Problem 1. Let be a topological space; let be a subset of . Suppose that for each , there is an open set containing such that . Show that is open in .
Solution. For each , choose an open subset such that . I claim that . Indeed, we have for each by definition, so . Furthermore, for each , one has , which gives the opposite inclusion. As each is open and arbitrary unions of open sets are open, the union is open in and hence is open in .