Mathematics - 2024-03-11 (page 1 of 2)

copper.hat

00:12

@XanderHenderson Right beside Berkzerkeley. I think SoCal is a different state now, ever since they adopted lesliecoin.

Balarka Sen

today i realized for any norm $|\cdot|$ on a vector space, $|a + b| = |a| + |b|$ implies $|\cdot|$ is linear along the entire line connecting $a$ and $b$

Proof: $|a + b| = |ta + b + (1 - t)a| \leq |ta + b| + (1 - t)|a| \leq t|a| + |b| + (1 - t)|a| \leq |a| + |b|$, but this was an equality! Hence everything is an equality

$|ta + b| = t|a| + |b|$

Xander Henderson

01:33

@copper.hat In that case, double yikes! You can't go 70 on San Pablo!

(I have an aunt who lives in Berkeley, about four blocks inland from San Pablo.)

copper.hat

02:14

no kidding. i still have been unable to find out what was happening. maybe on the news later, i can't imagine it ended well.

robjohn

06:58

We are down to a skeleton crew tonight.

leslie townes

💀💀

robjohn

Mr Bones

Daniel Donnelly

08:10

@Shaun finally got some more coding work :|

I'm quite jazzed

:D

Daniel Donnelly

08:44

Shaun, still interested in studying some group cohomology togeth?

My latest attempt on that famous problem has failed. Made a fatal algebric error

But that topology construction is still a topology. It just doesn't help prove what's needed. It's nice for framing it though. For example $0$ is a generic point for $Z/p_n\#$ in the topology.

The point is to prove that in the subspace $I=$ a large enugh interval, that $0$ is still a generic point, but I'm thinking no topological proof exists unless you delve further in but it has to be combined with the algebra in some way no matter what. We can't form the smallest topology making it a topological monoid under $\cdot$ I don't think if I recall correctly, because you lose some important properties that are needed to make the proof work.

For example the Furstenberg arithmetic progression topology is much too fine to use (too many open sets).

@Shaun

John Zimmerman

12:03

“What is the largest surface of revolution embedded in the unit cube?” Can this be considered a packing problem?

Xander Henderson

@JohnZimmerman Typically, a "packing problem" is "what is the largest number of $n$-balls I can fit into an $n$-dimensional volume?" Your question is not that.

So I wouldn't call it a "packing problem".

John Zimmerman

12:45

I think better to call it a maximization problem

Xander Henderson

It is certainly a kind of optimization problem.

John Zimmerman

Maximize the volume with the constraint that the surface is a surface of revolution

Soumik Mukherjee

Maximizing revolution...Hmm you should read Marx for that.

John Zimmerman

I want to prove some things about an extension of this problem in $N$ dimensions. Like is the sequence of volumes increasing

Ryder Rude

13:05

the elements $\sum n_i x_i$ form a subgroup, $x_i \in G, n_i \in Z$

the above result is only for Abelian groups $G$, right? (with summation being the group operation)

$x_i$ runs over some subset of $G$

Jakobian

@RyderRude what does that mean

Missing context or other details

Are $x_i$ fixed elements

The sum notation often doesn't make sense in commutative settings but this result stays true for any group, with proper interpretation

@RyderRude if $g_i\in S\subseteq G$ then $\{g_1^{n_1}...g_m^{n_m} : m\in\mathbb{N}_0, n_i\in\mathbb{Z}, g_i\in S\}$ is the subgroup of $G$ generated by $S$

That is, its the smallest subgroup in the sense of inclusion that contains $S$

Shashaank

13:27

Hello everyone, I want to find the power series expansion of $\frac{1}{1-2D^2+D^3}$ so I assume the power series to be $a_1+a_2D+a_3D^2+a_4D^3+a_5D^4......$, I call this summation P. These two are equal so, $P(1-2D^2+D^3)$ should be 1, I get the power series to be $1 + 2D + 4D^2 + 8D^3.....$

I don't understand where I am going wrong, Can someone correct?

Jakobian

@Shashaank what are you doing before saying you get 1+2D+4D^2+...?

Shashaank

The product of P and $(1-2D^2+D^3)$ is 1

Jakobian

I read that, but what are you doing to obtain 1+2D+...

Shashaank

I actually get the product as $a_1 + (a_2-2a_1)D + (a_3-2a_2)D^2......$ also $a_1$ is 1 and the other coefficients $(a_n-2a_(n-1))$ (n not 1) are zero

Jakobian

Well thats wrong, but you should be able to obtain recursion of some kind from this

Shashaank

13:35

So my product is wrong?

Jakobian

Was the original problem about solving a recursion?

Shashaank

The original problem was to find the power series of $\frac{1}{1-2D^2+D^3}$ and I thought of this method

Jakobian

@Shashaank yes you multiplied it wrong. For example coefficient next to D is a_1, its one next to D^2 thats a_3-2a_1

@Shashaank I see. Then its the wrong way to go about this.

Shashaank

Oh

How should I go about it? and why is it wrong?

Jakobian

You see, usually you are given a recursive sequence, which you want to solve explicitly so you obtain some formal series P which you then obtain explicit formula for

And to do this you expand it. You did this but backwards

Shashaank

13:39

I don't get you, Can you please explain

Jakobian

I mean the method is not really necessarily wrong but you will go through a lot more steps than necessary, while there is a simple solution

Shashaank

What is the simple way?

Jakobian

To expand your function into formal power series what you need to do is simple fraction decomposition, and then expand into geometric series

Shashaank

Also how can 1/(Something) be 1+.......

Oh

so D<1 to find the power series?

Jakobian

Not necessarily

Shashaank

13:43

oh

Jakobian

You don't need any bounds on $D$, treat it like formal variables, and forget about analysis for now

Shashaank

Oh, Btw I am learning this for solving differential equations using operators

Jakobian

By expanding into geometric series I mean to write your simple fractions as a/(1-bD) where a, b are some constants

Shashaank

Can I use this for complex numbers too?

Like I have $\frac{1}{(D-i)(D+1)}$

Can I write the power series?

nickbros123

14:02

@Jakobian my personal experience with LA, with even well regarded books like hoffman kunze is that a lot of proofs are vibes

i think the subject is like that. idk how to put this in correct words, its a bit mechanical

one potato two potato

So suppose $Z$ is a path-connected space with $\pi_1(Z) = \pi_2(Z) = 0$. Let $\Sigma$ be a surface. Let $f:\Sigma\to Z$ be a map. Then $f|_{\Sigma^{(1)}}:\Sigma^{(1)}\to Z$ is a map which is nulhomotopic as $\pi_1(Z) = 0$ (Here, $\Sigma^{(1)}$ is the $1$-skeleton of $\Sigma$ with a standard CW structure).

Using homotopy extension property on $\Sigma^{(1)}\times I\cup \Sigma\times\{0\}\xrightarrow{H\cup f}Z$ with $H$: nulhomotopy of $f$, I get a map $\tilde{H}:\Sigma\times I\to Z$ such that $\tilde{H}_1$ is a point. Hence, $\tilde{H}_1 = \Sigma/\Sigma^{(1)}\cong S^2\to Z$ which is nulhomotopic by assumption, so $f$ is nulhomotopic. Ta-da

Xander Henderson

Inspired by a low-quality question I saw on Math SE this morning, I am putting the following on my next calculus exam:

(In the section of more difficult questions which I expect only around (maybe) half the class to make real progress on. Though I think the hint makes it kind of trivial.)

SnoopyKid

Hello, this is an example of binary variable called as logical NOT https://www.fico.com/fico-xpress-optimization/docs/latest/mipform/dhtml/chap2s1.html?scroll=sseclognot

and this is the complement rule of probability https://brilliant.org/wiki/probability-by-complement/#:~:text=For%20two%20events%20to%20be,must%20always%20total%20to%201.&text=Equivalently%2C%20P%20(%20A%20c%20),1%E2%88%92P(A).

Are those things the same

Thorgott

@onepotatotwopotato correct, this also follows from general principle

one potato two potato

what is that general principle?

Thorgott

14:13

$\{\ast\}\rightarrow Z$ is a $2$-connected map, so the pushforward $\{\ast\}=[X,\{\ast\}]\rightarrow[X,Z]$ is surjective for any $\le2$-dim. CW-complex

one potato two potato

I don't know what that is

Thorgott

check 11.11.-11.13. in ch VII of Bredon

very foundational stuff

Xander Henderson

@Thorgott Oof. That book.

Things like that are what make me an analyst.

Thorgott

one of my favorite books

I have a copy right next to me

Xander Henderson

I also have a copy on my shelf. And, to be fair, it is better than a lot of books in the area. But... oof. I don't want to work in that area.

one potato two potato

14:19

@Thorgott Oh it's 2-equivalence in Bredon

I also have a copy...somewhere in my room

Ryder Rude

@Jakobian oh. so it's for general groups . thanks

one potato two potato

Interesting... I forgot many homotopy theories. But the point of the problem is to use HEP

Ryder Rude

why is closure guaranteed $(a^nb^m)(a^pb^q)=a^rb^s$

Jakobian

@Shashaank yeah

Shashaank

@Jakobian Thank you

Jakobian

14:31

@RyderRude huh?

$(a^nb^m)(a^pb^q) = a^nb^ma^pb^q$

Ryder Rude

@Jakobian i can see why this would be a subgroup if G is abelian, but i dont see how closure is satisfied if G is non abelian

Jakobian

So its in the desired form

Ryder Rude

@Jakobian oooh it already satisfies the form

Thorgott

@onepotatotwopotato right, it's different terminology for the same thing

Ryder Rude

so for general groups $G$, we cant say that $n_1 g_1+n2g_2$ , $n_i\in Z$ forms a subgroup, becuz it doesnt include elements like $n_1g_1+n_2g_2+n_3g_3+n_4g_4$? @Jakobian

but we can say that the set $g_k ^n$, $n\in Z$ span a subgroup, $g_k \subset G$

Jakobian

14:40

@RyderRude we usually don't use additive notation for non-commutative groups

Ryder Rude

yeah. this book is using the additive notation becuz theyre setting up homology group

it says theyre abelian

Jakobian

Homology group is commutative

Ryder Rude

yeah. so for these, we can say that $\sum n_i g_i$ , $g_i\subset G$ form a subgroup, where the same $g_i$ is not allowed to repeat in the sum

Jakobian

There is a difference between given indexed set $(g_i), i\in I$ and a set of elements $S$

Can you please slow down

Ryder Rude

@Jakobian yeah. in the latter, we r allowed to repeat

@Jakobian sorry

Mad

14:44

What does a "maximally" abelian Sub lie algebra mean? "spesifically refering to the maximality"

Jakobian

If $S$ is just a set, what I wrote works for non-commutative groups

Mad

IE all other abelian sub algebras are subalgebras of that? is that how you understand maximality in this context?

Jakobian

But if we are given an indexed set, $(g_i)$, then indeed this is somewhat specific to commutative groups

The reason being as you noticed, there is no reason for a product of elements of the form a^n b^m to be of the same form

Ryder Rude

yeah

Jakobian

This fails e.g. for the free group on two elements

Ryder Rude

14:47

thanks

Jakobian

@RyderRude I'm not exactly sure what you meant here but hopefully something like I just said

sharon_puthuparambil

On this post (math.stackexchange.com/questions/4695733/…): How can someone prove (1)=>(3)?

Ryder Rude

@Jakobian i just meant to write this there. the set of these products form a subgroup becuz repetition of the same $g_i$ is allowed in the product

Xander Henderson

@sharon_puthuparambil Ugh... that question is gross. I don't like the typesetting---makes it very hard to read. :(

Ryder Rude

i got confused becuz the book is using an indexed set but the statement they wrote isnt qualified by "abelian"

they have written just "group"

Thorgott

14:50

@Mad yup

Jakobian

About slowing down - I didn't mean to be rude but its slightly overwhelming, and additionally, I think slow ingesting of stuff means no mistakes in reasoning. But I get that you just want to, I am assuming here, learn this stuff as fast as possible.

Mad

@Thorgott Alright thanks

sharon_puthuparambil

@XanderHenderson One needs to prove that if f is sequentially lower semicontinuous, then f(𝑥)=sup𝑟>0inf𝑦∈𝐵(𝑥,𝑟)f(𝑦)

Ryder Rude

@Jakobian i just have this annoying habit to reply fast. i will work on it

sharon_puthuparambil

the inequality >= is trivial. I don't know how to prove the converse

Ryder Rude

14:52

it clouds the discussion

Jakobian

@RyderRude perhaps they meant what I wrote before then, without indexing the set $S$, then.

Or they decided it doesn't matter since homology groups are commutative

Ryder Rude

no, this is what they write : Take r elements x1,..., xr of G. The elements of G of the form
n1x1 +···+ nr xr (n_i ∈ Z , 1 ≤ i ≤ r) (3.7)
form a subgroup of G

Mad

@Thorgott i know you are well in lie theory, can you help me with my confusion about these two statements:
3. If g is semisimple, then any Cartan sub lie algebra h is maximally abelian. The converse is not true,however:
4. If h is abelian, and maximal among the Abelian subalgebras for which all ad_H are simultaneously diagonalizable, then h is cartan

But isnt being maximally abelian of ALL possible Abelian lie algebras a much stronger requirement than being maximally abelian of those sub lie algebras that fullfill the attribute that adh is diagonalizable, so shouldnt the converse be True…

Ryder Rude

@Jakobian yeah. i think they are somewhat careless about qualifying sentences by "abelian"

maybe becuz the general topic is about Abelians

Thorgott

@Mad sorry, I wasn't paying proper attention. this is wrong.

maximal just means there is no larger one

Mad

14:59

@Thorgott What do you mean larger?

Dimension wise?

Thorgott

with respect to inclusion

Jakobian

@RyderRude yeah. Not true in general

Mad

Thats the same as saying every other one is inside. isnt it not?

Brian C hall defines it like this: any X out of g such that [g,h]=0 then X is in h

Thorgott

@Mad no, consider the non-trivial 2-dimensional Lie algebra with basis x, y s.t. [x,y] = x

then x and y each span a 1-dimensional subalgebra that is maximally abelian, but neither is contained in the other

Mad

So can we define it like this: h is maximally abelian if there is no h' which is abelian and h \in h'? that would allow multiple "maximals"

Also my statement regarding Brian c hall is wrong i just noticed he is spesifically talking about Cartan sub algebras.

Thorgott

15:06

@Mad $h\in h^{\prime}$ doesn't make a lot of sense, it should be $h\subsetneq h^{\prime}$

Shashaank

What is (1/D)x = ? where D is the differential operator D := d/dx

Thorgott

and yes, there can be many maximally abelian subalgebras

Mad

yea i mean that. :D

So is that like completly a subset ? or can i have some elements of h be in h' but not all of them? these inclusion signs have always been amibgious. probably the latter

Xander Henderson

@Shashaank In what context are you asking this? Do you know anything about the spectral theorem or the functional calculus?

Shashaank

I am learning to solve differential equations using operators

I need relevant information to understand (1/D)x..

Thorgott

15:09

@Mad subset, but not equal

Xander Henderson

@Shashaank That only kind of answers the first question I asked...

Thorgott

or, put differently, any abelian subalgebra containing h is equal to it

Shashaank

@XanderHenderson functional calculus if that deals with solving differential equations..

Mad

@Thorgott Yes but i mean "completly" containing h right?

Thorgott

thats what "subset" means

Mad

15:11

I guess so. Thanks

Xander Henderson

@Shashaank (1) In what context are you asking the question? (Are you taking a class on functional analysis? an intro class on differential equations? What is your background?) Explain, in detail, where the question comes from. Give us some context.

(2) Are you familiar with the spectral theorem?

(3) Are you familiar with the functional calculus?

Shashaank

@XanderHenderson I am taking a class on solving differential equations ( intro ) also I am not aware of the spectral theorem or functional calculus, My prof just defined the Differential operator and I am solving some questions and I have come across (1/D)x which I am unable to understand

Mad

@Thorgott but that still doesnt really clear up the confusion, i mean even under this definition, i dont see how a lie algebra could be Maximally abelian and still not cartan.

@Thorgott wouldnt this imply that it is also maximally abelian under those that satisfy ad_h being diagonalizable and thus the other theorem would apply.

Xander Henderson

@Shashaank Perhaps you could provide the entire problem statement you have been given.

Shashaank

yeah

$(D^2 − 3D+2)y=xe^x$

sharon_puthuparambil

15:17

Can someone help me prove this?
If f is sequentially lower semicontinuous, then f(𝑥)=sup𝑟>0inf𝑦∈𝐵(𝑥,𝑟)f(𝑦)

Shashaank

I had to use exponential shift rule to solve this problem

Thorgott

@Mad consider the previous example. the subalgebra generated by x is maximally abelian, but not Cartan

@Mad well, you could be maximally abelian without every element h satisfying that ad_h is diagonalizable

Mad

@Thorgott Oh, so the condition of diagonalization is actually the stronger condition.

Thorgott

i dont think either is stronger a priori

Jakobian

15:44

@sharon_puthuparambil one inequality you havve for free

So the issue is to use definition of semicontinuity to prove the other inequality

Jakobian

16:13

@Shashaank $(1/D)x$ doesn't really make sense, but if you're looking for solution to Dy = x, then you can obtain one, modulo constant, by integrating

Shashaank

16:26

@Jakobian Oh

so is it x^2/2 ?

Dy = x solution?

Jakobian

+C for some C

A constant

Shashaank

Yeah Thank you, I got it @Jakobian

Jakobian

16:41

@Sahaj youtube.com/shorts/4t0kf_46460?si=mroxI4hGDoQfZblK

you

Pizza

16:59

hello!

Jakobian

@Thorgott context. I agree that the functor that assigns to a space its Alexandroff extension is important, but I don't fully agree with calling it a one point compactification of a space... just call it Alexandroff extension

Thorgott

I think it's fine

I've called non-compact spaces compactifications in some places

the meaning of that term is as contextual as anything else

but I do like Alexandroff extension

cause it also helps emphasize that the construction makes perfect sense in the non-locally compact setting when properly defined

Jakobian

well sure I wouldn't scrutinize someone for naming stuff, but I feel like this person is trying to force the name a bit too much when there are better alternatives

Thorgott

whereas people saying "compactification" usually subscribe to the idea that compact spaces are Hausdorff by deifnition

Jakobian

yes thats another reason why Alexandroff extension is a better name

Its odd that this is the only construction that I know of, that compactifies a space by one point

(in the sense that would be Hausdorff, and so unique, for Tychonoff spaces)

Xander Henderson

17:15

@Jakobian I have mixed feelings. On the one hand, I agree that simply calling something the "one point compactification" can cause problems, I also generally don't like to name things after people if a pithy descriptive name can be found. People-names can become obfuscatory. :/

Thorgott

you can always add a point whose only nbhd is the entire space to get a terrible "one-point compactification"

Xander Henderson

@Thorgott Right. And the one-point extension of a space that the Alexandroff extension provides needn't be compact.

Thorgott

no, the Alexandroff extension is always compact

it's not always Hausdorff

Jakobian

Alexandroff extension is always compact, I concur

Xander Henderson

Oh, sure.

It has been way too long since I've thought about basic topology.

Hausdorff. I live in a world where everything is Hausdorff.

I knew there was something that the Alexandroff extension didn't guarantee. But it is never relevant in any of the cases I care about.

Thorgott

17:19

the trick is that, in the non-Hausdorff case, the open neighborhoods of the point at infinity are not the complements of compact subspaces of the original space, but the complements of compact, closed subspaces of the original space

Jakobian

I think its better to live in a world where everything is Tychonoff. That way, everything has a Hausdorff compactification

Xander Henderson

If it's not Hausdorff, it's crap.

All Things Scottish - Saturday Night Live

►

Thorgott

I've recently been forced into taking a liking to weakly Hausdorff spaces

Jakobian

Other reasons to like Tychonoff spaces are 1. There is no loss of generality in considering rings of continuous functions from Tychonoff space to $\mathbb{R}$. And 2. regularity is very nice to have.

Pizza

The application $f_A : A \to T$ tells us that
$$f_A(x) = f (x) \quad∀x ∈ A$$ is called the restriction of f to the set A.

Is the restriction unique? Is the extension unique?

Xander Henderson

17:26

Well, suppose that there is another function $g : A \to T$ such that $g(x) = f_{A}(x)$ for all $x \in A$. Is it possible for $g$ to be distinct from $f_A$?

Pizza

@XanderHenderson wait with "distinct" you mean different?

Xander Henderson

Yes.

Pizza

ok my answer is no

Balarka Sen

@Thorgott whats your favorite non-LC hausdorff space

Pizza

right?

Sahaj

17:33

@Jakobian where did you find my videos !?

Thorgott

@BalarkaSen $S^{\infty}$

Balarka Sen

whats its one point compactification

>:)

Thorgott

quite ugly, I suppose

Balarka Sen

whats the one point compactification of $\Bbb Q$

easier question

Thorgott

not this space again

it haunted me just yesterday

Balarka Sen

17:35

lmao

nickbros123

how to correctly write statements like "$P(x)$ is true for utmost finite number of $x \in \mathbb{N}$", is that something like $P(x) \implies x \in F \ \text{where} \ \text{cardinality}(F) \in \mathbb{N} \cup \{0\}$

Thorgott

it's a compact space that is not the continuous image of any compact Hausdorff space

Balarka Sen

but what does it look like

Thorgott

also, its square is weakly Hausdorff, but not KC

Jakobian

@Sahaj you have some sick moves

Pizza

17:37

@XanderHenderson if the function $g(x)$ is always equal to $f_A(x)$, how can $g$ be different from $f_A(x)$?

so my answer is no

Xander Henderson

So what does that say about the uniqueness of the restriction?

Thorgott

the topology on $\mathbb{Q}^+$ is pretty awkward

Jakobian

$\mathbb{Q}^+$ is positive rationals?

Thorgott

no, Alexandroff extension

I'm not sure what exactly closed, compact subspaces of $\mathbb{Q}$ look like, probably all constructed (transfinitely?) from convergent sequences

in particular, all neighborhoods of the point at infinity should be dense

Pizza

@XanderHenderson it's unique

Jakobian

17:43

@Thorgott closed iff compact for $\mathbb{Q}^+$

3

A: Subspace closed iff compact implies Hausdorff?

Here's a counter-example. Let $X = \hat{\mathbb{Q}}$ be the one-point compactification of $\mathbb{Q}$. It's well-known that such compactification is Hausdorff iff the underlying space is locally compact and Hausdorff, so $X$ is not Hausdorff. If $Y\subseteq X$ is closed then it's compact as a cl...

In fact I've proved that this holds for Alexandroff extension of any metrizable space

Thorgott

yeah, it's a KC space, but its square is not lol

Xander Henderson

@Pizza Well, there you go.

Jakobian

of $\mathbb{Q}$... ah

they have empty interior for sure, they are countable

I can imagine if they all look sort of like $\{0\}\cup \{1/n : n\in\mathbb{N}\}$ iterated arbitrary (finite) many times

Thorgott

I think you can have a countable union of convergent sequences that's not expressable as a finite union of convergent sequences

Jakobian

You know, how I'd construct a set whose $n$th derived power is a point

Thorgott

17:48

but not every countable union of convergent sequences will be compact closed

robjohn

@Jakobian $\mathbb{Z}\cap\mathbb{Q}^+$

Jakobian

sure, $\{0\}\cup \{1/n : n\in\mathbb{Z}\cap \mathbb{Q}^+\}$

Pizza

oh wait

Jakobian

@Thorgott can you link me your question?

if you have one

(maybe you're just wondering about this)

Pizza

If $g : A \to T$ is a map of $A$ in $T$, every map $f : S \to T$ such that
$$f(x) = g(x) \quad \forall x \in A$$
it is called a prolongation of $g$ on $S$

Is the extension unique?

I forgot to write a part, sorry

this is it, anyway

Xander Henderson

17:56

@Pizza Think about it for a minute.

Work a concrete example.

Pizza

ok

Xander Henderson

For instance, suppose that $A = [0,1]$, $S = T = \mathbb{R}$, and that $g(x) = 47$.

Thorgott

I don't have a question, was just trying to respond to Balarka's comment

Pizza

@XanderHenderson ok ill try wait

Jakobian

@Thorgott I've actually never encounter this result in a book, the only time I ever seen it is when I figured out this could be a counter-example to that "compact iff closed" conjecture. So that's why maybe I take a little too much pride in it. Its not a new result so I shouldn't be

I guess this sort of thing appears in algebraic topology?

Thorgott

18:12

I've only ever seen it stated for $\mathbb{Q}$ either, it's nice tho

@Jakobian not really, I just happened to look for this counter-example yesterday

Jakobian

oh I see

it looks like one of those things that Patrick and people from pi-base would be interested in exploring (though knowing them its already there, along with 5 different variations)

Brian Scott has a lot of decent answers on the topic of topology, a lot canonical ones

@Thorgott The proof that product of $\mathbb{Q}^+$ with itself is not KC seems to just be this: Alexandroff extension of a metric space $X$ is Hausdorff iff $X$ is locally compact

So this actually has a quicker proof than what Brian proposes there

yeah PatrickR mentioned that too

So 1) Alexandroff extension of metrizable space is KC and 2) Alexandroff extension of $X$ is Hausdorff iff $X$ is Hausdorff and locally compact

If $X$ is not locally compact, then $X^+$ is not Hausdorff, so the diagonal $\Delta_{X^+}\subseteq X^+\times X^+$ is not closed.

so 2') Square of Alexandroff extension of a non-locally compact space is not KC

In particular taking any non-locally compact metrizable space $X$, $X^+$ is KC but $X^+\times X^+$ is not

Balarka Sen

18:35

@Thorgott whats a one-point compactification of a non-LC hausdorff space that you can see

Jakobian

maybe there is some connection to be drawn with compact-open topology as well, that would be nice

Alessandro Codenotti

@BalarkaSen is there such a thing?

Balarka Sen

@alessandro if not that should be a theorem

then one can put one-point compactifications of non-LC hausdorff spaces in the dustbin :P

Thorgott

@Jakobian nice

Alessandro Codenotti

What's the one-point compactification of $\ell^2$? I want to say it should be some kind of infinite dimensional sphere but that's probably wrong

Balarka Sen

18:46

that would be interesting

Thorgott

@BalarkaSen I don't have any, but that doesn't matter

it's a construction you make to have a certain property, not because it fulfills a geometric task

Balarka Sen

@Thorgott consider the set of twin primes, and define p + p' = (least twin prime occuring after p + p')

Jakobian

@AlessandroCodenotti You certainly can't connect it with vector spaces

Balarka Sen

this is a monoid, consider its grothendieck completion

that has a certain property

some constructions are inherently "meaningless"

even if they have a certain property

Thorgott

the Grothendieck construction certainly isn't meaningless

your monoid may be

Balarka Sen

18:49

same thing; one point compactification isnt meaningless

that of non-LC spaces may be

Jakobian

@AlessandroCodenotti I suppose if you take an injection $\ell^2\to S$ into the sphere $S = \{x\in \ell^2 : \|x\| = 1\}$ and replace topology of $S$ from subspace topology?

Then you would have some kind of picture of it

and by comparising topologies of $S$ and one-point compactification of $\ell^2$ you might understand this better

Thorgott

I think they're nice cause they allow us to make precise that a "sequence/net converging to infinity" is actually sequence/net that literally converges to a point called "infinity" in a larger space

Jakobian

this topology definitely won't be homogeneous though

a non-symmetric sphere

Pizza

@XanderHenderson the extension would be that if I compute f in the domain of g (f is different from the function g), do I find that the values of f are equal to those of g (the output values)?

Balarka Sen

Ok, sanity check. I have a norm on a vector space $V$ and three points $x, y, z$ such that the norm is linear along the segments $[x, y], [x, z], [y, z]$. Then its linear along the 2-simplex filling the triangle, yes?

EE18

18:58

OK the axiom of choice just HAS to be true, it just has to (tongue in cheek)

Jakobian

it might not be a problem with compactifying it using one point, it might be a problem with the definition of compactification not being symmetric with respect to all points in some sense, which turns out to be more intuitive in the non-locally compact space

EE18

I need AC to prove (b)!

Alessandro Codenotti

@BalarkaSen sorry, what does it mean for the norm to be linear along a subspace?

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