Let $n$ be a positive integer and consider the set $\{0,1\}^n$. What is this set usually called? I've seen some authors refer to it as the $n$-dimensional hypercube, whereas others typically use that name for the compact set $[0,1]^n$.
Now consider the set of all self maps $f : \{0,1\}^n \to \{0,1\}^n$. What is the cardinality of such set? Am I correct in thinking that it is simply $|\{0,1\}^{2n}|=4^n$?
Let $n$ be a positive integer and consider the set $\{0,1\}^n$. What is this set usually called? I've seen some authors refer to it as the $n$-dimensional hypercube, whereas others typically use that name for the compact set $[0,1]^n$.
I suppose it ultimately depends on what you want to do with it, and how one defines "cube" (the vertices? the faces? the inside as well?). One has a similar confusion between terms like "sphere" (often the boundary) and "disc" (often includes the inside).
Ultimately, so long as it's clear what you mean, I don't think it really matters. I'm preferential to "vertices of the $n$-dimensional hypercube" myself, which is perhaps the most literal (geometric) interpretation
Now consider the set of all self maps $f : \{0,1\}^n \to \{0,1\}^n$. What is the cardinality of such set? Am I correct in thinking that it is simply $|\{0,1\}^{2n}|=4^n$?
Yes. In general, we define $A^B$ to be all functions $f : A \to B$ and one has $$ \left| A^B \right| = |A|^{|B|} $$ and hence $$ \left| \{0,1\}^{2n} \right| = |\{0,1\}|^{2n} = 2^{2n} = 4^n $$ (You may interpret $n$ as the set $\{0,1,\cdots,n-1\}$, and $2n$ likewise as $\{0,1,\cdots,2n-1\}$.)
One could also argue this combinatorially since we're working in finite sets: $|\{0,1\}^{2n}|$ is the set of all $2n$-tuples $(a_1,\cdots,a_{2n})$ where $a_i \in \{0,1\}$ for each $i$, and since there are $2$ choices for each of the $2n$ members of the tuple, then you get $2^{2n} = 4^n$ possible tuples.
