Let $X$ be a compact space and $f_n: X \rightarrow \mathbb{R}, n \in \mathbb{N}$ are continuous and $f_{n+1}(x) \le f_n(x),\lim\limits_{n\rightarrow \infty} f_n=0, \forall x \in X $.
Prove that the sequence $\left\lbrace f_n\right\rbrace$ is uniformly convergent on $X$.
I'm trying to prove that with $G_n=\left\lbrace x \in X: f_n(x) < \epsilon\right\rbrace \Rightarrow \exists n_0: \forall n \ge n_0: G_n=X$ but I didn't succeed. Any solution is appreciated. Thank you.
$\varepsilon > 0$. $G_n := \{x\in X : \vert f_n (x) \vert < \varepsilon \}$. Since $f_n(x)$ is decreasing in $n$ we have $G_n \subset G_{n+1} $. Since $f_n$ is continuous $G_n$ is open. Let $x\in X$. Since $f_n(x) \to 0$, $x \in \cup_{n} G_n$. Thus $X\subset \cup_n G_n.$ Compactness of $X$ yields that there are $n_1 < \ldots < n_k$, such that
$$X\subset G_{n_1} \cup \ldots \cup G_{n_k} = G_{n_k}$$
Thus the convergence is uniform.