I am reading about the examples of vector space on wikipedia. One is the function space defined as the following,
Let $X$ be an arbitrary set and $V$ an arbitrary vector space over $F$. The space of all functions from $X$ to $V$ is a vector space over $F$ under pointwise addition and multiplication. That is, let $f : X → V$ and $g : X → V$ denote two functions, and let $α∈F$. We define
$(f + g)(x) = f(x) + g(x)$
$(\alpha f)(x) = \alpha f(x)$
where the operations on the right hand side are those in $V$. The zero vector is given by the constant function sending everything to the zero vector in $V$. The space of all functions from $X$ to $V$ is commonly denoted $V^X$.
If $X$ is finite and $V$ is finite-dimensional then $V^X$ has dimension $|X|(\text{dim} V)$, otherwise the space is infinite-dimensional (uncountably so if $X$ is infinite).
However wiki does not give any proof with above claim of the dimension of such function space. Anyone can help with a proof or a reference to some proof? Thank you!
Best Answer
Let $\{ e_i \mid i \in I \}$ be a basis for $V$.
You can check that a basis for $V^X$ is given by the set $$\{ f_{x,i} \mid x \in X,\ i \in I \}$$ where $f_{x,i} : X \to V$ is the function defined by $$f_{x,i}(y) = \begin{cases} e_i & \text{if } y=x \\ 0 & \text{otherwise} \end{cases}$$ for all $y \in X$.
This basis has $|X| \cdot |I|$ elements, and so $\dim(V^X) = |X| \cdot \dim(V)$.
This is true even when $V$ is infinite-dimensional (but not if $X$ is infinite).