I'm reading Advanced Linear Algebra by Steven Roman and in chapter two there is a theorem which is left for the reader to prove:
Theorem 2.7: If $n$ is a natural number, then any $n$-dimensional vector space over $F$ is isomorphic to $F^n$. If $\kappa$ is any cardinal number, and if $B$ is a set of cardinality $\kappa$, then any $\kappa$-dimensional vector space over $F$ is isomorphic to the vector space $(F^B)_0$ of all functions from $B$ to $F$ with finite support.
The first part is clear to me, but I can't understand the second one, specially the requirement for the functions to have finite support. Could anyone give me a proof of this second part of the theorem? Thanks.
I have finally been able to understand the isomorphism, and so I'll be posting an answer to my own question for anyone interested.
Let $\mathcal{B}$ be a basis for $V$ of cardinality $\kappa$. Let $B$ be a set of same cardinality as $\mathcal{B}$. Then there exists a one-to-one correspondence between these two sets.
We now prove that $(F^B)_0$ is indeed a vector space. It is easy to see that the set is closed under addition and multiplication by $\lambda \in F$. The set of functions $\{ f_b \} \subseteq (F^B)_0$, defined for any $c \in B$ as \begin{equation} f_b(c) = \begin{cases} 1 & c = b \\ 0 & c \neq b \end{cases} \end{equation} is linearly independent, since evaluating the zero expression at any $b_j$ gives: \begin{equation} 0 = \sum_i a_i f_{b_i} (b_j) = a_j \; . \end{equation} We can also see that $\{ f_i \}$ spans $(F^B)_0$, since any function with finite support from $B$ to $F$ is a finite linear combination of elements of $\{ f_i \}$.
Let $\tau \in \mathcal{L}((F^B)_0, V)$ be a linear transformation defined for any $f \in (F^B)_0$ as \begin{equation} \tau (f) = \sum_i f(h(c_i)) c_i = \sum_i f(b_i) c_i \; , \end{equation} where $b_i \in B$ and $c_i \in \mathcal{B}$. We use $h: \mathcal{B} \to B$ as a bijection which converts an element from $\mathcal{B}$ to an equivalent from $B$. Such bijection exists given the equality of the cardinalities of both sets.
We now prove that $\tau$ is an isomorphism. It is an homomorphism: \begin{equation} \begin{split} \tau (rf + sg) = \sum_i (rf + sg)(b_i) c_i = \sum_i (rf(b_i) + sg(b_i)) c_i \\ = r \sum_i f(b_i) c_i + s \sum_i g(b_i) c_i = r \tau (f) + s \tau (g) \; . \end{split} \end{equation} It is also injective, since $\tau (f) = 0$ implies \begin{equation} 0 = \sum_i f(b_i) c_i \; , \end{equation} and the linear independence of $\{ c_i \}$ implies that $f(b_i) = 0$ for all $b_i$ and so $f = 0$. Hence, $\operatorname{ker} (\tau) = \{ 0\}$.
Finally, it is surjective, since the output of all $f \in (F^B)_0$ contain all possible mappings with finite support to all possible values of $F$, therefore its functions produce all possible scalars in a linear combination in $V$. Hence, $\operatorname{im} (\tau) = V$.
We can then conclude that $V \approx (F^B)_0$.
Also, a shorter proof lies in the fact that each $\{ f_b \}$ in the basis of $(F^B)_0$ is defined for one $b \in B$, hence the dimension of $(F^B)_0$ is also $\kappa$. From there we have that two vector spaces with same dimension over a same field are isomorphic, as there is a bijection between their bases, which can be extended linearly.