Subnormality in \(\omega_1^2\) (Q1612258)

Let \(\omega_1\) denote the space of all countable ordinal numbers with the order topology. This paper proves that every subspace of the square \((\omega_1)^2\) is both subshrinking and collectionwise subnormal. Here are some definitions: A space \(X\) is subshrinking if for every open cover \({\mathcal U}\) of \(X\), there exists a cover \(\{F(U):U\in{\mathcal U}\}\) of \(X\) by \(F_\sigma\)-sets such that \(F(U)\subset U\) for each \(U\in{\mathcal U}\). A space \(X\) is collectionwise subnormal if for every discrete collection \({\mathcal F}\) of closed sets in \(X\), there exist countably many collections \({\mathcal G}_n=\{G_n(F):F\in{\mathcal F}\}\), \(n\in \mathbb{N}\), of open sets in \(X\) such that \(F\subset G_n(F)\) for each \(F\in{\mathcal F}\) and each \(n\in \mathbb{N}\), and such that for each \(x\in X\), there exists \(n\in \mathbb{N}\) with \(\text{ord}(x,{\mathcal G}_n)\leq 1\). A space \(X\) is subnormal if every pair of disjoint closed sets can be separated by disjoint \(G_\delta\)-sets. All subshrinking spaces are countably subparacompact and both countably subparacompact spaces and collectionwise subnormal spaces are subnormal. Thus, as a corollary, every subspace of \((\omega_1)^2\) is subnormal. It is, however, unknown whether every subspace of \((\omega_1)^n\) is subnormal for each \(n\in \mathbb{N}\).