Compact Convergence Implies Local Uniform Convergence if Weakly Locally Compact

Theorem

Let $T = \struct {S, \tau}$ be a weakly locally compact topological space.

Let $M = \struct {A, d}$ be a metric space.

Let $\sequence {f_n}$ 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 $f$.

$\blacksquare$

Also see

• Local Uniform Convergence Implies Compact Convergence