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**Opalg** The trouble there is that $\displaystyle (1,\ldots,1,0,\cdots)$ does not converge to $\displaystyle (1,\ldots,1,\ldots)$ in the space $\displaystyle l^\infty$ (because the norm there is the sup norm). To avoid this difficulty, you could modify the construction slightly by taking $\displaystyle y = \bigl(1,\tfrac1{\sqrt2},\tfrac1{\sqrt3},\ldots \bigr).$ Then $\displaystyle (1,\cdots,\tfrac1{\sqrt m},0,\cdots,0,\cdots)$ is in $\displaystyle \text{im}(T)$ since it is equal to $\displaystyle T(1,\ldots,\sqrt m,0,\ldots)$' and it does converge to $\displaystyle (1,\ldots,\tfrac1{\sqrt m},\ldots)$ in the $\displaystyle l^\infty$ norm.