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Let $X\subset l^\infty$ and $Y\subset X$ be a set of sequences with only finitely many nonzero terms. Show that $Y$ is not complete.

I was thinking about the following sequence of sequences: $[(1,0,\dots)$, $(1,1,0,\dots)$, $(1,1,1,0,\dots)$, $\dots$], which should converge to the infinite sequence containing only ones. Now the problem is that this sequence of sequences is not Cauchy, since for every $n>m$ we have that $\|(x_l)_n^l-(x_l)_m^l\|=\sup_\limits{l\in \mathbb{N}}|x_{n_l}-x_{m_l}|=1$, which does not hold for $\varepsilon < 1$.

How is this possible?

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The sequence you give is not convergent in the sense of $\ell^\infty$! $$||x_n-(1)_n||=1$$ for all n. Thus $(1)_n$ is not the limit of your sequence. Since its not a even a Cauchy sequence, it can't converge at all. For the incompleteness of $Y$, take a look at $x_n=(1,1/2,1/3,...,1/n,0,0,...)$.

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