TITLE: Cardinality of power set of $\mathbb N$ is equal to cardinality of $\mathbb R$
QUESTION [5 upvotes]: 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.

REPLY [19 votes]: 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 $B.$