Mathematics - 2016-11-29 (page 3 of 4)

Ted Shifrin

7:00 PM

@MikeM: Aren't you high in the air or something?

Mike Miller

I'm the air. The flight might be better if I was high, too.

Ted Shifrin

Oh, shelling out money for WiFi, eh? :D

Hopefully there's no one misbehaving as on the Delta flight yesterday (which cost Delta a bundle since they refunded all ticket purchases but for the troublemaker).

Hi @MartinS

Martin Sleziak

Hi!

Since I am trying to get the bounty room going, perhaps a reasonable thing could be to promote it here. So here is link to the relevant meta post and here is the room.

Ted Shifrin

A bounty room, huh? I've never played much with bounties, but maybe I should be less antisocial :)

Martin Sleziak

Well, so far not much content was posted there. But maybe if more people know about the room it will get going.

Ted Shifrin

7:04 PM

I'll check it out :)

Martin Sleziak

In short, the intention is that people could post there questions for which they offered bounty but did not receive sufficient answer.

Potentially, if enough people do this, the room might contain list of non-trivial (interesting ?) questions.

Ted Shifrin

I guess we would all have to go check in there and see if something interests us or is in our sphere of competence. Do the tags work as usual in there, I assume?

Martin Sleziak

Well, you can search the room for occurrences of the tag you're interested in. So far, there were two questions about topology. Nothing with tags containing the word geometry.

Adeek

@TedShifrin

hi

Martin Sleziak

As I said, only three questions have been posted there so far. If more people know about that room, maybe eventually it will have more content. (But maybe I am too optimistic.)

Ted Shifrin

7:08 PM

Hi Karim.

Danu

Hey @TedShifrin :)

Ted Shifrin

Hi, @Danu.

Adeek

I was wondering for theorems like this one. Let X be normeld space $Y \subset X$ then for all $f \in dual(X)$ $\| f|_Y \| = dis(f,perp(Y)).$

Ted Shifrin

@MartinS: I will have to figure out an easy way of getting to the bounty room to look at it frequently.

Adeek

I mean the proof I understand but how do people figure this stuff out ?

do they see geometry in infinite dimensional or what haha

Danu

7:09 PM

I'm indulging myself aesthetically with TikZ :)

Results will come soon!

Handlebody stabilization

Ted Shifrin

Karim: Normally, as with finite dimensions greater than $2$ or $3$, one derives some intuition (sometimes wrong) from pictures one can see and things in finite dimension. Is $Y$ a closed subspace of $X$ in your statement?

Adeek

Yeah

Ted Shifrin

@Danu: I've never progressed past Mathematica and Adobe Illustrator and \includegraphics :P

Adeek

yeah I mean finite dimensional picture makes sense

Ted Shifrin

Karim: So a linear functional is given by dot product, for example, and then it is just basic geometry in finite dimensions. Now you have to figure out why it generalizes.

Adeek

7:11 PM

I see

Danu

@TedShifrin Preview:

No, screw it

I'm only going to show the end product

Ted Shifrin

(In finite dimensions or in Hilbert space.) So you're moving to less structure. Perhaps a bit surprising then ...

I won't look, ever, @Danu.

Martin Sleziak

@TedShifrin There are various ways to access chat. One of them is to display your favorite rooms. You can add a room to your favorites by clicking on the star in the room info/room description.

Ted Shifrin

Have you given your Euler class talk yet?

Danu

No, next friday. But I prepared it now

Ted Shifrin

7:13 PM

Ah, thanks for the "adding to favorites" tip, @MartinS. :)

Danu

I'm actually going to prove that e is PD to generic zero section.

Martin Sleziak

I think that using favorite rooms is a relatively easy way if you want to know about several rooms whether something new was posted in them.

Danu

I'm afraid I'll lose all of the audience in that proof though. Cause it's a bit technical.

Ted Shifrin

Excellent, @Danu. I'm with MikeM that that is what people should come away remembering.

Danu

Yeah, me too. That's why I wanted to prove it.

Daniel Fischer

7:13 PM

Evening @Ted.

Ted Shifrin

Well, Thom class is technical. But it's all about intersection theory.

Danu

Also, I didn't know how to prove it a priori.

We already have the Thom class. But even from there, it is annoying to get to e

LeGrandDODOM

@Ted hey

Adeek

I see @TedShifrin it would be awesome to see the intuition for infinite dimension I mean in some particular cases I can see the intuition directly but mostly not.

Ted Shifrin

Before Bott/Tu did it in their book, the folklore result that cup product is Poincaré dual to transverse intersection appeared in no book. My algebraic topology prof spent a week trying to prove it (because I kept bugging him) and his proof collapsed the last day. @Danu

Danu

7:14 PM

Good times.... :P

Ted Shifrin

Karim: Often things fail in infinite dimensions, especially without Banach/Hilbert assumptions, so it's interesting that it generalizes, I suppose.

Danu

How do Bott & Tu prove it?

Ted Shifrin

@Danu: Thom class :D

Danu

The proof I am doing is a bit different from what appears in MS for instance

It of course uses Thom class

But in a different way

Ted Shifrin

You should look at Bott/Tu, btw. I should have mentioned that earlier.

Danu

7:15 PM

I have

Not much.

But I might do it more at some later point. Not now.

Ted Shifrin

They have much better proofs than MS.

Of course, they base things on deRham, not Steenrod squares (ICK).

Adeek

Is there a good book that talks about this kind of stuff like when things generalize to infinite dimension ? I am interested to read something like that ?

Ted Shifrin

I honestly don't know, Karim, since I don't think about that sort of thing much.

Danu

@TedShifrin But de Rham theory doesn't work with $\Bbb Z$ coefficients.

Ted Shifrin

Mostly I don't care, @Danu :D

Danu

7:17 PM

Meh

I do feel like it's preferable to keep everything topological rather than differential, from a first principles POV.

Ted Shifrin

Griffiths, Morgan wrote a really interesting book on rational homotopy theory (based on lectures in the 70s in Italy) using deRham.

Danu

Hmm

Ted Shifrin

I have it in the original handwritten notes form :) Dunno how the final Springer book looks.

Believe it or not, for some perverse reason, I scanned all those pages to .pdf when I was burning everything in my office.

Danu

Typewriter :\

Astyx

@Ted Hi

Danu

7:20 PM

hahaha :) Can't avoid sentimentality

Ted Shifrin

No, handwriting, @Danu :D

Salut, @Astyx.

Danu

You actually burned your stuff?

Ted Shifrin

Not literally. But it went to the recycle dump in the sky.

You can be sentimental when I'm gone, @Danu. I don't believe in it. :D

Danu

Hmkay

@TedShifrin When my grandfather died earlier this year, it turned out he had hoarded a lot of stuff. That was really not nice.

My father was severely disappointed, on a personal level

Ted Shifrin

You wouldn't believe how much got thrown out at both my house and my office, and I am not a hoarder. ... And I gave away many thousands of dollars' worth of books.

Danu

7:22 PM

Good on you, is what I'm trying to say.

Ted Shifrin

You can still be sentimental when I die :P

Danu

Speaking of sentimental, BEHOLD

3

fuck yes.

Fargle

Hello chat

Danu

Gotta bike home now but I'll be back later @TedShifrin!

I've actually got one small questoin about the Euler class left

Ted Shifrin

Damn good, @Danu. I could never draw that ...

hi @Fargle!!

Fargle

7:29 PM

@TedShifrin How goes it?

Danu

@TedShifrin 2 hours of work, even with already having the handles pre-defined (and considerable experience).

Ted Shifrin

My ex-colleague Will Kazez draws amazing topology diagrams using Illustrator. Absolutely amazing.

Danu

Nice. I'll check it out

Ted Shifrin

OK, @Fargle .. you learning anything this week?

Fargle

@TedShifrin Still going through Rudin, also going through D&F for some interesting exercises.

Ted Shifrin

7:32 PM

Well, I'm waiting for your next Rudin exercise set :)

Adeek

7:43 PM

hi @Fargle

Fargle

It's taking longer than I expected, haha.

And hello, @Adeek.

Adeek

I want to go through D&F and allufi chpter 0 problems once this semester ends.

Ted Shifrin

@Fargle: Focus. (Ha ha ...)

Adeek

@TedShifrin would you like to see my elliptic curve project?

Ted Shifrin

You can email it to me, Karim, but I don't have time to look at it for a while. Despite retirement, I'm still buried in grad school (and other) recommendations.

Adeek

7:49 PM

@TedShifrin at your email that is posted on mse ?

Ted Shifrin

Yup.

Adeek

sent it

if I have continous function such that $F(Span(A)) = 0$ does it mean that $F(clos(Span(A)) = 0$ ?

Xammm

Hello everyone, what are you talking about?

Adeek

@TedShifrin I am pretty sure this is correct that must be a different characterization of continuity in terms of closure.

Ted Shifrin

You should know that from basic analysis/topology. Go think.

Adeek

7:56 PM

I mean it is intuitively obvious. I keep forgetting some characterization of contuinity.

yeah ok

it is from the fact that $f(clos(A)) \subset clos(f(A))$

Semiclassical

@Danu oh hey, genus three. nice

Adeek

brb school

Mike Miller

@TedShifrin Ultimately, the best approach is just to calculate the cohomology of $BG$ and work with that... and if you like, you could do that with differential forms.

Ted Shifrin

@MikeM: Not.

Hi @Semiclassic.

Semiclassical

hi @ted

Mike Miller

8:04 PM

Your approach is literally contained in this, since an invariant polynomial in the curvature gives a differential form on $BG$, I rather believe.

Maybe even harmonic?

Tobias Kildetoft

@MikeMiller Maybe you know the answer to the question I asked Balarka yesterday (or maybe it was the day before). Take the unit cube in $\mathbb{R}^3$ and identify each pair of opposite sides with a $90$ degree twist. This gives $8$ "different" spaces depending on the choices of direction of the twists. How many of these are there when we identify them up to homeomorphism?

Also, what is the smallest $n$ such that the space embeds into $\mathbb{R}^n$ (and does this depend on the choice of twists).

Ted Shifrin

Isn't $BG$ infinite-dimensional, @MikeM?

Mike Miller

@TedShifrin Yeah. So?

Ted Shifrin

@Tobias: This sounds a lot like a problem in Hatcher. But not the counting different ones part.

Mike Miller

@TobiasKildetoft Do you always have the same fundamental group?

Ted Shifrin

8:07 PM

@MikeM: The construction I know for $BG$ isn't even giving a smooth manifold. I pass.

Tobias Kildetoft

@TedShifrin Yeah, it was inspired by recalling an exercise about that space from Hatcher (though I could not recall what the exercise was. Balarka reminded me that it was to calculate the fundamental group when all twists were counterclockwise).

@MikeMiller No idea

Ted Shifrin

So, as Mike suggests, different fundamental groups would tell you different homeomorphism type :)

Mike Miller

My guess is you'll probably get different fundamental groups but I don't have predictions on how many.

Trick Balarka into calculating them.

Tobias Kildetoft

@MikeMiller Good idea

Ted Shifrin

LOL ... He's got real school exams.

Mike Miller

8:10 PM

@TedShifrin If you have a free smooth action of $G$ on a Hilbert space, you're done. Certainly this is immediate for $O(n)$ and $U(n)$. Does every compact Lie group embed into one of them?

I could construct one in general if I worked harder but I won't.

Ted Shifrin

Yeah, sure. I guess the only construction I remember for $BG$ is the infinite join.

Mike Miller

Agreed that one's not so pretty.

Ted Shifrin

To be honest, I have no idea how to write down differential forms on Hilbert manifolds.

Much as my life is built on differential forms, they live on finite-dimensional creatures (sometimes with singularities).

Mike Miller

Nah, differential forms are fine. Still sections of $\Lambda^k T*M$. There's no top form anymore, but we're strong folks and will survive even this.

Ted Shifrin

Anything that depends on local coordinates computation is out the window. I'll have to think about my moving frames set-up and see what I think.

Heya, DogAteMy.

Akiva Weinberger

8:15 PM

Hey what's up

Ted Shifrin

Lunchtime for this bonzo.

Mike Miller

@TedShifrin Local coordinates should be fine on a Hilbert manifold, pick an ONB and expand in that. Perfectly good definition of $d$ this way. Alternatively work locally in each finite-dimensional slice. But I will admit that my preferred definition of $d$ is the invariant one (even if I never compute that way).

I've never actually worked out the story of harmonic forms on $BG$ but maybe I ought to.

Ted Shifrin

I have never thought about this, @MikeM. I don't know how much of the usual structure equations stuff (which is how I ordinarily do Chern-Weil, etc.) will make sense in this universality. If it does, I'm sure Bott already did it.

Mike Miller

Part of my current project is to do away with differential forms so I can use more general coefficients, anyway. So I'm maybe a ilittle biased.

Ted Shifrin

If you do away with differential forms, you never speak to me again! :D

Mike Miller

8:17 PM

In equivariant Floer homology, ya bozo.

Ted Shifrin

I didn't think there were any differential forms in that story.

Mike Miller

That's the only current kind of equivariant instanton homology - via differential forms (see Austin-Braam, the paper I once had that's now been torn to shreds via travel in my backpack at all times).

Semiclassical

homology via cohomology?

Ted Shifrin

interesting ...

Sometime you'll have to give me a lecture. Maybe on an actual blackboard.

Mike Miller

I can tell you the story when I actually have the freaking paper done.

Ted Shifrin

8:20 PM

Outta here for now. Bye all.

Semiclassical

later, @ted

Astyx

Bye

NaCl

8:32 PM

$A=\{x\in\mathbb{R}:x^k\in\mathbb{Z}\text{ where }k\in\mathbb{N}\}$
Is $A$ countable? I'd say yes, since it only contains numbers from $\mathbb{Z}$, so $A\subseteq\mathbb{Z}$ holds. Since there exists a bijection from $\mathbb{Z}$ to $\mathbb{N}$ it should be countable, shouldn't it?

SteamyRoot

Only contains numbers from $\mathbb{Z}$?

Semiclassical

$(\sqrt{2})^2=2...$

SteamyRoot

You'll find that $\sqrt{2} \in A$

NaCl

Nevermind, but then I'm unsure

Semiclassical

You might start with something simpler, then: $A_k=\{x\in \Bbb R:x^k\in\Bbb Z\}$

Mike Miller

8:35 PM

SemiC's hint is the way to go

SteamyRoot

^

NaCl

I'm clearly not made for this stuff, as I see literally no semantic difference

Semiclassical

With $A$, I have to consider every $k$ at once.

Astyx

It's countable is it not ?

SteamyRoot

$k$ is fixed

NaCl

8:36 PM

I know there are $\sqrt[3]{2}^3=\sqrt{2}^2$, does this yield to anything?

Semiclassical

that means that $\sqrt{2}\in A_2$ and $\sqrt[3]{2}\in A_3$.

NaCl

Yes

Semiclassical

So each $A_k$ is certainly countable.

NaCl

But we also find $-\sqrt{2}\in A_2$

SteamyRoot

You could also break up $A_k$ even further

NaCl

8:38 PM

But I can see why that doesn't matter

Semiclassical

Yeah. $A_k^+$ and $A_k^-$, for instance.

SteamyRoot

into sets $A_{k,n} = \{x \in \mathbb{R} \mid x^k - n = 0\}$.

NaCl

god, I'm so lost with this... I really wonder why we don't have a tutorial-course

SteamyRoot

where $n \in \mathbb{Z}$ is fixed.

Semiclassical

It'd be simpler had they said $x^k\in\mathbb{N}$.

Wait, nm

wrong statement

NaCl

8:40 PM

it doesn't matter

Semiclassical

Yeah.

NaCl

Since there is a bijection from $\mathbb{Z}$ to $\mathbb{N}$

SteamyRoot

How many $x$ can be in the set $A_{k,n}$, at most?

Semiclassical

That's a good point @SteamyRoot

NaCl

As for every $A_k$ it holds that $\vert A_k\vert=\vert\mathbb{Z}\vert$

Since we can just get the $k$th root

Semiclassical

8:42 PM

Well, there could be two $k$th roots. But that's not a problem.

NaCl

Yes, I was too lazy, sorry :P

Semiclassical

It'd be a bit more interesting if it was $x\in \Bbb C$ instead of $\Bbb R$

SteamyRoot

Yeah - if you want to, you can prove the following

Semiclassical

Then each $A_{k,n}$ would always contain $k$ elements.

Tobias Kildetoft

@Semiclassical But not that much different

NaCl

8:43 PM

That shouldn't change much, should it?

Semiclassical

That's probably right, but I wonder.

NaCl

As you'd have at most $k$-solutions

Which is still countable

Tobias Kildetoft

this is just a special case of the countability of the algebraic numbers

Semiclassical

Well, the reason I wonder is that if you now take a union---say, $A_{k,1}$ through $A_{k,n}$---then you'd end up with $\frac{k(k+1)}{2}$ solutions

SteamyRoot

damnit I was about to type that as an exercise :P

Semiclassical

8:44 PM

Wait, I'm still being silly

I meant $A_{1,n}$ through $A_{k,n}$.

NaCl

@Semiclassical Don't worry, probability is very high that I am much more silly than you

Semiclassical

So it'd grow like $k^2$ not like $k$.

Maybe that doesn't make a difference, though.

yeah, that shouldn't make a difference. $|\Bbb Z \times \Bbb Z|=|\Bbb Z|$.

NaCl

yep

Semiclassical

So it doesn't make a difference, no.

Aaanyways.

You should end up with $A$ being a countable union of countably infinite sets.

NaCl

In fact, as far as I understood, $\vert\mathbb{Z}\times\mathbb{Z}\times\cdots\times\mathbb{Z}\vert=\vert\mathbb{Z‌}\vert$ holds

Semiclassical

8:47 PM

For finitely many factors, yes.

NaCl

And you can replace $\mathbb{Z}$ with $\mathbb{Q}$ there as well

Semiclassical

Yeah.

SteamyRoot

Yup

Semiclassical

What I don't remember is if $|\Bbb Z^\Bbb N|=|\Bbb Z|$.

Mike Miller

@Semiclassical no, that's $|\Bbb R|$

Semiclassical

8:48 PM

okay.

Mike Miller

but that might not be what you mean

Alessandro

That's the set of functions Z to Z, it has the cardinality of R

Semiclassical

Hmm.

Mike Miller

the set of all countable sequences ($\Bbb Z^{\Bbb N}$ has the cardinality of the reals

Semiclassical

That fits with decimal representations being uncountable, sure.

Mike Miller

8:49 PM

the set of all countable sequences with all but finitely many zero is countable

SteamyRoot

Heh

Semiclassical

I'd have thought $\Bbb Z^\Bbb N$ is a countable union of countably infinite sets.

NaCl

@Alessandro What I didn't understand, how do I know how many functions exist from $\mathbb{Z}\to\mathbb{Z}$?

Alessandro

Is there some choice required in saying that the set of finite parts of an infinite set has the same cardinality of the set?

SteamyRoot

For any $z \in \mathbb{Z}$, how many images can it be given?

Alessandro

8:51 PM

There are ways to show that there are as many functions $Z\to Z$ as there are real numbers

NaCl

$\{1, 2\}\to\{a, b, c\}$ should have $\vert\{a, b, c\}\vert^{\vert\{1, 2\}\vert}=3^2=9$ possible functions, shouldn't it?

Alessandro

Hm, wait, nevermind my previous question, I'm pretty sure you need choice

NaCl

axiom of choice?

Alessandro

Yep, for my question about finite parts

Mitch

You guys have probably seen the ['Is This Prime Game'](http://isthisprime.com/game/)?
What are some other math ... things that could be treated like that?

Meigo62

8:53 PM

Hello

Semiclassical

I'm still a bit perplexed. Looking elsewhere, it seems like a union of countably many countable sets is countable. But then how is Z^N uncountable?

Mitch

The difficulty is that anything symbolic needs a symbolic math engine behind it (in the javascript).

NaCl

You correctly sorted 16 numbers. ouch

Mitch

So maybe stick with numbers and any binary feature?

NaCl

@Mitch basically any type of "special" numbers

Smith numbers

Taxicab numbers

Catalan numbers

Mersenne primes

Mitch

8:55 PM

but presumably not with small probability

Semiclassical

Triangular numbers

Meigo62

I have a question which I can´t find answer to :)
If the wave amplitude of a wire is 0,9mm, how much wire is there per 1m?

Tobias Kildetoft

@Semiclassical How is that a countable union of countable sets?

Mitch

so Mersenne is sorta out

Semiclassical

...derp

$\cup_k \Bbb Z$ is different than $\prod_k \Bbb Z$.

Mitch

8:56 PM

maybe restrict the possibilities so that they're 50-50 yes/no

Semiclassical

Could do it as something non-numerical as well. As a simple example: Show a polygon and ask "is this convex"

NaCl

@Semiclassical $\vert\bigcup^\infty_{k=1} A_k\vert=\vert\mathbb{Z}\vert$ with any $\vert A_k\vert=\vert\mathbb{Z}\vert$, but $\bigcap^\infty_{k=1}A_k=\emptyset$?

Semiclassical

That seems to be true, yes. (Not that I remember how it is proven)

Tobias Kildetoft

@Semiclassical Right (and for the latter one, if instead of a specific set we just took a collection of non-empty countable sets, we would need choice to even know that the product is non-empty)

NaCl

Basically, I should prove it

But I donÄt know how, so fuck it

Semiclassical

8:58 PM

Pretty much, yes.

You can definitely find that question on MSE, though.

SteamyRoot

Uhhh

Mitch

@Semiclassical good direction...easy to do javascript graphics... but the concepts seem ... basic.

SteamyRoot

wouldn't the intersection of all $A_k$ contain at least $\mathbb{N}$?

Semiclassical

Point.

SteamyRoot

or $\mathbb{Z}$ even

Semiclassical

9:00 PM

yeah. $\Bbb Z^k \subseteq \Bbb Z$, after all (usual multiplicative product here)

meow-mix

good afternoon

Astyx

@Semi you mean like $\{z^k \mid z\in \Bbb Z\}$ ?

meow-mix

@WillHunting yes, i have. i have not, however, studied any rigorous analysis, besides a little bit of complex analysis (cauchy's integral formula, laurent series, residue theorem)

Semiclassical

Yeah.

image of $\Bbb Z$ under $z\mapsto z^k$.

Complex analysis > real analysis

meow-mix

im interested in complex algebraic geometry :)

Semiclassical

9:03 PM

Though that may be because real analysis is a foundations course and complex analysis typically isn't.

meow-mix

hello, @AkivaWeinberger

Semiclassical

I always found that a bit weird, if I'm honest.

Calling it complex analysis rather than complex calculus.

NaCl

I'd argue with hilbert's hotel

meow-mix

calculus courses are usually a bit less rigorous

Akiva Weinberger

Hi

meow-mix

9:05 PM

i mean, it depends if its an undergraduate or graduate course, but either way, i think its just much more rigorous (even if not very rigorous) than the calculus sequence

Semiclassical

Yeah, but in complex analysis you do a lot of stuff that would fit more in the calculus side than the analysis side

e.g. contour integration, residue theorem

I guess you could argue that any calculus course after real analysis should be an analysis course

NaCl

Guys

Last time you said $\{y\in\mathbb{R}:\exists a,b,c\in\mathbb{Q}\text{ such that }ay^2+by+c=0\}$ is countable, because you can represent the coefficients as tuples

However, what if $a=b=c=0$?

Then it holds for any $x\in\mathbb{R}$

and thus it's uncountable

Astyx

That's true

SteamyRoot

They meant that at least one is nonzero

NaCl

9:20 PM

The task doesn't say anything about that, so I assume that all may equal zero

:)

SteamyRoot

I'm pretty sure that's an oversight :P

Sylent Nyte

hey guys, need some help with maths revision

well, further maths

how would you plot arg(z -a)=arg(z-b), like what does it look like?

because i completely forgot

Astyx

What's a ?

Kaj Hansen

bonjour toute le monde

meow-mix

hey kaj :)

would you care to play a game of chess?

Kaj Hansen

9:31 PM

Sure @meow-mix. Was that your friend request I got recently?

meow-mix

yes; finap13

Semiclassical

lazy questions are annouying

meow-mix

i don't even think he tried on that one.

Semiclassical

She, based on the name, but yes.

Kaj Hansen

Indeed @Semiclassical. It's funny. People have gotten so used to / tired of that, that if one writes a question with even a little effort, it typically gets showered with upvotes.

See my questions, e.g. haha

meow-mix

9:34 PM

sorry didn't realize

Semiclassical

No title, question text is in an image, no work shown

not even the tags are right

Kaj Hansen

LOL

meow-mix

lol once i saw a problem about solving the midpoint and one of the tags was "algebraic-geometry"

i actually dont blame them though, they probably thought it was literal geometry w/ elementary algebra

Semiclassical

Yeah, I can understand those (though ffs look at the meanings of the tags before you use them)

meow-mix

> representation-theory

Semiclassical

9:36 PM

^

Kaj Hansen

What're your preferred time controls @meow-mix ?

meow-mix

yeah, ok, missus.

@KajHansen i usually play 30 | 0

Kaj Hansen

Cool cool. That's fine with me

Balarka Sen

Nice, @MikeM. Have a +1 from me.

Semiclassical

Time controls for what?

meow-mix

9:36 PM

chess

Semiclassical

Ah.

Kaj Hansen

I sent a challenge @meow-mix

meow-mix

@Kaj received :)

NaCl

Any ideas how I could show that $A_1\times A_2\times \cdots\times A_n\text{ countable}\Leftrightarrow A_i\text{ countable, with }i\in\{1, 2, \cdots, n\}$?

Specifically, I have trouble showing $\Rightarrow$

Kaj Hansen

What happens if $A_k$ is uncountable for any $k$ @NaCl ?

NaCl

9:44 PM

@KajHansen That doesn't matter. I need to show that both holds iff the cross-product is countable, then $A_i$ is countable

Kaj Hansen

Playing chess atm @NaCl; I'll give it a closer look in a bit

NaCl

Thank you!

Danu

@TedShifrin Thanks for brining up Kazez: His paper on introductory contact topology seems nice!