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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?
 
I don't know any more than you do
 
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
 
Your textbook author is suggesting you might be in trouble, DogAteMy.
 
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.
 
It adds up the roots of $f(z)=a$, @anakhro.
 
12:19 AM
@TedShifrin Clearly
 
So that means I have to show that both the residue and the winding numbers are 1?
 
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 $\gamma$.)
 
If it is simple, then it is $a$.
err
the root
 
Right, the root.
 
So then I have to consider higher order roots?
And I suppose there is a formula for this I should find in a book
 
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.
 
Where $u$ is a change of variables?
 
The Laurent expansion is naturally going to be in $u$, right? That's easier than writing $(z-z_0)$ everywhere.
 
I am afraid I am lost. Maybe I need to read up on Laurent expansions again.
 
OK.
Or Taylor.
 
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.
 
12:34 AM
But use that $z_0$ is a double-root of $f(z)-a$.
Write $f(z)-a = (z-z_0)^2 g(z) = u^2 g(u+z_0)$.
 
Should I avoid expanding everything out? I took the liberty of also determining $f'$ with this
 
Yes, avoid. We're just trying to find the coefficient of the $1/u$ term.
 
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)
 
Hmmm ... Seems like a reasonable question. I don't have an automatic response.
 
I am not sure what I am trying to do algebraically. Certainly not just pull a term 1/u out.
 
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.
 
I have something that looks like $$\frac{(u+z_0)(2ug(u+z_0) + u^2g'(u+z_0))}{u^2g(u+z_0)}.$$
 
OK, so what is the coefficient of $u$ upstairs?
 
I can get rid of one u in the top and bottom
So it is $\frac{(u+z_0)(2g(u+z_0) + ug'(u+z_0))}{ug(u+z_0)}$
 
Now what's the constant term upstairs?
And what's the resulting coefficient of $1/u$ in that thing?
 
$2z_0g(u+z_0)$
 
12:46 AM
So we get $2z_0$. Bingo.
Double root.
I leave it to you to generalize to $m$th order root.
 
There's a reasonable "canonical answer" and some clever answers
 
How do you conclude $2z_0$ from that? How does the $g$ part disappear?
 
Look at the denominator.
Hmm @DogAteMy.
 
OH
I see.
 
You should practice doing computations like this ... if you're preparing for a complex analysis qualifying or something.
 
12:50 AM
Yeah, definitely. I 100% agree.
I just feel so pressed.
 
Practice is the only way to improve speed.
 
Yeah. I just wish I didn't have to take so many courses so I would have more time to practice.
 
DogAteMy: Something indexed by primes, maybe?
 
Thanks for your help, though!
I appreciate it ever so much.
 
Anyhow, you should only take a few seconds to do the $m$th order case, @anakhro, now.
 
12:53 AM
It should follow in a similar manner, or do you mean just guessing the answer?
 
Both.
 
Well I think I have the guess in a second, let me try the algorithm.
Yeah I did it
Heh
 
DogAteMy: What is my set $A_p$ is the set of all prime multiples of $p$?
See, @anakhro: Told you so :P
 
I thought I was in for some real hard thinking but the moment I started I realized what changes.
 
So you have $A_p=\{pq:q\text{ prime}\}$ for prime $p$?
 
12:55 AM
Yes.
 
That works
 
I take it that's not the canonical answer.
 
It's very close, it's like one step removed
 
Hmm ...
 
I never said these sets are subsets of $\Bbb N$
 
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.
 
Elaborate?
 
Oh, no, what I said is dopey.
I have to look at affine subspaces.
 
You know, since you already gave an answer I might as well tell you it
 
Well, don't ruin it for everyone else.
 
OK I won't then
Take "$A_p=\{pq:q\text{ prime}\}$ for prime $p$" and get rid of the number theory
 
1:06 AM
LOL, how do I get rid of number theory in that?
Oh, $y=n(x+1)$.
 
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
 
$A_n$ is the set of unordered pairs of integers containing $n$
 
@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.
 
@TedShifrin Don't they all contain $(-1,0)$?
 
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
 
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.
 
Do you wish you had done more in research?
Or do you like the balance you struck?
 
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.)
 
@TedShifrin I don't quite understand. Where do $A_1$ and $A_2$ intersect?
if not at the point you just removed
 
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
 
So where do $A_m$ and $A_n$ intersect
$m(x-m)=n(x-n)$
 
They're non-parallel lines, so they have to intersect.
And surely that point won't work for a different index :D
 
They intersect at $(m+n,mn)$, I think
which works
 
Or I can resort to primes again :P
 
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.
 
1:20 AM
Yeah, $(m+n,mn)$ is right.
 
That's basically what he just did I think @Thorgott
 
That's effectively what I've done, @Thorgott.
Oops.
 
These are the lines through $(n,0)$ with slope $n$
 
Oh lol, I didn't try to visualize what you were posting :p
 
Do I pass, DogAteMy? :D
 
1:21 AM
Yeah lol
 
Whew!
 
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)
 
You know that once you banned "number theory" I had to go "geometric." :D
 
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
 
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
 
1:23 AM
which can be done in a variety of ways (e.g. restrict their centers to those with rational positive coordinates)
 
Oh, that's nice, DogAteMy. Probably morally equivalent to mine if we blow up the origin.
 
Don't these usually intersect in two points?
 
@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)
 
But he's making one of them fixed and removing it.
 
Ah, that's nice
 
1:25 AM
Oh, I'm behind, I think.
 
Oh wait, you want the origin to be on the circle, not inside
 
No, @Thorgott, because in the tangent bundle fibers are distinct.
 
You can also do the graph of $y=\sin(x-t)$ over $x\in[0,\pi)$, for a countable collection of $t$
 
You mean the envelope of actual tangent lines.
 
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
 
Yeah, that's what I meant
 
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.
 
Actually, can't we even get an uncountable family using tangents of the curve?
 
Of course.
My example with the lines is easily made uncountable, too.
 
I would say my first answer is "set theoretic"
 
1:32 AM
I claim that DogAteMy biased us in the direction of $\Bbb N$ with the countable restriction.
 
and thus works with any cardinality
 
We're firing you.
 
I was being paid?!
 
You can be fired from a volunteer non-paid position.
 
1:34 AM
As can I.
waits eagerly
 
Once you have a countably infinite solution, you can take enough copies and get a solution of any cardinality
 
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)
 
A set is an unordered n-tuple.
Right.
 
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
 
You can just work with cartesian product minus diagonals (or not minus) ... modulo symmetric group.
 
1:39 AM
Sure
 
I'd write $[(i,j,k)]$, as in the equivalence class
 
I wonder if there's a plane through $(i+j+k,ij+jk+ik,ijk)$
er
 
I was about to type exactly that.
Lots of planes.
 
for fixed $i$ and variable $j,k$
What's $A_i:=\{(i+j+k,ij+jk+ik,ijk)\in\Bbb R^3:j,k\in\Bbb R\}$
 
It's a varying conic.
Put $s$ and $t$ in there instead of $j$ and $k$.
 
1:44 AM
$(x-i)i^2-yi+z=0$?
which is a plane
 
Ah, I was playing with quadratics. Is that correct?
 
So if $A_n$ is the plane $xn^2-yn+z=n^3$ then this solves the triple version of the problem geometrically
 
Are we sure no three intersect? Generically, three planes will intersect in a point.
Oh, no four.
"Never mind."
 
Compare this to the 2D version, $y=n(x-n)$, aka $xn-y=n^3$
 
Yeah, got it.
 
1:48 AM
I'm kinda curious what these look like
 
Mathematica can draw those.
 
You could also generalize the other version to tangent planes of a section of the 2-sphere
 
@Thorgott: You don't need a sphere, of course.
 
@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
 
1:51 AM
Are you restricting to a circle, DogAteMy?
 
Restricting what?
 
The tangent planes of the sphere along a circle?
Generically, Thorgott's right, I think, but if we specialize ...
 
I don't think any three intersect in a point then
 
The planes all intersect in a line.
Oh, no, that's wrong.
Agh. Enough for me.
 
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
 
1:57 AM
Oh right. even three of them need not intersect in a point. Would've been too nice.
 
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
> …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
 
 
1 hour later…
3:43 AM
@TedShifrin heya Ted
 
4:10 AM
Any advice appreciated
0
Q: How can I make a shift to applied mathematics?

sashasI 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...

 
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?
 
um computer science IS a field of applied mathematics? And I won't lie, Impact measured in quantity of people seems like a fallacy
 
 
2 hours later…
6:13 AM
Morning!
 
6:28 AM
What are you doing awake at dawn
 
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

maths studentSuppose 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.

 
7:50 AM
@Alessandro frantically reading about cyclotomic fields
 
Is there an exam coming up?
 
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
 
I see
I have one on the 10th and one the 11th I think, not sure about the second
And one in March
 
Nice, feeling prepared?
 
For the March one definitely not (that's why I postponed it lol)
 
Lol fair, I'm feeling okay for alg. number theory but modular forms is kicking my ass
 
For the other two I'm definitely feeling more prepared for set theory than global analysis, but there's still time to study
 
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
 
Makes sense
Do you only have those two exams?
 
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
 
@EdwardEvans sounds like a smart approach
 
Truuuuth
plus cyclotomic fields are super interesting and cool so I wanna learn them properly instead of cramming lol
 
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
 
8:26 AM
Nice :D
hahaha
I'm just looking at ramification phenomena in cyclotomic extensions
 
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
 
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 :(
 
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
 
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
 
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?
 
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
 
Applied meaning? Numerical/probability/mathematical physics/discrete?
@EdwardEvans I'd suggest Italian, but Russian is also cool :P
 
lol it's slightly blurred, functional analysis 2 counts as applied
but yeah I'll probs take some physics
 
@EdwardEvans what's in the syllabus?
 
errr
hang on
not a clue
but the lecturer is a physicist I think
 
That doesn't really tell me much hahaha
 
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
 
Uh Banach manifolds
Weird stuff
Looks like an interesting course though
 
maybe
idk what's being offered next semester tho
excited to take some alg. geo in 3rd semester tho lol
and algebraic topology
 
I was looking forward to AG but then didn't like it. I think you will though
 
yeah maybe
we'll see hehe
 
9:01 AM
Hello
Anyone worked through baby Rudin here?
 
Hello
 
9:29 AM
@AlessandroCodenotti wanna study an endgame with me?
 
I'm busy with lectures today
 
10:07 AM
4.5 hours of cyclotomic fields this morning already hahaha
 
10:35 AM
@EdwardEvans you need a break
 
10:52 AM
I'm just watching the proof of quadratic reciprocity tho
well that was nice
 
 
2 hours later…
12:52 PM
World's 3rd Superpower
 
 
1 hour later…
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
 
3:02 PM
0
Q: Is the Lebesgue Measure on $[a,b]$ perfect?

Subhasis BiswasDefinition: 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...

I am not sure whether I did it correctly.
 
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.
 
$G$ is an inner regular set w.r.t the Lebesgue measure on $\mathbb{R}$ only (by the arguments I provided), right?
 
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
 
@Thorgott Does it directly imply that the subset $A$ of $G$ can be approximated from within?
Since, $A \subset G \subset \mathbb{R}$?
 
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?
 
Hey chat! What textbook would you recommend as a good reading companion to Huffman/Kunze's Linear Algebra?
 
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
 
@StupidQuestionsInc, I would recommend Elementary Linear Algebra Applications by Howard Anton
 
@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?
 
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 $X$.
 
@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?
 
Yeah, though that's pretty much just the definition of subspace topology.
 
4:26 PM
Thank you very much. It is much more clear now.
 
All of what I said is essentially just carefully unraveling definitions.
 
Could you post that as an answer?
 
@CroCo will look into it
 
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.
 
@Thorgott I actually know nothing about measure theory. But I can sort of understand what you did.
 
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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