12:27 AM
@Krijn I'm sorry if you felt abandoned. I left soon after I posted and just got back.

4 hours later…
4:34 AM
@robjohn No new page will be opened when the bookmark starts with JavaScript. I meant I tried to find a solution for the problem so I added a hyperlink, as you did in your installation page, in the description of a room (I did that because the "Start ChatJax" hyperlink works well in your installation page). But, that way could not solve the problem because when you tap any link in a chat room, a separate window will be opened.

4:49 AM
@Later When I click the "Start ChatJax" bookmarklet in my browser's bookmark bar, it does not open a new window. The <a href="...">start ChatJax</a> is there only to provide something to drag to the bookmark bar. It does also render the MathJax on the page, but using it that way, locks it on a given page.

@robjohn I think you have some Apple's device. Can you please test the web browser?

@Later what do you mean "test the web browser"?

5:05 AM
@robjohn This is the link of the browser. Can you install it on your device to see whether the JavaScript code works well in a chat room or not?

5:16 AM
I have no idea whether that browser supports enough of JavaScript to run the "start ChatJax" bookmarklet. I use Safari on the iPhone and the bookmarklet runs there.

It was tricky, though, getting the bookmarklet installed on iPhone and iPad.
I use it in Safari and Chrome both.

I have asked people if the instructions are sufficient. If not, I welcome clarifications to the instructions.

As I said the JavaScript code works well in Safari and Chrome. The browser supports running JavaScript code in the installation page, but it does not work in chat rooms.

I remember that I originally figured it out and passed something on to you. I recently did it myself and remembered just to paste the bookmarklet "text" into an already set-up bookmark and it worked fine.
You're clicking on the bookmarklet once you're already in a chatroom, @Later?

@TedShifrin Yes

5:21 AM
Hmm, works perfectly for me on iPad and iPhone, both browsers.

@TedShifrin it also works perfectly for me with Safari and Chrome. I meant it does not work for this browser

Oh. Well, if you're determined to use a different browser, I guess you have a good programming puzzle to solve.

@TedShifrin That JavaScript code should not work for any web browser?

I do not know enough to address this.
Howdy, a @Balarka!

Hi @Ted

5:27 AM
@Later I imagine some browsers might disable some JavaScript if the authors of the browser think that JavaScript pose a security risk.

Heya nerds

Oh, howdy, a @Fargle.

Hi @Fargle

Just one? Thank goodness

I didn't want a Balarka to think he was the unique a.

5:30 AM
@robjohn But, with that browser the "Start ChatJax" link renders well the MathJax codes in your installation page, so it should support running JavaScript, right?

I am struggling with probability

Is it winning?

@Later I don't know how one would install the bookmarklet in that browser. I know how to do it in Safari, Firefox, and others are similar.
@TedShifrin The house ALWAYS wins.

Algebra, topology, those are one thing. It is probability where you truly learn the measure of a man
:^)

smacks a Fargle
I don't think his probability is housed, @robjohn.

5:32 AM
@TedShifrin there's a chance he's correct.

We should know a Balarka is immeasurable.

Hey, if I can't use my scant knowledge of mathematics to make pithy jokes on the Internet, what can I do with it?

You call that pithy?
sucky, maybe.

:(

@TedShifrin perhaps to a pithed head...

5:34 AM
I guess my work here is done.

Bullying Fargle: A Geometric Approach

Ah, back to bullying Ted.

I like the title. You hate my jokes.
The asymmetry is clear

Almost as bad as the way you-know-who is dismissing the eminent Dr. Fauci.

@TedShifrin He who must not be named?

5:37 AM
I have a splitting headache most of the time, @robjohn, so naming will only make it worse.

The first president not describable in under ten words

Really? One four-letter word should suffice.
I suppose this is inappropriate for this room. I might need to ban myself.

Look, it's 12:40 AM. I'm just here for the bad jokes.

Oh, I forgot you were Central.

To everyone's lives, yes.

5:39 AM
Hmm, delusional like ...

Someone who should go to bed, yes.

awful banter

Not arguing with that point.

When have you ever known me to have good banter?

As I recall, it all started when you announced your struggles.

5:43 AM
I am trying to dissect a proof to see if it applies to more general circumstances but I am confused on what the most difficult problem is
@TedShifrin Do you know the Heisenberg group?

Casually.

I never noticed it is a very natural sub-Riemannian manifold

Yes, I recall that from a seminar we did on sub-Riemannian geometry 8 years ago. But that's all I recall.

The Lie algebra of $H(\Bbb R)$ consists of upper triangular matrices with $0$'s along diagonals, and the three obvious basis vectors $X, Y, Z$ satisfy the relation $[X, Y] = Z$, $[X, Z] = 0$, $[Y, Z] = 0$, so the subalgebra generated by $X$ and $Y$ exponentiates to a contact distribution on $H(\Bbb R)$.

OK. What does this have to do with probability?

5:50 AM
Ah, OK, the probabilistic story is a little long, but I can tell you the setup. Let me say a bit more about the Heisenberg group first; consider the subgroup $H(\Bbb Z)$ of upper triangular matrices with $1$'s along the diagonal, integral entries. This is actually a "Lagrangian lattice" inside of $H(\Bbb R)$:

Warning: I am not here for very long.
Cool picture.

Notice the path $z = xyx^{-1} y^{-1}$ in the Cayley graph which climbs from the origin to the unit vector in the $z$-axis. These are like the Lagrangian paths in a contact manifold
Here's how all of this relates: Imagine you were Euclidean, and working with the lattice $\Bbb Z^3$ in $\Bbb R^3$. $\Bbb Z^3$ is to be thought as the graph whose vertices are integral, edges and nearest-neighbor edges, equipped with the graph metric $d$. The metric spaces $(\Bbb Z^3, d/n)$ converge to $(\Bbb R^3, \|\cdot\|)$ as $n \to \infty$.
$d/n$ being, scale each edge by $1/n$. The lattice becomes dense and denser and converges to $\Bbb R^3$ (in some appropriate sense, the notion of convergence of metric spaces you want is Gromov-Hausdorff convergence)
It's Pansu's theorem that same exact scenario happens in the Heisenberg group. $H(\Bbb Z)$, scaled by $1/n$, converges to $H(\Bbb R)$ with the subRiemannian metric.
That's all I had to say

Interesting.
I have a headache, so thinking isn't my strong suit right now. I'm surprised the limit of those metric spaces is complete.
I might have intuited $\Bbb Q^3$.
But goodnight for now!

@TedShifrin Yeah that's an accurate intuition. I think you only compare compact metric spaces using GH convergence, so you take closed balls around origin in $\Bbb Z^d$ and $\Bbb R^d$ and see how far apart they are
This is called pointed GH convergence
I think if $(X, d_X)$ and $(Y, d_Y)$ are compact metric spaces you say they are $\varepsilon$-close if there is a map $f : X \to Y$ which is a $\varepsilon$-isometry (meaning $d_Y(f(x),f(y))$ is bounded above and below by $(1 \pm \varepsilon)d_X(x, y) \pm \varepsilon$) which is $\varepsilon$-surjective (meaning every point of $Y$ is within $\varepsilon$ distance of some point in $f(X)$)
So in that sense, for example, $1/n \cdot \Bbb Z$ converges to $\Bbb R$

3 hours later…
8:43 AM
If I have got $$\lim_{x \to \infty} ln (y) =0$$ (where y is a function of x) then can I say $$\lim_{x\to \infty} y=0$$ ??

1, not 0

Okay, can you please show it

lim e^u as u -> 0 is 1
So lim e^(ln(y)) as ln(y) -> 0 is 1
So lim y as ln(y) -> 0 is 1

Yes

So lim y as x -> inf i 1

8:50 AM
Thank you

No problem, sorry it wasn't very well formatted, got a meeting in 10

Is the direct product of simple groups simple?

G x {1} is always normal in G x H

@Drathora Have a good meeting!

ah right. Thanks @BalarkaSen

8:55 AM
is the product of two primes a prime?

@BalarkaSen Is every proper non-trivial normal subgroup of this form?
(for the direct product)
Can't imagine there would be others...

Take the diagonal subgroup in G x G

diagonal subgroup?
is this a counterexample to my question statement?

No, sorry, nevermind that.
I'd have to think about what you said. Possibly it's true.

9:07 AM
@user Yeah it's true; if G and H are simple, all proper normal subgroups of G x H are {1} x H, G x {1}

What's the idea? Projection ? How did you prove it?
I guess Lagrange is helpful?

yeah just project

ok Thanks! Lagrange is only helpful if the groups are cyclic so nvm about that...
oh woops I meant prime order rather than cyclic

wHaT If G aNd H aRe ThE sAmE

simple means nonabelian to me to be clear
otherwise of course you can take G = Z_p, G x G and take diagonal subgroup

9:42 AM
Is there a preferred font for names of categories? Or do people just use text? rofl

\sf

nise
thanks

@EdwardEvans Comic Sans

Is there any way to ask a question to a particular person on the room ?

hahaha

9:52 AM
How ?

$\sf{my name jeff}$
@ronakjain you should just ask the question instead of pinging someone, unless you had already been speaking to someone

I am new to stackexchange and I am just a school student so please tell me can you solve my queries on maths ?
What happened Mr Edward Evans

Wait are you telling me I've been typing 4 extra characters with \mathsf the whole time?

@ronak idk man, at some point it all went wrong, I think it started with my father, he was often absent but we got along pretty wel-.. oh, sorry I misunderstood

How can we find the volumes of solids by deciding them into slices like sphere ?

9:58 AM
@Alessandro apparently so
idk I've just been using \text

Whoops
How did I delete that

How can we find the volumes of solids by deciding them into slices like sphere ?
What happened guys is there noone to response ?

I sometimes only respond when people @t me
can anybody make a torus no quotient?

10:29 AM
Can anyone recommend a good abstract algebra book for reference? Something useful to have on the shelf when one might need to be reminded about basic results/theorems.

10:41 AM
Had been revisiting some very elmentary stuff recently
and looked more closely in rotations in terms of complex eigenvectors
After fiddling a bit with the algebra to convince myself that it does work, it still amazes me why it works at all
like...
the rotation we are performing is a 90 deg rotation on the xy plane
so as the (2,0) vector rotate, its x component slowly reduces while its y component slowly increases, and eventually we have (-2,0)
Resolving this in terms of the complex eigenvectors (1,i) and (1,-i) with eigenvalues i and -i respectively, and then varying the angle of rotation $\theta$ so that the eigenvalues are $\pm e^{i\theta}$, then it becomes clear what is happening in the imaginary plane ixiy
the imaginary components are always conjugate, that is why there are no leftovers throughout the whole process of the rotation

Hey is there anyone to answer my questions ?

but why does that work, why does expressing an $\Bbb{R}^2$ rotation in $\Bbb{C}^2$ and the two extra components can always cancel out in conjugate other than algebraic reasons, so one never end up with any leftover imaginary stuff?

10:57 AM
Well think about it @ronakjain. If you take all the "sum" of the area of all of these slices you end with the volume
But the point is that these slices are so thin and there are "so many" of them that it doesn't make sense to write it as a sum
So this is where integration comes into play
Are you familiar with integration?

actually wait what am I doing
for a real vector to be expressible in terms of the span of complex eingevectors
it must be of the form $z+\bar z$
A rotation of the real components of the vector under $\Bbb{C}^2$ will thus be a rotation of all these complex eigenvectors in a 4 dimensional vector space
Since rotation preserve angle relationships, it follows that the conjugation in the sum $z + \bar z$ will be preserved, thus ensuring the irrelevant imaginary components to cancel out and never show up
Therefore, real space rotation formalised as rotations in a complex vector space results in:
1. Redundant imaginary components that never show up because of conjugation
2. A higher dimensional vector space to describe the rotation
3. The loss of 1,2 is offsetted by the huge gain in not need to deal with sin and cos directly, as these "shadows" are rationalised as rotations about planes in $\Bbb{C}^2$, which is described by complex multiplication

11:58 AM
Is the fact that the natural numbers are well ordered provable in ZF?

@S.D. Yes

@TobiasKildetoft How? Could you point me to a proof?

I don't recall how the proof goes, and I don't know of an online reference (though it may well have been covered on MSE at some point

any book on set theory should have this

I see. Anyway, you're sure that AC (or some equivalent version of it) isn't necessary in the proof?
@Thorgott Is it in Halmos' book?

12:03 PM
yes and yes

Ah, okay. And is the principle of mathematical induction also provable in ZF alone?

@S.D. depends on what you actually mean by that

yes

most formulations of that principle are not particularly interesting (at least for something that bears the name "principle")

induction is just a restatement of the fact that the natural numbers are the minimal set containing the empty set and the successor of any of its elements, which you get directly from a slight massaging of the axiom of infinity

12:07 PM
@TobiasKildetoft Let $S \subseteq \mathbb N$ satisfy $1 \in S$ and $n \in S \implies (n+1) \in S$ then $S = \mathbb N$
@Thorgott Oh, makes sense

@S.D. For set theory, that is an incorrect statement actually, since the set theoretic naturals contains $0$
But anyway, that statement is equivalent to the naturals being well-ordered and having $1$ as the smallest element
(or equivalently that all elements except $1$ have an immediate predecessor)

@TobiasKildetoft Heh, I didn't know this. Interesting
I should go through the construction

I think it is a mistake to focus on the naturals when discussing induction once we are in the realm on set theory anyway. Once we have gotten that far, people ought to be comfortable enough with abstraction to deal with the general version of induction

if you know the naturals are well-ordered, then regular induction is just a special case of transfinite induction

Hackathon
Prize amount 1/2 Million INR

12:34 PM
@Thorgott Do you know $\Bbb R^{n+1}$ has a division algebra structure iff. $S^n$ is parallelizable?

@feynhat Are they actually related, or is it just a coincidence of small numbers?

I actually don't know, that's what I am trying to work out.
Are you saying that we can't prove this without invoking Bott-Milnor's 1,2,4,8 theorem?

I am saying I have no idea

There's no easy way to prove S^n is parallelizable => R^{n+1} has a division algebra structure without going through classification of parallelizable spheres as I remember it
Ah, no, wait, I remember something.

the claim is definitely true
and the other direction should be easy, I think

12:42 PM
Other direction is trivial
Ok, here's the story, and this was the classical approach. Call a subspace $V \subset M_n(\Bbb R)$ of the $n^2$-dimensional vector space of matrices totally nonsingular if every nonzero element of $V$ is an invertible matrix.
The maximum dimension of such a $V$ is equal to the maximum number of linearly independent vector fields on $S^{n-1}$, plus 1.
This is actually not hard to see and is a good exercise. Mike helped me do this once upon a long time ago
For $n = 1, 2, 4, 8$, there are natural candidates for such subspaces which realize the maximal number. For example, if $n = 2$, consider the algebra spanned by $I, J$ in $M_2(\Bbb R)$ where $J$ is the complex structure on $\Bbb R^2 = \Bbb C$. If $n = 4$, consider the algebra spanned by $I, J, K$ the quaternion structure on $\Bbb R^4 = \Bbb H$.
Those are the candidate $V$'s
This doesn't actually help you see what you want because it's not trivial that you can find an $n$ dimensional totally nonsingular subspace of $M_n(\Bbb R)$ iff $\Bbb R^n$ is a division algebra. That's just an algebraic reformulation of Adams' theorem
But yeah at least this makes things transparent
Basically the other direction should not be trivial because Adams' theorem is hard while classifying for which n, R^n is a division algebra is really an easy theorem
Nah I shouldn't say an easy theorem. I don't actually know a proof, but I thought it was classical

it is
elementary is perhaps the more apt descriptor

I'm not sure though.

In General Topology, is closed nbhd of point x a family of closed sets where x belongs?

In mathematics, more specifically in abstract algebra, the Frobenius theorem, proved by Ferdinand Georg Frobenius in 1877, characterizes the finite-dimensional associative division algebras over the real numbers. According to the theorem, every such algebra is isomorphic to one of the following: R (the real numbers) C (the complex numbers) H (the quaternions).These algebras have real dimension 1, 2, and 4, respectively. Of these three algebras, R and C are commutative, but H is not. == Proof == The main ingredients for the following proof are the Cayley–Hamilton theorem and the fundame...
proof right there on Wikipedia

@BalarkaSen It is definitely classical. Probably it is an easy consequence of some of the deeper results on composition algebras and their possible dimensions

12:57 PM
OK cool

@flowian it's a subset of the space that is a) closed and b) a neighborhood of the point x

I kept thinking it was Hurwitz but he's the composition algebra guy yeah
I don't know a proof either way, without assuming Adams' work (for which I only vaguely know the rough idea)

@Thorgott thanks👍

@BalarkaSen Yeah, Hurwitz was who I was thinking of as well (without being able to remember the name)

In general the maximal dimension in the above problem are realized by the Clifford algebra of $\Bbb R^n$ if I remember correctly
Which sits inside the matrix algebra as a totally nonsingular subspace
anyway whatever too much algebra

1:18 PM
I'm utterly confused with probability

Can the number of connected components of a 2-manifold change upon removing a point?

No
If two points were joinable by a path earlier they are still joinable by a path, if the path goes through the point just nudge it a little bit around the chart of the deleted point

ok, that's the same argument as I had in mind, nice

by same argument connectivity of a smooth manifold is unchanged if you remove a codimension >= 2 submanifold
take a path, make it transverse

I know the Cartesian product of finitely many countable sets is countable but does the result also hold for countably many countable sets?

1:24 PM
interesting

@S.D. {0, 1}^N is in bijection with P(N), uncountable

@BalarkaSen Yes, thanks, I thought that but which part of the proof goes wrong? I thought the proof is similar to "countable union of countable sets is countable"

i dont know what "the" proof refers to
write down the proof and you'll probably immediately see what goes wrong

Say I'm referring to this one for example

what is the infinite table for {0, 1}^N
how do you enumerate it
the diagonals of the table A x B are finite, and there are only countably many diagonals since A, B are countable
the "diagonals" of {0, 1}^N are infinite so there's no enumeration like this
its an infinite 2x2x2x2x... cube i mean, how do you enumerate it

1:32 PM
@Balarka I'm trying to figure out the dR cohomology of a manifold upon puncturing in terms of the original manifolds cohomology. I think I've obtained it in case the manifold is orientable. Is there a good answer in the non-orientable case?

@BalarkaSen I'm trying to think about it. I still don't feel convinced. For example, if we can prove that the Cartesian product of two countable sets is countable shouldn't by induction we be able to prove that the Cartesian product of a countable number of sets is countable?

@Thorgott codimension 1 is the only problem

Or maybe induction fails somehow

that's not how induction works

@S.D. If we can prove that the union of two finite sets is finite shouldn't by induction we be able to prove that the union of a countable number of finite sets is finite?
It's pretty much the same situation

1:38 PM
@S.D. I don't know what it means to induct from finite to countable
I can induct from 1 to 2
1 to infinity is a lot

@Balarka hmm, I can't say anything about top cohomology in the non-orientable case either

top cohomology is totally zero

that comes as a surprise to me

@AlessandroCodenotti That's a good point but it's not immediately obvious how it relates to this situation. I mean if we assume that the Cartesian product of k countable sets is countable we just need to prove that the product of (k+1) sets is countable. Then that should prove that the product n countable sets for all $n \in \mathbb N$ is countable

If we assume that the union of k finite sets is finite we just need to prove that the union of (k+1) finite sets is finite. Then that should prove that the product n finite sets for all $n\in\mathbb N$ is countable.

1:43 PM
@Thorgott Just calculate I mean. M is a compact manifold, I hope. Run the Mayer-Vietoris sequence with a chart at the point and the complement of the point

What's the difference

Use H^n(M) = R if M is orientable =0 if M is nonorientable

I'm not assuming $M$ compact

if M is noncompact H^n(M) = 0

@S.D. Yes, the product of $n$ countable sets is countable for any natural number $n$. That was not what you started with

1:45 PM
just assume compact and calculate for now

hmm, should that be obvious?

I don't know. It follows from noncompact Poincare duality
H^n(M) = H_0^BM(M), the 0-th Borel-Moore homology, which is zero if M is noncompact
you can probably give some ad hoc de Rham proof

@AlessandroCodenotti Uh, I guess the problem with induction is that it only proves for all $n \in \mathbb N$ but $\infty$ is not in $\mathbb N$?

does that version of Poincaré duality hold for non-orientable manifolds too?

Yes

1:47 PM
damn

You're doing field coefficients, PD always holds

I only know Poincaré duality for orientable manifolds

If you work in Z/2 coefficients PD always holds because 1 = -1
everything is orientable

I don't know any of the coefficient stuff
my life is continuous suffering from not knowing alg top

Just assume compact man
It becomes needlessly technical for someone who just knows de Rham

1:49 PM
Knowing thorgott he probably isn't assuming second countable, hausdorff or paracompact either
2

LMAO

@TobiasKildetoft I see, so for my original claim I'd need some kind of transfinite induction (which fails)?

right

Why exactly does transfinite induction fail though :P

assuming compact won't help me if all the dualities I know need orientable though
so I'll just assume orientable

1:52 PM
assume compact and calculate H^n-1 for nonorientable, that's what I'm saying
you just need to run a Mayer Vietoris sequence, not a duality

I've already run M-V, but I can't extract much info from it without knowing the kernel of the boundary map
actually, I am gonna assume compact regardless

You get 0 -> H^n-1(M) -> H^n-1(M - {pt}) -> H^n-1(S^n-1) -> 0, right?

oh, I see your point, but I don't know the fact that nonorientable implies top cohomology zero
and I don't see an a priori reason for why the boundary map at the end should be the 0 map

@Thorgott this is doable. do you know the orientation double cover

no man, I don't know anything

2:00 PM
given any smooth manifold M there is a smooth manifold M' such that M' is orientable and M' -> M is a smooth 2-sheeted covering map
do you have a guess for how to construct such a thing

I don't even know what a 2-sheeted covering map is, give up on me

It's a covering map with fiber cardinality 2...?
All I'm saying is there is an elementary de Rham proof that if M is nonorientable H^n_dR(M) = 0 and we can work through it if you're willing to
I'll just do probability if you're not willing to

ah, now remind me what a covering map is. surjection and the preimage of an open set decompose into disjoint unions of homeomorphic (via the map) copies of that open set? so in this case, we just want a surjection such that the preimage of any open set consists of two disjoint open sets, each mapped homeomorphically to that open set?

@BalarkaSen D:
I still haven't told you why e(G)=2 implies G is virtually Z if you need to procrastinate on some probability

how do you prove whether the mapping class group is linear?

2:07 PM
@AlessandroCodenotti I'll read Bridson-Haefliger sometime this week so I'll probably take that as an exercise instead

@Thorgott not of any open set, but all points have nbhds whose preimages are two disjoint open sets, each mapped homeomorphically onto that nbhd

I've been tricked into learning geometric group theory in the name of probability, it's getting worse

ah, ok

@Thorgott diffeomorphically because everything's a smooth manifold but yeah in addition to what Alessandro said

probability

2:08 PM
@BalarkaSen Maybe you can explain me what's the precise version of "a random group is hyperbolic" then

makes sense
I'm being reminded of the fact that the Whitney sum of the Möbius bundle with itself is trivial

Do you have the drawing of $\Bbb R$ as a spiral over $S^1$ in mind? That and $S^2\to\Bbb R\Bbb P^2$ are the two drawings to imagine covering maps in my opinion

yeah, I remember that drawing from when I got like 10 pages into Hatcher
the latter is actually the 2-sheeted cover we want for RP^2

@Alessandro Didn't you tell me that once? Choose a free group on n things, choose m random words, and then build the presentation, or something like this

actually, scratch the "the", I don't know if it's unique, but it is one

2:11 PM
I'm doing percolations so I don't know random models for groups
@Thorgott yeah

@BalarkaSen Am I that old I'm forgetting stuff? I don't remember any of this lol. But what's the probability distribution? Like how am I choosing relators at random

ok, the statement sounds believe, I'll try figuring this out for a bit

I think you choose m random relators of length at most N uniformly, let N -> infty
n, m fixed.
then the probability of being hyperbolic goes to 1
So basically uniformly picking m elements from the ball of radius N in F_n, passing to the quotient and taking N large

That seems sensible

when you quotient a square by the sides to make a torus, is this the same as saying you are quotienting \Bbb R^2 by \Bbb Z^2

2:15 PM
yes geocalc

oh I was under the impression that first you quotient \Bbb R^2 by \Bbb Z^2 to get the unit square (where the integers are one point on the square) and then do the gluing process

but (1, 1/2) and (0, 1/2) already gets identified in R^2/Z^2
they are distinct points in the unit square

so is this any different than compactifying \Bbb R^2 to the unit square and then gluing sides? do you just get a smaller torus?

erm
i think ill go do probability

ok
was it a bad question?

2:23 PM
@TobiasKildetoft By the way, by that logic shouldn't the proof that "countable union of countable sets is countable" also fail?
There also they seem to be using some form of induction
Or does countable union only mean union for n many sets
Where n is a natural number

@S.D. Indeed, you need AC for that proof

I'm a bit confused. 1. So is the term "countable product" (or "countable union") by definition different from "product of countably infinitely many sets" (or "union of countably infinitely many sets")?

no, those mean the same thing

2. Is the result "product of countably infinitely many sets is countable" not true unlike "union of countably infinitely many sets is countable"? The latter I think needs countable AC but the former is straight-out false
@TobiasKildetoft Umm, but if you notice the proof here proofwiki.org/wiki/… they only prove it for all $n$....there is no transfinite induction of any kind

@S.D. No, that is not what they do at all

2:33 PM
So I'm now even confused how the union of countably infinitely many countable sets is countable
@TobiasKildetoft Could you elaborate a bit? From what I see they clearly write $S = \cup_{n \in \mathbb N} S_n$
That is the union is only for n many countable sets

no, it is not

@Thorgott they're writing $n \in \mathbb N$ no?
Oh but $\mathbb N$ itself is countably infinite
So why doesn't the same logic apply for the Cartesian product :[

indeed
because the proof doesn't work for the Cartesian product?!

I'm experiencing an unprepared seminar talk right now
and it's quite painful to watch

if you try drawing a similar picture, you will already need such an infinite square for just the cartesian product of two countable sets and one more dimension for each additional factor, so a countable product of countable sets will be pictured as an infinite-dimensional infinite cube and that won't be countable
@Edward yikes
some prepared ones already feel unprepared

2:44 PM
@Thorgott I get the idea of the infinite dimensional infinite cube but is there any way to rigourously write down why it's not countable?

@S.D. It all boils down to the usual diagonal argument

More simply how can we prove that $\times_{n \in \mathbb N} \mathbb N$ is not countable

@Thorgott this is supposed to be a proof of local class field theory and it's just being read straight out of a paper with some things slightly paraphrased

@TobiasKildetoft What is the diagonal argument? Like Cantor's diagonal argument?

2:48 PM
I only know Cantor's argument for two dimensions lol...how do I deal with an infinite cube

@EdwardEvans Does the speaker have the paper projected instead of slides?

(Sorry for being thick)

Well, first step is to note that it contains the set of all maps from the naturals to $\{0,1\}$

@Tobias not quite that bad, but I am reading the paper he's writing from and it's just quoted verbatim with things like "inverse image" replaced with "preimage"

then you do diagonal
heh
(I did a long time ago see a speaker actually have their talk consist of going through the paper, with scrolling and everything)
It was completely unwatchable

2:51 PM
Yeah this is painful, I'm just gonna not ask questions

Yeah, better to just let it end quickly

or bring out your inner sadist and ask especially difficult questions

lol or ask some simple questions so he can win a couple of points