# Compactness of open map

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?

• 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 $f$". – 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.