Let $f_n:X\to\Bbb R$ be a sequence of continuous functions with $f_n(x)\to f(x)$ for each $x\in X$. If $f$ is continuous and if the sequence $f_n$ is monotone increasing and if $X$ is compact, then the convergence is uniform.
Proof. For each $x\in X$, by the continuity of $f_n$, there is $U_x$ such that for any $y\in U_x$, $|f_n(x)-f_n(y)|<\epsilon/3$. Also by the continuity of $f$, there is $V_x$ such that for any $y\in V_x$, $|f(y)-f(x)|<\epsilon/3$. Let $W_x:=U_x\cap V_x$ which is nonempty open set as $x\in W_x$. Then $\{W_x\}_{x\in X}$ is an open covering of $X$ so by compactness, there is a finite subcover $\{W_{x_1},...,W_{x_n}\}$. Note that for each $x\in X$, there is a number $N_x$ such that $n\geq N_x$ implies $|f_n(x)-f(x)|<\epsilon/3$. So define $N = \max\{N_{x_1},...,N_{x_n}\}$.
Now let $y\in X$ be given. Then there is $i$ such that $y\in W_{x_i}$. For $n\geq N$, $$|f_n(y)-f(y)|\leq |f_n(y)-f_n(x_i)|+|f_n(x_i)-f(x_i)|+|f(x_i)-f(y)|<\epsilon.$$ I think I didn't use monotone increasing condition. I can't find the error. Could you help?
