Mathematics - 2020-01-20 (page 1 of 3)

12:16 AM

@AkivaWeinberger DogAteMy: I hadn't known about that. Wiki mentions his "deteriorating mental condition." Maybe it's a death wish to study stat mech?

Akiva Weinberger

I don't know any more than you do

anakhro

Ted, I hate to bother you again, but residues are killing me. I am wanting to determine what $\frac{1}{2\pi z}\int_\gamma \frac{z f'(z)}{f(z)-a}\,dz$ counts, as in the 5th question here: its.caltech.edu/~ivrii/practice-exam.pdf

Ted Shifrin

Your textbook author is suggesting you might be in trouble, DogAteMy.

anakhro

But this means I have to determine residues of solutions to $f(z) = a$ of that function $\frac{zf'(z)}{f(z)-a}$

But I don't know how to start approaching that.

Ted Shifrin

It adds up the roots of $f(z)=a$, @anakhro.

Akiva Weinberger

12:19 AM

@TedShifrin Clearly

anakhro

So that means I have to show that both the residue and the winding numbers are 1?

Ted Shifrin

Well, if $z$ is a simple root of $f(z)=a$, what is the residue (according to what I told you earlier)?

I'm assuming $\gamma$ is known to be a simple curve that winds once around all the roots. (Or else you add up the roots appearing inside $\gamma$.)

anakhro

If it is simple, then it is $a$.

err

the root

Ted Shifrin

Right, the root.

anakhro

So then I have to consider higher order roots?

And I suppose there is a formula for this I should find in a book

Ted Shifrin

12:23 AM

Now, if $z_0$ is a double root, say, then what will the Laurent series be there?

Hint: Let $u=z-z_0$.

Ignore all the stupid formulas with derivatives. I hate those.

anakhro

Where $u$ is a change of variables?

Ted Shifrin

The Laurent expansion is naturally going to be in $u$, right? That's easier than writing $(z-z_0)$ everywhere.

anakhro

I am afraid I am lost. Maybe I need to read up on Laurent expansions again.

Ted Shifrin

OK.

Or Taylor.

anakhro

So I let $u=z-z_0$ and write $\frac{zf'(z)}{f(z)-a} = \frac{(u+z_0)f'(u+z_0)}{f(u+z_0)-a}$

I feel like just naively filling it out has gotten me in a rut

Normally there is something to use the power series for.

But because of this f'/(f-a) stuff I am not sure how I would go about it in the normal way.

Ted Shifrin

12:34 AM

But use that $z_0$ is a double-root of $f(z)-a$.

Write $f(z)-a = (z-z_0)^2 g(z) = u^2 g(u+z_0)$.

anakhro

Should I avoid expanding everything out? I took the liberty of also determining $f'$ with this

Ted Shifrin

Yes, avoid. We're just trying to find the coefficient of the $1/u$ term.

Akiva Weinberger

Curious what your thoughts are on this. I found several not-obviously-equivalent solutions:

> Hardest problem on 1st year mid-term logic and sets was: "Find a countably infinite collection of sets such that any two intersect in exactly one element, and any three in none." What's the first solution you can think of (that fits in a tweet)? Students found several. — Andrej Bauer on Twitter

(Not a test I took, to be clear)

Ted Shifrin

Hmmm ... Seems like a reasonable question. I don't have an automatic response.

anakhro

I am not sure what I am trying to do algebraically. Certainly not just pull a term 1/u out.

Ted Shifrin

12:43 AM

Well, depends on the level of the students and course, I guess. I would never have put that on an exam for our Intro to Higher Math course ... unless we'd done it in class.

The denominator has a $u^2$, @anakhro, so, as usual, you want the $u$ coefficient upstairs.

anakhro

I have something that looks like $$\frac{(u+z_0)(2ug(u+z_0) + u^2g'(u+z_0))}{u^2g(u+z_0)}.$$

Ted Shifrin

OK, so what is the coefficient of $u$ upstairs?

anakhro

I can get rid of one u in the top and bottom

So it is $\frac{(u+z_0)(2g(u+z_0) + ug'(u+z_0))}{ug(u+z_0)}$

Ted Shifrin

Now what's the constant term upstairs?

And what's the resulting coefficient of $1/u$ in that thing?

anakhro

$2z_0g(u+z_0)$

Ted Shifrin

12:46 AM

So we get $2z_0$. Bingo.

Double root.

I leave it to you to generalize to $m$th order root.

Akiva Weinberger

There's a reasonable "canonical answer" and some clever answers

anakhro

How do you conclude $2z_0$ from that? How does the $g$ part disappear?

Ted Shifrin

Look at the denominator.

Hmm @DogAteMy.

anakhro

OH

I see.

Ted Shifrin

You should practice doing computations like this ... if you're preparing for a complex analysis qualifying or something.

anakhro

12:50 AM

Yeah, definitely. I 100% agree.

I just feel so pressed.

Ted Shifrin

Practice is the only way to improve speed.

anakhro

Yeah. I just wish I didn't have to take so many courses so I would have more time to practice.

Ted Shifrin

DogAteMy: Something indexed by primes, maybe?

anakhro

Thanks for your help, though!

I appreciate it ever so much.

Ted Shifrin

Anyhow, you should only take a few seconds to do the $m$th order case, @anakhro, now.

anakhro

12:53 AM

It should follow in a similar manner, or do you mean just guessing the answer?

Ted Shifrin

Both.

anakhro

Well I think I have the guess in a second, let me try the algorithm.

Yeah I did it

Heh

Ted Shifrin

DogAteMy: What is my set $A_p$ is the set of all prime multiples of $p$?

See, @anakhro: Told you so :P

anakhro

I thought I was in for some real hard thinking but the moment I started I realized what changes.

Akiva Weinberger

So you have $A_p=\{pq:q\text{ prime}\}$ for prime $p$?

Ted Shifrin

12:55 AM

Yes.

Akiva Weinberger

That works

Ted Shifrin

I take it that's not the canonical answer.

Akiva Weinberger

It's very close, it's like one step removed

Ted Shifrin

Hmm ...

Akiva Weinberger

I never said these sets are subsets of $\Bbb N$

Ted Shifrin

12:56 AM

Yes, I realize.

That seemed a natural (pun intended) starting place to me.

Oh. Axes in $\Bbb R^\omega$.

This is a great homework question, but I wouldn't have put it on an exam on an introductory course.

Akiva Weinberger

Elaborate?

Ted Shifrin

Oh, no, what I said is dopey.

I have to look at affine subspaces.

Akiva Weinberger

You know, since you already gave an answer I might as well tell you it

Ted Shifrin

Well, don't ruin it for everyone else.

Akiva Weinberger

OK I won't then

Take "$A_p=\{pq:q\text{ prime}\}$ for prime $p$" and get rid of the number theory

Ted Shifrin

1:06 AM

LOL, how do I get rid of number theory in that?

Oh, $y=n(x+1)$.

Akiva Weinberger

I'm thinking of $A_m=\{\{m,n\}:n\in\Bbb N\}$ for $m\in\Bbb N$

Bob

@TedShifrin I was wondering if you used LaTex for the "book" on differential geometry

Akiva Weinberger

$A_n$ is the set of unordered pairs of integers containing $n$

Ted Shifrin

@Bob, yes, for all my books.

DogAteMy, so the affine lines in the plane do it.

Ah, so you're sticking to integers, but maybe it's the same idea.

Akiva Weinberger

@TedShifrin Don't they all contain $(-1,0)$?

Ted Shifrin

1:10 AM

I see, unordered pairs, rather than ordered. Sneaky.

Oh, posh. How 'bout $y=nx+1$, and remove $(0,1)$ from the plane?

Bob

@TedShifrin thanks

Ted Shifrin

OR my example removing $(-1,0)$.

@Bob: In truth, had LaTeX not been around, I never would have written any books. Probably would have been a lot better for my research paper production, but oh well.

anakhro

Do you wish you had done more in research?

Or do you like the balance you struck?

Ted Shifrin

My one regret is that my work with two other guys (where we produced a bunch of papers) had one more giant paper on which we had worked for years. But we were all daunted by the realization that we'd have to write 3 50-60 page papers to incorporate what we'd done, and none of us had the energy to do that.

Each of us had multiple folders full of notes.

I enjoyed collaborative work a lot.

But in honesty I know my books (and YouTube lectures) are impacting orders of magnitudes more people than the research.

4

(Not to mention generating hatred.)

Akiva Weinberger

@TedShifrin I don't quite understand. Where do $A_1$ and $A_2$ intersect?

if not at the point you just removed

Ted Shifrin

1:16 AM

Oh crap @DogAteMy. I need to make parabolas instead.

Or make the intersection point move, which was my intention.

Duh. $y=n(x-n)$.

Yes, I was trying to make the intersection point move. I think I succeeded. :P

Akiva Weinberger

So where do $A_m$ and $A_n$ intersect

$m(x-m)=n(x-n)$

Ted Shifrin

They're non-parallel lines, so they have to intersect.

And surely that point won't work for a different index :D

Akiva Weinberger

They intersect at $(m+n,mn)$, I think

which works

Ted Shifrin

Or I can resort to primes again :P

Thorgott

My idea would be something like this: Put a line with positive slope through the origin in R^2, move a bit to the right, add a line with a steeper slope going through that point, continue. Don't have the time to work it out formally right now, but I think it should work if one makes the increases gradually smaller.

Ted Shifrin

1:20 AM

Yeah, $(m+n,mn)$ is right.

Akiva Weinberger

That's basically what he just did I think @Thorgott

Ted Shifrin

That's effectively what I've done, @Thorgott.

Oops.

Akiva Weinberger

These are the lines through $(n,0)$ with slope $n$

Thorgott

Oh lol, I didn't try to visualize what you were posting :p

Ted Shifrin

Do I pass, DogAteMy? :D

Akiva Weinberger

1:21 AM

Yeah lol

Ted Shifrin

Whew!

Akiva Weinberger

The answer I had was, $A_n$ is the set of unordered pairs of integers containing $n$, as $n$ ranges over the integers @Thorgott

(which geometrically looks like a vertical line starting at the x-axis and bouncing off the x=y line)

Ted Shifrin

You know that once you banned "number theory" I had to go "geometric." :D

Akiva Weinberger

One of my geometric answers was,

look at the set of circles of unit radius containing the origin

Delete the origin from each circle

Now we need a countable collection of them, taking care not to include two circles that are exactly opposite each other

Thorgott

I think if you decrease the distance between the x-intercepts of the lines and increase their slopes, you get something resembling the tangent bundle of a quarter circle

Akiva Weinberger

1:23 AM

which can be done in a variety of ways (e.g. restrict their centers to those with rational positive coordinates)

Ted Shifrin

Oh, that's nice, DogAteMy. Probably morally equivalent to mine if we blow up the origin.

Thorgott

Don't these usually intersect in two points?

Akiva Weinberger

@Thorgott Yeah I saw someone else had the set of tangents to a circle (again, countable collection, no two exactly opposite each other, can be solved in the same way eg restrict to quarter circle with countable slope)

Ted Shifrin

But he's making one of them fixed and removing it.

Thorgott

Ah, that's nice

Ted Shifrin

1:25 AM

Oh, I'm behind, I think.

Thorgott

Oh wait, you want the origin to be on the circle, not inside

Akiva Weinberger

Yeah

Ted Shifrin

No, @Thorgott, because in the tangent bundle fibers are distinct.

Akiva Weinberger

You can also do the graph of $y=\sin(x-t)$ over $x\in[0,\pi)$, for a countable collection of $t$

Ted Shifrin

You mean the envelope of actual tangent lines.

Akiva Weinberger

1:26 AM

(Notice that that's a half-period)

You could also do great circles on a sphere of a given inclination (latitude of highest point), provided you identify opposite points

Thorgott

Yeah, that's what I meant

Ted Shifrin

DogAteMy: I admit I'm now troubled by the fact that I first tried to give a discrete ("number theory" based) solution rather than a geometric one. I am getting senile.

Thorgott

Actually, can't we even get an uncountable family using tangents of the curve?

Ted Shifrin

Of course.

My example with the lines is easily made uncountable, too.

Akiva Weinberger

I would say my first answer is "set theoretic"

Ted Shifrin

1:32 AM

I claim that DogAteMy biased us in the direction of $\Bbb N$ with the countable restriction.

Akiva Weinberger

and thus works with any cardinality

Ted Shifrin

We're firing you.

Akiva Weinberger

I was being paid?!

Ted Shifrin

You can be fired from a volunteer non-paid position.

Akiva Weinberger

Fair

Ted Shifrin

1:34 AM

As can I.

waits eagerly

Thorgott

Once you have a countably infinite solution, you can take enough copies and get a solution of any cardinality

Akiva Weinberger

If we want any three to intersect in one point and any four to have empty intersection, we can let $A_n$ be the set of unordered triples containing $n$

and thus $A_i\cap A_j\cap A_k=\{i,j,k\}_3$

(…I don't actually know of any notation for unordered triple)

(…though it doesn't matter since we don't need to deal with repeated elements, so I can just use sets)

Ted Shifrin

A set is an unordered n-tuple.

Right.

Akiva Weinberger

In general you'd want $\{1,1,2\}_3\ne\{1,2,2\}_3$.

The natural framework for "roots of a polynomial" is unordered tuple, because of multiplicities

Basically unordered tuple = set with multiplicity

Ted Shifrin

You can just work with cartesian product minus diagonals (or not minus) ... modulo symmetric group.

Akiva Weinberger

1:39 AM

Sure

Thorgott

I'd write $[(i,j,k)]$, as in the equivalence class

Akiva Weinberger

I wonder if there's a plane through $(i+j+k,ij+jk+ik,ijk)$

er

Ted Shifrin

I was about to type exactly that.

Lots of planes.

Akiva Weinberger

for fixed $i$ and variable $j,k$

What's $A_i:=\{(i+j+k,ij+jk+ik,ijk)\in\Bbb R^3:j,k\in\Bbb R\}$

Ted Shifrin

It's a varying conic.

Put $s$ and $t$ in there instead of $j$ and $k$.

Akiva Weinberger

1:44 AM

$(x-i)i^2-yi+z=0$?

which is a plane

Ted Shifrin

Ah, I was playing with quadratics. Is that correct?

Akiva Weinberger

So if $A_n$ is the plane $xn^2-yn+z=n^3$ then this solves the triple version of the problem geometrically

Ted Shifrin

Are we sure no three intersect? Generically, three planes will intersect in a point.

Oh, no four.

"Never mind."

Akiva Weinberger

Compare this to the 2D version, $y=n(x-n)$, aka $xn-y=n^3$

Ted Shifrin

Yeah, got it.

Akiva Weinberger

1:48 AM

I'm kinda curious what these look like

Ted Shifrin

Mathematica can draw those.

Thorgott

You could also generalize the other version to tangent planes of a section of the 2-sphere

Akiva Weinberger

grapher.mathpix.com/…

Ted Shifrin

@Thorgott: You don't need a sphere, of course.

Akiva Weinberger

@Thorgott Can you? You can easily get 4 to intersect in a point

A point gives us an infinite collection of tangent planes through it

Ted Shifrin

1:51 AM

Are you restricting to a circle, DogAteMy?

Akiva Weinberger

Restricting what?

Ted Shifrin

The tangent planes of the sphere along a circle?

Generically, Thorgott's right, I think, but if we specialize ...

Akiva Weinberger

I don't think any three intersect in a point then

Ted Shifrin

The planes all intersect in a line.

Oh, no, that's wrong.

Agh. Enough for me.

Akiva Weinberger

In the original problem, we could do parabolas:

$y=(x-n)^2$

Right?

I think in the 3D case we can do horizontal translates of paraboloids

Wait

No never mind

Thorgott

1:57 AM

Oh right. even three of them need not intersect in a point. Would've been too nice.

Akiva Weinberger

2:14 AM

> Scott Aaronson has written about how easy it is to predict people trying to “be random”:

>> In a class I taught at Berkeley, I did an experiment where I wrote a simple little program that would let people type either “f” or “d” and would predict which key they were going to push next. It’s actually very easy to write a program that will make the right prediction about 70% of the time. Most people don’t really know how to type randomly. They’ll have too many alternations and so on. There will be all sorts of patterns, so you just have to build some sort of probabilistic model. Even a very…

@Secret This reminds me of you for some reason

(From section V of slatestarcodex.com/2019/06/04/…)

> …Being genuinely random is important in pursuing mixed game theoretic strategies. Henrich’s view is that divination solved this problem effectively.

I’m reminded of the Romans using augury to decide when and where to attack. This always struck me as crazy; generals are going to risk the lives of thousands of soldiers because they saw a weird bird earlier that morning? But war is a classic example of when a random strategy can be useful. If you’re deciding whether to attack the enemy’s right vs. left flank, it’s important that the enemy can’t predict your decision and send his best defend…

This is so neat I have to repeatedly stop myself from posting more quotes here

Let me just put one:

> Rationalists always wonder: how come people aren’t more rational? How come you can prove a thousand times, using Facts and Logic, that something is stupid, and yet people will still keep doing it?

Henrich hints at an answer: …

So much of this is quotable

Jacksoja

3:43 AM

@TedShifrin heya Ted

sashas

4:10 AM

Any advice appreciated

0

Q: How can I make a shift to applied mathematics?

I have done dual degree with bachelors in computer science. I am currently a software developer (India). I always have had interest in mathematics. But once was I got into college I did not work towards making myself mathematically mature. In college I just used to participate in coding contests...

danportin

4:23 AM

Is there a standard graduate-level textbook on regular languages and automata theory (e.g., one that introduces both algebraically) that is currently in print? Eilenberg is the standard reference, but his books run for $150 on Amazon?

Adam

um computer science IS a field of applied mathematics? And I won't lie, Impact measured in quantity of people seems like a fallacy

Edward Evans

6:13 AM

Morning!

Alessandro Codenotti

6:28 AM

What are you doing awake at dawn

maths student

7:10 AM

5. Suppose that $u(x, y)$ has a local maximum at $(0,0) .$ Show that for any direction $\vec{V}$
\[
\frac{\partial u(0,0)}{\partial \vec{V}}=0
\]

0

Q: Directional derivative at local maximum

Suppose that $u(x, y)$ has a local maximum at $(0,0) .$ Show that for any direction $\frac{\partial u(0,0)}{\partial \vec{V}}=0$. So I want to show that directional derivative is zero. So finally I get ${u(ah,bh) - u(0,0)}$ / $h$ as limit $h$ tends to zero ; how to proceed after that.

multivariable-calculus maxima-minima

Edward Evans

7:50 AM

@Alessandro frantically reading about cyclotomic fields

Alessandro Codenotti

Is there an exam coming up?

Edward Evans

8:04 AM

Yeah I have exams on the 6th and 10th of Feb, though I think I'm going to delay them until March

Alessandro Codenotti

I see

I have one on the 10th and one the 11th I think, not sure about the second

And one in March

Edward Evans

Nice, feeling prepared?

Alessandro Codenotti

For the March one definitely not (that's why I postponed it lol)

Edward Evans

Lol fair, I'm feeling okay for alg. number theory but modular forms is kicking my ass

Alessandro Codenotti

For the other two I'm definitely feeling more prepared for set theory than global analysis, but there's still time to study

Edward Evans

8:11 AM

Aye, I just feel like if I postpone until March I'll do really well, rather than cramming for the feb. exams and failing lol

Alessandro Codenotti

Makes sense

Do you only have those two exams?

Edward Evans

Yeah I only took two modules this semester so I could ease myself back into mathematics after a year out lol

unfortunately my poor mental health has hamstrung me somewhat, which is the main reason for postponing :P

Alessandro Codenotti

@EdwardEvans sounds like a smart approach

Edward Evans

Truuuuth

plus cyclotomic fields are super interesting and cool so I wanna learn them properly instead of cramming lol

Alessandro Codenotti

Fair

For my ANT class two years ago I gave a presentation on how to factor ideals in cyclotomic extensions based on Neukirch's 1.10

That's also how far I've ever gotten in Neukirch

Edward Evans

8:26 AM

Nice :D

hahaha

I'm just looking at ramification phenomena in cyclotomic extensions

Alessandro Codenotti

I used to know very little ANT, but now I forgot that too

I regret doing algebraic geometry instead of class field theory in my first semester in Bonn though

Edward Evans

Obviously I think it's amazing, but a lot of the formalities at the start of an ANT course can get kinda boring

Í'm hoping CFT will be offered next semester but I've heard on the grapevine that it might not be :(

Alessandro Codenotti

Oh, that sucks

On the other hand I like that here they offer different courses every semester instead of only having the usual big ones

Edward Evans

indeed, I think maybe some courses on lie theory will be offered and I'll be taking algebra 2 for the homological algebra lol

yeah that's cool

Alessandro Codenotti

You won't get bored

Can you take anything you want as long as you reach 120 credits or are you forced to do stuff from different areas of maths?

Edward Evans

8:34 AM

But I'm mainly interested in doing lots of number theory so I'll be slightly disappointed if CFT doesn't get offered during my time here lol

we have to do 2 applied mathematics modules and then some "interdisciplinary skills" course

so I'm gonna take Russian language next semester

Alessandro Codenotti

Applied meaning? Numerical/probability/mathematical physics/discrete?

@EdwardEvans I'd suggest Italian, but Russian is also cool :P

Edward Evans

lol it's slightly blurred, functional analysis 2 counts as applied

but yeah I'll probs take some physics

Alessandro Codenotti

@EdwardEvans what's in the syllabus?

Edward Evans

errr

hang on

not a clue

but the lecturer is a physicist I think

Alessandro Codenotti

That doesn't really tell me much hahaha

Edward Evans

8:39 AM

yeah idk lol

his website doesn't have any info about it

I. Differentialkalkül in Banachräumen und Satz von der impliziten
Funktion
II. Banachmannigfaltigkeiten und Fredhomabbildungen, Satz von
Sard-Smale
III. Der Abbildungsgrad modulo 2
IV. Der Abbildungsgrad von Brouwer, topologische Anwendungen
V. Der Leray-Schauder-Grad
VI. Lösungszweige und Verzweigungen von Lösungen
VII. Monotone Operatoren

I think that's it :P

Non-linear functional analysis

Alessandro Codenotti

Uh Banach manifolds

Weird stuff

Looks like an interesting course though

Edward Evans

maybe

idk what's being offered next semester tho

excited to take some alg. geo in 3rd semester tho lol

and algebraic topology

Alessandro Codenotti

I was looking forward to AG but then didn't like it. I think you will though

Edward Evans

yeah maybe

we'll see hehe

flowian

9:01 AM

Hello

Anyone worked through baby Rudin here?

Jacksoja

Hello

Leaky Nun

9:29 AM

@AlessandroCodenotti wanna study an endgame with me?

Alessandro Codenotti

I'm busy with lectures today

Edward Evans

10:07 AM

4.5 hours of cyclotomic fields this morning already hahaha

Leaky Nun

10:35 AM

@EdwardEvans you need a break

Edward Evans

10:52 AM

I'm just watching the proof of quadratic reciprocity tho

well that was nice

Secret

12:52 PM

australiandefence.com.au/news/…

World's 3rd Superpower

Secret

2:12 PM

@AkivaWeinberger You probably recalled my discussions about indeterminable sequences some time ago. Otherwise, that whole review, despite being cultural anthropology, does have some deep philosophical ideas that interests me for my philosopher hobby

For example, the central idea on how we are basically kept alive by randomness, and how that leads to traditions to evolve as a safeguard from killing us and to obey authority, the reason why there is power struggle, and coercion, all boils down to one reason:

> In giving humans reason at all, evolution took a huge risk. Surely it must have wished there was some other way, some path that made us big-brained enough to understand tradition, but not big-brained enough to question it. Maybe it searched for a mind design like that and couldn’t find one. So it was left with this ticking time-bomb, this ape that was constantly going to be able to convince itself of hare-brained and probably-fatal ideas.

This is a problem more ancient and much much harder than the millenium prize problems:

How to get rid of reason completely

and as we can see from history, nature still failed to solve it

Put it in another way. That this problem is still open and nature is picking a partial solution to is, is the origin of injustice

Subhasis Biswas

3:02 PM

0

Q: Is the Lebesgue Measure on $[a,b]$ perfect?

Definition: A measure space $(X, Σ, μ)$ is said to be perfect if, for every $Σ$-measurable function $f : X → \mathbb{R}$ and every $A ⊆ \mathbb{R}$ with $f^{−1}(A) \in Σ$, there exist Borel subsets $A_1$ and $A_2$ of $\mathbb{R}$ such that ${\displaystyle A_{1}\subseteq A\subseteq A_{2}{\mbox{ a...

measure-theory lebesgue-measure compactness solution-verification borel-sets

I am not sure whether I did it correctly.

Thorgott

3:31 PM

@Subhasis What you are saying is that $G$ is a measurable subset of $\mathbb{R}$ and $\mathbb{R}$ is an inner regular measure space, so $G$ is an inner regular set. This is not the same as saying that any subset of $G$ is an inner regular set with respect to the Lebesgue measure on $G$.

What you want to do is look at the subsets of $G$ and see if you can approximate them within $G$. Then use that the Lebesgue measure on $G$ is the restriction of the Lebesgue measure on $\mathbb{R}$ and that this measure is inner regular.

Subhasis Biswas

$G$ is an inner regular set w.r.t the Lebesgue measure on $\mathbb{R}$ only (by the arguments I provided), right?

Thorgott

It should also be noted that the Borel-algebra on [a,b] is the same as the trace algebra of [a,b] in the Borel-algebra on R

Yes

The idea of proof is similar, but it's important to be precise on the details in measure theory

Subhasis Biswas

@Thorgott Does it directly imply that the subset $A$ of $G$ can be approximated from within?

Since, $A \subset G \subset \mathbb{R}$?

Subhasis Biswas

3:54 PM

@Thorgott Approximating a set $G$ in $\mathbb{R}$ means that $\forall \ \epsilon>0$, $\exists$ an open set $U$ such that $\lambda(U \setminus G)< \epsilon$. Is that right?

Stupid Questions Inc

Hey chat! What textbook would you recommend as a good reading companion to Huffman/Kunze's Linear Algebra?

Thorgott

4:13 PM

Yes, that's the idea. Let $A\subseteq G$ be measurable with respect to the Borel-algebra on $G$. Then $A\subseteq\mathbb{R}$ is measurable with respect to the Borel-algebra on $\mathbb{R}$ (Why?). This measure is inner regular, so $\lambda(A)=\sup\{\lambda(K)\colon K\subseteq A\text{compact}\}$. Now all the $K$ the supremum is taken over are subsets of $A$, hence subsets of $G$.
Now note that this is the same as saying that $A$ is an inner regular set in $G$, because a) a compact subset of $G$ is the same as a compact subset of $\mathbb{R}$ that is contained in $G$ and b) the restriction of…

CroCo

@StupidQuestionsInc, I would recommend Elementary Linear Algebra Applications by Howard Anton

Subhasis Biswas

@Thorgott A finite measure $μ$ is inner regular if and only if, for all $ε > 0$, there is some compact subset $K$ of $X$ such that $μ(X \setminus K) < ε$. Let $X=[a,b]$. Consider $K=[a+\epsilon/4 , b-\epsilon/4]$. Evidently, $\mu(X\setminus K)=\epsilon/2 <\epsilon$

can the special case of the inner-regularity of $[a,b]$ be proved this way?

Thorgott

No, you are mixing the concept of inner regular measure and inner regular set up.

This just shows $X$ is an inner regular set (both within $\mathbb{R}$ and within itself).

But to show $X$ (with the inherited sigma-algebra and measure) is a perfect measure space, you need to show that every measurable subset of $X$ is inner regular in $X$.

Subhasis Biswas

@Thorgott EVERY

@Thorgott indeed. It is very confusing.

@Thorgott here we are using $C$ is compact in $Y$ iff $C$ is compact in $X$, where $Y \subseteq X$. Right?

Thorgott

Yeah, though that's pretty much just the definition of subspace topology.

Subhasis Biswas

4:26 PM

Thank you very much. It is much more clear now.

Thorgott

All of what I said is essentially just carefully unraveling definitions.

Subhasis Biswas

Could you post that as an answer?

Stupid Questions Inc

@CroCo will look into it

Thorgott

Actually, I misspoke at the end. I said this argument shows that restrictions of perfect measures are perfect, but that's not what I meant. What the argument shows is that the restriction of inner regular measures is inner regular. The statement about perfect measures is still true, but it requires a different argument. Also sure, I can.

Subhasis Biswas

@Thorgott I actually know nothing about measure theory. But I can sort of understand what you did.

Sha Vuklia

4:44 PM

heya guys, what's the reason(/motivation) behind taking $x,y$ and $x',y'$ distinct in the definition of 2-transitivity (where we define 2-transitivity to be a group action where for any distinct $x,y$ and $x',y'$ we have $\sigma_g(x)=x'$ and $\sigma_{g'}(y)=y'$ for some $g,g'\in G$, where $\sigma\colon G\to S_X$ is our group action)

I would think that if we don't force $x,y$ and $x',y'$ to be distinct, then we would also have transitivity

usually 2-transitvity is called "stronger", but by having $x,y$ and $x',y'$ distinct, it seems like it's not stronger, but just different from transitivity

or am I wrong, and is a 2-transitive action still transitive?

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