Show that the cardinality of $P(\mathbb{N})$ - the power set of $\mathbb{N}$ - is $2^{\aleph_o}$.
Hint: Show that $P(\mathbb{N})$ has the same cardinality as the set $N = \{f: \mathbb{N} \rightarrow A| f$ is a function$\}$, where $A = \{0, 1\}$.
My attempt:
Using an inductive argument with the statement that if $|X|=n$, then, $|P(X)| = 2^n$ for all $n\geq 0$, we have
Base: If $n = 0$, $X = \emptyset$. But the empty set is the only subset of itself, so $|P(X)| = 2^0$.
Inductive: Suppose for $m \geq 0$, if $|X| = m$, then $|P(X)| = 2^m$.
If $|X| = m + 1$, pick some $x \in X$ and let $Y = X - \{x\}$.
By assumption, $|P(Y)| = 2^m$.
There are two subsets of $X$: subsets with $x$ and subsets without $x$. For each subset, there are $2^m$ elements so $|P(X)| = 2P(Y) = 2^m + 2^m = 2^{m+1}$.
We have shown that the power set of a set $X$ has $2^{|X|}$ elements.
Let $X = \mathbb{N}$, then, as $|\mathbb{N}| = \aleph_o$, it follows that $|P(\mathbb{N})| = 2^{\aleph_o}$.
QED.
Is this proof correct? If so, what would a proof that uses the hint look like? Any insight is much appreciated.