If $X$ is a topological vector space(specially a Banach space with it's weak topology), Does $X$ have the following property:

A subset $A$ is closed $\Leftrightarrow $ every bounded net in $A$ converent to an $x_0$, implies $x_0\in A$.

In Metric spaces, this is true, because every converging net in a metric space has a bounded subnet(even has an eventually bounded subnet). But in a topological vector space, a net (according to Must a weakly or weak-* convergent net be eventually bounded?) may be doesn't have an eventually bounded subnet(although I don't know about a bounded subnet).

Can anyone help or guide me? I think this is a hard problem.

  • $\begingroup$ To be clear, you are asking about a characterization of closed sets in the weak topology on the dual space $X^*$, correct? $\endgroup$ – Math1000 Nov 28 '19 at 23:18
  • $\begingroup$ Closed sets in $(X,\tau_w)$. $\endgroup$ – Darman Nov 28 '19 at 23:20
  • $\begingroup$ Although in every T.V.S, I don't know the answer $\endgroup$ – Darman Nov 28 '19 at 23:22
  • $\begingroup$ Oh sorry, I misread the definition of weak topology. It indeed is a topology on $X$, not the dual space $X^*$. Please disregard my prior comment. $\endgroup$ – Math1000 Nov 28 '19 at 23:22

In general a topological vector space doesn't have that property.

An example in $\ell^2(\mathbb{N})$ in its weak topology:

For $m \in \mathbb{N}\setminus \{0\}$, let $$A_m = \bigl\{ x \in \ell^2(\mathbb{N}) : \lVert x\rVert_2 = m, \bigl(n \leqslant \tfrac{m(m-1)}{2} \lor n > \tfrac{m(m+1)}{2}\bigr) \implies x_n = 0\bigr\}$$ and define $$A = \bigcup_{m = 1}^{\infty} A_m\,.$$ Then the intersection of $A$ with every closed and bounded set is compact (even in the strong topology), thus all accumulation points of bounded nets in $A$ lie in $A$, in particular if a bounded net in $A$ converges to $x_0$, then $x_0 \in A$.

But $A$ is not weakly closed, we have $0 \in \operatorname{cl}_w(A) \setminus A$. For every weak neighbourhood of $0$ contains one of the form $$V(\varepsilon;\xi_1, \dotsc, \xi_k) = \{ x \in \ell^2(\mathbb{N}) : \lvert\langle x, \xi_j\rangle\rvert < \varepsilon \text{ for } 1 \leqslant j \leqslant k\} $$ where $k \in \mathbb{N}$, $\xi_1,\dotsc,\xi_k \in \ell^2(\mathbb{N})$, and $\varepsilon > 0$. And $V(\varepsilon;\xi_1,\dotsc,\xi_k) \cap A_m \neq \varnothing$ for all $m > k$.

This construction can be imitated in every infinite-dimensional normed space and yields a set that isn't weakly closed but whose intersection with every weakly closed bounded set is weakly closed.

The property holds (as you know) in every metrisable topological vector space.

For convex $A$ we have the equivalence $$A\text{ closed} \iff (A\cap B)\text{ closed for all closed and bounded } B$$ if $X$ carries the weak topology of an originally locally convex and metrisable space. In particular in the weak topology of a Banach space, closed convex sets can be characterised by the convergence of bounded nets.


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