# Surjectivity of Compact Operators on an Infinite Dimensional Vector Space

A pretty standard result says that if $$T$$ is a compact operator acting on a Banach space, then $$T$$ cannot be surjective. The proofs I've seen use the open mapping theorem and the fact that a topological vector space is locally compact if and only if it is finite dimensional. The open mapping theorem is usually phrased in terms of Banach spaces, although in can be phrased in terms of F-spaces and topological vector spaces. However, in both cases, completeness of the metric/norm seems essential. My question is,

Does there a compact operator on an infinite dimensional incomplete metric/normed vector space that is surjective?

• I guess no: we can take the completion of the space and extend $T$ by continuity.. – Berci Jul 8 at 17:18

The idea of the proof given there is as follows. Suppose that $$T:X \to X$$ is surjective. Then one can use the Baire category theorem to find a ball $$B(x,s)$$ contained in the unit ball of $$X$$ such that $$T(B(x,s)) \subseteq T(nK)$$ for some $$n$$, where $$K = \overline{T(B(0,1))}$$ is compact. Then if $$\tilde{T}: \tilde{X} = X/ \ker T \to X$$ is the operator induced by $$T$$ on the quotient space then one can check that $$\tilde{T}$$ is injective, compact and satisfies $$\tilde{T}(B_{\tilde{X}} (\pi x, s)) \subseteq \tilde{T}(\pi(nK))$$ where $$\pi: X \to \tilde{X}$$ is the usual projection. By injectivity, this in turn implies that $$B_{\tilde{X}} (\pi x, s) \subseteq \pi(nK).$$ But the right hand side is compact and so this would imply that $$\tilde{X}$$ contains a compact closed ball and hence is finite dimensional. This would in turn imply that $$X$$ itself is finite dimensional since $$T$$ onto implies that $$\tilde{T}: \tilde{X} \to X$$ is onto also.