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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.

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

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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.

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