction space" rather than merely adjunction

[whole gyro]

surely adjunction space has to be one of the worst terms ever

unless there is actually some adjunction involved im not aware of

[opaque scroll]

I think it's just because you're adjoining two spaces together

[whole gyro]

yeah

i think "attaching space" is just much better tho

in general I really dislike when terms with technical meaning are used in contexts devoid of any

[unreal stratus]

Real

[iron bolt]

I'm guessing the term adjunction space probably came first?

so it's a bad term now, but was not a bad idea back then

[rancid umbra]

is it really that easy to confuse?

i would think from context it’d be easier to disambiguate, but i guess not

[unreal stratus]

it would say it's not easy to confuse, just when you say "adjunction" as a form of colimit that seemed just wrong rather than another use (cause of this)

but it is chill

[rancid umbra]

oh i didn’t mean it any type of way

i just thought it was funny how much hate it was getting

[rancid umbra]

but yea i could it being interpreted like that causing issue

[solemn iris]

What does this notation mean actually

U is a set.

I’ve seen it for set/(equivalence relation) but here Rn/U where U is open nonempty and bounded, what does it mean?

[opaque scroll]

x~y iff x and y are both in U (or x=y)

[solemn iris]

ooh i see

thanks!

[scenic garden]

I'm not sure if this is the right channel for this question, but I am wondering... is it always possible in 3 dimensions to draw an arbitrary graph without any edges intersecting?

I know this might seem like a graph theory question, but I feel like it's actually a topology question

[rancid umbra]

yes. every graph can even be embedded into R^3 with straight lines.

see this mse post

the wikipedia page

and this paper also linked on the wiki page

however, if you ask that the graph be embedded with all edges having unit length, it fails. for example, the smallest euclidean dimension that K_n has a unit embedding into is n - 1.
see here, for example

[scenic garden]

Cool! I am thinking of making a graph visualizer program, and my intuition told me that 3 dimensions would be enough to avoid intersections, but I wasn't sure. Being unable to always have lines of unit length makes sense though

[rancid umbra]

yes. the paper gives an algorithm for finding an embedding, even

i think you could also get to the same result using like the whitney embedding theorem or some theorem about embedding CW complexes, but that seems overkill

[scenic garden]

My current plan is to build a primitive physics simulation that creates a force-directed graph rendering

[rancid umbra]

i’m not quite sure what that means

[scenic garden]

Basically it is where you model the nodes as objects that repel each other, and the edges as springs that are trying to keep neighboring nodes a specific distance from each other. You let the simulation play out, and the graph (hopefully) forms into a good shape.

Lemme find the video I watched yesterday...

https://youtube.com/playlist?list=PLubYOWSl9mIvtnRjCCHP3wqNETTHYjQex&feature=shared

[rancid umbra]

ah, i see. i think i simulated one of these a while ago in a physics class. it was fun to watch. that sounds like a cool project tho! gl

[cosmic bone]

pretty basic topology question, trying to self-learn analysis and of course some mandatory introduction to topology ensued

prove "If M is an uncountable subset of a topological space with a countable base, then some point of M is a limit point of M."

my idea is as follows: Consider our base G. If we restrict G to only members of M, call this new set G' (this isnt a base, per se, since M may not be closed (?)). Choose a point $x_n$ in each $G^\prime_n$. Then the set X = {$x_1$, $x_2$, ..., $x_n$, ... } is countable (or finite).
We will now show X must be everywhere dense in M. If X were not everywhere dense in K, then the open set L = T - [X] would contain no points of X. But this is impossible, since X is a union of some of the sets $G^\prime_n$ in our set $G^\prime$, and $G^\prime$ contains the point $x_n$ in X.

[gentle osprey]

stephenw

[cosmic bone]

What does this tell us? There must exist a countable or finite (due to the base being countable) everywhere dense set X whose closure [X] is equal to M

[ruby delta]

G is a collection of open sets, while M is a collection of points, so you'll have to explain what restricting G to members of M means

[cosmic bone]

I understand-- our base contains a set of open sets. I'm defining G' as G with all open sets containing elements not in M removed

[ruby delta]

well G' could potentially be empty then, since that's a really strong condition to be asking

[cosmic bone]

This work is kind of analogous to the proof that if a topological space has a countable base, it contains an everywhere dense subset

hm

thats a good point actually, the proof falls apart obviously in that case

[ruby delta]

I presume you mean G' to be the collection of all open sets having zero intersection with M?

[cosmic bone]

No, I think I didn't think about this properly

thank you for pointing out that very important flaw early on in the proof

[tender halo]

every space contains an everywhere dense subset, its called the entire space

[cosmic bone]

I think I came up with a less braindead proof (but its very possible theres some problems with it).

Let G be our countable base. Since G is a base for T, there is at least one base containing each x $\in$ M. In order for x $\in$ M to not be a limit point, G contains at least one set $G_x$ containing x along with a finite number of members of M.
However, in order for all x $\in$ M to not be limit points, there would need to be an uncountable number of sets in our countable base, which is a contradiction (as per the above argument prescribing a finite ratio between the number non-limit points and the number of bases) .

im sure this is not well exposited and perhaps it's even missing something, but hopefully it checks out?

[gentle osprey]

stephenw

[tender halo]

why do you need the "along with finite number of members of M"?

the proof checks out but i would formulate it as

[cosmic bone]

I included that because its possible of course to include {x, x_1, ... x_n} for some finite n as a set in your base, making x not a limit point

[tender halo]

if X is second countable, then M is second countable, a discrete uncountable space cannot be second countable

[cosmic bone]

what does it mean to say M is second countable since its not necessarily a topological space?

[tender halo]

but also there exists a member of the base that only touches x and doesnt touch anything else

otherwise x is a limit point

[tender halo]

it is a topological space as a subspace of X

[cosmic bone]

hmm

oh

yes if there exists a countable base for T it seems reasonable there is necessarily a countable base for any subset of T

[unreal stratus]

Yes, the definition of the topology on the subset gives you such a base

[cosmic bone]

one more silly question-- why is it obvious that a discrete uncountable space cannot be second countable?

or is that just a very basic result that I could write down in a proof?

[tender halo]

discrete topological spaces only have one base kinda

in a certain sense at least

[cosmic bone]

ah

I didnt think before asking that

that makes perfect sense

[unreal stratus]

If B is a base then every open set is a union of elements of B

[cosmic bone]

although isn't it not necessary that we imbue M with a discrete topology?

[unreal stratus]

So you can write each singleton as a union of base elements, which means that singleton must be in the base

[cosmic bone]

it needs some kind of discreteness (or you will get limit points) so the same argument follows but

[tender halo]

your condition is translated to "M is a discrete subspace"

they are the same thing

[cosmic bone]

okay

one last confusion I think you answered this pretty well but I still seem to be slightly confused

[cosmic bone]

why is this necessary?

[tender halo]

ease of counting i guess

[cosmic bone]

couldnt the only components of the base with x be {x, x_1, x_2}?

not T

[tender halo]

then x would be a limit point

[cosmic bone]

x is a limit point of a subset M if all of its neighborhoods contain infinitely many point of M

perhaps my textbook is using a nonstandard definition

a quick google search says it is a nonstandard definition

[tender halo]

that would be a condensation point

[cosmic bone]

interesting

[tender halo]

in Hausdorff spaces both are the same

[cosmic bone]

here they call contact points what you call limit points

[tender halo]

wait no sorry

[cosmic bone]

its a fairly old book by kolmogorov, but as I understand it the translator took some creative liberties

[tender halo]

condensation is when every nbhd has uncountably many points

[cosmic bone]

here its only specified that its infinite

weird

these seem like pretty old definitions its interesting to me that there is this unique phrasing chosen by this author

[tender halo]

this is called an \omega-accumulation point

[cosmic bone]

the proof mentioned above would work for accumulation points, I hope?

since a countable number of countable sets is also countable, we couldnt hope to contain every element of M in its own omega accumulation point (since M is uncountable)

[plush folio]

So this quotient of the unit square gives us the Klein bottle. What happens if the top and bottom arrows also point in the opposite direction, do we get something familiar?

[shy shoal]

I think we would get a real projective plane

[rancid umbra]

yes

[tender halo]

dont get it twisted

[plush folio]

ah, if we flip the bottom arrow then it's just the circle with the antipodal map

what if we flip the top arrow? then they don't all point clockwise around the square, do we still get RP1?

[rancid umbra]

RP2

RP1 is a circle

[plush folio]

ah, right 👍

[rancid umbra]

there are only four topologies you can get by identifying edges of a square

[plush folio]

torus, klein bottle, RP2 and ...?

[hidden abyss]

torus, rp2, klein, what's the fourth?

[rancid umbra]

i believe flipping the top is still RP2, but if not, try cutting down the middle or diagonal and gluing again until you get something familiar

[rancid umbra]

sphere

[hidden abyss]

I mean if you're counting all identifications you can get stuff like the dunce hat too

[rancid umbra]

the wiki page on the fundamental polygon has good information on what i mean

just pairs of edges

[hidden abyss]

oh i didn't even realize that you can get the sphere that way

I thought you were talking about collapsing the whole perimeter to a point

[rancid umbra]

nah, but that also works if you allow for more than just pairs of edges

then you get the dunce cap

and…

i think thats it?

maybe you can get a cone-thing?

[hidden abyss]

if you identify all sides via the same orientation, you get a space with H_1(X) = Z/4Z

[rancid umbra]

like the dunce cap right before it wraps around its bottom circle?

[hidden abyss]

or wait do you

[rancid umbra]

i think thats the sphere again

you identified all four sides

are you collapsing them? or leaving them

[hidden abyss]

leaving

[rancid umbra]

hmm

[hidden abyss]

so like RP2 but twisted once more

[hidden abyss]

this should follow via cellular homology

[rancid umbra]

weird

[hidden abyss]

but yeah, flipping a pair of arrows that gets identified doesn't affect the end result, so it's just the same ABAB shown on wikipedia

[rancid umbra]

but yea, its a circle with a disk glued on in a 1:4 ratio along the boundary

[plush folio]

I see, so ABAB = A^-1BA^-1B in that notation?

[hidden abyss]

yup

[rancid umbra]

don’t Riemann surfaces of z^(1/n) also have cyclic first homology? just guessing here, since they wind around in a weird way

[hidden abyss]

since the arrows only tell you how the identified sides are oriented wrt each other

[plush folio]

oh, I see, the equivalence classes are literally the same. I get it, thanks

[hidden abyss]

I tried looking for the definition for the Riemann surface of a function and found a handout with just loads of alternative definitions for a Riemann surface

[rancid umbra]

yea, i know nothing about the technical aspects of them

purely vibes here

[unreal stratus]

Vibes are fun

Riemann surfaces are 1 dimensional

[rancid umbra]

not getting the jokes is what pushes me lmao

[shadow frost]

yeah and the actual one - Etale space definition - doesn't make sense for a while

[warped bone]

Hi guys, i need help with this proof:
I have an hausdorff space X, A a compact subset of X and x a point of X that is not in A. I need to prove that there exists two open sets U and V such that

1. x is in U
2. A is subset of V
3. The intersection of U and V is empty

My idea was this:

Since X is hausdorff, i can build a finite collection of open sets that contains every element of A lets say that this is our V (since A is compact) and a collection of finite open sets that does not intersect any of the open sets of that cover lets say that this is U, since they both are union of open sets, they are open and they do not intersect. And obviously A is subset of V

My problem is on how to write this down in a formal way.
I think the idea is somewhat right, but i'm not sure on how to build the U that should contain x in the right way.

I thought the finite collection U to be made by the open set i can find using the hausdorff property, an open set V_1 that contains a_1 and since X is hausdorff i know that there exists a U_1 open set that contain x and does not intersect V_1

Hope i wrote everything in a comprehensible way

[plush folio]

You're on the right track, you start with using the Hausdorff property to build an open cover of A. Then you need to use compactness at some point. For the next step, think about what you can do with a finite collection of open sets that you cannot do with an infinite collection (which is why compactness is needed)

[warped bone]

From the open cover of A i take a finite open cover and every sets V_i of this finite open cover will have a corrispective open set U_i that contains x such that U_i and V_i does not intersect, then i take the intersection of all those U_i and that's my U since the finite intersection of opens is open?

(And the finite open cover is V)

[plush folio]

Yep, good job 💪 then U and V are disjoint, you can easily check this just to be sure

[warped bone]

Thanks!

[brazen sierra]

isnt E* clopen? then whats the difference between strictly convex and just convex?

[cosmic bone]

Let X be the [0,1]. Let $\tau$ be the system of sets consisting of the empty set and any subset of the closed unit interval obtained by deleting a finite or countable number of points from X. Show that (X, $\tau$) satisfies neither the second nor first axiom of countability, that its $T_1$, and not hausdorff.

My attempt at a sol'n:

To show it satisfies neither axiom of countability we can just show it does not satisfy the first axiom of countability.

The first axiom of countability dictates that there is a countable system of sets $H_x$ for any point x such that for any open set G containing x, there is a neighborhood of x O $\in$ $H_x$ such that O is a subset of G. Pick a point x and an open set $G_1$. Since any neighborhood of x $O_1$ is also an open set, we can remove a countable number of points from $G_1$ to make $G_2$ such that $O_1$ is not a subset of $G_2$. Since a countable union of countable sets is itself countable, for any countable system of sets H, we can always construct an open set that contains no $O_n \in H$ in a fashion similar to above. Thus, X is not first countable.

To show it's $T_1$ is trivial. Given any x,y, we can construct a $N_x$ by removing the point y from X, and $N_y$ by removing the point x from X. Then x is not in $N_y$ and y is not in $N_x$. If X were hausdorff, then for any x,y we can find a $N_x$ and $N_y$ such that their intersection is empty. Since X is uncountable and our open sets are made from removing at most countable points from X, in order for our space to be Hausdorff, we would need to remove uncountable points from $N_x$ to have an empty intersection with $N_y$. (really this boils down to you cant remove uncountable common points in the neighborhoods of x,y by removing a countable number of points)

[gentle osprey]

stephenw

[cosmic bone]

does this sound reasonable or have I made some grave error?

[queen prism]

#advanced-analysis but convexity of a normed space refers to convexity of the unit ball
strict convexity means that if you draw a line between any two (distinct) points on the unit sphere, the norm of any point inside the line segment is < 1, meaning that point is in the interior of the unit ball
convexity just means it's <= 1

[brazen sierra]

ah got it, thank you!

[prime elbow]

I want to show that the metric p(x,y) = | arctan x - arctan y | is not complete on R.

So, I am thinking of taking a sequence x_n = n for all n in N.

I have to show two claims,

Claim 1, the sequence x_n = n is cauchy in the given metric space

Claim 2, since graphically arctan n tends to π/2, and there is no such x in R such that arctan x = π/2.

But I want some rigours argument to show arctan n tends to π/2 when n tends to ∞.

And any hint to prove claim 1, i think I have to use the formula of arctan x - arctan y.

Great, thank you ❤️

So | arctan n - arctan m | = |n - m | / ( 1+ c^2) for some c in (n,m).

Then | arctan n - arctan m | ≤ |n-m| /(n^2)

Yes

Yes if we show claim 2 is true then it will be cauchy so claim 1 will be done

I want to generalize this question, say d(x,y) = | f(x) - f(y) |, where f :R -> R.

So when we can say (R,d) is complete metric space ?

[tender halo]

arctan is bijective and bounded and monotonic, its obviously cauchy just from that

[prime elbow]

Sorry I don't get it

[tender halo]

arctan n is monotonic and bounded

therefore it converges to pi/2 because of like stone weierstrass or something

[prime elbow]

Oh

[cosmic bone]

I think im using the same book as you

kolmogorov?

[prime elbow]

no carother

[prime elbow]

oh

[rancid umbra]

how is the unit interval given the CW complex structure with no 0-cells and one 1-cell?
im having issues understanding that if I_0 = ∅, then (I_1,I_0) is an adjunction space of 1-cells

like, how do you attach a 1-cell along nothing?

[quartz horizon]

The push out should work categorically

[rancid umbra]

how? if you want the pushout to be of ∅ <- ∅ -> I_0 = ∅, then the pushout is again ∅.
if you want it to beof D^1 <- ∅ -> I_0 = ∅, then the indexing set doesn't make sense

the base of the span should be S^0, not ∅ in that case

but the only way to have a map into ∅ is if the domain is ∅

is this just a special case we have to accomodate for?

[quartz horizon]

Hang on lemme draw it

@rancid umbra this is the pushout diagram right

the bottom-left represents the 0-skeleton

the top-right represents the single 1-cell

and the top-left represents what you're attaching it along

is that right?

[rancid umbra]

my issue is that the top left should be the boundary of [0,1]

[quartz horizon]

why is that?

oh hm i'm looking at the CW complex definition

right, the attaching maps need to have domain the boundary of the cells?

[rancid umbra]

here for example

and on the n-lab for CW complexes

[quartz horizon]

yeah yeah exactly

then i think i'm stumped too

cause you can't get a function from $S^0$ to the empty set

[gentle osprey]

Pseudonium

[rancid umbra]

right

im just going to treat it as a special case and move on

its a triviality and the natural thing that you want to do works

[quartz horizon]

mhm

[rancid umbra]

it messes this up too tho. the n skeleton can be thought of like that as long as the n-1 skeleton is non-empty. otherwise, it needs to be thought of as a disjoint union of closed n-cells

[unreal stratus]

Pretty sure a CW complex w no zero cells is empty

[rancid umbra]

i’m just going based off of this wikipedia article

[unreal stratus]

Sure, what's the issue

Oh lol

End of the paragraph

Eh it's wikipedia I would ignore this, just a mistake im sure

As it contradicts the definition and I have never seen someone talk about cw complexes without 0 cells lol

It would mess up many things

[rancid umbra]

i have been trying to examine the cell of X x I because of this and was curious what would happen if I had the cell structure mentioned on the wiki page

[rancid umbra]

yea, i can imagine

but uh. it’s kind of difficult to give an ad hoc argument. it is proven in topology and groupoids, however

[unreal stratus]

Wdym

[rancid umbra]

with regards to the issue i forwarded

[unreal stratus]

Oh sure

Yes this is a classic Hatcher thing lol

[rancid umbra]

kinda frustrating

he states it like it’s trivial

[unreal stratus]

yes i remember being annoyed about this proof

[rancid umbra]

i started examining it because of this exercise

but uh, i guess i got caught up in the details

[unreal stratus]

I guess here the point is that $Y:= X^{n-1} \cup A^n$ is just some subcomplex of X$^n$, and you can form $X^n$ by attaching n-cells to $Y$. Then if you think about $n$ cells of $X \times I$ as products of cells of $X$ with cells of $I$ then you arrive at this recipe

[gentle osprey]

Prismatic Potato

[unreal stratus]

I think it is helpful for the combinatorics / visualisation to do some reductions

Okay so I mean like

Let's assume X = X^n and A = A^n (to reduce notation - it makes no difference) and set Y = X^{n-1} u A as above

And for ease, say X is formed from Y by adding one cell, i.e. like

$X = Y \coprod_{\partial D^n} D^n$

[gentle osprey]

Prismatic Potato

[unreal stratus]

Then note $X \times I = (Y \times I) \coprod_{\partial D^n \times I} (D^n \times I)$ but we can cut this up as like $(Y \times I) \coprod_{\partial D^n \times I} (D^n \times {0})$ and then adding the remaining stuff

[gentle osprey]

Prismatic Potato

[unreal stratus]

And indeed you can finally attach D^n x I along this as Hatcher tells you

Hope that makes sense? @rancid umbra

[rancid umbra]

yea, that all makes sense and agrees with the visual i have. i guess trying to formalize the last part has been tricky for me

i’ll retry with this in mind tmrw

thanks

[rancid umbra]

for the general case, do you think a proof by transfinite induction would work?

or is that not the idea

[unreal stratus]

No induction should be necessary, just you take a big union instead of a single one

Which is fine as the cells are all "independent" in a sense

(Since you attach them to the n-1 skeleton)

Just it is notationally nicer for a single cell

[rancid umbra]

sure

i was thinking through it and i guess there is kind of a recursive structure to it, you keep growing the sub complex A_n

but the pushouts take care of it all since you can attach many cells at once and like you said, that they are all being attached to the n-1 skeleton

it would be different if i were trying to attach n-cells to n-cells

[cosmic bone]

Im slightly confused about this question

It seems to me we can construct an infinite subset X as the set of all ordinal numbers greater than 5. But then, for every ordinal number $\lambda$ $\in$ X, we can consider the open set that ($\lambda-1$, $\lambda + 1$) union (1,3). But then there is a neighborhood of $\lambda$ that contains one point in M, and is thus, not a limit point

[gentle osprey]

stephenw

[cosmic bone]

In other words, for any ordinal number in this infinite subset, it seems to me there should be an open subet that contains only one point in the original infinite subset

(and thus, no point is a limit point -> T is not countably compact)

I doubt Kolmogorov messed up so its clear I'm missing something with this counterexample

Even more of a simple example, for any point $\lambda$ in an infinite subset, we will have an open set ($\lambda$ - 1, $\lambda + 1$) union (x-1, x+1) for some arbitrary x, and this open set only contains two elements (and thus, any arbitrary point has a neighborhood that is very finite, and thus, this point is not a limit point)

I think I'm misunderstanding how our open sets are defined, but I don't see how

(Im using terrible shorthand, the ordinal number $\lambda$ is the order type of the set {$a_1$, $a_2$, ... , $a_{\lambda}$})

[gentle osprey]

stephenw

stephenw

[rancid umbra]

what happens if lambda has no immediate predecessor?

[rancid umbra]

the open sets are generated by intervals form [0,k) and (a,b)

this is just the order topology

[cosmic bone]

can't we construct an immediate predecessor for any lambda by removing one element from the end or the start of lambda?

[rancid umbra]

how do you construct an immediate predecessor for w_0?

[cosmic bone]

hm

I don't think I can

[rancid umbra]

right, so that’s at least one spot where your proof breaks

[cosmic bone]

hm

thank you!

[rancid umbra]

hint: ||given a collection of ordinals, show that the union of the collection of ordinals is again an ordinal, and that the supremum of the collection of ordinals is equal to the union of the collection of ordinals||
hint: ||show that a in a well-ordered set X, every infinite subset gives rise to a strictly monotone sequence||

[cosmic bone]

I think I lack the intuition one has for metric spaces or sets when it comes to compactness but I probably have to read a little

I will save this hint, hopefully I can come up with something but I may end up having to refer back to it at some point, thank you for your help

[cosmic bone]

can't you do this by taking the order type Z and removing the first element?

no

you cant

I didn't carefully read our order relation

[cyan roost]

I have a bit of trouble with compactness. I am not sure if I understand this correctly. The half open intervals like [a, b), a<b are not compact when in R with the standard euclidean topology. Would this also be true for half open intervals in all sets X with the standard euclidean topology. Like if X=[0,5] with the standard euclidean topology would a half open set/interval like [a, b) stil not be compact?

[ruby delta]

if an interval has any open end it is not compact (in R), try and prove why

[tender halo]

compactness is not a relative property

[cyan roost]

Is this for intervals in sets with the standard euclidean topology only? For couldn’t you have topology on for example [0,1) that would make [0,1) compact?

[tender halo]

[0; 1) is a certain space, it is both a subspace of X and R, but its compactness does not rely on either

[cyan roost]

It rely on the topology or am I wrong?

[tender halo]

yes, what i mean is that compactness just relies on the space itself and it's properties, it does not matter what the surrounding space is

if a space is homeomorphic to a half interval, then it is not compact

[cyan roost]

Ah yes okay, so in my first message, if X is a space with the standard euclidean topology then any half open intervals in X are always not compact because X has the standard euclidean topology?

[tender halo]

what is "a space with the standard euclidean topology'?

a subspace of R?

[cyan roost]

Yes let us say [0,9]

[tender halo]

what is a half open interval in X? if it an intersection of X as a subset of R with some half open interval as another subset of R?

or is it an intersection of X with a half-interval [a; b) where both a and b lie in X?

because the former can be compact, the latter cannot

[cyan roost]

If X=[0,9] with standard Euclidean topology and for example let’s say we have a subset A of X that is a half open interval let’s say A=[1/2, 4). Would A be not compact because X has the standard topology?

[tender halo]

A is not compact

but like

uhh

i mean sure from a certain point of view i guess

its a strange way to phrase the argument, to me it makes much more sense to state that A is a subset of R and we know how compact susbsets of R look

[quartz horizon]

Like there are topologies you can put on [1/2, 4) that make it compact

But yeah Heine-Borel classifies the compact subsets of R^n (with the usual topology)

In this case you’re giving A the subspace topology

[quartz horizon]

As a silly example, the indiscrete topology is always compact

[tender halo]

i hate when textbook writers put goofy topologies on R specifically

instead of like an arbitrary space of cardinality continuum

[ruby delta]

what, you don't like the topology generated by [x, y)

[cyan roost]

Yea so like if I have [0,8] with the standard euclidean topology then half open A=[1/2, 4) in this topology is not compact

[quartz horizon]

With the subspace topology, yeah

[tender halo]

sure

[quartz horizon]

It’s a convention that one puts the subspace topology on a subset of a topological space

[tender halo]

well sorgenfrey line makes sense because it makes use of the order on R

i mean more like cofinite topology

[ruby hinge]

How do I prove that R^n-{x1,x2} is simply connected?

[ruby delta]

what is x1,x2 here?

just two arbitrary points?

[ruby hinge]

Yeah

In R^n

Like okay look at the fundamental group

But thats not enough right

[ruby delta]

oh wait, simply connected

[ruby hinge]

you need to have trivial fundamental group

[ruby delta]

presumably you have n>=3 then?

[quartz horizon]

Yeah

It’s just false for n < 3

[ruby hinge]

Yes

[cyan roost]

Thanks for the help people🤝

[quartz horizon]

Are you sure

I remember this argument being harder than you’d expect even for 1 point removed

[ruby delta]

hmm, let me think about this more

[quartz horizon]

I.e. showing, say, R^3 \ 0 is simply connected

[hidden abyss]

A bit overkill, but you could also prove that its homotopic to the wedge of two spheres and then use seifert van kampen

[quartz horizon]

The issue is you can have space-filling curves

So e.g. a curve on S^2 which is a continuous surjection

You can still deal with these but you have to think a little more

[ruby hinge]

Like you have that the fundamental group of R^n-{a} is S^{n-1} right?

Is isomorphic to that of*

[quartz horizon]

It’s the same as that of S^{n-1}

Yeah

[hidden abyss]

What is the classical proof for spheres being simply connected anyway

[quartz horizon]

I don’t actually know

Haven’t done enough algtop~

[ruby hinge]

doesn't matter it's given lol

[ruby hinge]

So what about 2 points removing?

Is it still just S^n-1

[hidden abyss]

No its the wedge of two n-1 spheres

I.e. you glue two spheres of the same dimension together at a single point

[ruby hinge]

It's like figure eight or what?

[hidden abyss]

Yes though the spheres can be higher dimensional than 1

[ruby hinge]

yeah ok

[hidden abyss]

Do you know the seifert van kampen theorem?

[ruby hinge]

Yes

Like if you have 2 open sets and their intersection is nonempty

[hidden abyss]

Yes

[ruby hinge]

That cover the entire space

[hidden abyss]

So suppose the spheres are glued at the north poles. Then take U = the spheres \ south pole of first sphere and V = the spheres \ south pole of second sphere

[ruby hinge]

but how do we get 2 spheres

because that's removing 2 points?

[hidden abyss]

Explicitly writing the homotopy equivalence doesnt sound fun

The idea is basically the same as for R^n{x} ≃ S^{n-1}

You inflate the wedge of spheres outwards to fill all of R^n outside the spheres and inwards to fill all but a point in each of the spheres

[ruby hinge]

I dont't really have intuition how the seifert van kampen theorem would apply here...

[hidden abyss]

Once you have the wedge of spheres, you'll wanna do a decomposition like this

U, V and U ∩ V are all homotopy equivalent to spaces youre familiar with

[tender halo]

to prove it in a dumb way, there are two spheres around the points where the curve is absent (because compactness), we can flatten the curve to a plane + two spheres

[ruby hinge]

So in general you'd get a wedge of n spheres if you remove n points?

[unreal stratus]

ye

[ruby hinge]

you can pick open sets outside of the spheres to make it open right?

Like how do we know that U and V are open

[tender halo]

then for every segment of the path that goes through the surface of the sphere we can make it so it goes around the edge

and then we rotate and can just pull into a null loop

[hidden abyss]

Thats why I added those little "arms" to the individual spheres

You can define U = X \ a point so it will automatically be open in X

[ruby hinge]

what I thought was that you have that the fundamental group of R^n-{x1,x2} is just isom. to that of S^{n-1}-{x1} but then you really need some extra seifert van kampen like stuff righth?

But like how would removing 2 points in for example R^3 make 2 spheres?

Like that just does not make sense to me

[ruby hinge]

Ah alright

[hidden abyss]

This isnt true, S^{n-1}\x1 is isomorphic to R^{n-1}, thus contractible, but R^n {x1,x2} isn't, (intuitively, because those removed points are holes)

Or well the fundamental groups are isomorphic but it's not trivial

[unreal stratus]

It is sort of hard to visualise but basically point is you can expand the holes to get balls taken out of R^3 and then compress stuff to the boundary

As for R^2 with how you get two circles

In general, R^n minus k points is homotopy equivalent to a wedge of k (n-1)-spheres

[ruby hinge]

So that's basically just a crucial fact I guess?

What about taking U= R^n-x1 and V = R^n-x2 and then the inversection is most definitely non-empty, meaning we van apply seifer van kampen?

But \pi_1(U) is just S^n-1

Is this like an idea?

[rancid umbra]

viewing them as cw complexes is nice. there are some theorems in hatcher which deal with this easily

[rancid umbra]

U and V aren’t subsets of R^2 - {x1,x2}

[unreal stratus]

Adding to c squared, there are things like the "cellular approximation theorem", or for example you can use simplicial complexes (where the result is a bit easier to prove)

But yes the result that pi_k(S^n) = 0 for k < n [in particular you can take k = 1 here lol] is not easy and is an important input to a lot of things

[quartz horizon]

Interesting

[rancid umbra]

silly algebraic invariants. your were supposed to make life easier 🤦🏽‍♂️

now spheres are hard

[quartz horizon]

I really should learn algtop lol

Lots of cool things within it

[hidden abyss]

Wlog assume x1=(0,...,0), x2=(1,0,...0), you can now take U = (-∞,2/3) × R^{n-1} \ x1 and V = (1/3,∞) x R^{n-1} \ x2, then you can apply SvK directly

It will look something like this

The intersection is a strip which is contractible, and U and V are both isomorphic to R^n \ a point

[rancid umbra]

yea. this is the way to get around it without using the homotopy to a wedge of spheres

you can do a similar thing when more points are removed

[prime elbow]

I don't get the hint, I know I have to show p is complete metric space on G, but I think it is computational, so any hint?

[tender halo]

a conceptual hint is that we un-cauchy some cauchy sequences

[prime elbow]

Oh, how?

[ruby delta]

that's what the problem is trying to do

[tender halo]

we dont want sequences that go toward F to be cauchy, because they point "outside"

[ruby delta]

the only way cauchy sequences in G can go wrong is if they approach the edge, ergo if we can make a new metric that don't become cauchy if they approach the edge then we're golden

and so we're adding "distance to the edge" to the metric

[prime elbow]

What do you mean by edge?

[prime elbow]

Oh

[ruby delta]

the closure of G minus G

[prime elbow]

I see

[ruby delta]

e.g. (0,1) in [-1,2]

the only points where issues can arise is the boundary {0,1}

[prime elbow]

Yes

[ruby delta]

so, if we have a cauchy sequence approaching the boundary, we artifically inflate the distance between successive terms as it gets closer and closer to the boundary

so that it's no longer cauchy

[prime elbow]

I see

[ruby delta]

a concrete example would be if 1/n converges to 0, we could "invert" them, for instance p(1/n,0) = 1/d(1/n,0), which goes to infinity so the original sequence is no longer cauchy in this new metric

[prime elbow]

Oh

Great idea

[ruby delta]

that's what the problem is doing already

just investigate the given metric

[prime elbow]

So is that computational part?

[quartz horizon]

You should prove it’s actually a metric, for example

[prime elbow]

It is a metric function, yeah

[quartz horizon]

How do you know?

[prime elbow]

We can easily observe, p(x,y) = 0 iff x = y and symmetric property are easy to see and triangle inequalities come easily with the help of triangle inequality in R, right?

[quartz horizon]

Sure, just checking

[ruby delta]

in fact that's basically what the problem is doing

since the space is complete, the only way a cauchy sequence fails to converge is if it approaches the boundary

so, if we can "fold" the boundary to infinity such that the sequence is no longer convergent, period

for instance for (0,1) you can stretch the interval to get a homeomorphism to R, which is complete

[prime elbow]

Yes

[ruby delta]

converging to boundary bad => make boundary at infinity

[prime elbow]

I see

[hidden abyss]

The biggest hurdle in the exercise is probably checking that the identity map on G is an isomorphism

[prime elbow]

You mean isometry?

[hidden abyss]

no

[ruby delta]

they mean homeomorphism

[prime elbow]

You mean homeomorphic

I see

[hidden abyss]

Right mixed up the words

[ruby delta]

close enough

i mean

[hidden abyss]

Isomorphism in the category of topological spaces

[ruby delta]

yeah

[hidden abyss]

I went back through my intro to topology course's lecture notes and remembered that the prof called homeomorphisms topological equivalences

I dont think I've seen that name used ever again since then

[ruby delta]

hmmm yes it's a natrual transformation from itself to itself

[unreal stratus]

Equivalence is used for things like this often (not just between functors lol) but I have never heard someone say topological equivalence

Other than people trying to handwave away topology

[hidden abyss]

Might be relevant to add that the course was in german

But even then, Homöomorphismus should be the standard name

[thin tide]

yeah ngl i have heard people use topologically equivalent to mean homeomorphic but i think that's a consequence of people trying to do this cause that's such a bad name

like all it suggests is that it's some equivalence relation with topological information so for all u know it could be homotopy equivalence

[rancid umbra]

because homeomorphism is a bad name?

[thin tide]

nah i'm saying topologically equivalent is a bad name for homeomorphic

cause it's just unnecessarily ambiguous

[rancid umbra]

yea, i was just about to say something about that

but i misunderstood you

[thin tide]

no worries

[unreal stratus]

yh lol

[iron bolt]

homotopy equivalence? surely you mean homotopical equivalence? 😛

homotopeomorphism

[sudden flower]

If $(X,\tau)$ is a regular topological space, does there exist a maximal topology $\tau'$ containing $\tau$ such that $(X,\tau')$ is regular?

[gentle osprey]

mormore

[opaque scroll]

The discrete topology would do

[haughty jungle]

Yea this question is a little strange because in some intuitive sense the finer a topology is the like more regular it becomes so obv the finest topology works

[sudden flower]

Maybe its some kinda trick question as I was thinking along the lines of some Zorn's lemma type of stuff

But hey, the discrete topology works indeed

[tender halo]

that is not true

you can have a finer topology that is not regular

it only works for T1 and T2

[opaque scroll]

There's like a tricky balance. Finer topologies give you more open neighborhoods, but also more closed sets

[tender halo]

yea

[opaque scroll]

Both the discrete and indiscrete topologies should be regular, so you just need to get out of that scary middle ground

[tender halo]

as an example, let H = {1/n | n \in N}, and take the topology of R with the additional nbhds of 0 having the form (-a; a) / H

it is finer than R but not regular

[opaque scroll]

I guess similar question could be if tau is any topology is there a maximal regular topology contained in tau?

It's not totally clear to me if the supremum of a chain of regular topologies is regular

[tender halo]

it is

[opaque scroll]

Cool, same for intersection?

[tender halo]

doesnt even need to be a chain i think, regular topologies just need to be cofinite

same with any other separation axiom

even T6 somewhat surprisingly

for intersection doesnt sound true to me at all

except T1 obviously

[opaque scroll]

What do you mean that they need to be cofinite?

[tender halo]

in the chain

for any x in the chain there needs to be a regular a in the chain with x <= a

hmm actually no

maybe its not true

[opaque scroll]

Ah cofinal, I see.

Hmm, did you have an argument in mind?

[tender halo]

i forgot that supremum of the chain is not the union

[opaque scroll]

Yeah, it's almost though, so maybe one can work with it

[tender halo]

ah of course dont know how i got those mixed up

[opaque scroll]

Yeah if they're cofinal in the chain, then they form their own chain with the same supremum

[sudden flower]

If Q(a,X) denotes the quasi-component of a in X, then is it true that Q(a,X) x Q(b,Y) ⊂ Q((a,b), X x Y)?

[languid patrol]

what's a quasicomponent?

[sudden flower]

So there seems to be a few different definitions, but I guess this is what I'm working with (https://proofwiki.org/wiki/Definition:Quasicomponent)

One is that Q(a,X) is the intersection of all clopen subsets of X containing a

[languid patrol]

what have you tried to prove this?

have you used the definition of the product topology? (generated by a basis of boxes)

[sudden flower]

Not much yet, I initially proved that if you replace the quasi-components with just components you get this inclusion for free from the fact that the components are connected and the product of two connected sets is connected, but I can't find any results I could use with the quasi-components.

[cyan roost]

I know that the half open interval (a,b] is not compact with the standard euclidean topology, but what if we have the topology generated by half open intervals (a,b] with a<b in R. Would (a,b] then be compact?

[tender halo]

no

it has an even worse chance of being compact because its finer than the standard topology on R

if there is a cover of (a; b] with the standard topology with no finite subcover, then that same cover is also a cover for (a; b] with the upper limit topology

and therefore also doesnt have a finite subcover

[cyan roost]

so the cover is the same in both the standard topology and in the upper limit topology? is this also true if we take [a,b) with the lower limit topology?

[tender halo]

there is a cover that disproves compactness for both

[languid patrol]

I think if one of X or Y is compact this is true by the tube lemma, but I'd suspect in general it's false

[tender halo]

(a; b] with upper limit topology is homeomorphic to [a; b) with lower limit topology

[cyan roost]

ah yes, I didn't think of that

[unreal stratus]

Lol fun story is that one time I got really sidetracked by the tube lemma

[cyan roost]

Is there an example for this, I am just trying to get my intuition on this. Is it always true that if there is a cover of (a,b] in the standard topology then that same cover is also a cover for (a,b] with the upper limit topology?

[unreal stratus]

Basically someone pointed out that most proofs that the product of two spaces is compact use the axiom of choice without acknowledgement

Which is especially funny considering that it is very well advertised how Tychnoff is equivalent to choice

Anyway so like I remember learning the harder non choicy proof and getting obsessed with this instead of working on other stuff rip

[tender halo]

how?

also wdym constructive proof?

[tender halo]

i mean yeah

any set open in the standard topology is also open in the upper limit topology

so a cover stays a cover if you go from the former to the latter

[cyan roost]

could it be something like (a, b-(b-a)/(2n)] with n natural number, or is that not a cover for (a,b].

[tender halo]

no, because it's not a cover in the standard topology

also doesnt include b

[cyan roost]

oh I see hmmm

[tender halo]

(a + (b - a)/2n, b] is a cover in the upper limit topology though

[cyan roost]

so if for example we had (0,1] with the upper limit topology, then (1/2n, 1] is a cover for (0, 1] with n natural number

[tender halo]

yea

[cyan roost]

and this doesn't have any finite subcollection that covers (0, 1], because if we take a finite subcollection from (1/2n, 1] then we would miss elements near 0? or is my intuition wrong on that

[unreal stratus]

Okay I should not say constructive lol. I more mean without choice

[gritty widget]

(a, b] is not compact in the standard topology, so there is some open cover, say U_i, with no finite subcover.

(a, b] in the upper limit topology is finer than the standard, every open set in the standard topology is also open in the upper limit topology. Hence, U_i is also an open cover in the upper limit topology, and it still has no finite subcover.

[unreal stratus]

Well usually you pick a bunch of finite subcovers for each point or smth and use the tube lemma

But you pick finite subcovers possibly infinitely many times

Uhhh I cannot remember details as it has been a while

[gritty widget]

Oh beaver said it first.

Smh.

[tender halo]

sure that works

[tender halo]

ah i see

my preferred proof of the finite is due to Kuratowski

[unreal stratus]

How does that go?

[tender halo]

a space X is compact iff forall Y the projection X x Y -> Y is closed

[unreal stratus]

Ah yes that's good

The way I know to prove that is from tube lemma aha but ye

[tender halo]

so if you have a product of compact spaces X1, X2, ... Xn, then X1 x X2 x ... x Xn x Y -> Y is a composition of X1 x (X2 x ... x Xn x Y) -> X2 x (X3 ... x Xn x Y) -> .. -> Xn x Y -> Y

which is closed as a composition of closed maps

[unreal stratus]

This is very good and nice cause lol motivation for things in algebraic geometry

[cyan roost]

what if we took something like [a,b) with the upper limit topology, then is it also the same as before, that since [a, b) with standard topology has a cover with no finite subcover and then not compact, then [a,b) with the upper limit topology is not compact since we can take the same cover in the upper limit topology

[tender halo]

that is true yeah

[cyan roost]

what if we took [a,b) as a subspace of R with the upper limit topology, then could the same argument be made for [a,b) with the subspace topology?

[tender halo]

its the same topology?

upper limit topology on [a; b) is the same as the [a; b) subspace of (R with upper limit topology)

[cyan roost]

you are right, I think I just confused myself there for a moment.

[blazing jackal]

Can anybody link me to a reading that describes how to compute the similarity between two simplicial complexes?

[haughty jungle]

Oo good catch ty

[rancid umbra]

what do you mean by similarity?

[blazing jackal]

Looking for ways to compute a similarly metric, like a correlation coef, between simplicial complexes

Would like to work with the simplex data and not something like a persistent homology

[cosmic bone]

Im realizing I misunderstood our open sets

oh wow

or at least I really didnt get why this problem was asking about ordinals in particular

idk how you saw that right away but thank you

[sudden flower]

Is the space {(x,y) | y = 0 or x^2 + y^2 < 1 } locally compact?

I somehow managed to conclude that it is, but my classmate told me that it isn't

[quartz horizon]

I don’t think it is

[sudden flower]

Is the issue on the points (1,0) or (-1,0)?

[quartz horizon]

That’s what I’d think

[alpine nest]

Same

I'm assuming this is in the standard Euclidean metric on R^2?

[sudden flower]

Yesss

If you take a ball around (1,0) then in the subspace topology it has like a bit of the disc and a bit of the x-axis

[quartz horizon]

yeah

[sudden flower]

One needs to argue that the closure of this is not compact, but I don't see how

[quartz horizon]

There are some equivalent formulations of compactness for metric spaces that’d be useful here

[sudden flower]

In terms of sequences?

[quartz horizon]

Yeah

[sudden flower]

Not sure I see how that would be helpful?

[quartz horizon]

It depends on you knowing these

[sudden flower]

I mean there are bunch of them? Its the image of a continuous function from the Cantor set?

[quartz horizon]

Oh huh I had no idea that was one lol

The one i mean is compact <-> sequentially compact <-> complete and totally bounded

[alpine nest]

Sequential compactness is what I'd invoke here.

But I've always favoured the caveman methods.

[quartz horizon]

Caveman methods?

Oh i guess there’s a faster way to do this

[languid patrol]

a compact subset of R^n with the subspace topology has to be closed and bounded as a subset of R^n

for any open or closed ball around (1, 0) the intersection with your thing isn't even closed in R^2

[alpine nest]

I mean, I didn't really stop to think of a faster way, that's my point.

I see a metric space, I start picking points and hope for the best.

[quartz horizon]

Wdym

[alpine nest]

Just that when I do topology in a metric space, my first instinct is to think in terms of points, balls and sequences.

[quartz horizon]

Oh sure same here

I thought you were saying you found a better method than sequential compactness

[sudden flower]

Wait but aren't we concerned about the closure of that which is closed?

[languid patrol]

the closure within the subspace is not closed inside of R^2

because it will never contain any points of S^1

except for (1, 0)

[median sand]

In the definition of uniform structures what do we need the inversion condition for (if U is an entourage, then so is U^-1)? All the basic stuff "a uniformity induces a topology", "uniformly continuous maps are continuous", etc., all don't require this, only the condition VV<=U. I realise it is motivated naturally by considerations involving the metric and group topologies, but I've yet to see where it actually enters a proof.

[prime elbow]

What is caveman methods?

[warm hedge]

I have a question. Lets say X is a Polish space with a countable basis B. If I close B under finite unions, finite intersections, and complements to obtain a countable Boolean algebra A, and then take the topology generated by A, is the resulting topology still Polish?

[warm hedge]

could i use this corollary to say that. Because A will be a collection of borel sets, so there is a finer topology which is polish and make all the sets of the boolean algebra open, so we could say the boolean algebra can produce this finer polish topology

or i am wrong ?

actually it doesnt mean that the new topology is the one generated by the A, right ? Its finer ?

[tender halo]

yeah its some topology finer than the one generated by A

[warm hedge]

so the corollary doesnt help, right ?

[queen prism]

what does sigma-open mean?

[warm hedge]

its open in the new topology

[queen prism]

o lol

I thought that was like some weird countable set operation

[warm hedge]

no no just open in the σ topology

[warm hedge]

I have an other question. Lets say we have a topology τ and its basis B. Also lets say F is a close set. The topology generated by τ union {F} is the same with the topology generated by B union {F} ?

[rancid umbra]

any topology containing T U {F} will contain B U {F} and vice versa

[warm hedge]

and if we say B isnt a basis but B is a collection of sets which generated T. Can we say the same about T U {F} and B U {F}?

[tender halo]

yes

F being closed doesnt matter

[warm hedge]

yeah lets say its just not open. Cause i guess if its open it doesnt change the topology anyway

[warm hedge]

so i guess now i can say that the topology is still Polish. I will try to explain why and if someone wants to check if my thoughts will be correct i will appreciate it

[tender halo]

i can see why its metrizable and separable, not sure how to prove completeness tbh

[warm hedge]

i dont know either how to prove completeness so i will try to use a result that if i add a close set in the topology the one generated by this collection of sets is still polish

I start with a countable base B for a Polish topology τ.
My goal is to enlarge B step by step until it becomes a Boolean algebra.

First step.
I add one set that is missing from B.
• If the new set is open in τ, the topology does not change, so the space remains Polish.
• If the new set is closed in τ, I replace τ by the smallest topology τ₁ containing both τ and that closed set. By known result τ₁ is still Polish.

So now when i will add the next set, it will be either open or close (there is no other choice cause it will be either finite union/intersection or complement of sets in our collection which are opens in τ₁) in the new τ₁ topology so it is the same as step 1. And again we have a Polish topology.

So we create a sequence of polish topologies that their intersection is Hausedorff (its the first polish topology) and by known result the union of all these topologies is Polish.

the known results i use

am i correct or i am missunderstanding things

[tender halo]

sounds right i think

[warm hedge]

thank you

[neat mantle]

I'm flipping through Engelking as a refresher before picking up Hatcher. I used Munkres for Topology back in the day. What is this symbol, the inverted triangle. I tried finding it in the earlier pages unsuccessfully. I thought it was a "join" of maps, I searched and I don't think "join" is the right word for it; incidentally couldn't find anything. What is this symbol? Are two functions being pasted together?