hm

Because for every given Union of such, you can, according to the Axiom of union, get another set which has all the sets in that set, and all the sets in another given set that we will construct from two sets (or more - axiom of union on the axiom of pairing) -

The clock on MSE seems to be 40 seconds slow.

@Studentmath But that doesn't prove that you actually have the desired set.

And we go on and on.

What was your idea?

Hm, how come?

I guess you could do transfinite induction to complete the process.

Not sure if I am allowed to though.

My idea is this:

$$\bigcup\{\{\{x\}\}:x\in\mathbb N\}$$

Hm

Huh

It's really nice

That's better.

Now that's even better -

The use of axioms to show such a thing may exist is with just three axioms, that way, which I think is really lovely.

Axiom of union, axiom of comp...whatever and axiom of pairing. Correct?

That is true.

@Mike We took over the chat once again

Oh Karl, there's another way I just thought of, think it might be nicer.

Say we assume N exists (we have to), then according to the axiom of the power set, P(N) exists. we define $\phi$[P(N)] where $\phi$ takes only the sets A where for every x1,x2,...EA, x1=x2=x3=....

We used only the axiom of power set and the axion of compherence

comprehense*

@KarlKronenfeld

yeah, that works too

You should not define $\phi$ that way though (with numbers as indices)

I suggest, $\phi(A)\iff \exists a,\forall x(a\in A\land x\in A\implies a=x)$.

Yep, much better. Thanks!

Oh, an equivalent but better sounding form is $$\phi(A)\iff \forall x\forall y(x\in A\land y\in A\implies x=y)$$

This is exactly what you wrote above, but we removed indices.

I see, yes.

I have to go. Bye @Studentmath

See you @KarlKronenfeld !

Damn you, scary nature!

@IanMateus "Damn nature, you scary."

@PedroTamaroff poetic license

@anon

yes

@anon This.

@PedroTamaroff Suppose $N$ and $M$ are distinct subgroups of size $2^3$. Then $NM$ is $2$-torsion so it is also a $2$-subgroup; what can its size be?

Purrfect.

Do you want to add an answer or shall I edit that into mine?

you can have it

I gave you an upvote in another answer.

good, dent my 0 streak

I have so many answers with zero upvotes.

@Mike iZone?

@TedShifrin

Pete hasn't answered my mail. =(

Mr @Pedro

Can I forward it to you and you tell me what you think about it?

Next time I see him I'll ask

Ok ... Might take me a while. Busy next days.

Sent.

does anyone know what an elliptical distribution is in a deep way?

i have seen the definition on wikipedia and am trying to reconcile it with a different definition

I have a grad school question.

Is it normal for a professor in a phD level class to not give a syllabus?

@JessyCat It seems not uncommon to me.

$$\sum_{p\leq x}_{p=n^2+1}1\approx \text{Li}(x^{1/2})(\prod_{p\equiv 1 \text{ mod } 4}\frac{p-2}{p-1})(\prod_{p\equiv 3 \text{ mod } 4}\frac{p}{p-1})$$

How are you supposed to know what all you're planning to cover, how grades are calculated, how many exams and when, etc.? Just basic course policies.

Or are we all supposed to be so freaking brilliant by this point that we don't need such things as mere mortals need..,

hola

Got another interesting one. Have to prove for given set A that P(P(A)) exists using only the axiom of Pairing, axiom of empty set, and axiom of union. Hola mike.

I am thinking towards proving it with induction + for A being the empty set of course. But it's gonna be a bit tough, the n-->n+1 step. Any other ideas?

@Mike Wat.

@pedro Que?

@Studentmath Something like $\{A,\{A\}\}$?

@Mike Just saying hi. Don't be an ass to me man.

My god. That's so simple.

I am inlove @PedroTamaroff

@Pedro Sorry, I didn't know I was being an ass...

But no way it's possible..

@Mike HA, JK.

I got troleld there.

@Studentmath I probably did something stupid, I'm no set theorist.

Oh nah wait. Yeah, it won't work. You need all subsets too, not just that.

@Studentmath Well, unions give you maximal elements with respect to some property.

@Pedro I'm gonna guess that's using more than "the axiom of pairing, empty set, and union"

That would be invoking the axiom of comperhence, which I am not allowed to :)

@Studentmath axiom schema of comprehension, yes?

Correct

I have only three axioms at my disposal. Pairing, empty set and union.

I am sure though there is a prettier way than the ugly induction, with the terrible step between n and n+1 in here.

OK, empty set tells you $\varnothing$ exists, yes?

Union tells you $\cup \{A,B\}$ exists given $A,B$ sets?

Correct and correct.

Pairing tells me that for two given sets/items x,y the set of {x,y} exists too.

Oh.

OK.

Should I write my idea down?

OK.

It's ugly though.

Alright..

If the given set A is the empty set, then according to the Axiom of Empty set $\varnothing$ exists. According to the axiom of Pairing, the set of {$\varnothing$} exists too, and again according to the axiom of pairing, the set {{$\varnothing$},$\varnothing$} exists.

That is precisely the set of P(P(A)) if A is the empty set.

KAY:

I'll be right back, you keep writing and @Mike will help,

@Mike , always comes back to the set-Theory...

Hey @ethan!

hey

Let A be the given set, and let A have one item in it, a. According to the axiom of pairing, the set {a} exists. According to the axiom of empty set, $\varnothing$ exists. again invoking the axiom of pairing, the set {{a},$\varnothing$} exists. Again we invoke the axiom of pairing few times, we get to: {{{a},$\varnothing$},{a},{$\varnothing$},$\varnothing$} which exists. That is precisely P(P(A)) if A has one item in it.

We will now use induction. We assume it is right that for given set A with n items in it, P(P(A)) exists, and we will prove that for B with n+1 items, P(P(B)) exists too.

Now it gets ugly.

@Studentmath I can live with it :)

Let y be the 'new' item, as in the n+1 item, in B. Invoking the axiom of pairing, we pair it up with every given subset of P(A), and invoke the axiom of union on it. All these sets exists due to the axiom of pairing and axiom of union of course.

We take each of these sets and pair it with the set without the y. This is possible as we showed that such a set exist, and due to teh axiom of pairing.

Each of two of these -new- sets we add up invoking the axiom of union.

Then we pair up again each two.

And union -new- two.

@Ethan I am going out for a walk now, let me know where you eventually go later in the year!

Eventually we will reach P(B). But I think it's just.. too ugly.

And too hard to show.

ideas/comments/suggestions willb e appreciated by all :)

1-800-ASAF

xD

But, how do you deal with infinite sets?

Yeah, I might just post it on the website..

Well, I basically don't. Since the moment I go n+1 I covered all possible sets, finite or infinite.

blarg

Clean your mess, Ethan.

Well, website worthy isn't it? @Ethan what's up?

Yes, I think it is.

alot

ORLY?

i got so much paper

Dafaq is that?

old military papers

ww2 or ww1

i think

Where from?

Yours?

mother's father's father's

etc

its his:

$(r(e(m(o)v)e)d)$ ;-)

old

It's wrong?

yea

Wha...?

Why?

It says D Doster, not D Foster.

oh crap

well the last guy I emailed died 3 months later

STAAAHP!!!

Hi @chris

How are you?

sup folks

why do you constantly delete stuff?

ocd

did you read the last line

what the author did with the O notation mike

nope

proofwiki.org/wiki/Prime_Number_Theorem

@Ethan Do you try to control yourself?

i saw the link but wheres the issue?

and then he has the nerve to attribute the garbage to newman

@skullpatrol yea

Good

The proof is wrong and even if it was right he assumes that $M(x)=o(x)$ which in itself is enough to prove the pnt

so basically its wrong and circular

Every time I post a question in set theory

looks like loads of fun

How long ago were you first diagnosed @ethan?

I imagine Asaf staring at it and going "this guy oughta be stupid.."

2-3 years ago

oh asaf is the logic guy

:D

im so bored

Oh snap

just figured easy way to do it

@PedroTamaroff your idea of going around was right.

how old are you student math

Or nope, it wasn't.

18

You?

same here

Nice, going to college now I assume?

yea waiting on admission decisions

a month and a half from now

I am sure it'll go well

Any preferred one?

they make you write alot about yourself over here

good but a pain i guess

dont you have to do like conscripted military service over there

or are you in an Atuda

i gota sign off for ss, but that's not mandatory lol

you had to sign it b4 Vietnam

@Studentmath Explain?

I saw that Ted was here?

@PedroTamaroff I thought we could indeed go P(A)={{A},emptyset}

but we can't go around like that..

Dang Ramanujan

(Not you @Ethan, this )

ramanujan.sirinudi.org/Volumes/published/ram12.pdf

@Ethan Are you onto computational NT?

idk thats sort of vague

Hmm.

I am usually in analytic more than algebraic, so computational jumps from the window time to time.

@Ethan From when you are doing mathematics? 17?16?15?

Higher math, I mean

uhm

a while

Just started in 18?

no

Oh.

19

Wat?

20

Then you are 81?

41

That's my age, 41.

Darn these questions. I am so tired, I can't focus.

I hate getting red marks on my proof papers. geez even if I did right I still face rejection ughhh wtf so irritating

@usukidoll Why are you facing rejections?

like if it's not enough words it's too many words... if there's too many symbols, prof goes bat shit crazy

I CAN'T TAKE IT ANYMORE!

If one defines $\dim V$ to be the minimum cardinailty of a set that spans $V$, how does one prove the existence of linearly independent sets $E$ such that $E \in V, |E|=\dim V$?

Actually, must such an $E$ necessarily exist?

@Alyosha Existence of a basis for non-finite dimensional spaces requires axiom of choice, if my sleep-deprived brain remembers correctly. Look up "Hamel basis".

@BalarkaSen Are you 41 or 14?

@usukidoll Relax, relax.

@daniel You look very much like one of my lecturers who is a German, lol.

@JasperLoy 14 =D

No, seriously, I am 14.

Oh, hey, by the way, Banach Analytic Manifolds.

I mean @Jasper.

I cannot feel my fingers

It is so frigging cold.

And seriously, who the hell cares how old @Balarka is, the little Doogie Howser wannabe whippersnapper.

Age is stupid.

@JessyCat How rude.

Don't let age, or lack thereof, define your accomplishments

@Balarka, sorry.

Yay! 80s sitcoms!

Sorry, I'm having a bad day.

Try week.

@JessyCat That's ok. I heard worse than that.

=D

@BalarkaSen: I'm 16 mod 14 compared to you.

@Nick That means 2.

Yeah probably what I meant

@BalarkaSen: So, how are you?

how are things going?

Fine. Having a hard time doing a nice little problem. Wanna try?

... does it have calculus?

No. Factorization and irreducibility.

No thank you then, I just ate.

But share anyway :D