Define the term closure \(\overline A\) of a subset \(A\) of a topological space \(X\). Prove that if \(B\) is a closed subset of \(X\) such that \(A\subset B\) then \(\overline A\subset B\).
Solution
1. Define the closure of \(A\) by\[ \overline A =\bigcap\{F\subseteq X:F\text{ is closed in }X\text{ and }A\subseteq F\}. \]Thus \(\overline A\) is the smallest closed subset of \(X\) containing \(A\).
The family being intersected is nonempty because \(X\) itself is closed and contains \(A\). An arbitrary intersection of closed sets is closed, so \(\overline A\) is closed. Every member of the family contains \(A\), hence their intersection contains \(A\). Finally, if \(F\) is any closed subset of \(X\) containing \(A\), then \(F\) is one of the sets in the defining family, so \[ \overline A\subseteq F. \] This proves the stated minimality.
2. If \(B\) is closed and \(A\subseteq B\), then\[ \overline A\subseteq B. \]
1. An intersection is contained in each set being intersected.Therefore\[ \overline A =\bigcap\{F\subseteq X:F\text{ is closed and }A\subseteq F\} \subseteq B. \]