Local Uniform Convergence Implies Compact Convergence

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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







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.

$\blacksquare$


Also see