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I am trying to prove that zero is in the weak closure of set $\{ x\in l^1 \,:\, \lVert x\rVert =1\}$ . And I need example too. Is there any countable subset of this set which has $0$ in weak closure?

Nina
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  • The key observation (which follows from the definition of weak neighbourhood) is that all weakly-open subsets in and infinitely dimensional normed space contain a straight line. –  May 09 '16 at 12:59
  • For the second part: let $A={x\in l^1,:, \lVert x\rVert=1}$. There is a countable subset $S\subseteq A$ whose strong closure is $A$. The weak closure of $S$ in $l^1$ contains the strong closure of $S$ in $l^1$... –  May 09 '16 at 13:08
  • Sassatelli has given some pretty strong hints regarding the right answers. It might be a good thing to realize that the following true fact does not imply that there's no countable set with $0$ in the weak closure: There is no sequence in $A$ that tends to $0$ weakly. (In fact, curiously, weak convergence of a sequence in $l^1$ implies convergence in norm!) – David C. Ullrich May 09 '16 at 13:32
  • http://math.stackexchange.com/questions/153889/prove-the-weak-closure-of-the-unit-sphere-is-the-unit-ball – Tomasz Kania May 09 '16 at 15:10

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