The number of subsets of a set of cardinality $n$

Please help with this question. Show that for a finite set $A$ of cardinality $n$, the cardinality of $P(A)$ is $2^n$, where $P(A)$ is the power set of $A$. Thank you in advance for any help that is given.

Think of how you could go about constructing a subset of $A$. For each element, you choose to either include it or exclude it from the subset you are building. That gives you 2 choices for each of the $n$ elements of $A$. Multiplying your choices together, you get $2^n$ total possibilities. That is, there are $2^n$ different subsets you can build from $A$.

HINT:

Suppose I am choosing elements to put in some subset of the power set. Then each element can either be in, or not be in, my subset. So this means that altogether...

Number the elements of $A$ as $a_1, a_2, \dots, a_n$. Consider the map $\chi:P(A) \to \{0,1\}^n$ given by $\chi(X)=(x_1,x_2,\dots,x_n)$, where $x_i=1$ iff $a_i \in X$. Then $\chi$ is a bijection.

Each subset $S$ of $A$ can be formed by considering each element of $A$ and deciding whether or not that element is to be in the subset.

There are two choices for each element $a$ of $A$ -- either the element $a$ is in $S$, or it is not. Each distinct set of choices (one for each element of $A$) determines a unique subset of $A$. So, since there are $k$ elements with $2$ possible choices each, there is a total of

$$\underbrace{2\times2\times\cdots\times 2}_{k\textrm{ factors}}=2^k$$ different subsets.

Multiplication is used because you are using "and" to combine the choices. For example, "$a_1$ is in $S$, and $a_2$ is in $S$, and $a_3$ is not in $S$, and etc".