# Prove that the sequence $\{f_n\}_{n \in \mathbb{N}}$ is uniformly convergent on a compact space $X$ [duplicate]

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.

## marked as duplicate by Robert Z, Martin R, TheSimpliFire, YuiTo Cheng, José Carlos Santos general-topology StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jul 28 at 11:25

$$\varepsilon > 0$$. $$G_n := \{x\in X : \vert f_n (x) \vert < \varepsilon \}$$. 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
$$X\subset G_{n_1} \cup \ldots \cup G_{n_k} = G_{n_k}$$
• Why $f_n(x) \rightarrow 0$ then $x \in \bigcup_n G_n$ ? – Nguyen Thy Jul 28 at 8:36
• $f_n(x)\to 0$ means that for every $\varepsilon>0$ there exists some $N\in \mathbb{N}$ such that $|f_n(x)|\leq \varepsilon$ for every $n\geq N$. In particular, $x\in G_N$. – WoolierThanThou Jul 28 at 8:41