Mathematics - 2012-03-21 (page 1 of 4)

Jonas Teuwen

00:00

@user1123950 That's not the main goal, no.

N3buchadnezzar

@user1123950 If it can be answered within a single post =)

user1123950

It can be

But I just don't know how to solve this problem

Asaf Karagila

@tb Can you inspect it for possible mistakes?

Jonas Teuwen

@AsafKaragila Aha. So you accept it after all?

Asaf Karagila

It's a lot shorter than usually.

@JonasTeuwen I accept nothing. (I don't even accept this fact, which I am not accepting anything; now reiterate).

Mariano Suárez-Alvarez

00:01

how do you people flag questions as "invalid"?

Asaf Karagila

@MarianoSuárezAlvarez Invalid? At most I flag the flagging invalid.

Mariano Suárez-Alvarez

hm?

can you flag a flagging?

Asaf Karagila

Yes.

Mariano Suárez-Alvarez

!!

can regular users see the flags on a post?

Asaf Karagila

When a user/automatic process flags something as not answer or low quality 10k'ers can review this flag and support it or vote against it.

Then you can mark a flag as invalid.

Mariano Suárez-Alvarez

00:02

ah

cool

Asaf Karagila

For example if there is a good discussion in the comments the software will raise a flag. I can mark it invalid if I think that there is no need for a moderator intervention.

Wow, I really had fun with this question!

Mariano Suárez-Alvarez

good to know

Asaf Karagila

I learned a lot about topological properties of Hamel spaces and whatnot.

t.b.

@MarianoSuárezAlvarez It is described here. Only the standard or autogenerated flags are shown, not those including typed messages.

Hi, btw.

Mariano Suárez-Alvarez

it would be useful if sufficiently many flags as invalid would resolve the flagging automatically

Asaf Karagila

00:04

@MarianoSuárezAlvarez I think that five does that.

Mariano Suárez-Alvarez

mods need to actually clear the flag independending of the flagging on the flagging

Asaf Karagila

Maybe four or three even.

Mariano Suárez-Alvarez

I've just cleared a few with 5

Asaf Karagila

I'm guessing some of the invalids were flagged by me.

t.b.

@AsafKaragila Does the result on continuity of linear maps on Banach spaces extend to Fréchet spaces? Or how do you want to apply it to $\mathbb{R}^\mathbb{N}$?

Asaf Karagila

00:06

@tb If I recall correctly $\mathbb R^\mathbb N$ is a Bananach space...

Mariano Suárez-Alvarez

@AsafKaragila, a few :D

t.b.

@MarianoSuárezAlvarez If there are sufficiently many uncleared flags, there is a yellow alert drawing attention to that flagging tab (as for suggested edits). This alert lighted up a lot recently.

Asaf Karagila

@MarianoSuárezAlvarez What I hate about this method the most is that if I voted a flag as invalid, and it is indeed invalid I'm not getting any flag weight for it.

Mariano Suárez-Alvarez

@tb, I was in a conference for the last two weeks, so was mostly offline

so there were too few mods around

t.b.

Oh, that explains it. I didn't mean to complain :) I just tried to explain why there might have been an increase in flags and valid/invalid flags.

Mariano Suárez-Alvarez

00:08

math.stackexchange.com/questions/122657/missing-number is quite annoying as a whole... I wish it would have been closed

t.b.

@AsafKaragila what is the norm? yellowness?

Asaf Karagila

@tb Sure. Why not. Well, Gerald remarked on my confusion as well, Polish group need not imply Banach...

I should really look into that whole measure and category thing.

Either way, is there a nice property of Polish groups with the Baire property?

Jonas Teuwen

Well guys. Good night!

t.b.

@AsafKaragila yes, homomorphisms are continuous :)

Pettis theorem.

Asaf Karagila

@tb So essentially I need to change that answer to BP (which takes away the inaccessible, huzzah) and then a homomorphism from $\mathbb R^\mathbb N$ to itself has to be continuous but by cardinality games there would have to be more?

Jonas Teuwen

00:15

@tb Huh? Strongly measurable <=> weakly measurable and a.s. separately valued?

t.b.

@AsafKaragila Yes, I think this should work. Reference: Kechris, 9.C

@JonasTeuwen same Pettis, other theorem.

Asaf Karagila

Thanks!

Jonas Teuwen

Hmm... Okay. Good night.

t.b.

Good night!

N3buchadnezzar

scratches head

@JonasTeuwen sleep tight¨

Jonas Teuwen

00:18

Dream about Prokhorov's theorem?

I'll try. Thanks.

N3buchadnezzar

@JonasTeuwen about the skullpatrol in the hut theorem =)

t.b.

Try and visit a Prokhorov space. Compactness is nice!

Asaf Karagila

@tb Do you have a prepared mSE-bib entry for Kechris? :-)

t.b.

just a sec. librarian runs

@AsafKaragila is that one good enough? desired section starts on page 60 by the way

Asaf Karagila

@tb Yes.

t.b.

00:24

Theorem 9.10

Asaf Karagila

Indeed. :-)

t.b.

On a completely diferent note: seen this?

Asaf Karagila

So how does my answer stand now?

Please Read: Chat Rules

t.b.

@AsafKaragila looks good :)

Asaf Karagila

Huzzah.

t.b.

00:27

Very nice.

Asaf Karagila

Also, this all night brought me to a very impressive note which seemed to have been missed before...

N3buchadnezzar

@AsafKaragila oh ?

t.b.

@AsafKaragila Garnir's dream?

Asaf Karagila

Assume all Banach spaces have only continuous functions from them to normed spaces. Consider the space $\ell^\infty$, its algebraic dual has only continuous functionals and therefore is its continuous dual, which is $\ell^1$ under these assumptions.

Now it begs the question, it is very clear how $\ell^1$ embeds into $\ell^\infty$, but it is also obvious that this embedding is not surjective.

So we have two new things to share:

1. There is a space which is strictly smaller than its dual; and

2. There is a vector space which has an infinite dimension, and is isomorphic to its algebraic dual.

Am I right?

t.b.

I don't know what exactly you mean by embedding into its dual.

Asaf Karagila

00:34

Well, since $\ell^1$ is also a Banach space, and $\ell^\infty$ is a normed space we have that any two mappings between the two has to be continuous.

t.b.

So?

Asaf Karagila

Can there be a continuous linear bijection between $\ell^1$ and $\ell^\infty$?

t.b.

I don't think so.

Asaf Karagila

There you have it.

t.b.

$\ell^1$ is separable and $\ell^{\infty}$ isn't. At least if you have the Baire category theorem at hand, you'd get a contradiction to the open mapping theorem.

Asaf Karagila

00:37

Well, every set has the Baire property...

t.b.

You essentially reproduced Väth's theorem now.

Anyway I should go now. See you soon!

Asaf Karagila

Wait, no!

@tb What do you mean by that?

t.b.

@AsafKaragila this

Asaf Karagila

So there is no continuous linear bijection between the spaces.

Hooray! I win! :-D

Oh, wait. I'm not winning yet.

Drats.

t.b.

What's up?

Asaf Karagila

00:42

Why is it that we can drop the topology and not add more functions?

t.b.

What do you mean?

Asaf Karagila

I mean, under AC $\ell^1$ and $\ell^\infty$ are algebraically isomorphic: both are real vector spaces of dimension $\frak c$.

I keep forgetting about the norms.

t.b.

Problem is that in Pettis theorem you need the target to be Polish which $\ell^\infty$ isn't.

Asaf Karagila

Oh well, I will find that space which is isomorphic to its double dual someday...

N3buchadnezzar

Double duality all the way!

what does it mean ?

Asaf Karagila

00:45

Wait wait wait wait.

t.b.

@AsafKaragila I really have to go now. We can discuss tomorrow.

Asaf Karagila

We will.

I got totally confused.

Some REM sleep will help.

t.b.

Me too :)

Asaf Karagila

Hi @azarel.

azarel

Hi there

Asaf Karagila

00:46

I was wondering, what do you study?

azarel

I did my PhD on set theory

Asaf Karagila

Oh? I'm intrigued.

azarel

I'm in my first year as a postdoc

Asaf Karagila

Where? (if you don't mind sharing.)

azarel

I did my PhD at the University of Toronto

Asaf Karagila

00:49

With Stevo?

azarel

yes

Asaf Karagila

Whenever someone would talk about "Walks on ordinals" I would ask if this walk is a silly walk...

azarel

rofl

Asaf Karagila

Once I was going to a seminar with my advisor and I explained to him the meaning of this repeated question which no one really hears me say, and told him I'd love to ask Todorcevic this question sometime. He said that he won't get it...

azarel

you should ask anyway

Asaf Karagila

00:53

Of course I should.

I wish I could come to YST2012 and do that.

Alas, I have no funding and my advisor has no grants this year...

azarel

That sucks

I just got my ticket for Luminy

Asaf Karagila

You're giving a talk, aren't you?

azarel

yes, I'm

Asaf Karagila

So it makes a lot of sense that you could actually arrive as well! :-D

azarel

I guess

well, It was nice talking to you but I need to go

Asaf Karagila

00:58

I should hit the hay as well.

3am...

azarel

maybe we can chat another day

Asaf Karagila

Well, be seeing you.

azarel

take care

Fortuon Paendrag

@AsafKaragila 3 Am! 0_o. Where in the world are you? Its only two here in hungary.

Asaf Karagila

Israel.

Fortuon Paendrag

01:01

Ah! That would explain things.

Asaf Karagila

Like my fluent Hebrew? :-)

Fortuon Paendrag

I was thinking more like your last name, but that works too ;)

Asaf Karagila

:-)

What would you make of that name?

I mean, it's considered a bit weird in Israeli standards... :-)

Fortuon Paendrag

It sounds vaguely Middle-Eastern to my indian ears..

And Asaf definitely Is, right?

Asaf Karagila

Yeah.

N3buchadnezzar

01:06

Kari gunnlaugsdottir

Mariano Suárez-Alvarez

02:56

Is Karaglia at all common in .il?

Kannappan Sampath

Hi @Mariano @Akhil

Mariano Suárez-Alvarez

hi

user19161

Hi @kan!

Kannappan Sampath

Hi @Will

user19161

@KannappanSampath You know how to use synctex on TeXmaker right?

Kannappan Sampath

03:01

@WillHunting Yes, I do. : )

user19161

@KannappanSampath Good, people might not be aware of this cool feature!

Kannappan Sampath

@WillHunting But, I use that from time to time.

user19161

@KannappanSampath I thought you can only change your username once in 30 days? How did you manage to change it again from profile to be deleted?

Kannappan Sampath

@WillHunting I had to take help from a moderator...

I posted it as a question on Meta.

user19161

@KannappanSampath Ah I see. Hehe. I think I will change back to myself in a couple of weeks.

Kannappan Sampath

03:07

@WillHunting Oh, sure. It was good. : )

user19161

Once I ate some really hot chilli in an Indian food outlet. It was unbearable for a whole day.

user19161

No matter what I drank or ate it was still terribly hot.

user19161

I think the name of the place is Muthu's curry.

user19161

Anyway I like tosai quite a lot!

Kannappan Sampath

I am not sure what that curry is! : (

user19161

03:10

It is just the name of the owner!

Kannappan Sampath

@WillHunting Me too. (It is spelt dosai (dosa, often) BTW).

user19161

@KannappanSampath How is it read? Doh-sah-ee?

Kannappan Sampath

@WillHunting ee looks a bit long. Just Doh-sa-i

Jeff

03:24

Hi folks. I'm just poppin' in for a quick question. Can anyone explain the $\frac 12* \frac 14$ term on the top in this answer: math.stackexchange.com/a/122774/22544?

$P(A \cap B)$ equals a 50-50 chance that both are white ($\frac 12*1$) plus _____ ($\frac 12*\frac 14$) (fill in the blank pls).

Kannappan Sampath

Look at EMS's answer. His notation makes it transparent. :-)

Jeff

@KannappanSampath you're right. thx

Kannappan Sampath

Np

David K.

Can someone help with a quick topology question?

Kannappan Sampath

What does it concern?

David K.

03:33

Homeomorphisms

Kannappan Sampath

Let me try then. But I am not sure I know them thoroughly.

David K.

I'm trying to show that $f^{-1}(Y-V)$ is closed in $X$ where $f:X\to Y$ is a homeomorphism, and $V\subset Y$ is open. I have that $f^{-1}(Y-V)=f^{-1}(Y)-f^{-1}(V)=X-f^{-1}(V)$. Since $X$ is both closed and open, and since $f^{-1}(V)$ is open, can I conclude that $X-f^{-1}(V)$ is closed so that $f^{-1}(Y-V)$ is closed in $X$?

Kannappan Sampath

Fine. Good!

David K.

@KannappanSampath Yes. Sorry for the edits and ambiguity.

@KannappanSampath Ok, thanks! I thought so, but it kinda seemed too easy.

Kannappan Sampath

Yes, You're right. :-)

David K.

03:38

@KannappanSampath Great. Thanks!

Kannappan Sampath

04:26

Hi @TheChaz Sorry. I did not see you posted. Hope you got notified of my later message.

The Chaz

I did. Like I said, I'm not that active on chat : )

Kannappan Sampath

Should this be voted to close? As what?

The Chaz

as retarded

Kannappan Sampath

:-) As too localized? Because retardation cannot be a global phenomenon?

The Chaz

It's surprisingly ubiquitous...

t.b.

04:32

Hi guys

Kannappan Sampath

Hi @tb

The Chaz

Howdy. I was hoping to find you here!

t.b.

What's up?

The Chaz

Just the Catalan conjecture

t.b.

But first I need to get something off my chest: interesting how a comment on SE can become an upvoted answer on MO :)

@TheChaz I thought Mihailescu proved it some 10 years ago...

The Chaz

04:34

Hah. And hah.

Hopefully my first answer on MO will be more original...

Kannappan Sampath

@tb Holy! You can sue him for plagiarism, I think! : )

t.b.

Well, Alex Eskin is a cool guy, he knows what he's talking about :) I find the guy who needed to Google that and came up with a crappy reference much more interesting.

The Chaz

Right... I saw that. 10 minutes after Alex's answer

t.b.

scratches head

So now the guy accepted the answer already? Interesting. If he was satisfied by Wikipedia then why on earth did he crosspost to MO.

Mariano Suárez-Alvarez

that MO question is answered by pretty much any minimally related google search one may attempt :/

t.b.

04:38

Mysteries over mysteries over mysteries

@TheChaz So what was this with Catalan? You mean the follow-up-comment-posted-as-answer-now-question?

What about it?

The Chaz

Yes. Maybe it's the liquor, but many things on MSE have struck me as funny tonight.

Mariano Suárez-Alvarez

Dio-equation... sigh

The Chaz

I don't know... how bad the question is, on many levels ??

Kannappan Sampath

@MarianoSuárezAlvarez I left a comment for the OP.

math.stackexchange.com/questions/122798/…

Mariano Suárez-Alvarez

thanks!

t.b.

04:45

@TheChaz I saw your alter ego advertisement just before something was deleted :)

The Chaz

Oh, yeah! Too bad he deleted it... my comment could have been an answer on Math Underflow

*Undertow

t.b.

That was one for the gallery. But the one on chess wasn't bad either.

The Chaz

Chess?

t.b.

Oh, you missed that one?

Aside Will Hunting is cycling in and out.

The Chaz

Something @Mariano might know about...

t.b.

04:50

The first one was right.

The Chaz

Nice.

t.b.

@TheChaz let's delete that

Will Hunting stabilized!

The Chaz

How many rep until I can see deleted questions?
I don't know if my mathematical career will last that long...

t.b.

10k

Another thing for the gallery

The Chaz

; )

Hmm... strange!

Kannappan Sampath

05:02

Can Someone tell me if I should I add proofs for all my claims, please?

The Chaz

@t.b. I added a comment on meta about how we have trained certain users to have a 100% accept rate, yet they continue to ask the worst questions...

Is there a consensus about naming names?

t.b.

Yes, it is probably better not to. There were a few kerfuffles due to that.

By the way: the link went to chess.com

The Chaz

Ah.

t.b.

I'd have sent him to FIDE

The Chaz

A purist :)

t.b.

05:09

No, but they will react to such questions as we do to proofs of Collatz and Fermat

@KannappanSampath No, certainly not. I would assume they proved in class that $A_5$ was simple.

The Chaz

Charles usually deals with the quacks who post proof of FLT on the other forum

t.b.

Here we have Gerry

Kannappan Sampath

@tb We were assigned to prove that a simple group of order $60$ is $A_5$ and my god, I proved it after a week!

t.b.

(and Charles)

Kannappan Sampath

@tb And, most certainly, a good news to my laziness. : ) Thank you!

t.b.

05:13

@KannappanSampath Oh, sure it isn't that easy. I would guess that either you know that fact or you need to do some Sylow stuff for the other ones.

Kannappan Sampath

@tb Let me get back to this after class. I don't understand this remark though. : (

t.b.

@KannappanSampath well, you only need to think about 60 and 98. I would assume that a little fiddling should exclude the 98.

Kannappan Sampath

@tb You did not read my answer. : (

t.b.

No I didn't.

Yes, you confirmed my hunch :)

This strikes me as a terribly bad idea

The Chaz

TL;DR

Because I'm about to pass out.

See you guys later.

t.b.

05:23

See you!

Kannappan Sampath

@tb Can you help me see why the kernel of nilpotent operator is not just $\{0\}$. I don't see this at all.

(On a finite dimensional vector space.)

t.b.

What is the definition of nilpotent? :)

Kannappan Sampath

$T^r \equiv 0$ for some $r \in \Bbb N$.

t.b.

So, can $T$ be injective?

Or: take the minimal such $r$. If $r=1$ done. Otherwise $T^{r-1}v \neq 0$ for some $v$.

Kannappan Sampath

Yes, right!

Thank you. I seem to be missing out a lot !

t.b.

05:29

It's just a matter of practice :)

Did you get both solutions?

Kannappan Sampath

Meaning, " why it cannot be injective? " I got the other one.

t.b.

Yes.

Kannappan Sampath

OK. If $v \neq u \implies T(v) \neq T(u)$, then, we have $T^r(v) \neq T^r(u)$. A contradiction, right?

(For some $r$ greater than the minimal $r$ for which $T$ is identically $0$.)

@tb Is the above thing fine?

t.b.

That's also right. A third option would be to argue that in finite dimensions a linear operator is injective if and only if it is surjective.

Hence it would be bijective and thus $T^r$ can't be zero.

Kannappan Sampath

Ah! Right.

I have proved this using dimension formula.

t.b.

05:35

I don't like the last one (third option) because it needs finite-dimensionality. The other ones go through without it.

Kannappan Sampath

But, the dimension formula is true irrespective of dimensions right?

t.b.

That's true.

Jonas Teuwen

07:00

Hmm...

Matt N.

07:16

@tb 3.5 hours is not what I had in mind.

Kannappan Sampath

07:29

@MattN Hi!

Matt N.

Hi Kannappan.

Kannappan Sampath

Did you look at the CA room?

Matt N.

No, I need to get some additive combinatorics done. But I will soon, hopefully on Friday.

Kannappan Sampath

@MattN Alright. Good luck with Combinatorics. I like that so much but we have no course now. : )

Matt N.

Thanks : )

Benjamin Lim

07:38

@KannappanSampath Hey

Kannappan Sampath

@BenjaminLim Hi!

Benjamin Lim

I went to see my supervisor today. I told him about B. Sury

he was stunned !!!!

There was a big smile on his face. Indeed you and I do have a connection!!

Did you tell your supervisor something like that too?

Kannappan Sampath

Oh, Nice. What did he have to say?

Benjamin Lim

he smiled and stuff and asked how I met you and things

Kannappan Sampath

@BenjaminLim I went to meet him but he did not come to his office that day for some reason. So I could not meet him! : (

Benjamin Lim

07:40

Did you tell your supervisor??

oh

Kannappan Sampath

I hate it when smilies split but not exact sequences. :P

Hi @Zhen

Zhen Lin

hi

Benjamin Lim

@ZhenLin I found out today a short proof of how $A \otimes M \cong M$ using universal property

Zhen Lin

Very good. Do you understand tensor products better yet?

Benjamin Lim

yeah

For example

I came up with an example of how tensoring does not usually leave a sequence exact

even though the original one was

Zhen Lin

07:46

Good, good...

and are you convinced yet that category theory is worth learning? :p

Benjamin Lim

hahaha

too abstract for me for now

Zhen Lin

You haven't even tried learning it.

Benjamin Lim

well learn be do algebraic topology first

Zhen Lin

Universal properties are part of the language of category theory. If you can understand those, then you are ready enough to learn category theory.

Kannappan Sampath

08:09

Hi @anon

anon

yo

Kannappan Sampath

Oh, I did not notice that I just had to get 10 upvotes for breaking 5k! : )

So far So good.

anon

puns puns puns

Kannappan Sampath

08:24

:-))

anon

that joke took way too much time

Zhen Lin

I don't even know what an adele is, and I took a course in local fields!

Asaf Karagila

Isn't that the reduced product of all $p$-adic integer rings or something?

anon

plus R, yes (I'm familiar with the term "restricted" product)

Asaf Karagila

Well, one of my office-mates works with the adeles and he told me that you always assume that $\mathbb R$ and $\mathbb Z_2$ are wiped out in the reduction. :-)

anon

08:30

well, restricted product of the p-adic fields $\mathbb{Q}_p$ - restriction means cofinitely many are p-adic integers.

I'm reading about the character group of $\mathbb{Q}$ right now.

Asaf Karagila

Is it fun?

anon

Yes, but short. $\widehat{\mathbb{Q}}\cong \mathbb{A}_\mathbb{Q}/\mathbb{Q}$!

Asaf Karagila

Where $\mathbb A_\mathbb Q$ is?

anon

$\mathbb{A}_\mathbb{Q}$ is in New York maybe? I don't understand your question.

3

Asaf Karagila

I haven't had my coffee yet. :-D

In fact just this moment I brought something to drink at all.

I meant to ask "What is $\mathbb A_\mathbb Q$?"

anon

08:44

It's $\displaystyle \mathbb{R}\times \prod^{res}_p \mathbb{Q}_p$, where the restricted product means cofinitely (all but finitely) many of the $a_p$ components are p-adic integers i.e. in $\mathbb{Z}_p$.

Asaf Karagila

Why do you have the need to explain to me what "cofinitely many" mean? :-)

anon

Because that's the first time I've used the word :P

Asaf Karagila

You have used it a few lines earlier.

anon

Okay, $*$this conversation$*$ is the first time.

Asaf Karagila

Ah. :-)

anon

08:47

Also, $\mathbb{Q}$ diagonally embeds in $\mathbb{A}_\mathbb{Q}$ as $r\mapsto (r,r,r,\cdots)$.

Asaf Karagila

Clearly.

anon

Because the only primes $p$ for which $r\not\in\mathbb{Z}_p$ are those dividing the denominator in $r$'s reduced form.

Asaf Karagila

Yeah, this is why $\sum\frac1n$ doesn't converge on any completion of $\mathbb Q$. Either it's the real numbers or $\frac1n$ doesn't approach zero. :-)

Ilya

@anon who are them?

anon

Adele.

Ilya

08:51

Do you love this name especially?

anon

I think it's cool, I guess, why?

Ilya

It's cool, yeah - I just wondered why did you post it

Asaf Karagila

Because there is an algebraic structure called "Ring of Adeles".

anon

Because "I'm reading about the character group of Q" which is $A_Q/Q$, a quotient of the ring of adeles.

Asaf Karagila

What anon did was to construct a nonstandard model of that ring. :-)

Ilya

08:52

@AsafKaragila icic

Asaf Karagila

Let's count how many puns I made in that previous message of mine...

I think about $\pi$.

Zhen Lin

I'm not sure that "reduced product" in his sense is the same as in your sense...

Ilya

time to be polite. Good morning @Asaf, @Zhen. Good night @Anon

Zhen Lin

Good morning.

Asaf Karagila

@ZhenLin Reduced product just means that you divide by a filter rather than an ultrafilter. The cofinite subsets of $\omega$ form a filter, well they did that the last time I checked.

@Ilya If you're being polite I guess I'll have to be rude... bite me! :-P

Zhen Lin

08:56

Eh, why divide by a filter? I thought the point of taking a quotient was to get a 2-valued model.

Asaf Karagila

Well, sometimes you want more.

The usual product topology is a reduced product.

Ilya

@AsafKaragila trfying to kfeep tfhe bfalance, aren't fwe? :) [a little accent due to biting Asaf]

Transcript for

$Mathematics$ Mathematics