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Let $T:X\to Y$ be a linear operator between two normed vector spaces. My question is:

If $T$ is a closed map (sends closed sets to closed), then is the graph of $T$ a closed set of $X \times Y$?

It seems to be the case according to this question, even if the converse is not true. I haven't found a counter-example.

Thank you for your help!

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    $\begingroup$ It is true for every continuous map between separated space, the converse of this assertion is the closed graph theorem. $\endgroup$ – Tsemo Aristide May 29 '16 at 13:14
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    $\begingroup$ @TsemoAristide : thank you. 1. Ah yes, for instance here. But if $T$ is not assumed to be continuous, but only linear? $\tag*{}$ 2. For the converse, don't we need $X$ and $Y$ to be Banach spaces? For me, the closed graph theorem tells us that the graph of $T$ is closed iff $T$ is continuous. In particular, I would get that $T$ is closed? $\endgroup$ – Alphonse May 29 '16 at 13:22

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