Cantor's Theorem for $\Bbb N$.

Let $S$ be the set of positive integers. Suppose there is a bijection $f : S \to P(S)$. Then every subset of $S$ is equal to $f(s)$ for some $s \in S$. For any $s\in S$, $f(s)$ is a subset of $S$, and it is certainly the case that either $s\in f(s)$ or $s\notin f(s)$. [For example, there exists $s_1$ such that $f(s_1)= S$, and then $s_1 \in f(s_1)$; likewise, there exists $s_2$ such that $f(s_2) = \emptyset$, and then $s_2 \notin f(s_2)$.] Define $A$ to be the set of all elements $s$ of $S$ such that $s \notin f(s)$; symbolically, $$A = \{s\in S\,|\,s \notin f(s)\}.$$

(In the above notation, $s_1 \notin A$ but $s_2 \in A$.)
Certainly $A$ is a subset of $S$; that is, $A\in P(S)$. Therefore, as $f$ is a bijection, $A = f(a)$ for some $a \in S$.
We now ask the question: does $a$ belong to $A$?
If $a \notin A$, then $a \notin f(a)$, so by definition of $A$, we have $a \in A$. This is a contradiction. And if $a \in A$ then $a \in f(a)$, so by definition of $A$ we have $a \notin A$, again a contradiction.
Thus we have reached a contradiction in any case. So we conclude that there cannot be a bijection from $S$ to $P(S)$.

Hint: Let $(b_1,b_2,...,b_i,...)$ denotes a new bit string. And let $(a_{i1},a_{i2},...,a_{ii},...)$ denotes the $i$th string that corresponds to the $i$th integer. Take $b_1 \neq a_{11}$, $b_2 \neq a_{22}$, ..., $b_i \neq a_{ii}$, and so on. It's called Cantor's diagonal argument.