If $E$ is a Banach space, $A$ is a subset of $E$ and is compact with the weak topology $\sigma(E,E')$, that is the most coarse topology which make every $f\in E'$ continuous, is it true that $A$ is weakly sequentially compact? As we all know, in $C_1$ spaces, compact concludes sequentially compact. So, we should show that $E$ is a $C_1$ spaces with the topology $\sigma(E,E')$. Some known conclusions:$\forall x_0\in E$ , the basis of neighborhoods of $x_0$ constitutes of $V$ with the below form $$V=\lbrace x\in E;|(f_i,x-x_0)|\lt\epsilon, i\in I\rbrace$$
where I is a finite set. Can we prove $E$ is a $C_1$ spaces? Now, for every such $V$, we can find a $\delta>0$, such that $B(x_0,\delta)\subset V$.

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    $\begingroup$ What is weak self sequence compact? Do you mean weakly sequentially compact? $\endgroup$ Jul 13, 2012 at 7:03
  • $\begingroup$ I think so, too. $\endgroup$ Jul 13, 2012 at 9:51
  • $\begingroup$ oh,yes, it is weakly sequentially compact $\endgroup$ Jul 13, 2012 at 10:10

1 Answer 1


Yes, it's the Eberlein–Šmulian theorem

  • $\begingroup$ I don't think so. When $E=E'$, one can apply Ebelein-Smulian theorem. $\endgroup$ Jul 31, 2012 at 14:54

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