Rudin PMA p.150
This is the theorem;
Suppose
(i)Let $X$ be a metric space and $K$ be a compact subset.
(ii)Let $\{f_n\}$ be a sequence of continuous functions such that $f_n:K\rightarrow \mathbb{R}$
(iii)$\forall x\in K, f_n(x)≧f_{n+1}(x)$
(iv)$f_n \rightarrow f$ pointwise on $K$.
(v)$f$ is continuous.
Then $f_n\rightarrow f$ uniformly on $K$.
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I think the condition (iii) is not essential.
I think (iii) can be replaced to "$\{f_n(x)\}_{n\in \mathbb{N}}$ is monotonic for all $x\in K$" and i actually proved it.
Here's what i tried;
Define $g_n=f_n - f$.
Let $A=\{x\in K|\{g_n(x)\}_{n\in\mathbb{N}} \text{ is monotonically decreasing}\}$ and $B=\{x\in K|\{g_n(x)\}_{n\in\mathbb{N}} \text{ is monotonically increasing}\}$
Let $K_n=\{x\in K|g_n(x)≧\epsilon \bigvee g_n(x)≦-\epsilon\}$.
Then $K_n$ is compact.
Also, it can be shown that$\forall x\in A \forall n\in \mathbb{N}, g_n(x)≧0$ and $\forall x\in B \forall n\in \mathbb{N} g_n(x)≦0$.
Thus $K_{n+1} \subset K_n$.
Since $g_n\rightarrow 0$ pointwise on $K$, $K\cap (\bigcap K_n)=\emptyset$.
Thus, $\exists N\in \mathbb{N}$ such that $n≧N \Rightarrow K_n=\emptyset$. Q.E.D.
Is my argument correct?
