Let \(X\) be a topological space and \(A \subseteq X\) an arbitrary subset. Prove that the topological closure \(\operatorname{cl}_X(A)\) (or \(\overline{A}\)) is the smallest closed subset of \(X\) containing \(A\): \[\overline{A} = \bigcap \{F \subseteq X \mid F \text{ is closed and } A \subseteq F\}.\]
Solution
Let \[ K=\bigcap\{F\subseteq X:F\text{ is closed and }A\subseteq F\}. \]
1. The set \(K\) is closed and contains \(A\).
Arbitrary intersections of closed sets are closed, and every set in the intersection contains \(A\).
2. The closure \(\overline A\) is closed and contains \(A\).
Certainly \(A\subseteq\overline A\). If \(x\notin\overline A\), some open neighborhood of \(x\) misses \(A\); the same neighborhood lies in \(X\setminus\overline A\). Thus \(X\setminus\overline A\) is open.
3. Since \(\overline A\) is one of the closed supersets occurring in the definition of \(K\),\[ K\subseteq\overline A. \]Conversely, every closed set containing \(A\) contains \(\overline A\), by minimality of closure under the neighborhood characterization. Hence\[ \overline A\subseteq K. \]Therefore\[ \overline A=K. \]