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11:00 PM
@MeAndMath Eu nunca soube se aplicada é só aplicada e bacharel é pura e aplicada ou se bacharel é mais pura.
 
@t.b. I read that as "weak-ass" topology
 
@GustavoBandeira é ,a "aplicação".
 
@t.b. Yes.
I should do something else, I am so unproductive for weeks! 8-).
I hate to be ill.
 
@GustavoBandeira tem bacharel em matemática,bacharel em matemática aplicada,licenciatura em matemática,bacharelado em estatística e bacharelado em ciência da computação.
 
@JonasTeuwen so what's the problem?
 
11:02 PM
There probably is none. The problem is Kopf.
 
@MeAndMath Ah, tem bacharel em matemática e bacharel em matemática aplicada. Aqui só tem bach de matemática.
 
Oh, I know that problem. Too well.
 
I spent five hours figuring out why $P(\mathbf N) = \mathbf R$. Apparently.
 
@GustavoBandeira estranho
 
Ah, yes. Is it mathematicians thingie...?
 
11:03 PM
@JonasTeuwen Does that mean "powerset"?
 
I don't care if it does.
 
@JonasTeuwen What do you mean by $P(\bf N)$???
 
Perhaps the Cech-Stone compactification? Or the ultrafilters?
 
11:04 PM
Mas deve ser normal, isso aqui é Nordeste.
 
dá uma olhada
 
É completo demais. Tem até:
Matemática Aplicada - Bacharelado - Habilitação em Pornografia quântica.
Matemática Aplicada - Bacharelado - Habilitação em depilação estocástica de pelo de axila de primata.
 
Menschen geben mir Kopfschmerzen.
3
 
@GustavoBandeira rsrsrsrsrs
 
@MeAndMath Eu quero comparar os cursos.
Parece que num é muito diferente não.
 
11:11 PM
é?
 
Pelo que eu me lembro das matérias, parece não ser muito diferente.
Cê já teve topologia?
 
@GustavoBandeira LOL
 
@GustavoBandeira estou estudando por minha conta agora...
 
@MeAndMath wait, how do you guys laugh?
 
@PeterTamaroff LoL is acceptable.
 
11:13 PM
@PeterTamaroff hahahah rrsrsrsrsrsr kkkkkkkk LOL sahushaushaushuash
 
@MeAndMath huehuehuehue is also acceptable
 
@MeAndMath why "rsrsrsrsrsrs"?
 
@PeterTamaroff rs = risos
risos = laughs
 
@GustavoBandeira look at that
 
@MeAndMath Ah é? Idem, mas tu deve saber MUITO mais que eu.
@PeterTamaroff What?
 
11:15 PM
@GustavoBandeira I don't know...
 
@MeAndMath Cê sabe o que é um anel?
@MeAndMath É uma coisa que põe no dedo. =D
@MeAndMath Eu fui na faculdade falar com uns professores... O cara me fez essa pergunta: Me segurei pra não responder issso.
 
@JonasTeuwen can you explain the notation?
 
@t.b. In terms of cardinalities.
 
@MeAndMath Como foi? Tu fazia o curso e saiu?
 
@JonasTeuwen And P(N) is what?
 
11:18 PM
Just $2^{\mathbf N}$.
Uh, wait.
No, I mean...
Or wait, I do mean.
That's the Cantor set right? Decimal expansions.
 
yes.
ternary expansions, though.
 
So, for sets it is $\mathbf R$. I was trying to figure out that if $Y$ is a basis for a Hausdorff (compact?) topology if the space could maximally be $P(P(Y))$ in terms of cardinality. I think so, through filter bases, but... Kopf.
 
I'm still at a complete loss about what you're trying to do.
 
Oh, working on some game theory. One moment, trying to recall. Need some fuel.
 
@JonasTeuwen Do you mean this?
 
11:22 PM
@GustavoBandeira eu sei ,sim.não ,eu não saí.
 
@JonasTeuwen Does the $P$ stand for powerset? (Why won't you tell me?)
 
I want to figure out what the additional conditions on $X$ other than compactness should be for the following to hold: The induced topology on $\mathcal M_+(X)$ by the vague topology is locally compact (yes), separable and metrizable.
@t.b. Ah, cool, The Ilya!
 
@MeAndMath Tá fora por causa da greve?
 
Last week he was complaining he had was so unproductive, I asked a bit. Sounds like me 350 days per year. 8-).
 
@GustavoBandeira não,não tem greve aqui não!
 
11:24 PM
@MeAndMath Então o que o faz estudar por conta própria?
 
só estou estudando sozinha mesmo.
 
I know the answer to his question.
 
topologia não faz parte da grade e eu prefiro.
And because Peter is here to help!
 
We happen to have all kinds of crazy compactification guys here... van Mill, KP.
With his ultrafilters on $\mathbf N$.
@PeterTamaroff Sorry, it is the powerset. Sorry, it is not personal that I did not reply.
 
@MeAndMath I can't follow portuguese =D
 
11:27 PM
@PeterTamaroff He was asking me why was I studying topology by myself and I told him because I wanted and because you can help me !
 
@MeAndMath Oh, yes. math.SE is here to help.
 
@PeterTamaroff I'm happier because of SE!I'm learning more!
 
@JonasTeuwen The map $x \mapsto \delta_x$ is a homeomorphism onto its image in quite some generality. So I'm pretty sure that it is necessary for $X$ to be metrizable and second countable.
 
Wait, what is a non-complete separable metric space? How does the separable set relate to the completion? It just is...?
 
@MeAndMath It is a cool page, yes.
 
11:28 PM
@t.b. Yes, I was afraid it was. (same argument)
The diracs always fuq things up.
And there is no good way to exclude them as they are Radon.
 
@PeterTamaroff People help me!Answer my questions!We talk about math (mostly) all the time!AND I LOVE IT!
 
@JonasTeuwen well, you could try to pick a measure on $X$ and look at only those measures which are absolutely continuous with respect to it.
 
no one has been able to answer my question
 
@t.b. Yep, I will try that next.
 
I need to know how to solve these types of questions
 
11:30 PM
@JonasTeuwen so you'll get $L^1$ :)
 
@Khromonkey What question?
 
Also I had stuff like $f \in L^p(X, \mu_i)$ for a collection of $\mu_i$ probability measures.
 
@t.b. Is $$\left(\sum_{i=1}^n (x_i+y_i)^p\right)^{1/p}\leq \left(\sum_{i=1}^n x_i^p\right)^{1/p} + \left(\sum_{i=1}^n y_i^p \right)^{1/p}$$ hard to prove?
 
@t.b. Yeah, but that is also quite ugly...
My beloved Ornstein-Uhlenbeck. Is degenerate on $L^1$. That broke me.
 
11:33 PM
@Khromonkey Is that $p$ something like probability?I still didn't understand tha $p$.
 
@PeterTamaroff no, not really.
 
That's just Hölder.
 
no, i just used it to say that all of the elements of a are $\leq 100$ and $\geq 1$
 
I like the integral inequality more as mixed norm inequality.
 
user19161
Pedro is trying to prove every theorem himself perhaps, which is good.
 
11:34 PM
At this harmonic analysis conference people had a hard time recalling the order. "But it is just a mixed norm inequality!".
 
@t.b. I only know how to prove Cauchy-Schwartz, and the case $p=2$ follows from that.
 
@Khromonkey Ok.in a set of 16 distict elements.
 
@PeterTamaroff Well, try for $n = 2$ and $p$ not $2$.
 
yes
 
user19161
@JonasTeuwen What, they don't know the holder or minkowski?
 
11:35 PM
@Khromonkey Give me a sec.I will think.
 
@JonasTeuwen And then generalize?
 
@PeterTamaroff Then try to see what is the problem.
That is what all non-Grothendieck mathematicians do. If you cannot prove it add more conditions until you can.
 
@JonasTeuwen What do you mean?
 
user19161
if they can't remember holder or minkowski @jonas they have no business at a harmonic analysis conf
 
@JasperLoy No.
Damn it.
No energy for discussion.
 
11:38 PM
I have to use all of those for my math olympiad
 
user19161
seriously, what do these conference people know these days?
 
The best in the world.
 
user19161
@JonasTeuwen time for bed then
 
@PeterTamaroff It's Cauchy-SchwarZ (no t please). Why didn't Jasper correct that already?
 
It is mental energy, not tired.
@t.b. And also not Ornstein-Uhlbeck?
 
user19161
11:39 PM
time for coffee then or teae
 
@JonasTeuwen you're the expert :)
 
why dont you all try to solve my problem instead of fight?
 
user19161
@t.b. I was thinking which it was. I got confused with the distributions.
 
@t.b. Yes, I know.
 
@t.b. I wish. But apparently I am since the other experts are lacking a bit.
 
11:40 PM
@PeterTamaroff then why don't you apply your knowledge?
 
@t.b. Just a slip of the mind =)
 
@JasperLoy I mean the mixed norm Minkowski.
For general measures.
 
As Garling quipped: The most difficult thing about Cauchy-Schwarz is how to teach the students how to pronounce Cauchy and how to spell Schwarz.
 
Did you read his inequalities book (Garling)?
 
@Khromonkey if you have 100 numbers and 16 that form a set ,it gives a possibility of 1345860629046814650 different sets.Is that it?
 
11:42 PM
Cauchy, that's not so hard is it?
 
@JonasTeuwen Parts of it. It's nice.
 
@t.b. Hahhaa. I know how to pronounce Cauchy!
But you'd die if you hear how people generaly pronounce it here
 
user19161
My favourite proof is a one line proof that tb said was from the master Hardy himself.
 
@t.b. I think I took it from my advisor. I don't know how it got on my book shelf.
 
user19161
It just involves completing the square.
 
11:42 PM
Cauchy is just French - easy.
Schwarz have to say it like Goebel (in the speeches).
Wollt ihr...
 
@JonasTeuwen They generally sound like this Just press the speakerphone
 
@PeterTamaroff O.o
 
@PeterTamaroff What on earth?! The english one is comparably bad though..
 
@FortuonPaendrag What on **E**arth what?
 
The worst thing ever is that I heard somebody from Bordeaux pronounce "Hermite" like... an Australian.
 
11:44 PM
@MeAndMath i think there are $\frac {100!}{(100-16!)(16!)}$ but how does that solve the question?
 
And... then a German was commenting on that.
 
@Khromonkey It doesn't.I was just thinking.well,at least one of them satisfies the condition...
AFK.BBL
 
@t.b. Would it be too crazy raise the ineq to the power of $p$? and use the binomial theorem?
(Twice?)
 
@JasperLoy I thought the one you like particularly is the original one by Cauchy: part 1 part 2.
 
How did people typeset integrals and so on preTeX?
@t.b. Jan van Mill's thesis was on a typewriter. I have seen it. In the introduction he does not say "thanks to my advisor van Douwen" but thanks to my coauthor...
 
11:47 PM
there were other apps available
before that, typewritters
there is a MO question on the subject
 
Cauchy? Looks better than a typewriter.
 
can someone recommend me a book on combinatorics that might help me answer my problem?
 
user19161
What did they use before typewriters?
 
@PeterTamaroff start with $\lVert x + y \rVert_{p}^{p}$ on the left.
 
11:49 PM
from gutenberg onwards that's pretty much it
 
@MarianoSuárez-Alvarez hehee
 
user19161
I wonder if LaTeX will ever be replaced by something else in a few decades.
 
user19161
Maybe Microsoft Word would have gotten so advanced that it has a proper superset of TeX functionality.
 
@t.b. OK.
 
@PeterTamaroff then use that $|x_i + y_i|^p \leq (|x_i| + |y_i|) |x_i +y_i|^{p-1}$. Split up into two sums, use Hölder.
 
11:52 PM
@JasperLoy Then... that would be the end of the world.
 
user19161
@JonasTeuwen Which some say is 2012.
 
@t.b. I don't know Hölder... maybe I should prove that first?
 
Yes. But that is Young's inequality.
 
@PeterTamaroff sounds like a good idea. It's a substitute for Cauchy-Schwarz when $p \neq 2$.
 
@t.b. Oh, OK.
 
11:55 PM
How can i contact brian scott
??
 
user19161
@Khromonkey What do you mean by contact?
 
@Khromonkey Just wait till he shows up.
 
have him read my text
 
user19161
@Khromonkey If it is a question you can just ping him on the main site.
 
he was helping me solve my problem
 
user19161
11:56 PM
@Khromonkey But why only him? Because he stopped halfway?
 
@t.b. Is the absolute module necessary in Hölder? Cause CS uses no absolute modules.
 
He is the only person who is interested in it
 
user19161
@Khromonkey Then it must be something quite advanced?
 
@Khromonkey Link?
 
@PeterTamaroff well, it's sharper, as $\left| \sum x_i \right| \leq \sum |x_i|$
so, no.
 
11:57 PM
Its going to dissapoint you but here it is math.stackexchange.com/questions/186866/…
 
user19161
@Khromonkey Just ping him on the post itself.
 
@t.b. What do you mean by sharper?
 
@JasperLoy Is that equivalent to " Your problem sucks im not wasting my time on it"?
 

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