Continuous selections and prime numbers (Q2052567)

Let \(X\) be a Hausdorff space and \(\mathcal{F}(X)\) the space of all nonempty closed subsets of \(X\) with the Vietoris topology. For a subspace \(\mathfrak{F}\) of \(\mathcal{F}(X)\), a map \(f : \mathfrak{F} \to X\) is called a continuous selection for \(\mathfrak{F}\) if \(f\) is continuous and \(f(A) \in A \) for every \(A \in \mathfrak{F}\). For a positive integer \(n\), let \(\mathcal{F}_n(X) =\{S \subset X: 1 \leq |X|\leq n \}\) and \([X]^n = \{ S \subset X : |X| =n\}\) be equipped with the relative topologies of \(\mathcal{F}(X)\). Let \(\mathrm{Sel}_{\leq n}^c(X)\) and \(\mathrm{Sel}_{n}^c(X)\) denote the set of all continuous selections for \(\mathcal{F}_n(X)\) and those for \([X]^n\), respectively. The following problem was posed by \textit{V. Gutev} and \textit{T. Nogura} [Appl. Gen. Topol. 5, No. 1, 71--78 (2004; Zbl 1064.54014)]: Does there exist a Hausdorff space \(X\) such that \(\mathrm{Sel}_{2}^c(X)\ne \emptyset\), but \(\mathrm{Sel}_{\leq n}^c(X) = \emptyset\) for some \(n>2\)? \textit{M. Hrušák} and \textit{I. Martínez-Ruiz} [Fundam. Math. 203, No. 1, 1--20 (2009; Zbl 1170.54008)] proved that the answer to this problem is negative for every separable space. But the problem for more general spaces is still open even in the case \(n=3\). Concerning this problem, \textit{V. Gutev} [Topology Appl. 157, No. 1, 83-89 (2010; Zbl 1186.54014)] proved the following theorems: If \(X\) is a Hausdorff space such that \(\mathrm{Sel}_{\leq n}^c(X) \ne \emptyset \ne \mathrm{Sel}_{n+1}^c(X)\) for \(n \geq 2\), then \(\mathrm{Sel}_{\leq n+1}^c(X) \ne \emptyset\). If \(X\) is a Hausdorff space such that \(\mathrm{Sel}_{\leq 2n+1}^c(X) \ne \emptyset\) for \(n \geq 1\), then \(\mathrm{Sel}_{2n+2}^c(X) \ne \emptyset\). In this paper, the author generalizes the latter theorem of Gutev by proving the following theorem: If \(X\) is a Hausdorff space such that \(\mathrm{Sel}_{\leq n}^c(X) \ne \emptyset\) for \(n \geq 2\) and \(m\) is an integer such that \(n\leq m\leq 2n\) and \(m\) is not a prime number, then \(\mathrm{Sel}_{m}^c(X) \ne \emptyset\). It is also proved that, for any Hausdorff space \(X\), \(\mathrm{Sel}_{n}^c(X) \ne \emptyset\) for every \(n\geq 2\) if and only if \(\mathrm{Sel}_{p}^c(X) \ne \emptyset\) for every prime number \(p\).