Consider $X=[0, 1]^{\mathcal{P}(\mathbb{N^*})}$ (where $\mathbb{N^*}=\{1,2,3,\ldots$} and $\mathcal{P}(\mathbb{N^*})$ its power set) with the standard product topology. It is known that $X$ is Hausdorff and compact but not sequentially compact.

Let's define a sequence $(x_n)_{n \in \mathbb{N^*}}$ of points in $X$ by specifying the values of all the functions $x_n : \mathcal{P}(\mathbb{N^*}) \to [0, 1]$ on each $S \subseteq \mathbb{N^*}$ :

if $S=\{n_{1}, n_{2}, n_{3},\cdots\}$ with $n_1<n_2<n_3<\ldots$, then $x_{n_{1}}(S)=x_{n_{3}}(S)=x_{n_{5}}(S)=\ldots=0$ , $x_{n_{2}}(S)=x_{n_{4}}(S)=x_{n_{6}}(S)=\ldots=1$ , and $x_{k}(S)=\frac{1}{2}$ for any $k \in \mathbb{N^*} \setminus S$.

In other words $x_m(S) = \left\{\begin{matrix} 0 & \exists k \in \mathbb{N^*} : & m=n_{2k+1} \\ 1 & \exists k \in \mathbb{N^*} : & m=n_{2k} \\ \frac12 & \not\exists k \in \mathbb{N^*} : & m=n_{k} \end{matrix} \right.$

This sequence $(x_n)_{n \in \mathbb{N^*}}$ does not have any convergent subsequence (known result) but does have a cluster point, since $X$ is compact.

My question is whether it is possible to find this (these) point(s) $x \in X$, cluster for the sequence $(x_n)_{n \in \mathbb{N^*}}$, by specifying the function $x : \mathcal{P}(\mathbb{N^*}) \to [0, 1]$. I would really appreciate your help. Thanks.



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