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I know that it would be true if $X$ and $Y$ both would be Banach spaces AND assuming that $T$ maps bounded and closed subsets of $X$ onto closed subsets of $Y$.

But what happens in the case that $X$ is just a normed vector space, $Y$ is Banach and one does NOT have the assumption about $T$?

My intuition says that in this case the range is not closed. But is there any example where I can see it clearly?

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    $\begingroup$ Let $X$ be an incomplete normed space, $Y$ its completion and $T$ the inclusion. $\endgroup$ – Daniel Fischer Oct 24 '16 at 12:05
  • $\begingroup$ Is the range in this case not closed because otherwise $X$ = $Y$ which is not true? $\endgroup$ – TigerLa Oct 24 '16 at 12:14
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    $\begingroup$ The range of a bounded linear transformation from a Banach space to itself need not be closed. $\endgroup$ – GEdgar Oct 24 '16 at 12:27
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    $\begingroup$ Related: Examples of bounded linear operators with range not closed $\endgroup$ – Martin Sleziak Oct 24 '16 at 12:58
  • $\begingroup$ Thanks @MartinSleziak that helps me to understand it better! $\endgroup$ – TigerLa Oct 24 '16 at 19:40

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