Suppose $K$ is compact.
(a){$f_n$} is a sequence of continuous functions on $K$
(b){$f_n$} converges pointwise to a continuous function $f$ on $K$
(c) $f_{n+1}(x) \le f_n(x)$ for all $ x \in K$, $n=1,2,3,...$ Then $f_n \rightarrow f$ uniformly on K.
Proof:- Put $g_n=f_n-f$. Then $g_n$ is continuous, $g_n \rightarrow 0$ pointwise, and $g_{n+1} \le g_n$
Let $\epsilon >0$ be given. Let $K_n$ be the set of all $x \in K$ with $\epsilon \le g_n(x)$.
Since $g_n$ is continuous, $K_n$ is closed by theorem(*), hence compact.
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Theorem(*)
A mapping $f$ of a metric space $X$ into a metric space $Y$ is continuous
iff $f^{-1}(C)$ is closed in $X$ for every closed set $C$ in $Y$.
In here, I have no idea how to apply that theorem to the proof.
$K_n$ is $f^{-1}(C)$ in that theorem?