Assume that $T:\ell_1\to\ell_2 $ is bounded,linear and one-to-one. Prove that $T(\ell_1)$ is not closed in $\ell_2$
2026-03-31 08:43:45.1774946625
Example of open operator but not closed
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If $T$ is one-to-one then by closed graph theorem $\ell_1$ may be regarded as a infinite dimensional closed subspaspace of $\ell_2$. Any closed infinite dimensional subspace of $\ell_2$ is $\ell_2$. Thus we conclude that $\ell_1$ is isomorphic to $\ell_2$. This is impossible because $\ell_2$ is reflexive while $\ell_1$ is not.