2
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ 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
3
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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

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.