12:00 AM
I was looking at your questions. Sobolev spaces, all sorts of measure theory.

@manooooh yes, you want to conclude $X\in\{\emptyset\}$, which is the same as $X=\emptyset$. is this the same as $X\in\mathcal{P}(A)$ and $X=\emptyset$?

Math is just a hobby for me, so i have studied on my own multiple things, i dont have the formal part

Weird that you've done measure theory and Sobolev spaces and are asking this question.
What exactly are you confused about?
@Thorgott: Do you not agree that what he's trying to prove is wrong?

Im starting to see some things on sigma-algebras

But when you wrote those semicolons, user76284 appropriately asked what you were writing. I interpret it as three separate questions that should not have been combined in the way you wrote it. Am I correct? @Alek
@manooooh: Oh, I see, you just had containment. So there are lots of subsets on the right-hand side as I described that are not subsets of $A-B$, but you just claimed containment. My mistake.

12:03 AM
I really dont know how u interpret that, for me, i'm trying to write the inverse image ofsome collection of subsets of a set $Y$

@TedShifrin yes: take $U=\{1,2\}$, $A=\{1,2\}$ and $B=\{2\}$. Then $A-B=\{1\}$ so $P(A-B)=\{\emptyset,\{1\}\}$, and $P(A)=\{\emptyset,\{1\},\{2\},\{1,2\}\}$, $P(B)=\{\emptyset,\{2\}\}$. So $P(A)-P(B)=\{\{1\},\{1,2\}\}$, and hence $P(A-B)\subseteq(P(A)-P(B))\cup\{\emptyset\}$

I dont think that I'm trying to see that as 3 different inverse images

So are you asking three separate questions, as I said, @Alek, or are you taking the inverse image of the union of those subsets?

@AlekMurt The inverse image is defined for each of those subsets, not a collection of them.

I don't think it's wrong

12:04 AM
Are you trying to get the collection of inverse images?

No, I was thinking it claimed equality, not containment. I apologized above. @Thorgott, @manooooh.

No problem

I guess there's no way to write it other than $\{X\in\mathcal P(A): X\cap B = \emptyset\}$.

ah, ok

If you love using masochistic notation then $f^\star \circ ([0,1],(1,4],(4,\infty))$ might work. $f^\star$ is the preimage with the pullback notation, and I'm exploiting this notational convenience. Don't try this at home though.

12:06 AM
No, I'm not talking about the union of those subsets is more of something like @user76284 says

@Thorgott showed me that $X\in B'$ implies $X\subseteq (P(B))'$ or $X=\emptyset$, but I can't figure out how to use this to finish the proof here: chat.stackexchange.com/transcript/message/53348033#53348033

I think that never in my 45+ years as a professional mathematician have I encountered this notation/notion. @Alek, @user76284.

the proof is essentially finished
you just need to realize that the last statement is equivalent to the conclusion

@Thorgott do you agree we have $(X\in P(A)\wedge X\in(P(B))')\vee(X\in P(A)\wedge X=\emptyset)$?

12:08 AM
What does $X\subseteq (\mathcal P(B))'$ mean?

So let's enter in context, because for me is also super weird and i would like to know if I'm writing this ok

You mean element, @manooooh? And complement in $\mathcal P(A)$?

@TedShifrin $(P(B))'$ is the set not containing elements of the power set of $B$. i.e. $(P(B))'=U-P(B)$, where $U$ is the universe

I'm complaining about $\subseteq$ instead of $\in$.

I didn't write subset but $\in$

12:11 AM
I have a $\sigma ([0,1],(1,4],(4,∞))$ and i have the function that i wrote above, and i would like to know $f^{-1}\sigma ([0,1],(1,4],(4,∞))$

@manooooh: You did write it and that's why I asked what it meant.
Look up 12 "chats."
I still don't know what $\sigma()$ with a bunch of commas means, @Alek.

@TedShifrin $X\in P(B)'$ is a relationship like $x\in A=\{1\}$. I write $x$ in capital letter since $P(A)$ contains sets, and sets are commonly denoted with a capital letter

So i think that i have to look for $\sigma (f^{-1}([0,1],(1,4],(4,∞)))$, that why I ask for that inverse image, where $\sigma ()$ is the sigma algebra generated for those elements inside ()

Commas, semicolons, whatever. Unless $\sigma$ is a function on the set of ordered triples of subsets ...
oh, crazy.

If $X$ is an element of $P(A)$ then $X$ is a subset of $A$

12:14 AM
So $f^{-1}$ of the $\sigma$-algebra is again a $\sigma$-algebra, I guess, but you apply $f^{-1}$ to each element of that $\sigma$-algebra, one at a time.

and the converse also holds

@manooooh: Yes, I know this stuff. I'm saying you wrote the equivalent of $X\subset P(A)$.
All I said was it should have been $\in$.
You can find what you typed if you count 12 things up from where I put that.
Anyhow, I'm done with this discussion.

Initially that was my doubt, and my thoughts were that the only thing that makes sense is to apply inverse image for ever subset, but this topic has been really confusing for me, so i wanted to make sure my thoughts were correct

Where did I write, with other words, $X\subset P(A)$?

Hey there guys!

12:16 AM
Hello!

@manooooh: I already told you where a while ago.

Yes, I agree we have that

Demonark!!! Where the hell have you been?

hey @Daminark

I guess that since Demonark moved to Wisconsin he became too good for us :P

12:17 AM
Lmao, in Chicago fine but Madison???
Nah but yeah thing is last semester I felt I wasn't really getting any math done because the volume of immediate deadlines dropped like a rock

Ted here ^^^^^^^^^^? But I didn't write $X\subset P(A)$. I wrote $X\in P(A)$

And I was like uh okay I need to pull myself together a bit, gonna try to get offline a bit more and focus a bit

No, not there. Geez.
Is it going better, Demonark?

In a way no, I just found other distractions. This semester I'm taking classes that actually all assign psets and I'm gonna try to be a part of some reading groups
At least if I can artificially place some deadlines on my head then maybe that'll help a bit

Oh, now I see Ted, yes you are right again. I should have write $X\subset B'$ implies $X\in P(B)'$ or $X=\emptyset$

12:22 AM
Or not artificially so much as like, okay with reading groups obviously it's not like the due dates are quite as "over your head" but hopefully this will at least get me to focus a bit

Demonark: In general, faculty expect graduate students to be more grown-up and responsible. Although I knew better and always assigned weekly or bi-weekly problem sets in every graduate course I ever taught. :D

That makes sense, yeah. This semester I'm going a bit harder on the algebra though eventually I'll wanna get back to some analysis. I've got homological algebra, elliptic curves, and AG2 (starting with Riemann Roch, then doing some basic scheme theory)

LOL, all algebra.

What've you been up to, Ted and Lukas?

are you saying that graduate lectures usually do not have assignments?

12:25 AM
@Thorgott oh, yes since we know that $p\wedge q\to p$, so $X=\emptyset$ and $X\in P(A)$ implies $X=\emptyset$, which is the same as $X\in\{\emptyset\}$. And we are done!

@Daminark I've been giving seminar talks on group schemes/algebraic groups and rigid-analytic geometry and I'm taking lectures on adic spaces and L functions

(On the side I've got a few people and we're gonna read some K-Theory, I'm thinking using Atiyah?)

and I'm TAing intro abstract algebra

It costs me to understand why if $X$ is the empty set then $X$ belongs to $\{\emptyset\}$

Ah dope

12:27 AM
oh and I'm in the process of finishing my masters application

are you saying that $p\land q\rightarrow p$ generally?

I've written CV, motivation letter, asked for recommendation letters and all that

Nice, still going with the plan you mentioned that one time? ALGANT?

yeah

@Thorgott yes of course. It is one law from Logic

12:28 AM
but I probably can't go to LA :/

I was wondering what happened with that, Lukas.

due to illness, I won't finish my bachelor as early as I planned

$X=\emptyset\Leftrightarrow X\in\{\emptyset\}$ holds qua definition

I certainly don't like writing $\{A\}$ when $A$ is a subset.
But I give up on this stuff.

@Thorgott oh, I see, it is pretty obvious

12:30 AM
But yeah on my end I'm TAing Calc 2 again

i agree with Ted in that this is too heavy on symbols

When I took point-set topology in college from Munkres, he literally would not allow us to write any symbols other than $\in$, $\implies$, $\subset$, $\cup$, $\cap$ (and I'm sure I've forgotten a few) and the abbreviation s.t. for "such that."

Let me clear it for you and me: \begin{align*}\text{PROVE: $P(A-B)\subseteq (P(A)-P(B))\cup\{\emptyset\}$}\\\hline X\in P(A-B)=P(A\cap B')&\to X\subseteq A\wedge X\subseteq B'\\&\to X\in P(A)\wedge(X\in(P(B))'\vee X=\emptyset)\\&\to(X\in P(A)\wedge X\in(P(B))')\vee(X\in P(A)\wedge X=\emptyset)\\&\to(X\in P(A)\cap(P(B))')\vee(X=\emptyset)\\&\to X\in P(A)-P(B)\vee X\in\{\emptyset\}\\&\to X\in(P(A)-P(B))\cup\{\emptyset\}\end{align*}
Thanks @Thorgott and @TedShifrin for your huge help!

@TedS I see that it left a stain on you

I buy that, at least for psets
On boards I tend to be a bit more symbolsy
$\forall$ and $\exists$ in particular

12:40 AM
One can use symbols or can't use symbols. Is up to you. I prefer to use symbols in this particular situation because does not include quantifiers and the expressions are not long enough to get lost on the road

So, it's officially up to you in the sense of, we all have free will. But if you want other people to read your stuff then that's something you wanna take into consideration

Erm... Is up to you to read my question or not...

I mean sure, just like, if there over comes a time where you ever do need others to read what you're doing, and it will likely come, be ready to make some concessions

1:02 AM
@Thorgott Not at all. I found that every mathematician from whom I took courses or with whom I studied used hardly any symbols at all. If you're going to work in formal logic, then have at it. But if you ever are going to write a paper for publication, mathematics (not logic) journals want readable mathematics, i.e., words and not symbols.
And it's important for pedagogy to instill an understanding of concepts, not of symbols.
In the meantime, I'll avoid reading what manoooh chooses to write.

Let's do like they used to do and use words for operations too.
No more equals sign.
NO LONGER WILL WE BE BOUND TO MATHEMATICAL NOTATION.
LANGUAGE WILL TAKE BACK WHAT IS RIGHTFULLY THEIRS.

2 hours later…
3:06 AM
Can the product of a transcendental number and an algebraic irrational number be algebraic?

products of algebraic numbers are algebraic and reciprocals kf algebraic numbers are algebraic so no

showing that the sum of an algebraic number and a transcendental number is transcendental is actually an instructive exercise in logic

3:35 AM
@LukasHeger thank you! Also, @BalarkaSen, @Thorgott and @Ted: thank you all :)

also can an integer to the power an irrational be transcendental?

41

The fact that there exists irrational number $a,b$ such that $a^b$ is rational is proved by the law of excluded middle, but I read somewhere that irrationality of $\sqrt{2}^{\sqrt{2}}$ is proved constructively. Do you know the proof?

Take a look at the accepted answer
In mathematics, the Gelfond–Schneider theorem establishes the transcendence of a large class of numbers. == History == It was originally proved independently in 1934 by Aleksandr Gelfond and Theodor Schneider. == Statement == If a and b are algebraic numbers with a ≠ 0, 1, and b irrational, then any value of ab is a transcendental number. === Comments === The values of a and b are not restricted to real numbers; complex numbers are allowed (they are never rational when they have an imaginary part not equal to 0, even if both the real and imaginary parts are rational).In general, ab ...

4:25 AM
If $a$ is algebraic, $b$ is irrational and $a^b$ trancendental, then is $b$ always algebraic?

4:54 AM
does the Gelfond–Schneider theorem imply the above?

5:05 AM
nope, not trivially if true

5:15 AM
which book has set operations defined in terms of logical operations, For example union defined as $\or$

any formal set theory book
Paul Halmos "Naive Set Theory", probably

5:58 AM
@AkivaWeinberger what does ELO as in ELO rating system stand for?

6:52 AM
@LeakyNun It's Elo, it's a guy's name

7:08 AM
@AkivaWeinberger nice

3 hours later…
9:44 AM
@LeakyNun Hi

10:23 AM
@AlessandroCodenotti Hi
@AlessandroCodenotti I have a question

11:05 AM
7

11:39 AM
Sample 3 binary strings uniformly of length 30. What is the exact expected minimum distance between the closest pair?

12:05 PM
0

A pregnancy test is accurate 97% of the time when someone is pregnant and 98% accurate when someone is not. Assuming that 60% of people who take the test are pregnant, and that someone tests positive twice in a row, what is the probability they actually are pregnant? I think this is application ...

12:21 PM
Is it true that $(A/B) \cup C= A \cup (B \C)$?
sorry
that should be
$(A/B) \cap C = A \cap (B/C)$
or better $(A/B) \cap C = A \cap (C/B)$

12:56 PM
So I ask is it true that $(A/B) \cap C = A \cap (C/B)$

1:25 PM
draw a diagram

1:38 PM
I think it is true
by basic properties of sets
I am just sanity checking, because it makes a question rather too trivial
@Thorgott

2:04 PM
it is

hi @user, @Thorgott

2:26 PM
hi Lucas

Yo chat

4 hours later…
5:57 PM
@LukasHeger hey

math.stackexchange.com/questions/3523121/… has been closed but for no reason that I can see. Is it possible to have it reopened?

No one in here can help with that. You need a moderator. The question seems clear. It would have been nice if you'd shown some of your attempts or an approach other than just having a numerical hunch. But I personally do not even understand the question, so I can't help.

@TedShifrin Hey Ted

Howdy @Jacksoja.

@TedShifrin need some hint Ted ! Been stuck for days

6:13 PM
Days, not weeks?

Want to show order of any element in a finite cyclic group has to divide order of the element of highest order.
No, going crazy as it is

Oh, there's a version of that question that's much harder, but this isn't hard.

What version

Leave out cyclic, just say abelian.

I tried alot of ways

6:15 PM
So a cyclic group has a generator. What is the order of the element of highest order?

Oh yes
That is my question haha
Gonna prov it is cyclic

Huh????

@TedShifrin it is p-1 but am not alowed to use that

You told me the group was cyclic, and now you're saying you're gonna prove that.
What is the correct statement?

My question is the hard version

6:19 PM
So you have a finite abelian group.
Nothing more.

Yes

What tools do you have? Do you know that such a group can be decomposed as a direct sum of cyclic groups?

No
I tried using

Do you know something about the order of $ab$ if the orders of $a$ and $b$ are relatively prime?

Euclid algorithm ,sorry typing from phone
Bit slow

6:21 PM
Yes, that might be relevant.

I tried that

So my hint (this exercise is actually in the algebra book I wrote, so I remember it) is to answer the question I wrote 5 lines up.

Product of their order
Ok thanks

OK, find a way to use that for your question. You have to be slightly tricky.

On diff geo
What are prereq ?

6:24 PM
Multivariable calc and some linear algebra (like linear maps and eigenvalues/eigenvectors in 2D).
The YouTube lectures are for the most part much more advanced than the diff geo material.
You don't need analysis for the diff geo.

So the task is to show that the order of an element in a finite abelian group divides the order of the element of highest order?

Don't give it away, @Thorgott or DogAteMy!

Ok thanks so much !
Yes let me try the hint

6:38 PM
Ah, I see it now. Yeah, that's a nice exercise.

@Jacksoja oh, lol. I came here to ask the same

@LucasHenrique haha good luck
Tell me if you solve it
You taking algebra as well ?

Kind of. I'm studying Galois theory

7:15 PM
The same @Lucas?!!

Ok. Let $a$ be one of the elements with maximum order and let $b \not \in <\!a\!>$ (otherwise it's trivial)
(I can't typeset this properly)

Oh, because of finite fields this could show up in Galois theory (although you have more ways around it, I think).
@Jacksoja didn't ask for more hint, but I would suggest proceeding by contradiction ... using the beautiful fact I asked him above.

I'm trying to show that if $\operatorname{ord}b \not \mid \operatorname{ord}a$, we can choose $r,s$ such that $\operatorname{ord}(a^r b^s) > \operatorname{ord} a$
Which looks like Bezout's lemma

Yes, that's correct, but it's a bit sneaky, because you need relatively prime orders for our lemma above.

@Ted, I thought about that direct sum of cyclic groups thing. Is that done by finding cyclic subgroups of maximal order, taking quotients (since each subgroup is normal cuz commutativity) and using induction?

7:21 PM
Whoa. I haven't thought about it in a while, but I think it's easier than that.

@TedShifrin oh. :(. I'm not sure of how I'd get relatively prime orders.

That's why it's a bit sneaky ;P

Could you give me a hint, please?

Well, first of all, what is your set-up if you're doing a contradiction argument?

I'm not sure if I understand your question, but I'd try to show that... my hypotheses are contradictory. :P Specifically, I'd try to show that there's an element with strictly bigger order than the one we supposed to have maximum order

7:27 PM
Yes, but with what starting hypothesis?

There's an element with order not dividing the maximum order

So can you phrase that as a gcd statement?

I was literally writing that down haha

That's gonna be my hint, then.

If $m$ is the maximum and $n$ is the other one, $\operatorname{gcd}(m,n) < n$

7:33 PM
Yup, so you want to play with that gcd to come up with your $r,s$.

Ohh
So that $\operatorname{gcd}(m,n) \geq n$

I'm not saying any more. You're more clever in algebra stuff than I am, anyhow :P

Thank you @Ted. I'll play with these symbols for a while

That last sentence is too confusing, because we already had an $m$.
BTW, do you think that lemma generalizes to lcm?

Sure

7:35 PM
Nope. Find a counterexample.

smirks
I said the same thing (when I was writing my algebra book) and then realized it was wrong.

That question seemed too innocent for the answer to be positive

LOL @Thorgott ... particularly if one knows Ted.

$2$ and $6$ over $\mathbb Z_{12}$?
$\operatorname{ord}(2+6) < \operatorname{lcm}(\operatorname{ord}2, \operatorname{ord}6)$

7:41 PM
Also, if you are willing to use the structure theorem, the question is trivial, because you can explicitly classify the orders of all elements, but that's not necessary here

OK. I had a slightly different version of the same example.

But couldn't you have $a^rb^s$ with order being the lcm?

Yes, in fact, that's always possible

In this case, $1*2 + 3*8$
@Thorgott that's what I'll be trying then
(I think I should, at least)

I think you can do it with even less work, but that definitely suffices

7:44 PM
OH. So that's the subtlety

losted

$a^{\operatorname{gcd}(m,n)}b$
the first one's order is $\frac m{\operatorname{gcd}(m,n)}$ which is relatively prime to $n$

I think I made a mistake like that, too.

it looks I shouldn't be hahaha

I just noticed I made a mistake like that too lol
But it's easily fixed

7:49 PM
@Thorgott: Many years ago it took me a while to figure out how to fix it.

that's wrong
oh hell. for $4, 6 \in \mathbb Z_{12}$, $4^1+6 = 10$ has order $6$ as expected
idk what I'm doing wrong
but these orders are relatively prime, duh

nods

Is it about the chosen orders?

Come back to your false claim. If $m/d$ and $n$ aren't necessarily relatively prime, how do you fix that? Forget the group theory for a while.

@TedShifrin okay, found a counterexample. I'd take $m/d$ and $n/d$ but it I don't think this choice will work

8:02 PM
Now comes the tricky stuff, then.
Heya, a @Balarka.

Hi @Balarka @Ted

@Balarka, with regard to a question on main on which I'm commenting ... there's no elementary way to see that the commutators of the generators of $\pi_1(\Bbb R^2-p-q)$ are nontrivial, right?
hi, demonic @Alessandro (will I get flagged yet again?).
3

Hi guys, I'm trying to solve the limit $\lim_{x\to 0} \frac{\sqrt{1+x}-\frac{1}{2}\sin(x)-\cos(x)}{\arctan(x^2)\sin(\frac{\pi}{3}\cos(x))}$. I can't seem to find the trick in order to solve it. Any suggestions please?

I always use Taylor polynomials, @vesii.

8:07 PM
@TedShifrin there's only one way to find out

Unfortunately, I did not learn taylor yet :(

The $\sin(\frac{\pi}3\cos x)$ term is sort of a red herring (i.e., irrelevant to the issue). Can you do it without that?
There are some super-zealous people voting to close everything. I've seen a few where they're over-doing it.
And I haven't heard of most of the people so voting, either.
Too many people have vote-to-close privileges now.
@vesii: So you have no choice but to use L'Hôpital, but I truly hate that.

@TedShifrin $m$, $n/d$?
I tested it a few times, it worked and I don't know why, haha

You can make counterexamples for that one, too, just like for $m/d,n$, right?

Not sure if the restriction $m > n$ implies anything.

8:12 PM
Oh, true, it isn't totally symmetric.
What about $m=12$, $n=9$?

@TedShifrin, I'll try, thanks :)

I guess I'll have to change both orders simultaneously

@vesii: Factor out the $\sin(\pi/3)$ limit of that one term. It's a mess otherwise.
@LucasHenrique Bingo.
Now I'm done!
Lunchtime for Ted.

bon appétit

@TedShifrin Doubtful. I at least need to draw the covering space and invoke the lifting results.
That's the easiest proof I know

8:21 PM
Thank you, @Ted :). Bon appétit!

9:20 PM
@BalarkaSen Yup, me too.

@TedShifrin I suppose one way to argue could be to show that $T^2$ is not homotopy equivalent to $S^1 \vee S^1 \vee S^2$. This you can conclude via ring structure in cohomology.
This proves the commutator is not nullhomotopic, because that's what the $2$-cell in $T^2$ is attached to the 1-skeleton by.

$a^r b^s$, if $m/r$ and $n/s$ are relatively prime, has order $\frac{mn}{rs}$ which we want to be $>m \implies n > rs$

I wouldn't call this easy by any means, just interesting. You could potentially write a purely de Rham theoretic proof maybe.
Replacing cell structure by handlebodies to make everything a manifold
Fun fact: $\Sigma T^2$ is nonetheless homotopy equivalent to $\Sigma(S^1 \vee S^1 \vee S^2) = S^2 \vee S^2 \vee S^3$
Suspending completely "untangles" the cell structure, thanks to higher homotopy groups being abelian (commutator becomes nullhomotopic after suspending)

9:47 PM
@Balarka: This was an exercise (?) in a complex variables book, however :P
@LucasHenrique You'd better specify something about $r$ and $s$ ...

$0 \le r < m,\ 0 \le s < n$?
I feel dumb compared to you guys, lol

10:08 PM
@Balarka speaking of de Rham, there's a swiss theoretical physicist in London called de Rham who funnily enough works in cosmology

@LucasHenrique I don't think that'll do it.
Don't you need something like $rs|n$ (and $rs<n$)?

@Edward in Germany we usually say De Shane

w h a t
I assume this is a joke but I don't get it

it's not good anyway

From my experience in Germany thus far I didn't expect it to be
lol

10:19 PM
@Edward, you have my permission to smack @Lukas.

lol

I was thinking of proposing a mathunderflow site to handle the periodic surge of homework questions :-)

Periodic? It seems periodic with period 0.
But I have to run off to a concert. Hi/bye @copper.hat :)

Bye :-)

@edward do you still not get it?

10:21 PM
Indeed I do not
hahaha
ah okey
ich habs
duh

can anyone help me to prove this integral?
$\int_0^{\pi/2} (\tan x)^{1/3}dx = \frac{\pi}{\sqrt{3}}$

@TedShifrin why do I need $rs \mid n$?
I mean, $s\mid n$ is sufficient (I think)
Maybe I'll have to appeal to prime factorization

Just write down your proof completely and then we'll see whether it works

I don't have one and this problem looks too dumb to waste hours
I'm sad to have spent so much time on it, to be honest

11:31 PM
Jeez, finally
Let $\ell = \operatorname{lcm}(m,n)$ with $m = \prod_{i=1}^k p_i^{\alpha_i}$ and $n = \prod_{i=1}^k p_i^{\beta_i}$. Break $\ell$ down in two relatively prime products $m' = \prod_{\alpha_i \geq \beta_i} p_i^{\alpha_i}$ and $n' = \prod_{\beta_i > \alpha_i} p_i^{\beta_i}$
$a^{m/m'}b^{n/n'}$ is our element
Could anyone check this out, please?

11:47 PM
Yup, that works
And also gives you the general procedure to find an element whose order is the lcm (granted the elements commute)
Here's what I did: Let $d=\mathrm{gcd}(m,n)$. Then $a^d$ has order $m/d$ and $b^d$ has order $n/d$, which are coprime, hence the order of $a^db^d$ is $mn/d^2=l/d$. But, by commutativity, this is $(ab)^d$ and that implies the order of $ab$ is $l$.