let's consider $\mathscr{U}$ a free ultrafilter on the natural numbers and consider its corresponding ultrapower \begin{align*} \widetilde{X} = (\ell ^{\infty}(X _{i})/\operatorname{ker}\mathcal{N},\Vert \cdot \Vert), \end{align*} where $\widetilde{x}$ is the equivalence class formed by the sequence $(x _{n})$ and the norm $\Vert \cdot \Vert$ is defined as \begin{align*} \Vert \widetilde{w} \Vert = \lim _{n, \mathscr{U}} \Vert w _{n} \Vert. \end{align*} We define \begin{align*} D(\widetilde{w}) = \lim _{m, \mathscr{U}} \left( \lim _{n, \mathscr{U}} \Vert w _{m}  w _{n} \Vert \right). \end{align*} Let $T: C \to C $ a nonexpansive function and $C$ a set of $X$. For the minimal set of \begin{align*} \mathscr{A} = \{ K \subseteq C: \ K \ \mbox{is not empty, weak compact, convex and $T$invariant}\} \end{align*} called $C _{0}$. We know that $C _{0}$ is closed, convex and $T$invariant. We define the set \begin{align*} \widetilde{C _{0}} = \{ \widetilde{w}: w _{n} \in C _{0}, \ \mbox{for all} \ n \in \mathbb{N} \} \in \widetilde{X} \mbox{.} \end{align*} Let \begin{align*} \widetilde{W} = \left\{ \widetilde{w} \in \widetilde{C _{0}}: \ \Vert \widetilde{w}  \widetilde{x} \Vert \leqslant \frac{1}{2} \ \mbox{and} \ D(\widetilde{w}) \leqslant \frac{1}{2} \right\} \mbox{.} \end{align*} Why $\widetilde{W}$ is closed?.
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$\begingroup$ What is $\tilde x$ in the definition of $\widetilde W$? $\endgroup$ – Matthew Daws Jan 11 at 21:59

$\begingroup$ is the equivalence class formed by the sequence $(x _{n})$, where each $x _{n} \in C _{0}$ $\endgroup$ – Amanda Gael Jan 12 at 4:02

$\begingroup$ I know that. What I mean is: $\tilde x$ is not a "bound" variable in the definition of $\widetilde W$. We know what $\tilde w$ is: it varies, and we select those which satisfy the condition. But $\tilde x$ appears nowhere else, so I don't know, at present, what $\widetilde W$ means. $\endgroup$ – Matthew Daws Jan 12 at 10:04