# Compact open operator between Banach spaces

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.

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.