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This question has some context, which I explain below.

Let $(X, d)$ be a metric space, and fix some $x_0$ in $X$. Define $$BC(X) = \{ f : X \rightarrow \mathbb{R} : f~\text{is continuous and bounded}. \}.$$ Equip $BC(X)$ with the supremum metric, that is, $$\rho(f, g) = \sup_{x \in X} |f(x) - g(x)|.$$ It's easy to see that $(BC(X), \rho)$ is a metric space. Now consider the map $\Phi : X \rightarrow BC(X)$ which sends $x$ to $\Phi_x$, where $$\Phi_x(y) = d(x, y) - d(x_0, y).$$ The map $\Phi$ is an isometry. I've got no problem up to this point.

Now, suppose $K \subseteq X$ is a compact set and $\{x_n\}_{n \geq 1}$ is a sequence in $X$ (not necessarily in $K$). My question is:

How can I show that there is a subsequence of $\{\Phi_{x_n}\}_{n \geq 1}$ that converges uniformly on $K$?

The first thing that popped in my head was, of course, Arzelà-Ascoli, but the family $\{\Phi_{x_n}\}_{n \geq 1}$ is not pointwise bounded, so I'm stuck. Any help is appreciated. Thanks.

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1 Answer 1

up vote 2 down vote accepted

Try $X = \mathbb R$ with the usual metric, $x_0 = 1$, $K=[0,1]$ and $x_n = n$. Then $\Phi_n(x) = |x - n| - |x-1| = n-1$ on $K$. There is no subsequence converging on $K$.

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you're right, that was a silly question. I got this from a qual exam and never stopped to think about the veracity of the statement. Thanks! –  student Jul 24 '12 at 20:06

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