Cardinality of power set of $\mathbb N$ is equal to cardinality of $\mathbb R$ How to prove that;

The cardinality of power set of $\mathbb N$ is equal to cardinality of $\mathbb R$

I need a reference or proof outline. I looked up but didn't find results
thanks.
 A: Hint: think of an element of $P(\Bbb N)$ as corresponding to a binary expansion-for a given subset $A$, let $x=\sum_{i \in A} 2^{-i}$ This gives you (almost) a bijection between $P(\Bbb N)$ and $[0,1]$
A: I. There is an injection $f:P(\mathbb N)\to\mathbb R,$ that is, $|P(\mathbb N)|\le|\mathbb R|.$
Proof. For $S\subseteq\mathbb N$ define $f(S)=\sum_{n\in S}10^{-n}.$
II. There is an injection $g:\mathbb R\to P(\mathbb N),$ that is, $|\mathbb R|\le|P(\mathbb N)|.$
Proof. Fix an enumeration $\mathbb Q=\{q_n:n\in\mathbb N\}$ of the set $\mathbb Q$ of all rational numbers, and define $g(x)=\{n:q_n\lt x\}.$
The equality $|P(\mathbb N)|=|\mathbb R|,$ that is, the existence of a bijection between $P(\mathbb N)$ and $\mathbb R,$ follows from I and II by virtue of the Cantor–Schröder–Bernstein theorem. Since I can't tell from your question if you are familiar with that theorem, I will now give a proof.
III. Lemma (Knaster–Tarski Fixed Point Theorem). If a function $\varphi:P(A)\to P(A)$ satisfies the condition
$$X\subseteq Y\implies\varphi(X)\subseteq\varphi(Y)$$
for all $X,Y\subseteq A,$ then $\varphi(S)=S$ for some $S\subseteq A.$
Proof. Let $\mathcal F=\{X:X\subseteq\varphi(X)\}$ and let $S=\bigcup\mathcal F.$ First note that, if $X\in\mathcal F,$ then $X\subseteq S$ and $X\subseteq\varphi(X)\subseteq\varphi(S).$ Since every member of $\mathcal F$ is a subset of $\varphi(S),$ it follows that $S=\bigcup\mathcal F\subseteq\varphi(S),$ that is, $S\in\mathcal F.$ From $S\in\mathcal F$ it follows that $\varphi(S)\in\mathcal F,$ whence $\varphi(S)\subseteq S,$ and so $\varphi(S)=S.$
IV. Theorem (Cantor–Schröder–Bernstein). If there are injections $f:A\to B$ and $g:B\to A,$ then there is a bijection between $A$ and $B.$
Proof. Define a function $\varphi:P(A)\to P(A)$ by setting
$$\varphi(X)=A\setminus g[B\setminus f[X]]$$
for $X\subseteq A,$ and observe that $X\subseteq Y\subseteq A\implies\varphi(X)\subseteq\varphi(Y).$ By the lemma, there is a set $S\subseteq A$ such that $\varphi(S)=S,$ that is, $g[B\setminus f[S]]=A\setminus S.$ Thus, $f$ maps $S$ bijectively onto $f[S],$ and $g$ maps $B\setminus f[S]$ bijectively onto $A\setminus S,$ so there is a bijection between $A$ and $B.$
A: Imitate/replicate Cantor's diagonal.
Assume P (N) is countable and that we can index each subset as $A_i $ for some natural $i $.
Now consider the set $B =\{n \in \mathbb N| n \not \in A_n  \} $.  In other words create a new set by  going through the sets $A_i $. if $i \in A_i $ exclude it.  If $i \not \in A_i$ include it.  
The resulting set will be distinct from all $A_i $ which means our indexing was not surjective which means no injective map from the naturals to P (N) exist, and P (N) is not countable.
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As Asaf Karagila points out this answer oversimplifies and merely shows that P(N) is uncountable but not that it has the cardinality of the reals. 
On the other hand, I wouldn't say my answer doesn't address the question "at all".  
As my argument replicates Cantors diagonals we can continue to conclude the same results as Cantors diagonal would conclude.
$P(\mathbb N) \cong \times_{\mathbb N}\{0,1\}$ via mapping $A \subset \mathbb N \rightarrow (x_1, x_2, x_3,..... )$ where $x_i = 1$ if $i \in A$ and $x_i = 0$ if $i \not \in A$.  (And by Cantor's diagonal both are uncountable-- although that was never asked to be shown apparently.)
And $|  \times_{\mathbb N}\{0,1\} | = |\mathbb R|$ is a standard analysis result.
$ \times_{\mathbb N}\{0,1\} \cong \{\sum_{k=1}^{\infty}\frac{x_k}{2^k}|(x_1,x_2,x_3,...,x_k,...) \in \{0,1\}^{\mathbb N}\}$ is clear when we show no two different sequences of {$x_i$} yield equal sums.  (Which is straightforward to show).
We have to then show $\{\sum_{k=1}^{\infty}\frac{x_k}{2^k}|x_k \in \{0,1\}\}= [0,1]$ which .... is tedious but the completeness/least upper bound property of the reals leads directly to that.
And finally we must show $[0,1] \cong \mathbb R$ with is a standard result.  $(0,1) \cong \mathbb R$ via $x \rightarrow \arctan x$.  And with tedium we can show for infinite $S$ and finite $I$ that $|S| = |S \cup I|$.
And that does it.  Although as no-one ever asked to show $P(\mathbb N)$ was uncountable, I could have skipped Cantor's diagonal altogether and just gone to the postscript.   
But if I had done that, the clear parallel between $A \subset \mathbb N$ and $x \in [0,1]$ wouldn't have been immediately clear.
