# Find a cluster point of a sequence in a compact but not sequentially compact space.

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 $$n_1, then $$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.