I have a "proof" for the following problem:

Suppose $X$ is a Banach space, the operator $T \in L(X,X)$ is open, and let $X_0$ be a closed subspace of $X$. Further the restriction $T_0$ of $T$ to $X_0$ is continuous. Is $T_0$ necessarily open?

I found a counterexample here.

What went wrong in my "proof" then? --

$T_0$ is continuous $\implies$ $X_0 \times T(X_0)$ is closed by closed graph theorem $\implies T(X_0)$ is closed because graph is closed so it must be closed in each of its components $\implies T_0$ is open by open mapping (because range is closed)

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    $\begingroup$ Between which spaces does $T_0$ map? $T_0 : X_0 \to X$? $\endgroup$ – gerw Apr 10 '18 at 6:30

$X_0\times T(X_0)=\{(x,Ty)\vert x,y\in X_0\}\neq \{(x,Tx)\vert x\in X_0\}=\mathrm{Graph}(T\vert_{X_0})$. (Otherwise every operator with closed graph would have to have closed range, which is certainly not true.)


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