Dimensionality of a function space

Question

Let $\mathcal{S}$ be a finite set belonging to the set of natural numbers $\mathbb{N}$, and consider the product set $\mathcal{S}^k$ for some positive integer $k \geq 2$, i.e., $\mathcal{S}^k = \{\underline{s} \in \mathbb{N}^k \,|\, \underline{s}_i \in \mathcal{S}, \forall 1 \leq i \leq k\}$. Then consider the set $\mathcal{F}$ of real-valued functions over $\mathcal{S}^k$, i.e., \begin{equation} \mathcal{F} = \{f \,|\, f: \mathcal{S}^k \to \mathbb{R} \}. \end{equation} Let $S$ be the cardinality of $\mathcal{S}$. My question is: Is the "dimension" of $\mathcal{F}$ equal to $S^k$ or infinite-dimensional? Also, is there a "natural" way to introduce an inner product and a basis in this setting? Thanks a lot.

Answer 1

The dimension of your space (as a vector space over $\Bbb R$) is $ S^k.$ There's a natural basis $B$ for $\mathcal F: B= \{ f: \mathcal S^k \to \Bbb R \vert f(p) =1 \text{ for some } p \in \mathcal S, 0 \text{ otherwise}\}.$ I'll leave it to you to show that $B$ is linearly independent and spans $\mathcal F$.

Answer 2

I think $S$ is isomorphic to $\{1,2,...,|S|\}$, so S^k is isomorphic to $\{1,2,...,|S|\}^k$ which is isomorphic to $\{1,..., |S|\cdot k\}$. So basically you have $\{f \mid f : \{1,..., |S|\cdot k\} \to \mathbb{R} \}$. So a function $f$ is really nothing else then a $|S|\cdot k$-tuple of real numbers, so it is isomorphic to $\mathbb{R}^{|S|\cdot k}$.