Consider the set of of all functions of the form $F = { f: {0, 1}^n \rightarrow \mathbb{R} }$ Let us consider this over the field of real numbers. We define vector addition and multiplication as such: \(h = f + g \implies h(x) = f(x) + g(x)\) \(h = \alpha g \implies h(x) = \alpha \cdot g(x)\) Then this forms a vector space.
Consider the dimension of the simple case where $n = 1$, then $F = { f: {0, 1}^n \rightarrow \mathbb{R} }$.
In order to find the dimension, we must find a basis for the space.
Let the basis be defined by two functions $f_1, f_2$, such that
\(f_1(0) = 0,\ f_1(1) = 1\)
\(f_2(0) = 1,\ f_2(1) = 0\)
Then given $a, b \in \mathbb{R}$, we can represent any function in $F$ as \(a \cdot f_1 + b \cdot f_2\) Thus, the dimension of the vector space must be 2 in this case.
Similarly, for greater values of n, we can consider the possible values of $f_i$ at $x = 0$ and $x = 1$, which can be either $0$ or $1$ again, and use this as one of the basis vectors.
We can reason that thus, there must be $2^n$ such basis vectors and as a result, we can say that the dimension of the vector space $F$ is $2^n$.