# Does every linear injective open map between Banach spaces map closed sets to closed sets?

Let $$X$$ and $$Y$$ be Banach spaces and $$T \in L(X,Y)$$, i.e. $$T$$ is a linear continuous map from $$X$$ to $$Y$$. Further let $$T$$ be injective and open. Does $$T$$ map closed sets to closed sets?

I know this doesn't hold if we drop the injectivity requirement ($$X := \mathbb{R}^2$$, $$Y := \mathbb{R}$$, $$T(x,y) = x$$, closed set = graph of $$\frac{1}{x}$$) but am struggling to find a counterexample or proof.

• If a bounded linear map is injective and open, then it must also be surjective and hence a homeomorphism Jun 2, 2019 at 19:33

If $$T:X\to Y$$ is an open linear map, then it is surjective, because by assumption the set $$A=\{Tx: \|x\|<1\}$$ is open in $$Y$$, and since it contains zero, it must also contain an open ball around zero, say $$B(0,\epsilon)\subset A$$ for some $$\epsilon>0$$. Now given any nonzero $$y\in Y$$, we have $$(\epsilon/2) y/\|y\|\in A$$, so there exists some $$x\in X$$ such that $$Tx=(\epsilon/2)y/\|y\|$$, so $$T$$ is surjective.
Up to this point, we only needed $$T$$ to be linear and open. If you also add the assumptions that $$T$$ is continuous and injective, then, as was commented above, $$T$$ is a topological homeomorphism between $$X$$ and $$Y$$, and in particular, a closed map.