# Properties of a set in $\ell_p$ spaces

$\ell_p = \{ x = (x_n) : x_n \in \mathbb C \ \ \forall \ \ n \geq 1 \ \ \text{and} \ \ (\sum_{n=1}^{\infty} |x_n|^p)^{\frac{1}{p}} < \infty \}$ and let $e_n \in \ell_p$ be the sequence whose $n$-th element is 1 and all other elements are zero. Then the set $S=\{ e_n : n \in \mathbb N \}$. Then which of the following are true ?

• $S$ is closed

• $S$ is bounded

• $S$ is compact

• $S$ contains a convergent subsequence

we know that if any $e_i , e_j$, then $||e_i - ej|| = 2^{\frac{1}{p}}$, if $i \neq j \in \ell_p$ , so $S$ contains no convergent subsequence and $S$ is a bounded sequence.

we know that if any closed and bounded subset of a normed linear space is compact then this space is finite dimmensional, but $\ell_p$ is infinite dimensional, So $S$ is not closed and compact. we know that every normed linear space is a metric space which is Hausdorff and every compact space in a Hausdorff is closed. If $S$ is compact then $S$ is closed, we got a contradiction. So $S$ cannot be compact.

I am unable how to check whether $S$ is closed or not. Any help would be appreciated. Thank you.

• Your proof that $S$ is not compact is completely wrong. With the same reasoning, you prove that $\ell^p$ contains no compact subsets, which is obviously false. – Giuseppe Negro Jun 1 '16 at 14:06
• I does not mean that $\ell_p$ has no compact subspace. I have proved that $S$ is cot compact. – Struggler Jun 1 '16 at 17:19
• @ Henno : Thanks for editing – Struggler Jun 1 '16 at 17:21

Suppose $x\in\overline{S}$ (the closure of $S$). Then there exists $\{x_n\}\subset S$ such that $x_n\to x$. Because of your remark: $\|e_i - ej\| = 2^{\frac{1}{p}}$, the sequence can only converge if it is definitively constant, which means $x\in S$. Therefore $\overline{S}=S$, that is $S$ is closed.