This question already has an answer here:

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 Jul 28 at 11:25

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


$\varepsilon > 0$. $G_n := \{x\in X : \vert f_n (x) \vert < \varepsilon \}$. 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.

  • $\begingroup$ Why $f_n(x) \rightarrow 0$ then $x \in \bigcup_n G_n$ ? $\endgroup$ – Nguyen Thy Jul 28 at 8:36
  • 2
    $\begingroup$ $f_n(x)\to 0$ 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, $x\in G_N$. $\endgroup$ – WoolierThanThou Jul 28 at 8:41

Not the answer you're looking for? Browse other questions tagged or ask your own question.