Local Uniform Convergence Implies Compact Convergence
Theorem
Let $X$ be a 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 locally uniformly to $f : X \to M$.
Then $f_n$ converges compactly to $f$.
Proof
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Recall that a sequence $f_n$ converges compactly to $f$, if $f_n$ converges uniformly to $f$ over any compact subset $K \subset X$.
Consider a sequence $f_n$ which converges locally uniformly to $f : X \to M$.
Let $K$ be an arbitrary subset $K \subset X$.
For $x \in K$ there exists an open neighborhood $U_x$ such that over $U_x$ the sequence $f_n$ converges uniformly to $f$.
Observe that $\set {U_x : x \in K}$ is an open cover of $K$.
Since $K$ is compact, there exists a finite collection of points $x_1, \ldots, x_\ell$ such that $\set {U_{x_1}, \ldots, U_{x_\ell} }$ is an open subcover of $K$.
Fix $\epsilon > 0$.
Consider $i \in \set {1, \ldots, \ell}$.
Then there exists $N_i \in \N$ such that:
- $\forall n \ge N_i: \sup \set {\map d {\map {f_n} z, \map f z} : z \in U_{x_i} } < \epsilon$
Take $N = \max \set {N_1, \ldots, N_\ell}$.
Then for $\ds z \in \bigcup_{i \mathop = 1}^\ell U_{x_i}$:
- $\sup \set {\map d {\map {f_n} z, \map f z} : z \in U_{x_i} } < \epsilon$
But since $\ds K \subset \bigcup_{i \mathop = 1}^\ell U_{x_i}$, and $\epsilon$ is arbitrary we conclude that $f_n$ converges uniformly over $K$.
The fact that $K$ is an arbitrary compact subset finishes the proof.
$\blacksquare$