Mathematics - 2022-10-03

if i am solving a problem that requires serious thought i need quiet, but for most things some non invasive tune is nice. wellerman in repeat for example :-).

Obliv

3:14 AM

I like music when doing chill/low effort work. I prefer not to be under massive pressure generally because I typically cave and run away

But yeah as an engineer you probably have some pretty important and complicated stuff you work on.

copper.hat

important no. complicated sometimes.

Obliv

What did you mean by if I learn my prof's name and how to pronounce it properly it'll help me focus?

I think I will put the effort in to do it, out of respect for him now, you had the right call.

copper.hat

i was kidding really, you wrote "my comprehension on the other hand"

but in general people appreciate being treated with dignity

this has been my experience even in some pretty delicate situations

Obliv

unless they're depraved, i tend to agree.

(in a consensual adult way)

i probably shouldn't have elaborated.

copper.hat

:-)

@Koro did you see my comment?

Koro

3:29 AM

Copper: yes, thanks for the comment. E need not be measurable.

The question probably needs a revision.

copper.hat

measurability seems to have featured a lot in the last few days...

Ted Shifrin

Two more months to endure, copper.

copper.hat

:-).

there was that conditional probability thingy the other day

Ted Shifrin

That too

Koro

4:05 AM

@copper: It's proposition 3.11.2 in this book amazon.in/Introduction-Measure-Integration-Inder-Rana/dp/….

This is only for the record.

I think that the second part of the theorem is not correct. Because as you said that would imply E is measurable, which need not be the case. @copper.

copper.hat

4:41 AM

Thanks @Koro

porridgemathematics

6:36 AM

i have a feeling this has a simple answer: math.stackexchange.com/questions/4544085/… , anyone want to take a look?

basically what is stumping me is I dont know how to go from the jordan forms of two such matrix logs being equal implies the matrix logs themselves are

under circumstances where we are not dealing with matrix logs of the same fixed matrix $C$ this is obviously false

so I am guessing there is some relationship between change of basis matrices for the jordan blocks of two matrix logs of $C$ and ta change of basis for $C$ turning it into a jordan block

or put another way, I think my question boils down to, assume $C$ itself is already in jordan form (or even more simply just assumes its a single jordan block), now can you show $M$ has to be unique under the eigenvalue condition?

Koro

7:04 AM

I'm confused again.

Suppose that $E$ is Lebesgue measurable, then given $\epsilon\gt 0$, there exists an open set $G_\epsilon\supset E$ such that $\lambda^*(G_\epsilon\setminus E)<\epsilon$.

If $\lambda ^* (E)<\infty$, then OK.

If $\lambda^*(E)=\infty$, then $\lambda^*(E_n)<\infty$ for all n, where $E_n:=E\cap (n, n+1]$.

So given $\epsilon>0$ there is $G_\epsilon\supset E_n$ such that $\lambda^*(G_\epsilon\setminus E_n)\lt \epsilon$.

Since $E_n\subset E$, we have $G_\epsilon\setminus E\subset G_\epsilon\setminus E_n$, whence $\lambda^*(G_\epsilon\setminus E)\le \lambda^*(G_\epsilon\setminus E_n)<\epsilon$

My question is: is my proof for $\lambda^*(E)=\infty$ correct? I ask because the book proves it a complicated way.

porridgemathematics

its not correct because your $G_{\epsilon}$ only includes $E_n$, it doesn't include $E$

you're using $G_{\epsilon}$ twice, and in both cases they mean different things

and so your penultimate line also is incorrect

Koro

@porridgemathematics oh yes. Thanks a lot.

:-)

Koro

7:27 AM

To fix this, I do the following: Write $E=\cup E_i, \lambda^*(E_j)<\infty, E_i\cap E_j=\emptyset$ for all $i\ne j$. Fix j and $\epsilon>0$. There is open set $G_j\supset E_j$ such that $\lambda^*(G_j- E_j)<\frac \epsilon{2^j}$. $\lambda^*( G- E)=\lambda^* (\cup G_j- E)\le \lambda^* (\cup_j (G_j-E_j))\le \sum \lambda ^* (G_j-E_j)\le \epsilon$

si-LV-er_and_b-LA-ck

1:37 PM

5

Q: Why does math need to be practiced and exercised, when L1 Linguistic Competence is subconscious?

If 'linguistic knowledge is largely subconscious'3, why isn't math? Most math instructors sermonize solving exercises and problems. But a student challenged why students need practice — because most students and adults don't need to know, or practice, any linguistics or socio-pragmatics to speak ...

exercises

leslie townes

interesting question. i note that human languages allow for some degree of 'technically incorrect but nevertheless understandable' and math sorta doesn't

or if it does some of the time, it doesn't enough of the time to prevent someone who can't be exactly correct from attaining what we would recognize as competence

Koro

3:04 PM

Given $E\subset U, F\subset V$. How is it 'super' obvious and highly intuitive that: $V\cap (x+U)\subset (x+U-F)\cup (V-E)\cup (E\cap (x+F))$?

correction: Given E⊂V,F⊂U

leslie townes

3:26 PM

koro in my very humble opinion anything that requires writing things down is not super obvious and highly intuitive. but i maybe also don't see much of a reason for vetting third party claims that things are super obvious and highly intuitive. a person who says that is probably far more careless with their words than i am.

and i say all kinds of irresponsible, goofy stuff

Thorgott

take $v\in V\cap(x+U)$ and $u\in U$ s.t. $v=x+u$. if $v\in V-E$, $v$ is in the RHS. if not, then $v\in E$. if $u\in F$, then this means $v=x+u$ is in $E\cap(x+F)$, whence in the RHS. and if not, then $u\in U-F$ and $v=x+u\in x+U-F$, whence in the RHS.

so I'll grant obvious, but I don't see what's intuitive about this. it's just symbol-pushing. granted, I also don't know the context.

leslie townes

stuff that invokes a mental picture of a venn diagram is not intuitive. i will die on this hill.

Koro

The book uses this in proving that $x\mapsto \lambda^*(E\cap (F+x))$ is continuous, where $\lambda^*(F)<\infty, \lambda^*(E)<\infty$.

leslie townes

weirdly, i find that intuitive.

Koro

E and F are given to be measurable (Lebesgue).

leslie townes

3:31 PM

even though i would not find the proof intuitive.

Koro

Leslie: the set inequality follows from Venn diagram?

:(

I find Venn diagram intuitive in the sense that it gives a picture to show the truth quickly.

leslie townes

i meant venn diagram in a generalized sense. if i have to consider cases based on where something is, that's venn diagram-ish to me.

and i do insist that they are not intuitive.

Koro

Then, the above theorem is used to prove Steinhaus theorem as a corollary. In fact, the book calls the above theorem Steinhaus theorem.

leslie townes

3:45 PM

nothing intuitive about that.

robjohn

@Koro By $(x+U-F)$, do you mean $(x+(U-F))$ or $((x+U)-F)$? I think it is probably the former.

the former

see? non-intuitive.

$\ddot\smile$

4:13 PM

you need better books

leslie townes

5:19 PM

twitter.com/CihanPostsThms/status/1576972957678587908

Akiva Weinberger

Define $\dot+$ to be the operator $x\dot+y:=\dot x+\dot y$

Xander Henderson

5:52 PM

@AkivaWeinberger No. I refuse. Never.

Alessandro Codenotti

6:24 PM

Hey @Akiva any news on your distinguishing R^n's in the first order language of posets?

Ted Shifrin

6:36 PM

Hi, demonic @Alessandro.

Fargle

Oh hey, it's the usual suspects

Akiva Weinberger

@AlessandroCodenotti No

Also I'm no longer confident that my original idea for singling out R^3 works

but I do think inductive dimension is probably workable

Ted Shifrin

Heya @Fargle

Fargle

How goes it @Ted?

Alessandro Codenotti

6:56 PM

Hi @Ted

I think I can distinguish R^n+1 from R^n with 2n+some constant quantifiers more or less by writing "having covering dimension n"

But that requires figuring out what's the smallest size of a cover of R^n+1 that witnesses dim>n (I think it might be as low as n+2)

Ted Shifrin

Still alive, @Fargle, and you?

Fargle

Likewise. Struggling with a physics problem at the moment, but of my own volition, at least.

I'm finding that it's possible that this textbook is in error, but I'm not actually too certain---it's also possible I'm not understanding what the question is trying to ask.

Ted Shifrin

8:03 PM

I doubt I should ask the nature of the problem.

Overtherainbow

I have the following definition of separable: We say that a space X is separable if there exists a countable subset S such that $\bar S=X$.

Using this I want to prove that $l^p$ is separable. Would it work to take $S=\{(x_n)_n: x_n\in \Bbb{Q}: \sum_n|x_n|^p<\infty\}$? I only want to know if I'm on the right track or if I'm totally wrong. So my idea is to use that $\Bbb{Q}$ is countable, that's why I get my S.

anak

@Overtherainbow I'd be careful since that's not "obviously" countable.

Overtherainbow

@anak so clearly I need to check this, but would this S work or is there a thinking error and I need to chose another S?

copper.hat

Try something similar but more 'finite'

anak

e.g. $\{(x_n)_{n\in\mathbb N} \mid x_n\in\{0,1\}\}$ is the set of binary sequences, and it can be shown to be in bijection with the subsets of $\mathbb N$ (using characteristic functions). What is the cardinality of the power set of $\mathbb N$?

Xander Henderson

8:12 PM

@copper.hat "more 'finite'"...

Like... 1?

Koro

did you try considering sequences with rational terms with tail 0 ?

anak

@Koro a little bit too heavy of a hint, I think copper.hat's would have gone far enough. :P

copper.hat

working on my subtlety

Overtherainbow

@copper.hat okey since I now read koros hint I will try it with sequences such that $x_k=0$ for $k\geq N$ for some $N$

Ted Shifrin

@Koro You need to learn to shut up.

Koro

8:16 PM

"all theorems seem to be saying the same thing in measure theory" read a post on reddit.

copper.hat

link?

Akiva Weinberger

@AlessandroCodenotti I think I can refer to the complement of an open sets's boundary (given that those are always closed)

Koro

@anak I was not completely sure if that would work so I thought I gave only a wild guess while writing my comment.

@Ted

Overtherainbow

@TedShifrin but can I ask where my original S would fail?

Akiva Weinberger

Basically I can expand the language to include closed sets and then translate back at the end by indentifying closed sets with their complements @AlessandroCodenotti

and that's good enough for inductive dimension I think?

That's a good question - what's the most elegant definition of the boundary complement $\partial'$ in this language

copper.hat

8:21 PM

@Overtherainbow it is not countable

Akiva Weinberger

(write $\partial'X$ for ${\rm all}\setminus\partial X$)

Koro

@copper.hat reddit.com/r/learnmath/comments/jl5ccg/comment/gan8bb1/…

copper.hat

thanks!

gee, real analysis is hard. why would one expect measure theory to be any easier?

Akiva Weinberger

@Koro I'm curious what specifically they're doing in their class

It might be stuff I didn't learn in my real analysis class

My real analysis class was spring 2020 and so partway through the semester we had an abrupt shift to virtual

and my ability to focus on the class plummeted

Overtherainbow

@copper.hat but I mean the only thing which changes between my original S and the one Koro speaks is that the sequences have only finately many terms. But doesn't countability means that the set is either finite or there is a one to one correspondence to $\Bbb{N}$? But then I don't see why this depends on the "lenght" of a sequence.

Akiva Weinberger

8:24 PM

but then the university went universal pass/fail so it was OK in the end for me, grades-wise

copper.hat

@Overtherainbow trun on your music and do a little thinking. also note @anak's comment above.

Overtherainbow

@copper.hat okey thanks!

Koro

and also note anak's comment.

copper.hat

@a equivalence class :-)

Koro

or monotone class :-)

copper.hat

8:26 PM

that is where measure theory gets a bit opaque for me...

its like universal properties...

i have some intellectual limit on generality

ok, back to paying work :-(

Ted Shifrin

@Overtherainbow show that the set of sequences of 0s and 1s is uncountable.

Koro

what is low dimensional topology?

Alessandro Codenotti

I think it's easier to do covering dimension @Akiva I'll write some details tomorrow. It's time to sleep now for me

anak

Topology in lower dimensions, @Koro

Usually 3-5

In particular, 3 and 4 are the most interesting ones.

Akiva Weinberger

@AlessandroCodenotti Interesting. I will be curious to see it

Incidentally, I think I have a characterization of the boundary complement

Alessandro Codenotti

8:35 PM

I don't think you can write "the space has covering dimension n", but you can definitely write "no cover of size k witnesses that the space has dimension more than n" for all pairs of k,n, and this is enough to distinguish the various euclidean spaces

Akiva Weinberger

Suppose we have a topological space $S$ with an (open) subset $X$. Let $\phi(Y)$ be the property that for all open sets $A\subsetneq S$ and $B\subseteq S$, if $Y\subseteq A$ and $A\cup B=S$, then $B\cap X\ne\emptyset$. Then I believe $\phi(Y)$ holds iff $Y=S\setminus\partial X$

@AlessandroCodenotti Interesting. Sounds plausible

Jakobian

@Koro that's the solution, yep

Alessandro Codenotti

I'm leaving now but I'll write more tomorrow

Koro

thanks anak. So it's something that I don't yet know.

Akiva Weinberger

@AlessandroCodenotti It's interesting to imagine the corresponding Ehrenfeucht–Fraïssé game

anak

8:40 PM

@Koro The idea behind it is that in lower dimensions, certain phenomena don't behave as nicely as they do when the dimension is large. For example, knots in 3-manifolds, and exotic structures in 4-manifolds, as well as the h-cobordism theorem and stuff like that.

Akiva Weinberger

@AlessandroCodenotti which you can learn about here:

Introducing Model Theory with Ehrenfeucht-Fraïssé Games on Linear Orderings #SOME2

►

Overtherainbow

@copper.hat would you show that S is countable using the cardinality or finding a bijection?

Because I mean if I fix $N>0$ and chose $(x_n)_n$ to be a sequence such that $x_n=0$ for $n> N$ then I know that $(x_n)_n=\{x_n:n\in \Bbb{N}\}$ is a finite set. But then since $\Bbb{Q}$ is countable is it correct to deduce that then S is countable?

anak

9:00 PM

What do you know about the cardinality of unions of families, @Overtherainbow?

Overtherainbow

@anak there is like a long inclusion exclusion formula. So I mean for two sets it is $|A\cup B|=|A|+|B|-|A\cap B|$ and this can be generalised for countably many unions

so in particular for example in case with two sets $|A\cup B|\leq|A|+|B|$

Fargle

@TedShifrin Simple orbital mechanics problem. I think I figured it out.

anak

9:16 PM

@Overtherainbow try thinking about infinite families

More Anonymous

Hey physics person here. Could someone tell me what's going on in wiki? in this section?

Congruence (manifolds)

In the theory of smooth manifolds, a congruence is the set of integral curves defined by a nonvanishing vector field defined on the manifold. Congruences are an important concept in general relativity, and are also important in parts of Riemannian geometry. == A motivational example == The idea of a congruence is probably better explained by giving an example than by a definition. Consider the smooth manifold R². Vector fields can be specified as first order linear partial differential operators, such as X →...

The example

Overtherainbow

@anak shouldn't I think about countably infinite families?

anak

9:31 PM

@Overtherainbow That's a great start!

Overtherainbow

@anak but the inclusion-exclusion formula does only hold for countably finite families right?

anak

@Overtherainbow yeah, so you have to do something different, so you can toss that idea out for now.

Overtherainbow

@anak So I still have that $S=\cup_{x_n\in \Bbb{Q}} \{(x_n)\} and I want to find $|S|$

anak

Sorry, I don't quite understand your S there.

Overtherainbow

9:46 PM

so I defined my S from the beginning to be $\{(x_n):x_n\in \Bbb{Q}, x_j=0 for j>N and \sum_n |x_n|^p<\infty\}$.

geocalc33

10:07 PM

How do you find the solutions (streamlines and integral surfaces) for $W(x,y,z)=\big(x\log x,-y\log y, -z\log z\big)$ for $0<x,y,z<1?$ I know that it's a system of seperable equations and I can solve the system in one lower dimension i.e. for $W(x,y).$

but with the third component it's throwing me off

Ted Shifrin

10:39 PM

What you mean solve for $W(x,y)$? You’re solving for the glow/integral curves. There’s nothing different with having the third component.

FLOW … stupid typing.

anak

11:23 PM

@Overtherainbow So $S$ depends on $N$?

copper.hat

11:36 PM

@Overtherainbow Think of $\mathbb{N}^2$ as an countable number of copies of $\mathbb{N}$. It is still countable.

Ted Shifrin

@MoreAnonymous I’ve never in my 50 years as a geometer heard this term. It sounds like a foliation by curves (perhaps with a few singular points thrown out). Can you say specifically what your question is?

leslie townes

solving for the glow, i like that

Ted Shifrin

11:55 PM

glowers at leslie

copper.hat

should be flowers at leslie

no easy pickin's for my jump suit rep hunting...

i need to answer an $-{1 \over 12}$ question, i think.

Ted Shifrin

I ain’t gotten nada!

copper.hat

:-). the use of double negative always throws me.

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$Mathematics$ Mathematics