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.