Let $F$ be a field, let $X$ be a set, and let $F^X$ denote the set of functions from $X$ to $F$. Then $F^X$ forms a vector space in a natural manner. What can we say about the dimension of $F^X$? This question is easy when $X$ is finite, so from here on, suppose $X$ is infinite.
The set $\{\chi_{\{x\}} \, \vert \, x\in X\}$ (where $\chi_S$ denotes the indicator function of $S$) is a linearly independent subset of $F^X$ with cardinality $\vert X \vert$, and so we can say that $\dim F^X \geq \vert X \vert$. In general, we don't necessarily have equality, since $\dim \mathbb{R}^\omega = \mathfrak{c}$. In fact, I think that $\dim F^\omega \geq \mathfrak{c}$ for any field $F$, which should be enough to show that $\dim F^X > \vert X \vert$ if I'm not mistaken.
Is there anything more that we can say about $\dim F^X$? It would be ideal if there were a formula for $\dim F^X$ in terms of $\vert F \vert$ and $\vert X \vert$, but I doubt that's true.
