Cardinality of set of all bijections $\mathbb{N}\to\mathbb{N}$; is my proof correct? I need to find cardinality of a set containing all bijections $\mathbb{N} \to \mathbb{N}$.
My proof goes like that:
Let $S$ be the set containing all bijections $\mathbb{N} \to \mathbb{N}$. There exists trivial injection $S \to {\mathbb{N}}^{\mathbb{N}}$ (identity). Therefore $S\subset\mathbb{N}^\mathbb{N}$. We define $$F: \{0,1\}^\mathbb{N} \to S$$ as $$F(f)=\lambda x.\text{if } f(x)=0 \text{ then }2x\text{ else }(2x+1)$$
$F$ is injective. Proof: Assume $F$ isn't injective. Then $\exists f_1, f_2 \in \mathbb{N}\to\{0,1\}. f_1 \ne f_2, F(f_1) = F(f_2)\implies\exists x\in\mathbb{N}.f_1(x) \ne f_2(x)\implies F(f_1)(x) \ne F(f_2)(x) \implies F(f_1) \ne F(f_2)$ Contradiction. 
It implies $\{0,1\}^\mathbb{N}\subset S$. $$|\{0,1\}^\mathbb{N}| = 2^{\aleph_0} =\mathfrak{c}  \\
|\mathbb{N}^\mathbb{N}| = \mathfrak{c} \\
\{0,1\}^\mathbb{N}\subset S\subset\mathbb{N}^\mathbb{N}\\
\text{Cantor-Bernstein:} \\
|\{0,1\}^\mathbb{N}| \le |S| \le |\mathbb{N}^\mathbb{N}| \\
|S| = \mathfrak{c}
$$
Is this proof correct?

@Eric Troy: I think I got your idea. You break the set of naturals into pairs. Now you flip the pair $n$ or not according to $a_n=1$, or  $a_n=0$. You get for every infinite sequence of $0$ and $1$ $(a_n)$ a bijection of $\mathbb{N}$.
A compact form:
$$(a_n) \mapsto \prod_n (2n-1,2n)^{a_n}$$
