In this example $H$ is a infinite-dimensional Hilbert space and let $(e_n)_n$ be an orthonormal sequence in $H$, and $S = \{\sqrt{n}e_n : n \in \mathbb{N}\} \subseteq H$. It was said in the example that: $S$ is weakly sequentially closed because every weakly convergent sequence in $S$ has to be bounded, so the sequence has to have at most finitely many distinct terms, and hence it stabilizes to an element of $S$. Now this argument seems to work because it has this term $\sqrt{n}$ that cannot go to infinity ( please correct me if i got it wrong), however if set S were defined by $S = \{e_n : n \in \mathbb{N}\} \subseteq H$ how could I show that he is weakly sequentially closed? If necessary include the null vector $S \cup \{0\}$ is weakly sequentially closed?
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Assume that you have a sequence $x$ of elements of $S \cup {0}$ that converges weakly to some $v$. If $x_n$ is infinitely often some given vector of $S$, or zero, then $v$ is said vector and we are done. Else, then for each $n$, $\langle x_p,,e_n\rangle=0$ for large enough $p$ so that $v$ is orthogonal to $S$. Thus for each $p$, $\langle x_p,, v\rangle=0$, so that $|v|^2=0$ and thus $v \in S\backslash {0}$. – Aphelli Sep 02 '20 at 20:52
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would you also know if this is true in banach spaces in general? – Ilovemath Sep 02 '20 at 23:14
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What exactly is your question in the case of Banach spaces? – Kavi Rama Murthy Sep 02 '20 at 23:20
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I saw that he argued in the end using the inner product, but if $S$ were a set in which all the elements have norm $1$ and the space was just Banach, with no inner product, I could also conclude that it is weakly sequentially closed, if so how can I argue that? – Ilovemath Sep 02 '20 at 23:24
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Probably not. For instance, in $\ell^1$, the sequences with norm $1$ and rational coefficients are countable and strongly dense. You’d need some sort of assumption of well-separedness of your vectors. But even then I’m not sure how to go about it. – Aphelli Sep 03 '20 at 07:08