Theorem. Let $X$ denote an arbitrary set such that $|X|=n$. Then $|\mathcal P(X)|=2^n$.
Proof. The proof is by induction on the numbers of elements of $X$.
For the base case, suppose $|X|=0$. Clearly, $X=\emptyset$. But the empty set is the only subset of itself, so $|\mathcal P(X)|=1=2^0$.
Now, the induction step. Suppose $|X|=n$; by the induction hypothesis, we know that $|\mathcal P(X)|=2^n$. Let $Y$ be a set with $n+1$ elements, namely $Y=X\cup\{a\}$. There are two kinds of subsets of $Y$: those that include $a$ and those that don't. The first are exactly the subsets of $X$, and there are $2^n$ of them. The latter are sets of the form $Z\cup\{a\}$, where $Z\in\mathcal P(X)$; since there are $2^n$ possible choices for $Z$, there must be exactly $2^n$ subsets of $Y$ of which $a$ is an element. Therefore $|\mathcal P(Y)|=2^n+2^n=2^{n+1}$. $\square$
From the above explanation, I don't understand why the set that contains $\{a\}$ will contain $2^{|n|}$ elements when it should clearly be $2^{|1|}$.
The construction of a new set $S$ is the union of the old set with cardinality $n$ and a new element $\{a\}$, therefore the set that does not contain $\{a\}$ still has cardinality $n$ and the set that contains $\{a\}$ is just $\{a\}$, one element.
Can someone please elucidate?
