Proposition 7.2.6.label Let $X$ be a topological space, $\kappa \subset 2^{X}$ be the collection of all precompact sets in $X$, and $(Y, \fU)$ be a uniform space, then the $\kappa$-open topology and $\kappa$-uniform topology on $C(X; Y)$ coincide.
Proof. By Definition 7.2.2, the $\kappa$-uniform topology is finer than the $\kappa$-open topology.
Let $K \in \kappa$ be compact, $U \in \fU$ be symmetric, and $f \in C(X; Y)$. For each $x \in K$, there exists $V_{x} \in \cn_{X}(x)$ such that $f(V_{x}) \subset U(f(x))$. Since $K$ is compact, there exists $\seqf{x_j}\subset X$ such that $K \subset \bigcup_{j = 1}^{n} V_{x_j}$.
Let $g \in \bigcap_{j = 1}^{n} M(V_{x_j}\cap K, U(f(x_{j})))$ and $y \in K$, then there exists $1 \le j \le n$ such that $y \in V_{x_j}$. In which case, since $f(y), g(y) \in U(f(x_{j}))$, $(f(y), g(y)) \in U \circ U$. Therefore
and the $\kappa$-open topology is finer than the $\kappa$-uniform topology.$\square$