2

Let $X$ a Banach space. Let $\mathcal T$ the weak topology on $X$ (the thickest topology s.t. linear form are continuous).

  • (1) $A\subset X$ is weakly compact if for all covering $\mathcal U\subset \mathcal T$ there are $U_1,...,U_n\in \mathcal U$ s.t. $$A\subset \bigcup_{i=1}^n U_i.$$

  • (2) $A\subset X$ is sequentially weakly continuous if for all $(x_n)\subset A$ there is a subsequence that converge weakly, i.e. there is $(x_{n_k})\subset (x_n)$ and $x\in A$ s.t. for all $\varphi\in X^*$ (where $X^*$ denote the topological dual) $$\lim_{k\to \infty }\varphi(x_{n_k})=\varphi(x).$$

Are (1) and (2) equivalent ?

copper.hat
  • 172,524
user352653
  • 832
  • What are your thoughts? – mathworker21 Aug 23 '18 at 23:18
  • @mathworker21: I know that compacity and sequentially compactness are equivalent for metrizable spaces. I also know that weak topology is not metrizable... so I have doubt that is true. – user352653 Aug 23 '18 at 23:21
  • Compact always implies sequentially compact. I'm curious, though, if there is any non-metrizable topological space such that any sequentially compact subset is compact. – mathworker21 Aug 23 '18 at 23:29
  • @mathworker21: In my OP, I'm talking about "weak sequentially compactness" not sequentially compactness... – user352653 Aug 23 '18 at 23:34
  • I'm fixing the underlying topology as the weak topology. Then what we are talking about is compactness and sequential compactness. – mathworker21 Aug 23 '18 at 23:35
  • @mathworker21: I'm really not used to this, please don't confuse me more than what I am. Is (1) $\iff$ (2) or not ? Just yes or no. If no, a counter example would be great, if yes, I'll try to make a proof. – user352653 Aug 23 '18 at 23:42
  • lol I'm not messing with you. Calm down. I don't know the answer. What I was saying is that (1) $\implies$ (2) since, in any topological space, compactness implies sequential compactness, and here we have the weak topology. – mathworker21 Aug 23 '18 at 23:53
  • 3
    Look up the Eberlein Šmulian theorem. TL; DR: Yes they are equivalent. – copper.hat Aug 24 '18 at 01:46

0 Answers0