A space \((X,\mathcal{T})\) is called extremally disconnected if the closure of every open set is open; that is, whenever \(U \in \mathcal{T}\) we have \(\operatorname{cl}(U) \in \mathcal{T}\). Show that in an extremally disconnected space, if \(U,V \in \mathcal{T}\) and \(U \cap V = \varnothing\), then \[\operatorname{cl}(U) \cap \operatorname{cl}(V) = \varnothing.\]
(Hint: first show that \(U \cap \operatorname{cl}(V) = \varnothing\).)
Solution
Let \(U,V\in\mathcal T\) with \(U\cap V=\varnothing\).
1. \(U\cap\operatorname{cl}(V)=\varnothing\).
Each \(x\in U\) has the neighborhood \(U\), which misses \(V\), so \(x\notin\operatorname{cl}(V)\).
2. Q.E.D.
1.Since \(\operatorname{cl}(U)\) is the smallest closed set containing \(U\), \(\operatorname{cl}(U)\subseteq X\setminus\operatorname{cl}(V)\), that is, \(\operatorname{cl}(U)\cap\operatorname{cl}(V)=\varnothing\).