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I'm working on a problem which amounts to showing that if $X$ is a compact metric space and $C(X)$ is the space of continuous functions $X\to\mathbb{R},$

$$\forall f\in C(X)\ \ f(x_n) \to f(x)\implies x_n \to x$$

I've done a fair amount of functional analysis (with weak and weak* convergence) but I'm not sure that this result is true.

If it isn't, please let me know, otherwise, I would appreciate a hint to get me started.

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    $\begingroup$ @Randall you beat me by one second. $\endgroup$ – Thomas Jan 17 '20 at 20:07
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    $\begingroup$ Of course this is possible if $X\subset\mathbb{R}$, but it doesn't make sense otherwise. $\endgroup$ – AlephNull Jan 17 '20 at 20:07
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    $\begingroup$ Let $f(t)=d(t,x)$ and ignore that $X$ is compact? $\endgroup$ – Hagen von Eitzen Jan 17 '20 at 20:25
  • $\begingroup$ As Hagen pointed out, you don't need compactness. $\endgroup$ – peter a g Jan 17 '20 at 20:38
  • $\begingroup$ Note that since $f$ can be any continuous function this is a lot stronger than weak convergence. $\endgroup$ – copper.hat Jan 17 '20 at 22:17
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Let $X$ be a compact metric space. Assume there exist a sequence $(x_n)$ in $X$ not converging to a point $x\in X$ having the property:

We consider the following continuous (help) function $h:X\to \Bbb R$, defined for $y\in X$ by $$ h(y):=d(y,x)\ . $$ It is continuous because of $$ |h(y)-h(y')|=|d(y,x)-d(y',x)|\le d(y,y')\ . $$ (The last is true, because it is equivalent to $-d(y,y')\le d(y,x)-d(y',x)\le d(y,y')$, and each $\le$ reduces equivalently to a triangle inequality.)

It is given that $h(x_n)\to h(x)$, i.e. $d(x_n,x)\to 0$. This implies $x_n\to x$.

Contradiction...

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  • $\begingroup$ I asked for a 'hint' but you spoilt it with the full solution... something like 'choose f in terms of the metric' would've been enough. Thanks for the answer anyway. $\endgroup$ – AlephNull Jan 17 '20 at 20:37
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The statement is true. Assume by contradiction that $x_n \nrightarrow x$. This implies that for some $\epsilon > 0$, there exists a subsequence $x_{n_k}$ such that $d(x_{n_k}, x) > \epsilon$. Now, choose $f: X \rightarrow \mathbb{R}$ where $f(y) = d(y, x)$ and note that $f\in C(X)$. Finally, note that $f(x_{n_k}) > \epsilon > 0 = f(x)$ so the subsequence $f(x_{n_k})$ is not convergent to $f(x)$ which implies that $f(x_n)\nrightarrow f(x)$; a contradiction

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