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.

  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.

  • 2
    $\begingroup$ I'm sure this came up in many other threads before, but it seemed easier to write an answer than to find a duplicate. $\endgroup$ – t.b. Aug 11 '12 at 11:21
  • 1
    $\begingroup$ Note also that completeness doesn't enter the argument. $\endgroup$ – t.b. Aug 11 '12 at 11:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.