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Let $X$ be an infinite dimensional Banach space. I want to show that any compact subset $\varnothing\neq A\subset X$ is nowhere dense.

I've been able to prove the statement for $X=(C[0,1],\|\cdot\|_\infty)$ by using an Arzelà-Ascoli argument. But this is not easy to generalize to an arbitrary infinite dimensional Banach space.

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  1. In an infinite-dimensional normed space the (closed) unit ball is always non-compact: using Riesz's Lemma you can construct a sequence of unit vectors $(e_n)_{n=1}^\infty$ such that $\lVert e_n - e_m \rVert \geq 1/2$ for all $m \neq n$. This sequence can't have a convergent subsequence. It follows that no closed ball is compact. More details on the linked Wikipedia page.

  2. Assuming the set $A \neq \emptyset$ is closed and not nowhere dense, it must contain an open ball $B_r(x_0) \subset A$ by definition of “nowhere dense”. Thus, if $X$ is infinite-dimensional, $A$ cannot be compact: otherwise the closed ball $\bar{B}_{r/2}(x_0)$ would have to be compact as a closed subset of $A$, contradicting 1.

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I'm sure this came up in many other threads before, but it seemed easier to write an answer than to find a duplicate. –  t.b. Aug 11 '12 at 11:21
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Note also that completeness doesn't enter the argument. –  t.b. Aug 11 '12 at 11:39

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