Reading Axler's Linear Algebra Done Right. He defines $F^S$ to be the set of functions from $S$ to $F$.
So for example, $\mathbb R^{[0,1]}$ is the set of all functions from $0$ to $1$ that map $\mathbb R$. So does this mean for all elements in $[0,1]$, these elements can have functions assigned to them, (infinitely many functions I assume), and by doing so, this will map out all of $\mathbb R$?
For example if you fix the function $f(x)= x + 1$ and make the domain $[0,1]$ you would get the image $[1,2]$, which would be some values of $\mathbb R$, but not all. Then you would find infinitely many other functions that map out the rest of it? And all of these would go into the set? Let me know what you guys think.
Best Answer
A simple example: \begin{align} & \{0,1\}^{\{a,b,c\}} \\[10pt] = {} & \left\{ \left[ \begin{array}{l} a\mapsto0 \\ b\mapsto0 \\ c\mapsto0 \end{array} \right], \left[ \begin{array}{l} a\mapsto0 \\ b\mapsto0 \\ c\mapsto1 \end{array} \right], \left[ \begin{array}{l} a\mapsto0 \\ b\mapsto1 \\ c\mapsto0 \end{array} \right], \left[ \begin{array}{l} a\mapsto1 \\ b\mapsto0 \\ c\mapsto0 \end{array} \right], \right. \\[10pt] & \qquad \left. \left[ \begin{array}{l} a\mapsto0 \\ b\mapsto1 \\ c\mapsto1 \end{array} \right], \left[ \begin{array}{l} a\mapsto1 \\ b\mapsto0 \\ c\mapsto1 \end{array} \right], \left[ \begin{array}{l} a\mapsto1 \\ b\mapsto1 \\ c\mapsto0 \end{array} \right], \left[ \begin{array}{l} a\mapsto1 \\ b\mapsto1 \\ c\mapsto1 \end{array} \right] \right\}. \end{align}
It's the set of all functions whose domain is $\{a,b,c\}$ and whose codomain is $\{0,1\}.$