@Sean I might be able to help, but I do not understand what you are asking...

The main problem with your proof by induction is that you leave out the process of getting to $\omega$ or any other limit ordinal.

You account for all finite ordinals, but that is insufficient to be able to assume $|\omega| <|2^{\omega}|$.

You are so kind peoplepower, thank you

For instance, it is easy to verify $n<2n$ on the finite numbers by induction, but that does not allow you to conclude $|\omega|<2|\omega|$, since the latter is false.

I have come to an understanding that I cannot do induction on countably infinite sets

But now I have a hard time understanding this prove by contradiction

problemhere.wordpress.com/2011/11/02/…

It is a tricky proof, indeed.

if you dont mind

can you read it and help me?

We are constructing $A\subset S$ to be the set of all $x\in S$ such that $x$ does not lie in $f(x)\in P(S)$.

i suspect my notation in this answer math.stackexchange.com/a/286460/1284 could be improved somehow, any suggestions?

@Sean I.e. we collect all points $x$ such that the complement of $f(x)$ rather than $f(x)$ itself contains $x$.

peoplepower, does it make sense to you?

@DanBrumleve its alright I think.

btw are you using LaTeX? I saw "$f(x)$"

i've been using \pmod in a relational context but i'm not sure about a functional one. i'd rather say % but i think that is wrong for this audience.

Does anyone know why some sources say that $N$ is a natural number and others say that $N$ is a real number in the definition of the limit of a sequence? I am quite confused.

I don't see why $N$ has to be a natural number.

Wikipedia seems convinced that it is.

@Sean Yes, I am using latex, see here to render it on chat.

@Limitless U are still awake?????

@Limitless it must be natural since it is the subscript of a sequence

aside from that it doesn't really matter

@DanBrumleve Oh, right.

To peoplepower: I dont understand the line "A=f(a)"? I thought A is the domain and f(a) is an element of the range.

@DanBrumleve If it's the limit of a function, it can be a real number.

@math101 Yes. I am working.

@Sean $S$ is the domain of $f$.

btw thank you peoplepower, but I'm using a school lab computer haha

@Limitless, rather, n must be natural, but it doesn't matter if N is natural or real.

So...

@DanBrumleve For an epsilon-n proof, should I just do nearest integer if I have a weird N value (e.g. $\frac{5\epsilon+4}{\epsilon^2}$?

Ahh, I took it as something else, never mind.

If A is a subset of S, why would "A=f(a)"? Consider f(a) is the range and not the domain

@Limitless use ceiling function to be sure

Mmmk.

actually since the def is \gt rather than \ge it doesn't matter.

@Sean Well, $f:S\to P(S)$ is surjective, so every subset $T\subseteq S$ can be written as $f(t)$ for some $t\in S$.

In particular, the subset we constructed can be written in the form $f(a)$.

You meant "f:S->P(S)" is assumed to be surjective, right? Since I have to prove that it is NOT surj

It is given, yes.

OK

Let me explain my thoughts

@DanBrumleve Some of these details are absolutely frustrating.

on second thought my last comment is wrong

certainly they can be :) :(

wave your hand past them :)

for the line "x belong to A iff x does NOT belong to f(x)" does not make sense

ROFL. I wish.

@Sean That is just the description defining $A\subset S$.

It is the set of all elements of $S$ which do not lie in their $f$-image.

since S={a, b}, P(S) has {empty set, {a}, {b}, {a,b}}

oh I forgot

ok let me move on to next lines, thank you ;))

is e^e^e^79 an integer?

Since f is surjective, there exists a belongs to S st A=f(a)

this doesn't make sense to me, either

shouldnt it be P(s)=f(a)?

No, remember, $f$ maps into (assumed to be onto) the powerset of $S$, so $f$ maps elements of $S$ to subsets of $S$.

Since x is an element of A and A is a subset of S and S is the domain, then f(a) should map x to P(A)

What is lowercase a?

sorry, x

wait, no

the line is in the link btw

problemhere.wordpress.com/2011/11/02/…

probably easier if you identify the line for me

wow, who was it that got in the flag war?

We are picking an element $a$ based on the property $f(a)=A$.

It will not be the powerset of $A$, since $f(a)$ is just a subset of $S$, not a set of subsets.

where is the property f(a)=A came from?

$f$ is assumed to be surjective, so if we have a subset $X\in P(S)$, we can find $x$ such that $f(x)=X$.

Just like any other surjective function.

I thought the definition is: For all p, exists q st f(q)=p

That is, if $p$ is the codomain of $f$.

The codomain of $f$ here is $P(S)$.

sorry about the wording, I'm using a school library and there is no LaTeX installed

@MarianoSuárez-Alvarez Jonas was one it seems. I don't know others.

kids these days

Sorry, guys.

what happened here? 4.bp.blogspot.com/-CkCkR0qZJ0U/TbdnHArWfkI/AAAAAAAACoI/…

@Sean I did not mean literally "we can find", sorry, it is technically "there exists".

@MarianoSuárez-Alvarez :P The most interesting thing was, after initial flagging, there was a lot of revenge flagging.

"since $f(a)$ is just a subset of $S$, not a set of subsets." why is f(a) a subset of S?

@Sean you don't meed LaTeX installed for viewing the site. You just need a ChatJAX bookmark. (See right, on the pinned message on the starboard. $\LaTeX$ support for chat.

@Sean It is in the codomain of $f$, which is the set of all subsets of $S$, i.e. the powerset of $S$.

Where is pinned message?

@Sean see this

@peoplepower: I see that now

@Orange Harvester: thank you

@Sean Great, so since $A$ is a subset of $S$, there exists $a\in S$ with the desired property.

wait a sec....I really need the TeX support, what should I do with the bookmark agaihn?

You made it a bookmark, right?

i actually appreciate reading tex in its raw form here without any plugin

I saw the java script

pastebin.com/UYgNvcVU

You can right click on the link "Start ChatJax" and there will generally be an option to create bookmark linking there.

if it's too hard to parse it shouldn't be in chat anyway

Why is the notion of p-adic numbers useful?

@Ethan it is useful as an alternative completion of the rationals at least

rather as an example of such

I cannot figuer it out, lol

can't build airplanes with it

where is the start chatbox

@Sean Oh, sorry I was going ahead. First scroll to the update section and click the link "all versions here"

well to start with ....99999 = -1 right?

what?

...9999 is a 10-adic

yes

where's the update section?

wait

I thought p had to be a prime

it doesn't really

then why define it to be?

....999 is an n-adic

good question

@Sean Just to make sure we are on the same page. Go here. The word "Update:" is in bold.

probably for nice multiplicative properties

at least for the additive ones it doesn't matter

Are there any useful theorems that can only be proved with p adics

ahh

I clicked all versions here

@Ethan no airplanes

@Ethan, one problem which is of interest in general is the factorization of primes in number field.

lol

then "start ChatJax"?

@Sanchez how does that relate to p-adics?

@Sean Right click it and there should be an option to add it as a bookmark.

@Ethan, for example, $\mathbb{Z}[i]$ is something that behaves like integers, it has primes, it has unique prime factorization and so forth.

yes

I still don'

@Ethan, you may want to understand how $p$ factors there.

and you need p-adic analysis for that?

@Ethan, but there are many primes in $\mathbb{Z}$. $p$-adic numbers is one way to let us focus on only one prime.

I added it to my bookmark bar

@Sean Or if you have a bookmark bar you can drag it to the bar.

Ok.

then?

Now at this page you just click the bookmark.

@Ethan Google throws up this @anon This is why NASA might be interested?

wow work like a charm

Thank you peoplepower

@Ethan, as for $n$-adic numbers. Well we really have to be careful here. When you say $p$-adic numbers, there are two things that come to mind: $\mathbb{Z}_p$, the "integers" and $\mathbb{Q}_p$, the "rational numbers"

though I cant type in TeX so you'll have to bear with me lol

@Ethan, there's no problem to use $n$ if we want to talk about $n$-adic integers, but you don't have the analogue of "rational numbers" in this case.

@Sean You're welcome. Do you understand the second line of the proof, now?

@Ethan, that's why people focus on primes in general. Also, to understand $n$-adic integers it suffices to understand $p$-adic ones.

let me go back a bit

@Sanchez thanks that helps me too

@Dan, sure :)

Consider the subset A of S which satisfies x belongs to A but not f(x)

That is a confusing way to word that line.

@Sanchez What do you mean, if we chose the n to not be prime, it just wont satisfy as many nice properties as it did when it was prime?

isnt that redundant since f(x) produces the powerset of A, which contain all the subsets of A

@Sanchez Like its n adic order being additive

@Sean The action of $f$ is not to produce powersets.

for example, A={a} then P(A)={0,{a}}

@Sean I will give an example $f$, it that helps.

wait what?

@Ethan, additive is not a problem. The problem is that $n$-adic integers is not an integral domain, i.e. there can be nonzero $n$-adic integers whose product is 0.

sure! thank you!

Let $S=\{a,b,c\}$. Then $f$ might be defined by $f(a)=\{a,b\}$, $f(b)=\{a,c\}$, and $f(c)=\{a\}$.

one question

@Sean Notice that the notation $f:S\to P(S)$ means that $f(x)\in P(S)$ rather than $f(x)=P(x)$.

P(A) = {0,{a},{b}...etc} so of course a is not in P(A), that why I think the line doent say anything more

I see

In the case I gave, what is the set $A$?

@Sanchez How can I show that $$\sum_{k=1}^nv_p(k)M(\frac{n}{k})=[log_p(n)]$$ where $M(x)$ is the mertens function and $v_p(x)$ is the p-adic order of x?

A={a,b,c}?

No, $a\in f(a)=\{a,b\}$.

then how did you plug "a" into f like f(a)?

@Ethan that is an interesting formula is it really true??

Yes I found it, yesterday im just interested how one might go about proving it using other methods

If you multiply both sides by ln(p) and sum over the primes less then or equal to n

you get

@Sean $f$ is a correspondence between the elements of $S$ and (some of) the subsets of $S$. In general functions do not need to be defined by formulas like $f(x)=x+1$. I defined the function $f$ in my example in a different way.

$$\sum_{k=1}^nln(k)M(\frac{n}{k})=\sum_{p\leq n}ln(p)[log_p(n)]$$

I see what you meant now, but I still don't know why A is not {a,b,c}.

Why is A={a,b}