Use the Principle of Mathematical Induction to prove that the number of subsets of a set containing $n$ distinct elements is $2^{n}$,for all $n \in N$.

(N/A) $P(n):$ The number of subsets of a set having $n$ elements is $2^{n}$,where $n \in N$.
For $n=1$:
Let $A$ be a set with one element,$A = \{x\}$.
The subsets of $A$ are $\phi$ and $A$.
The number of subsets of $A$ is $2 = 2^{1}$.
Thus,$P(1)$ is true.
Assume that $P(k)$ is true for some $k \in N$,i.e.,a set with $k$ elements has $2^{k}$ subsets.
Now,we prove for $n = k+1$.
Let $A = \{a_{1}, a_{2}, \ldots, a_{k}, a_{k+1}\}$.
The subsets of $A$ can be divided into two types: those that do not contain $a_{k+1}$ and those that do contain $a_{k+1}$.
The number of subsets not containing $a_{k+1}$ is the number of subsets of $\{a_{1}, a_{2}, \ldots, a_{k}\}$,which is $2^{k}$ by the assumption $P(k)$.
The number of subsets containing $a_{k+1}$ is also $2^{k}$ (each subset is formed by adding $a_{k+1}$ to each of the $2^{k}$ subsets of $\{a_{1}, a_{2}, \ldots, a_{k}\}$).
Therefore,the total number of subsets of $A$ is $2^{k} + 2^{k} = 2 \cdot 2^{k} = 2^{k+1}$.
Thus,$P(k+1)$ is true.
Hence,by the Principle of Mathematical Induction,$P(n)$ is true for all $n \in N$.