Let $\mathbb{R}^i$ be equipped with their Euclidean topologies (for $i=n,m$) and consider the following topologies on $C(\mathbb{R}^n,\mathbb{R}^m)$:
Topology of Pointwise convergence In Nagata's book, it is shown that the topology of pointwise convergence is equivalent to the point-open topology, which is generated by the sub-basic sets $$ \left\{f \in C(\mathbb R^n, \mathbb R^m) : f(x) \subseteq O \right\}, \quad x \in \mathbb R^n,\ O\subseteq \mathbb{R}^m \text{ open}. $$
Topology of Compact Convergence Similarly, in the same book, it is shown that the topology of compact-convergence (a.k.a. uniform convergence on compacts) is well-known to be equivalent to the compact-open topology on $C(\mathbb{R}^n,\mathbb{R}^m)$ which can be described by the sub-basic sets $$ \left\{f \in C(\mathbb R^n, \mathbb R^m) : f(K) \subseteq O \right\}, \quad K \subseteq \mathbb R^n \text{ compact, } O\subseteq \mathbb{R}^m \text{ open}. $$
Topology of Uniform Convergence From these equivalences one immediately has that compact convergence refines pointwise convergence. Now, it is also clear that uniform convergence refines compact convergence. Is there a sub-base of the topology of uniform convergence of the form $$ \left\{ f \in C(\mathbb{R}^n,\mathbb{R}^m):\, f(A) \subseteq O \right\}, \quad A \in \mathcal{A}, \ O\subseteq \mathbb{R}^m\mbox{ open}, $$ where $\emptyset \neq \mathcal{A}\subseteq 2^{\mathbb{R}^n}$ is a collection of non-empty subsets of $\mathbb{R}^n$?