Find the cardinality of $\{f: \mathbb{N} \rightarrow \mathbb{P}(\mathbb{N}) \mid ( \forall n ) ( n \in f(n) ) \}$ 
Find the cardinality of $A=\{f: \mathbb{N} \rightarrow \mathbb{P}(\mathbb{N}) \mid (\forall n ) ( n \in f(n) ) \}$.

I know $A\subseteq \mathbb{P}(\mathbb{N})^{\mathbb{N}}$, so ${\#}A \leq \mathfrak{c}$.
I want to show the other inequality by defining a function $\phi: \mathbb{P}(\mathbb{N}) \rightarrow A$. If $K \subseteq \mathbb{N}$ then $\phi(K)=f_{K}$ where $f_{K}: \mathbb{N} \rightarrow \mathbb{P}(\mathbb{N})$ is defined by $$f_{n}=
\begin{cases}
K, &\text{if }n \in K \\
\{ n \}, &\text{if }n \notin K.
\end{cases}
$$
Is this correct? If so how can I prove this function is injective. If I prove that then I think I prove ${\#}A=\mathfrak{c}$.

martini seems to have aptly answered the immediate problem, so I'll give an alternate method of showing the inequality $\mathfrak{c} \leq |A|$.
Consider the following subset of $A$: $$A^\prime = \{ f \in A : ( \forall n \neq 33687 )( f(n) = \{ n \} ) \}.$$
Note that the mapping $\psi : A^\prime \to \mathcal{P} ( A )$ defined by $\psi(f) = f(33687)$ is one-to-one, and the range of $\psi$ is $\{ K \subseteq \mathbb{N} : 33687 \in K \}$. As this set can easily be put into one-to-one correspondence with $\mathcal{P} ( \mathbb{N} )$ it follows that $\mathfrak{c} = | \{ K \subseteq \mathbb{N} : 33687 \in K \} | = | A^\prime | \leq |A|$.
