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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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up vote 3 down vote accepted

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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Is this all wrt $l_\infty$ norm? – Simon Sehayek Oct 15 '12 at 5:04
Yes, it is all wrt $\ell_\infty$ norm. – Robert Israel Oct 15 '12 at 5:04
So have I already shown the first part? $l_1 \subseteq A$ follows from the condition on absolutely converging series? – Simon Sehayek Oct 15 '12 at 5:07
I don't know what you mean by "the condition". It's the fact that the terms of a convergent series must go to $0$. – Robert Israel Oct 15 '12 at 5:39
So I need to fix a sequence in $A$ (this sequence is bounded) ${x_n}$ and show that I can construct a sequence ${y_n}$ in $l_1$ s.t. $\max|x_n-y_n|<\epsilon$, and if each component of the sequence is to converge I have $|x_n^j-y_n^j|<\epsilon$ – Simon Sehayek Oct 15 '12 at 6:24

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