Weak closure points not attainable as limits of sequences

Where in this post we discover an uncanny property of the weak topology: the points of the weak closure of a set cannot always be attained as limits of elements of the set. Naturally, the w-closure of a set is weakly closed. The so-called weak sequential closure of a set, on the other hand, is the set of cluster points of sequences made with elements from that set. The new set is not, however, weakly sequentially closed, which means that there may arise new cluster points; we may, in fact, have to take the weak sequential closure trans-finitely many time to obtain a set which is weakly sequentially closed – still, this may not be weakly closed.

Let $X$ be a Banach space. The weak topology $\tau_w$ in $X$ is the coarsest topology in $X$ which makes the elements of $X^*$ continuous. In infinitely dimensional spaces, the weak topology is not metrisable, while in finite dimensions the properties of $X$ coincide with the ones of $X^*$ as the strong and the weak topology coincide.

A basis of open neighbourhoods of $x_0$ of the weak topology comprises of the sets $B_{\Gamma, \epsilon}(x_0)$ for some $\epsilon >0$ and $\Gamma=\{x_i^*\}_{i\in \mathcal{I}}$ being a finite collection of elements of $X^*$

$\begin{aligned} B_{\Gamma, \epsilon}(x_0) = \{x: |x_i^*(x) - x_{i}^{*}(x_0)| < \epsilon, \text{ for all } i\in \mathcal{I}\}. \end{aligned}$

For example, in Hilbert spaces, these sets can be written as

$\begin{aligned} B_{\Gamma, \epsilon}(x_0) = \{x: | \langle x_i^*, x \rangle - \langle x_{i}^{*}, x_0\rangle| < \epsilon, \text{ for all } i\in \mathcal{I}\}. \end{aligned}$

We may use these sets to tell whether a sequence $(x_n)_n$ converges weakly to a point $x\in X$ ; we need to check whether for every finite collection $(x_i^*)_i \subset X^*$ we have that for every $\epsilon>0$ there is a $n_0 = n_0(\epsilon) \in \mathbb{N}$ so that for all $n \geq n_0$ ,
$|x_i^*(x_n) - x_{i}^{*}(x)| < \epsilon$ .

Alternatively, we may say that $(x_n)_n$ converges weakly to $x$ if

$\begin{aligned} x^*(x_n) \to x^*(x), \text{ for all } x^* \in X^*. \end{aligned}$

Instead of testing weak convergence against arbitrary finite collections $\Gamma\subset X^*$ we may instead use a total set $(e_k)_{k\in \mathcal{K}}$ in $X^*$ . Then $(x_n)_n$ converges weakly to a point $x$ if and only if $(x_n)_n$ is $\|\cdot\|$ -bounded and for every $k \in \mathcal{K}$ , $e_k(x_n) \to e_k(x)$ .

A prime example of a sequence which is weakly convergent, but not $\|\cdot\|$ -convergent is the sequence $(e_n)_n$ of $\ell_2(\mathbb{N})$ . Since $\|e_n\|_2 = 1$ , this is a sequence of elements on the unitary sphere of $\ell_2$ and so it does not converge strongly. However, for every $k\in \mathbb{N}$ , the sequence $|\langle e_n, e_k\rangle| = \delta_{nk}$ converges to 0, so $e_n \to 0$ weakly. Notice that the sequence is of course bounded.

The closure of a set $A \subseteq X$ in the strong (norm) topology of $X$ is the smallest closed set which contains $A$ . This is exactly the same as the set of all (strong) limit points of sequences made with elements from $A$ . This is not true in the weak topology the main reason for which being that it is not metrisable. The set of weak sequential cluster points of sequences in $A$ is the so-called weak sequential closure of $A$ which we will denote by $\mathrm{w{-}scl}(A)$ .

$\begin{aligned} \mathrm{w{-}scl}(A) = \{x: \exists \{a_n\}_{n\in\mathbb{N}}\subseteq A, a_n \overset{w}{\to} x\} \end{aligned}$

This is smaller that the weak closure of $A$ :

$\begin{aligned} \mathrm{w{-}scl}(A) \subseteq \mathrm{w{-}cl}(A). \end{aligned}$

As an example take the set $A = \{e_n\sqrt{n+1}\}_n$ in $X = \ell_2(\mathbb{N})$ . We may show that $A$ is closed and weakly sequentially closed, i.e., $\mathrm{ws{-}cl}(A) = \mathrm{cl}(A)$ , but not weakly closed.

Indeed, the origin is not a weak cluster point of $A$ : let $B_{\mathcal{I},\epsilon}$ be a weak basic neighbourhood of the origin. We may, therefore, find elements of $A$ in $B_{\mathcal{I},\epsilon}$ for every $\epsilon$ and $\mathcal{I}$ , so $0$ is in the weak closure of $A$ , but it cannot be attained as a limit of a sequence of elements of $A$ .

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Weak closure points not attainable as limits of sequences

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