Compact Convergence Implies Local Uniform Convergence if Weakly Locally Compact

Theorem
Let $T = \left({S, \tau}\right)$ be a weakly locally compact topological space.

Let $M = \left({A, d}\right)$ be a metric space.

Let $\left\langle{f_n}\right\rangle$ be a sequence of mappings $f_n : X\to M$.

Let $f_n$ converge compactly to $f:X\to M$.

Then $f_n$ converges locally uniformly to $f$.

Proof
Because $T$ is weakly locally compact, each point of $S$ has a compact neighborhood in $T$.

Because $f_n \to f$ compactly, each point of $S$ has a (compact) neighborhood on which $f_n \to f$ uniformly.

Thus $f_n$ converges locally uniformly to $f$.

Also see

 * Local Uniform Convergence Implies Compact Convergence