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Question: Let $(E,d)$ be compact metric space. Let $(f_n(x)_{n\in \mathbb{N}}$ be sequence of continuous functions from E to $\mathbb{R}$ such that $\forall n \in \mathbb{N}$ we have $f_{n+1}\leq f_n$. Suppose that $\forall x \in E$, the sequence $(f_n(x))_{n \in \mathbb{N}}$ converges to $f(x)$. Show that $\forall \epsilon>0, \exists N \in \mathbb{N}$ such that for all $n\geq N$ and $\forall x \in E$, we have $|f(x)-f_n(x)|< \epsilon$.

I am not sure how to tackle this problem. To me by the convergence of $f_n(x)$ to $f(x)$ proves this result without any work but im not sure how to prove this. Any hints will be helpful.

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A hint given by my professor was to define $\forall \epsilon >0$ and $n \in \mathbb(N)$, the set $X_n={x \in E: f_n(x)-f(x) \leq \epsilon}$. Show that these are closed. What is $\cap X_n$ –  Mathstudent Nov 5 '12 at 4:09
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This is called Dini's theorem if $f$ is assumed to be continuous. If $f$ is not assumed to be continuous, then this is false. Note the difference between the assumption and the conclusion, which is between pointwise and uniform convergence. The assumption that $\forall x\in E$, $(f_n(x))\to f(x)$ says that for each $x$, for each $\varepsilon>0$, there exists $N\in \mathbb N$ such that .... This $N$ may depend on $x$. However, in the conclusion here $N$ is picked first, and then $x$. –  Jonas Meyer Nov 5 '12 at 4:12
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It is implicit in your professor's hint, mentioned in a comment, that $f$ is assumed to be continuous, but that is not stated in your question. –  Jonas Meyer Nov 5 '12 at 4:17
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