The uniform convergence Let $u_n$ be a decreasing sequence of continuous functions on a compact Hausdorff $X$ and assume $u_n(x)\to0$ for all $x\in X$. Then I want to prove $u_n\to0$ uniformly.
I am trying to use Ascoli-Arzala Theorem. The uniform boundedness is quick but I can not get uniform continuous....
Also, by using "monotone" convergence, I have $\lim_{n\to\infty}\int_X u_n=0$, but it is not helpful... I need $L^\infty$ convergence but not $L^1$ convergence...
Please help!
 A: define
$$
U_n(\epsilon)=\{x \in X|u_n(x) \lt \epsilon\} 
$$
note that
$$
m \lt n \Rightarrow U_m(\epsilon) \subset U_n(\epsilon)
$$
because $u_n$ is a decreasing sequence. also each $U_n(\epsilon)$ is open (continuity) and
$$
 \cup_{n=1}^{\infty} U_n(\epsilon) = X
$$
now invoke compactness to show
$$
\forall \epsilon \gt 0 \exists n_{\epsilon}.\forall x \in X, n \gt n_{\epsilon} \Rightarrow u_n(x) \lt \epsilon
$$
A: Fix $\varepsilon>0$. Then for every $x\in X$, there is an $n_x\in\mathbb N$, such that
$$
0\le u_{n_x}(x)<\frac{\varepsilon}{2}.
$$
But as $u_{n_x}$ is continuous at $x$, there exists a $U_x$ open, where 
$\lvert u_{n_x}(y)-u_{n_x}(x)\rvert<\frac{\varepsilon}{2}$, and so
$$
0\le u_{n_x}(y)<\varepsilon, \quad\text{for $y\in U_x$}\tag{1}
$$
Now, as $\{U_x\}_{x\in\ X}$ is an open cover of a compact space, there are $x_1,\ldots,x_k$, such that
$$
X=U_{x_1}\cup\cdots\cup X_{x_k}.
$$
So, if $n\ge n_0=\max\{n_{x_1},\ldots,n_{x_k}\}$, and then
$$
0\le u_{n}(x)<\varepsilon, \quad\text{for $x\in X$}.
$$
For, if $x\in X$, then $x\in U_{x_j}$, for some $j=1,\ldots,k$, and as $n\ge n_{x_j}$,
then 
$$
0\le u_n(x)\le u_{n_{x_j}}<\varepsilon,
$$
due to $(1)$.
