Vector space over F (set of functions)

$\mathbb F^{n}$ is a set of tuples of size $n$ which is obtained by Cartesian product of $\mathbb F$ $n$ times, where $\mathbb F$ is a field. It's clear. I can imagine that in my mind ))).

But I don't understand how the same approach applies to $\mathbb F^{S}$ where S is a set.

This is the definition:

If S is a set, then $\mathbb F^{S}$ denotes the set of functions from S to F.

Here $\mathbb F^{S}$ is a vector space over $\mathbb F$.

For instance, $\mathbb R^{[0,1]}$ is vector space where elements are real-valued functions on [0,1].

I understand what vector space is and that $\mathbb F^{S}$ is a set of functions, but I don't understand how it correlates with $\mathbb F^{n}$. Is it a Cartesian product of $\mathbb F$ $S$ times (set times)?

• If you take $S = \{1,2,3,\ldots, n\}$ then $F^S$ is essentially the same as (isomorphic to) $F^n$. This is because an ordered tuple $(x_1, \ldots, x_n) \in F^S$ is equivalent to a function $f: \{1, 2, \ldots, n \} \rightarrow F$ given by $f(i) = x_i$. – Jair Taylor Jan 8 '18 at 9:16
• @JairTaylor, Thanks for the answer. Not it makes sense. Can you post it as answer so I can accept it? – Nurjan Jan 8 '18 at 9:50
• Sure, I've elaborated slightly and posted an answer. – Jair Taylor Jan 8 '18 at 19:04

The notation $F^S$ comes from the fact that if $S = [n]$ (using the notation $[n] = \{1,2,3, \ldots, n\}$) then $F^S$ is isomorphic to $F^n$ as a vector space. The isomorphism is $T: F^{[n]} \rightarrow F^n$ given by $T(f) = (f(1), f(2), \ldots, f(n))$. Since they are isomorphic, they have the same cardinality (number of elements). If $F$ is finite, say $|F| = q$, this is reflected in the fact that $|F^n| = q^n$, since there are $q$ choices for the first element of the tuple, $q$ choices for the second, etc., up to the $n$th element. Similarly, $|F^{[n]}| = q^n$ since there are $q$ choices for $f(1)$, $q$ choices for $f(2)$, etc.