Compact open operator between Banach spaces

Question

Let $X,Y$ be Banach space, $Y$ infinite dimensional. Show that no $T \in \mathcal{K}(X,Y)$ is open. By definition $T$ is open if and only if $\exists r >0$ such that $B_Y(0,r) \subset T(B_X(0,1))$ and I also know that the closed unit ball in $Y$ is not compact, since $Y$ is infinite dimensional.

Answer 1

You have all of the right pieces, you just need to assemble them.

Since the closed unit ball is not compact in $Y$, neither is $\overline{B}_Y(0,r)$ (since $y \mapsto ry$ is a homeomorphism for $r > 0$). Suppose $T$ is open and compact. Then $$B_Y(0,r) \subseteq T(B_X(0,1)) \subseteq \overline{T(B_X(0,1))}$$ where the last set is compact, since $T$ is a compact operator. This means that $\overline{B}_Y(0,r)$ is a closed subset of a compact set and hence is compact which is a contradiction.