# Why is $\widetilde{W}$ closed?

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 non-expansive 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?.

• What is $\tilde x$ in the definition of $\widetilde W$? – Matthew Daws Jan 11 at 21:59
• is the equivalence class formed by the sequence $(x _{n})$, where each $x _{n} \in C _{0}$ – Amanda Gael Jan 12 at 4:02
• 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. – Matthew Daws Jan 12 at 10:04