# Closure and Sequence.

I dug this example out of Chen's free online source notes.

I am very puzzled by the last comment

but the limit $\mathbf{a}$ does not belong to $\ell_0$.Hence the linear subspace $\ell_0$ is not closed in $\ell^\infty$

I don't understand the idea. To show a subset $\ell_0$ isn't closed in $\ell^\infty$, should you find a sequence in $\ell_0$ that doesn't converge in $\ell^\infty$? Not the other way around? Or generally speaking isn't this true for any metric space?

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The following result is used here:

Let $(X,d)$ be a metric space, and $F$ be a subset of $X$. If $F$ is closed, and $\{x_n\}$ is a sequence of elements of $F$ which converges to some $x\in X$, then $x\in F$.

(so by closed it means that "we can't escape by limit")

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Then what about my definition? I am guessing I am wrong then? – Hawk Feb 10 '13 at 22:27
Oh wait if the limit point isn't in the set, it can't be in the subset. – Hawk Feb 10 '13 at 22:30
But it also works to find a sequence that also converges outside of $\ell^\infty$ right? Abeit it may not be the easiest thing to do – Hawk Feb 10 '13 at 22:46
How could a sequence converge outside $\ell^\infty$? The limit will necessarily lie in $\ell^\infty$ (use the definition with $\varepsilon=1$). – Davide Giraudo Feb 10 '13 at 22:52
Can't the nth term of the sequence lie just on the boundary of the set? – Hawk Feb 10 '13 at 22:55