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Let $T$ be closable. I am trying to show $\Gamma(\overline{T}) \subseteq \overline{\Gamma(T)}$. I can already show the reverse inclusion. Any ideas?

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    $\begingroup$ Did you try to use the sequence definition of the closure? I mean, if $x_n\in D(T)$, $x_n\to x$ and $Tx_n\to y$, then $x\in D=(T)$ and $Tx=y$? $\endgroup$ – abatkai Mar 26 '13 at 19:16
  • $\begingroup$ Can someone clarify this notation? $\endgroup$ – Norbert Mar 27 '13 at 15:46

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