I consider $C(X,\mathbb{R})$ the space of continuous functions from a compact metric space $X$ to $\mathbb{R}$ with the supremum norm. I would like to prove Dino’s theorem
Let $(f_n)_n\subset C(X,\mathbb{R})$ be an increasing sequence of functions that converges pointwise to $f$ with $f$ continuous. Then $f_n$ converges uniformly to $f$.
Here is my attempt that should be false because I do not use the fact the sequence is increasing :
Let $\varepsilon>0$ and define $A_{n,\varepsilon}=\{x\in X : \lvert f(x) -f_n(x)\rvert<\varepsilon\}$. We have that $A_{n,\varepsilon}$ is open as the pre image of $[0,\varepsilon)$ (open in $\mathbb{R}_{\geq 0}$) by $g=\lvert f-f_n\rvert$ and $g$ is continuous. Let $x\in X$, we have by pointwise convergence the existence of an integer $N(x)$ such that $\lvert f(x)-f_{N(x)}(x)\rvert < \varepsilon$ such that $x\in A_{N(x),\varepsilon}$ so $X=\cup_{n\in\mathbb{N}}A_{n,\varepsilon}$. Since $X$ is compact, there exists a finite open cover $A_{i_1,\varepsilon},…, A_{i_k,\varepsilon}$. Let $N=\max(i_1,…,i_k)$, then for all $n\geq N$ we have that $\sup_{x\in X}\lvert f_n(x)-f(x)\rvert<\varepsilon$.
Do you see where is the mistake please ?
EditMistake found I think : in the definition of $A_{n,\varepsilon}$, the $n$ does not depend on $\varepsilon$ and is by no mean related to the very definition of pointwise convergence of $f_n(x)$ to $f(x)$. Hence, it is possible that by taking a rank $N$ we have $\lvert f_N(x)-f(x)\rvert<\varepsilon$, and then at rank $N+1$ we have that $\lvert f_{N+1}(x)-f(x)\rvert>\varepsilon$.