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Prove that the range $R(T)$ of a bounded linear operator $T:X\to Y$ need not be closed in $Y$.

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    $\begingroup$ Hi, welcome to Math.SE! Please add your thoughts on how to approach the problem and we will be glad to give some hints. $\endgroup$ – gt6989b Mar 17 '14 at 20:09
  • $\begingroup$ Take a compact operator from an infinite dimensional space with infinite dimensional range. $\endgroup$ – David Mitra Mar 17 '14 at 22:37
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Consider the identity map $$\mbox{id} :\ell^1 \longrightarrow \ell^2 .$$

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  • $\begingroup$ but $l^1$ is complete , so must be closed in $l^2$ as a subspace , then how is the range space not closed ? $\endgroup$ – user228169 Feb 19 '16 at 10:56

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