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As Aaron pointed out, "von Neumann's" example is really a non example. To salvage the problem, restate it as: construct a sequence in $\ell_2$ which has $0$ in its weak closure, but no subsequence converges weakly to $0$. First note that such a sequence must be unbounded (by Eberlein-Smulian). Secondly, observe that it is enough to have for each $\epsilon > 0$ a (necessarily bounded) subsequence that converges weakly to a point whose norm is at most $\epsilon$ (and, of course, no subsequence that converges weakly to $0$). With these "hints", it is easy to construct an example: Let $x_{nm}(k)$ be $1/n$ if $k=1$, $n$ if $n=m>1$, k=m>1$, and$0$otherwise. With the "obvious" definition,$0$is in the$2$-weak sequential closure of$x_{nm}$but not in the$1$-weak sequential closure. From this beginning it is natural to define for each countable ordinal$\alpha$the$\alpha$-weak sequential closure and to state an obvious problem. Another (not very difficult once you understand the example above) problem is to build a sequence in$\ell_2$whose norms tend to infinity and yet$0$is in the weak closure of the sequence. Another striking example of the phenomena sought by the OP is the following. Take a dense sequence in the unit sphere of$\ell_1$. Then$0$is in the weak closure of the sequence but no subsequence converges weakly to$0$because$\ell_1$has the Shur property. 2 Corrected typo. As Aaron pointed out, "von Neumann's" example is really a non example. To salvage the problem, restate it as: construct a sequence in$\ell_2$which has$0$in its weak closure, but no subsequence converges weakly to$0$. First note that such a sequence must be unbounded (by Eberlein-Smulian). Secondly, observe that it is enough to have for each$\epsilon > 0$a (necessarily bounded) subsequence that converges weakly to a point whose norm is at most$\epsilon$(and, of course, no subsequence that converges weakly to$0$). With these "hints", it is easy to construct an example: Let$x_{nm}(k)$be$1/n$if$k=1$,$n$if$n=m>1$, and$0$otherwise. With the "obvious" definition,$0$is in the$2$-weak sequential closure of$x_{nm}$but not in the$1$-weak sequential closure. From this beginning it is natural to define for each countable ordinal$\alpha$the$\alpha$-weak sequential closure and to state an obvious problem. Another (not very difficult once you understand the example above) problem is to build a sequence in$\ell_2$whose norms tend to infinity and yet$0$is in the weak closure of the sequence. Another striking example of the phenomena sought by the OP is the following. Take a dense sequence in the unit ball sphere of$\ell_1$. Then$0$is in the weak closure of the sequence but no subsequence converges weakly to$0$because$\ell_1$has the Shur property. 1 As Aaron pointed out, "von Neumann's" example is really a non example. To salvage the problem, restate it as: construct a sequence in$\ell_2$which has$0$in its weak closure, but no subsequence converges weakly to$0$. First note that such a sequence must be unbounded (by Eberlein-Smulian). Secondly, observe that it is enough to have for each$\epsilon > 0$a (necessarily bounded) subsequence that converges weakly to a point whose norm is at most$\epsilon$(and, of course, no subsequence that converges weakly to$0$). With these "hints", it is easy to construct an example: Let$x_{nm}(k)$be$1/n$if$k=1$,$n$if$n=m>1$, and$0$otherwise. With the "obvious" definition,$0$is in the$2$-weak sequential closure of$x_{nm}$but not in the$1$-weak sequential closure. From this beginning it is natural to define for each countable ordinal$\alpha$the$\alpha$-weak sequential closure and to state an obvious problem. Another (not very difficult once you understand the example above) problem is to build a sequence in$\ell_2$whose norms tend to infinity and yet$0$is in the weak closure of the sequence. Another striking example of the phenomena sought by the OP is the following. Take a dense sequence in the unit ball of$\ell_1$. Then$0$is in the weak closure of the sequence but no subsequence converges weakly to$0$because$\ell_1\$ has the Shur property.