@TedShifrin what do you mean?

@BalarkaSen Wouldn't group theory be dealing with the set of bijections between a set and itself (i.e. permutations)?

Also hello

@akiva those would be automorphisms, i believe

Self-bijections are not really different from bijections between two distinct sets of equal cardinality.

I thought things called automorphisms generally have more structure. Like a self-isomorphism

A set $X$ is infinite $\iff$ we can find a subset $A \subset X$ with $|A| = |X|$

These are automorphisms in the category of sets.

@BalarkaSen Yeah, but harder to give a group-theoretic operation

@KajHansen …assuming Choice

Ok, @Akiva, fair enough :)

Also I really meant finite sets, not infinite. Those are messed. I don't think permutation representation is a thing for infinite groups.

Without choice, "infinite" just means "not finite" — that is, not in bijection with $\{1,\dots,n\}$ for any $n$

I take it back, sure that's a thing.

Choice ruins everything :/

one just needs to look at the symmetric group on the underlying set of the group we're looking at.

@KajHansen But here it's the lack of choice that ruined it

@BalarkaSen which are a torsor over the automorphisms of either

haha, yeah I suppose so

@MikeM that's what infinity categories does to an honest topologist

soon you'll write articles in nlab and actively use the n-point of view in conversations

n

o

haha

Balarka....so tsundere towards nlab

You'll come around some day :)

@AkivaWeinberger aren't there a few different concepts of finiteness without choice that all become equivalent with AC?

yea

exists non-surjective self-injection, non-injective self-surjection, are popular

but finite literally means in bijection with 1,...,n

@Kaj I sure hope not

@KajHansen i have now a crazy bug. I can't visit my universities course site, but via proxy no prob. And it's not only my courses, any course. Something is really messed up.

hm, what I meant is like "every collection of subsets of $A$ has a $\subseteq$ maximal element" and "$A$ is Dedekind finite" (where Dedekind infinite is just what you said, existence of a non surjective self injection) are equivalent properties in ZFC which classify finite sets, but they're not in ZF

You've been IP banned from your own uni's website LOL

its funny, because I can visit my university site

only the coursematerial sites not

(they are on another server i believe)

You've been IP banned from your own uni's website course catalog LOL

I guess that'd be even more hilarious

But really, @Null, report it.

@MikeMiller At a glance that paper looks very nice. I'll bookmark it for reading.

@Alessandro Right, but only one of them is actually called "finite" (and only one concept is actually called "infinite")

lol, this is really funny

> Show that if $n > 1$, there is a bijective correspondence of $A_1 \times .... \times A_n$ with $(A_1 \times ... \times A_{n-1}) \times A_n$

can someone click on this pls? reh.math.uni-duesseldorf.de/~koehler/Lehre/2016-17/… and tell me if it's alright

you can get banned from mathworld for trying to download their entire site

(experience)

It looks like I'm also IP banned @Null

It opens fine for me

@arctictern it's not worth the odwnload!

@arctictern lol you actually tried that?

seemed like it in high school

@KajHansen do you have by chance irefox?

I guess I'll be enrolling in Dusseldorf since I'm the only one not banned then

Chrome

always fun with the elecronics :)

@Alessandro Same.

the right version of mathworld you'd want to download is CRC Encyclopedia of Mathematics, not the website

My attempted proof (of my earlier question) : Put $\alpha = A_1 \times .. \times A_n = \{(a_1, .., a_n) | a_1 \in A_1 \text{ and .... } a_n \in A_n\}$ and $\beta = (A_1 \times .. \times A_{n-1}) \times A_n = \{(a_1, .., a_{n-1}), a_n) | a_1 \in A_1 \text{ and .... } a_{n-1} \in A_{n-1} \text{ and } a_n \in A_n\}$.

Then define $f : \alpha\to \beta$such that $f(a_1, ... a_n) = ((a_1, ...a_{n-1}), a_n)$. From the definition of $f$ one can easily prove that is both surjective and injective, and thus bijective. Thus proving the bijective correspondence between the sets. $\square$

I had to google street slang I heard outside, is this what getting older feels like :(

What Your Teen Is Really Doing When She Says She’s Going To A Warehouse To Get All Holes Stuffed

do we get something interesting if we plugin the vectors of the standardbasis in a function?

...

O_o

@Null jesus, they really don't like your question i guess

@Perturbative Looks good

WAT @MikeMiller ?

@Kaj Lindsey was going to a place of industry to have anonymous group sex

it's a clickhole reference, surely

@MikeMiller That's an old Clickhole article, right?

googled. yup.

@Null This is the only interesting part of a linear map

@MikeMiller Maybe it's the tired delirium but I can't stop laughing at your deflection

Thanks for the verification! @AkivaWeinberger

@Krijn but what about nonlinear functions?

(or maps)

@MikeMiller POS is surely the mildest translation ;)

@MikeMiller It's so hard to keep up with teen speak nowadays, wow

Oh, yeah, pretty sure POS is something else

presumably you know firsthand

What are you implying

you are a teen

Oh, sorry, I thought you were calling either Null or me a POS

a mathemathician wouldn't do that, I suppose

@Null You'd be surprised.

I will have no part in this discussion.

(that is in no way a dig at anyone here)

@AkivaWeinberger well I mean, that too

I want to read a math paper that has random curse words sprinkled throughout just for the lolz

you seem like the type that would always have a parent over your shoulder

hey @BalarkaSen

"$N+1$ isn't divisible by any $p_i$, bitch."

hi

@AkivaWeinberger "Let V be some fucking, I dunno, algebraic variety or some shit."

If you do it sparsely, cursing can be an effective teaching tool

https://xkcd.com/66/

so I proved the following result that any isogeny between elliptic curves takes the form $(u(x) / v(x), g(x)y)$ and the degree of isogeny is just the maximum degree between u and v.

@MikeMiller As my linear algebra professor once said, "Learn the damn definitions."

@AkivaWeinberger I had a prof that defined convergence of a sequence to 0 as "this thing shrinks fast as f*ck" (best translation I could do)

Is that the equivalent of RTFM, @Fargle?

@BalarkaSen one can easily see that the frebenius map has degree p.

("Read the fucking manual")

just using definition.

@AkivaWeinberger I'd say so--or RTFC if you come from a background of collectible card games.

@AkivaWeinberger a paper that rhymes would tickle my fancy more

but what about the multiplication map why does it have degree $m^2$ ?

i.e the map given by $\phi_m : E \rightarrow E$ given by $\phi(P) = mP$.

Oh, I knew this once

@Fargle This would fit perfectly into an episode of Rick and Morty (for those who watch it)

@Krijn your talking to me ?

@Adeek Yes

@Adeek Ehh

@Perturbative Ah, yes. I have some problems with the show but on the whole I enjoy it.

cool

I've never thought about this I guess

have you tried writing down $mP$ explicitly?

@BalarkaSen seems like a horrifying suggestion

Is this the elliptic group thing

yeah :/

mP is just P + P + P ...

m times

no I meant in local coordinates

Ah yes, found it @Adeek

oh

@Adeek The point is that the degree of an isogeny is also the number of points in the preimage of a point

Aww, @Mike ninja'd me

oh

why's there m^2 many points satisfying $mP = O$?

oh duh

because the group is isomorphic to $\Bbb R^2/\Bbb Z^2$

@Krijn My analysis professor, concluding the proof that the subsets of the Cantor set are Lebesgue measurable "so we conclude that there's a shitload of Lebesgue measurable sets"

just because its Z/n times Z/n

oh ok

I was trying to cook up something geometric, so no fair

@Alessandro Lol. Not in the same vein, but also funny was a comment by a prof "We have way too many poles here, let's get rid of some"

It was unintentional

@Alessandro Even if that Cantor set has nonzero measure?

Cantor set does not usually mean the thick ones

this algebraic geometry stuff is very cool

@AkivaWeinberger He's referring to the standard one

punk

the usual Cantor set has measure 0

:P

Isn't a subset of a 0-measure set always measurable, though?

yep, because it must have measure 0 as well and sets with measure 0 are measurable

So why prove it specifically for the Cantor set

Oh, was it just the example of an uncountable measure-zero set?

@Alessandro You should teach me Lebesgue measures at some point. I only vaguely know the theory.

I wonder if there's a nonstandard version of measure theory (a la nonstandard analysis)

Like maybe the measure, if it exists, would be equal to the number of multiples of $\epsilon$ in the set divided by $\epsilon$ for all infinitesimals $\epsilon$

we considered the Cantor set as an example of a set which isn't Borel yet it's measurable

Cantor set isn't Borel? Huh

wait no, that's wrong

(Remind me: Borel is open sets plus unions and intersections, right?)

Cantor is Borel, but subsets needn't be

Cantor set is Borel

Ah

I guess I don't know an explicit way to see that other than cardinality considerations

uhm, now that I look at this again I'm confused and I don't follow my prof's reasoning

Right, it's the intersection of $C_n$ where $C_{n+1}$ is $C_n$ without the middle thirds and $C_0$ is the unit interval

Yea

as in the usual construction of the Cantor set

we wanted to show that there are Lebesgue measurable sets which are not Borel, so we showed that there are uncountably many measurable sets

couldn't we just pick the intervals?

err, the Cantor set is closed

so... it's Borel

Oh. Derp.

@Alessandro No, the point is there are $2^{|\Bbb R|}$ subsets of the Cantor set

Right.

yep, it is, I wrote it above after catching my mistake

But there are $|\Bbb R|$ Borel sets

Ah, right of course

I got confused, I was missing an exponent in my cardinalities

ok, so we took the Cantor set to show that there are Lebesgue measurable sets which aren't Borel, makes sense now

But I'm quite sure there are more concrete ways of deriving an example

Wait, hold on

As you can see I don't know much measure theory either :P @Balarka

It's not really my thing.

But I'll have an exam on it in June so I hope I'll know it by then

Point is, there's a shitload of Lebesgue measurable sets.