Definitions. Let $U$, $F$ be Banach spaces, and $f:U \rightarrow F$ a homeomorphism.

  1. $f$ is an open map, if it maps open sets into open sets.

  2. $f$ is compact on $U$, if it maps weakly convergent sequences in $U$ to strongly convergent sequences in $F$

Question. Is it true that if $f$ is open, then $f$ is not compact? And why?

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    $\begingroup$ Many authors take "$f(A)$ is pre-compact whenever $A$ is bounded" as the definition of a compact $f.$ And some say " completely continuous $f$" for "compact $f$". $\endgroup$ – DanielWainfleet Nov 5 '19 at 23:45

It is not true. Consider, e.g., $U = \mathbb R$ and $F = \{0\}$. Then the zero map $f$ is open (because $F$ is discrete) and it is compact (any sequence converges because the topology on $F$ is trivial). More generally, if $F$ is finite-dimensional and $f$ is surjective, then $f$ is compact and open.

Your statement becomes true if you assume that $F$ is infinite-dimensional. The reasoning behind this is as follows. Suppose $f: U \rightarrow F$ is compact. Let $B \subset U$ be the unit ball. Then any sequence in $f(B)$ has a sequence of pre-images that has a weakly convergent subsequence (because $B$ is bounded). This gets mapped back to a convergent subsequence, so $f(B)$ is pre-compact. Thus, a compact map maps bounded sets into precompact sets. Since no compact set in $F$ contains an open set (this is because in infinite-dimensional vector spaces, no ball is compact), $F$ cannot possibly be compact if it is open.

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    $\begingroup$ For a proof that any bounded sequence has a convergent subsequence see this question. $\endgroup$ – Levi Nov 5 '19 at 21:44
  • $\begingroup$ Yes, I understood $\endgroup$ – Alex Nov 5 '19 at 21:54

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