For a set \(A\subset X\), \(\overline A\) is defined to be the intersection of all closed sets containing \(A\). Using this definition, show: \(x\in\overline A\) if and only if every open set \(U\) containing \(x\) intersects \(A\).
Solution
Suppose first that \(x\in\overline A\). If some open set \(U\ni x\) were disjoint from \(A\), then \(X\setminus U\) would be a closed set containing \(A\) but not \(x\). This contradicts the definition of \(\overline A\) as the intersection of all closed sets containing \(A\). Hence every open neighborhood of \(x\) meets \(A\).
Conversely, suppose every open neighborhood of \(x\) meets \(A\). If \(x\notin\overline A\), then by the defining intersection there is a closed set \(F\supseteq A\) with \(x\notin F\). Its complement \(X\setminus F\) is an open neighborhood of \(x\) disjoint from \(A\), contradiction. Therefore \(x\in\overline A\).