Mathematics - 2024-08-15

nickbros123

2:25 AM

In a metric space, does countable basis imply separability?

Jakobian

2:38 AM

@nickbros123 countable basis always implies separability

If you have a basis consisting of sets $U_n$, assuming all of them are non-empty, just take some $x_n\in U_n$ for each $n$. Then $\{x_n : n\in\mathbb{N}\}$ is your dense countable set

Its the other direction that only real holds for metric spaces.

nickbros123

4:59 AM

@Jakobian ah, I see.

Jakobian

9:16 AM

@monoidaltransform by the way, there is a finite metric space which doesn't embed in any Hilbert space. And a finite space is an ANR

my example has this advantage that you can't have an isometry $f:L^p\to H$ into any normed space $H$ linearly homeomorphic to a Hilbert space

since Mazur-Ulam theorem shows that we can assume $f$ is linear, and so $L^p$ is linearly homeomorphic to a Hilbert space

Type and cotype of a Banach space

In functional analysis, the type and cotype of a Banach space are a classification of Banach spaces through probability theory and a measure, how far a Banach space from a Hilbert space is. The starting point is the Pythagorean identity for orthogonal vectors ( e k ) k = 1 n {\displaystyle (e_{k})_{k=1}^{n}} in Hilbert spaces...

but if you look at the type and cotype of $L^p$, you'll notice that one of them is not $2$, and so $L^p$ is not linearly homeomorphic to a Hilbert space

porridgemathematics

Suppose I have a function of the form $r = \sum_{k} \frac{c_k}{(z-z_k)^2} + \frac{c_k'}{(z-z_k)} $ with $c_k , c_k'$ real and $ \sum_k |c_k| > 0$, also all sums here are finite and $z_k$ are distinct points on the unit circle. I’m not sure if it’s always true that the supremem of $(1-|z|^2)|r|$ over the unit disk is achieved at a boundary point, I.e at one of the poles.

I want to believe it’s true because if I were to increase the exponent $2$ to something only slightly larger then my super mum becomes infinite at a pole of $r$ on the boundary… but I don’t see a way to directly argue using this that the same is true for $r$ itself

I also think it may not be true though, but I can’t think of any simple counterexamples…

*supremum

Mr Gubbo

Can someone suggest a Delta calculator online, for a two degree curve? (Delta as in a^2 + b^2 + c^2 + 2fgh -af^2 -bg^2 -ch^2)

Jakobian

@porridgemathematics yeah, why is the supremum not infinite in general here?

porridgemathematics

@Jakobian it is finite because the double poles get cancelled by the (1-|z|^2) term

That cancellation fails when 2 is say 2.00002 though, and in that case it’s not finite

I basically want to know when we can say the supremum is still achieved at a pole

Jakobian

I don't think it gets cancelled, but it would if it were (1-|z|)^2 instead

porridgemathematics

9:31 AM

Wait crap, I meant (1-|z|^2)^2

That was a typo, apologies

Jakobian

that explains it

porridgemathematics

I thought naively we can figure this out by just computing hessians… to show there can’t be local maxima in the interior but I think that would be a parameter nightmare

There are some special cases where I think it’s feasible there could be interior maxima, like if the z_j are evenly spaced, and if all the coefficients are equal..

Then my gut feeling is 0 is an interior maxima

I want to say “in most cases” the supremum is achieved at the boundary, because it seems like we would require very symmetric configurations of c_k and z_j to get interior maxima in general

But I’m struggling to make this precise

Jakobian

@porridgemathematics by achieved, you mean that it can occur at the interior as well?

or is that something you are wondering too

If you let $r = \frac{1}{(z-1)^2(z+1)^2}$ then $(1-|z|^2)^2 |r|$ achieves maximum at $\text{Re}(z)\in [-1, 1], \text{Im}(z) = 0$

so it can occur in the interior as well

is your question: can we always find maximum on the boundary?

porridgemathematics

9:56 AM

It’s if we can say that for almost all z_k and c_k, c_k’ under the constraints given, the supremum is achieved on the boundary (not whether it is only achieved on the boundary)

Or more broadly under what constraints would the supremum be achieved somewhere on the boundary

Also thanks for that example!

psie

10:12 AM

I have read a proof that the closure of open balls in a normed linear space are closed balls, however, is this also true for $\mathbb R^n$ equipped with the discrete metric? It seems like the discrete metric is used to counterexample this very statement, and hence I am not sure. After all, $\mathbb R^n$ is a normed linear space.

oh wait

the discrete metric is not derived from a norm

ok, so $\mathbb R^n$ equipped with the discrete metric is not a normed space.

that answers my question, thanks

My book writes that the closure of open balls being closed balls holds in $\mathbb R^n$. Is this true? It seems like it depends on which metric we equip $\mathbb R^n$ with.

Shaun

10:34 AM

I've been targeted with serial downvoting for the last three days. I think it's because I ran for moderator. Anyway, the system hasn't corrected for it yet.

FShrike

@psie Under the discrete metric, an open ball $B(x,a)$ is either $\{x\}$ (if $0\lt a\le1$) or the whole space, if $a\gt1$. In either case, this is a closed ball as well. We have the closed ball of radius $1/2$ or the closed ball of radius $1$, say, as options to witness this claim.

And every set is closed in the discrete metric, so the closure of these open balls are the same thing again, and can also be written as closed balls

Oh! Or did you mean, you want $\overline{B(x,a)}$ to be the closed ball of radius $a$ specifically? In which case you’re right the discrete metric is a counterexample

psie

@FShrike yes, but your comments were helpful nevertheless. $\mathbb R^n$ can be both a metric and a normed space, right? so the closure of open balls are only closed balls if $\mathbb R^n$ is viewed as normed linear space. Is this observation correct?

I am basing my claim on that the closure of open balls are closed balls in a normed space on this answer.

FShrike

Every normed space is a metric space too, yeah. $\Bbb R^n$ is both (it is many things). But I disagree with the next claim. You know if you use the metric induced from a norm that the closed ball claim is true; you don’t know, and have not tried to prove, that if your metric is not induced from a norm the closed ball claim is false. So I disagree with the “only if”

Maybe it is true but I suspect not. At any rate, no argument has been presented for jt

psie

ok, I see👍

@FShrike do you know what property of a norm is needed to prove the closed ball claim? I have read the proof in the link I sent above, but I can not tell what key property about normed spaces was used that makes the closure of open balls closed balls.

porridgemathematics

10:55 AM

Ok @Jakobian I have a new more concrete question, is it safe to say that if z_j are distinct points on the unit circle, then it is very unlikely that sum_j z_j^3 u_j , where 0<u_j<1 sum to 1 , remains on the unit circle?

Seems to be intuitively true, is it easy to show that actually outside a null set of n tuples of points on the unit circle, the convex combination does not stay on the unit circle?

Sorry ignore that question

What I wanted to ask is harder: how likely is it for sum z_j^3 u_j and sum_{j,k} u_j u_k (z_j z_k)^{-1} to coincide, with the same setup

Shaun

12:32 PM

Hi :) I'm doubting myself a lot lately. My 2nd year progression is coming up, which decides whether I go into my 3rd year of my PhD. My supervisor thinks I'm doing alright but I have no big theorems yet; just four small ones that I don't think are very good :/

monoidaltransform

@Jakobian woah that is very weird. what's the idea behind this? I find it really bizarre.

FShrike

@psie It is more the overall structure of a normed vector space. You can draw lines, and by scaling vectors, and use homogeneity of norm to give these lines a specific length. In this way you can construct points arbitrarily close with an open ball, on the boundary. In a general metric space you actually have no control on what points do and don’t exist; normed spaces have much more structure

ephe

How do you actually align something like this?

FShrike

@Shaun I know PhD students who had “nothing” until their final year. I imagine your situation is not uncommon. But best wishes.

2

ephe

I mean the part with |\delta_i^n|. I am about to go mad trying it

Shaun

12:35 PM

@FShrike Thanks :)

Thorgott

1:01 PM

@ephe what do you mean by "align"?

ephe

@Thorgott you know how |\delta_i^n| ends right where the (t_0,) etc. begins at the lower level and how the (t0)'s are aligned with the domain codomain

When I try it, I just get weirdly seperated pieces

Thorgott

1:41 PM

try what? what is the actual question?

ephe

@Thorgott It's a latex question :\

Thorgott

ohhh

sorry, I totally misunderstood

I'd assume they used an align environment

ephe

@Thorgott That is what I figured but the naive way of just putting a & for everything you want to be aligned like a table just gives huge seperations between some characters because latex tries to span the whole thing evenly

Sahaj

1:57 PM

Happy independence day to all my Indian friends. @nickbros123 @SoumikMukherjee

Ryder Rude

2:39 PM

suppose Q is a propositional consequent of P

does this mean that, in the truth table, $P\to Q$ is always True, or that $P\to Q$ has the entries of the implication operand?

hi

nvm it needs to be a tautology. So always True

Minsky

Imagine an arbitrary function f and a neural net with a single hidden layer but infinite neurons in that layer (and use ReLU or max(0,x) as activation).
Now think of f as broken up into small linear segments that approximate it locally.
If the NN has infinite neurons in that single layer, then it can approximate each of the little line segments.
This is very clear since the ReLU makes all the rest 0 besides the new weight, which corrects the previous ones.
Is this correct? (at least directionally)

Ryder Rude

@Minsky hi.. long time

Minsky

@RyderRude hi there

Ryder Rude

@Minsky u r not on the physics chat anymore..

Minsky

no i was blocked :-(

Ryder Rude

2:53 PM

what?? Who blocked u?

Minsky

i don't remember, it was quite harsh, i felt really bad and decided not to participate anymore; however i do like the challenging discussions in SE forums so im here and there, but not on PSE

Ryder Rude

really sad....

@Minsky but good that u r on SE :)

Minsky

thanks ! yeah always learning the basics XD

Ryder Rude

Did they ban u on the main PSE site? can u no longer access the physics chat?

@Minsky great :D same

Minsky

@RyderRude it was only in the chat, but i remember it was completely undeserved; it was for 30 days i think, but dont wanna participate anymore now

Ryder Rude

2:57 PM

@Minsky okay :( wont insist then

really wild they blocked u...

@Minsky are u still learning physics?

Minsky

@RyderRude not so much, back then it was mostly from a random book, now just doing deep learning, i find it interesting; what about you?

Ryder Rude

@Minsky i am learning propositional logic..

@Minsky yeah u have posted a question about it

i will star it so it doesn't get buried

Minsky

@RyderRude sounds interesting but not gonna lie, i don't not what it is (apart from the very basics.)

thank you :-)

Ryder Rude

@Minsky I'm at the very basics rn :P

just truth tables

Minsky

nice

Ryder Rude

3:03 PM

but this topic is always basic i guess... it is like a foundation to logic

i just started learning it.. hopefully, i can finish it

Minsky

@RyderRude yep, good luck !

ill be off now; it's good to see u around :-)

Ryder Rude

same to u. see ya !

nickbros123

3:35 PM

@Sahaj wish you the same!!

Jakobian

@monoidaltransform take 3 points, all distance 1 away from each other. Then take a 4th point distance 1/2 from those 3 points. If there were such an isometry, then the 3 points would form an equilateral triangle. But then there can be no point which is distance 1/2 apart from the 3 points

This happens because if $v_1, v_2, v_3$ are the $3$ points and $w$ is the $4$th, then strict convexity implies that $w$ is the unique element in $\overline{B}(v_1, 1/2)\cap \overline{B}(v_2, 1/2)$ so that $w = (v_1+v_2)/2$ and so on. Which then implies $v_1=v_2=v_3$ which is a contradiction.

So this metric space can't be isometrically embedded in any strictly convex space

@porridgemathematics again, grouping the $z_i^3$ and $z_j^3$ such that $z_i^3 = z_j^3$, we obtain from strict convexity that for at least four points it's in the interior, and for $3$ or less that it can only happen to be on the boundary when all the points third powers are equal

So unless of course there's just one point in the sum, it happens almost surely

Soumik Mukherjee

4:06 PM

@Sahaj sorry but I don't celebrate it. I don't think the country is independent yet. (However I do acknowledge the sacrifices made by our freedom fighters)

Jakobian

@porridgemathematics I expect it to hold a.e. as well but no argument for now

nickbros123

7:12 PM

@SoumikMukherjee did u buy any chance give the NBHM exam?

psie

7:39 PM

I am studying the Baire Category Theorem, and why the assumption of completeness is essential. Consider $\mathbb Q$. Arrange the rationals in a sequence $(s_n)$ and put $U_n=\mathbb Q\setminus \{s_n\}$. Then it is claimed that $U_n$ are dense in $\mathbb Q$, but $\bigcap_n U_n$ is empty. I feel silly for asking, but why is it that $U_n$ is dense in $\mathbb Q$?

Ok, I am silly. The closure is the union of the set and its boundary point. Clearly, the boundary point of $U_n$ is $\{s_n\}$.

And $U_n\cup \{s_n\}=\mathbb Q$.

leslie townes

yes. any rational you like is the limit of a sequence of other rational numbers, e.g. for any fixed n, s_n is the limit of the sequence a_k = s_n + 1/k and all of these example a_k's are (by the hypothesis that the sequence s enumerates all of Q and the fact that each of the a_k's is, by its defining formula, in Q and not equal to s_n) in Q \ {s_n}

psie

ah, true

Soumik Mukherjee

@nickbros123 Are you asking if I have given the exam?

Then yes, I wrote the exam twice, first time I couldn't clear the written phase. Second time I qualified the written test but couldn't clear the interview.

Jakobian

7:58 PM

@psie because rationals have no isolated points

psie

👍

Jakobian

By the way the theorem of Baire holds more generally

It's only needed that your metric space contains a dense subset which can be given a complete, equivalent metric

Equivalent in the topological sense

psie

interesting

Jakobian

what's interesting is that there are subsets of $\mathbb{R}$ which are Baire spaces, but don't contain a dense, (topologically) complete subset

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