Let \(X\) be a topological space; let \(A\) be a subset of \(X\). Suppose that for each \(x \in A\) there is an open set \(U\) containing \(x\) such that \(U \subset A\). Show that \(A\) is open in \(X\).
Solution
For each \(x\in A\), choose the open set \(U_x\) supplied by the hypothesis. Then \[ A=\bigcup_{x\in A}U_x. \] Indeed, every \(x\in A\) belongs to \(U_x\), while every \(U_x\) is contained in \(A\). Since arbitrary unions of open sets are open, \(A\) is open.