Don't get why argument of power sets being uncountable works Statement: $\mathcal P(\mathbb N)$ is uncountable. 
Proof by contradiction: 
Assume $\mathcal P(\mathbb N)$ is countable.
Let $A_{i} \in \mathcal P(\mathbb N)$. Now define a set $B = \{i\in A: i \notin A_{i}\}$. However, $B \subset \mathcal P(\mathbb N)$, which means $B = A_{j}$ for some j. 
This is a contradiction. 
If $j \in B = A_{j}$, then $j \notin B$ by definition of set B.
If $j \notin B = A_{j}$, then $j \in B$ by definition of set B.
Therefore by contradiction, powerset is uncountable.This is the generic proof.
When we say let there be a set $B = \{i\in A: i \notin A_{i}\}$. This set B is an element of power set, but by definition B cannot be in powerset, thus by contradiction power set of A is uncountable.
I understand what is at contradiction, but I don't get how we can just assume such a set B exists in the first place to be able to form a contradiction?
 A: The simplified proof is this: Let $A$ be any nonempty set.  Let $f:A\rightarrow 2^A$ be a function.  Assume $f$ is surjective (we reach a contradiction).  Define $$B = \{i \in A : i \notin f(i)\}$$ 
Since $B$ is a (possibly empty) subset of $A$, by surjectiveness there must be an element $a^* \in A$ such that $f(a^*)=B$.  Is $a^* \in B$?  Both "yes" and "no" lead to a contradiction. 
A: It might help to think of this more constructively. I claim:

Any time you give me a function $f:\mathbb{N}\rightarrow\mathcal{P}(\mathbb{N})$, I can give you a set $A_f\subseteq\mathbb{N}$ with $A\not\in ran(f)$.

This simplifies things by removing the unnecessary contradiction aspect.
So let's prove my claim. Suppose you give me an $f:\mathbb{N}\rightarrow\mathcal{P}(\mathbb{N})$. I'll build my set $A$ as follows: it's just the set of $i\in\mathbb{N}$ such that $i\not\in f(i).$ More succinctly, we're defining $A$ by saying $$i\in A\iff i\not\in f(i).$$ Note that you've already supplied the relevant function - there's no circularity here, since the definition of $A$ comes after you've already provided your $f$.
It's clear that if $A$ is a set then $A\not\in ran(f)$. So we only have one question: why is this a valid definition of a set? 
Well, ultimately this will come down to the specific axiomatic framework you're using, but the idea in the usual set theory is that "$i\not\in f(i)$" is a well-formed statement which is either true or false for each specific $i\in \mathbb{N}$. For example, if your particular $f$ has $$f(7)=\{2,4,6,8,10,...\},$$ then we'd have $7\in A$ since $7\not\in\{2,4,6,8,10,...\}$.
And now we go to our precise axiomatic system. One of the ways we build sets is via formulas: given a set $A$ and a property $p$, we can form the set $\{x\in A: x$ has property $p\}$. Here our property $p$ is "is not an element of $f$ applied to it."
Does this mean you can block Cantor's argument if you weaken your axioms enough? Well, sure, practically anything can be blocked if you are willing to throw away axioms willy-nilly. But the definition of $A$ here is so simple that (so far) there isn't any known useful set theory which is able to construct powersets but can't prove that $\mathcal{P}(\mathbb{N})$ is uncountable.
