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Consider the metric space $(X, d)$ given by $$X = \{\text{all continuous functions}\,f:[0,1]\to\Bbb R\}$$ with $$d(f,g)=\sup_{t\in[0,1]}|f(t)-g(t)|.$$ Find with proof a set $A \subseteq X$ with $\operatorname{diam} A < \infty$ which is not compact in $(X, d)$.

I can gather that $A$ must be infinite, because any finite set is compact. I can't however reconcile this with $\operatorname{diam} (A) < \infty$ (where diameter is defined as the supremum of all distances). I can't get the faintest picture in my head of what this would even look like, let alone a formal example.

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Set $f_n(x):=\frac{1}{n}$ for every $x\in[0,1]$ and every $n\geq 1$. Then the set $A:=\{f_n\;;\;n\geq 1\}$ has $\mbox{diam}\;A\leq 1$. Since $f_n$ converges uniformly to $0$, $0$ is in the closure of $A$, but not in $A$. So $A$ is not closed. A fortiori not compact.

Note: this is probably the simplest example when $A$ is not required to be closed. But Davide Giraudo's example is more interesting. Indeed, from his example, you see that the closed unit ball of your space is not compact, as his sequence does not admit a converging subsequence. And of course, the closed unit ball is closed and bounded.

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We can take a countable set: consider a sequence $(f_n,n\in\Bbb N)$ such that the support of $f_n$ is contained in $(4^{-n},4\cdot 4^{-n})$ whose graph is the polygonal interpolation of $(4^{-n},0),(2\cdot 4^{-n},1),(4\cdot 4^{-n},0)$. This means that $f_n(x)=0$ if $0\leqslant x\leqslant 4^{-n}$ or $x\geqslant 4\cdot 4^{-n}$, and $f_n$ is piecewise linear.

This sequence converges pointwise to $0$, but if $n\neq m$, then $d(f_n,f_m)=1$.

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  • $\begingroup$ First of all, thanks for the help. Wikipedia was helpful with the definition of 'support' but doesn't turn up anything at all for 'polygonal interpolation.' I'm starting to feel the professor asked a question about which he did not teach. Could you elaborate? $\endgroup$ Apr 28, 2013 at 13:48
  • $\begingroup$ I've edited. Hoping it's clearer. $\endgroup$ Apr 28, 2013 at 13:59

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