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Suppose we have a set $A$ which is the set of all sequences that satisfy $|x_n|\xrightarrow{} 0$. If we consider $l_1$ to be a subset of $l_\infty$. Show that the closure of $l_1$ in $l_\infty$ equals $A$.

I started by showing $l_1$ is closed since it is complete (even stronger condition is that it is compact in $l_\infty$ since it is closed in $l_\infty$). Then I can see why $l_1 \subseteq$ $l_\infty$ since this is a condition on converging (absolutely) series. This doesn't seem to make sense in the other direction... I think I'm seeing this wrong.

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No, $\ell_1$ is not closed in $\ell_\infty$. It is complete in the $\ell_1$ norm, but not in the $\ell_\infty$ norm. It is also not compact: no unbounded set can be compact.

To show that the closure of $\ell_1$ is $A$, you might proceed as follows.

a) show that $\ell_1 \subseteq A$.

b) show that any member of $A$ is the limit of a sequence of members of $\ell_1$.

c) show that $A$ is closed in $\ell_\infty$.

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    No, $A$ is the set of all sequences that converge to $0$. Hint: take $y_n = 0$ for sufficiently large $n$.2012-10-15