Let \(A\), \(B\), and \(A_\alpha\) denote subsets of a space \(X\). Prove the following:
If \(A \subset B\), then \(\overline{A} \subset \overline{B}\).
\(\overline{A \cup B} = \overline{A} \cup \overline{B}\).
\(\overline{\bigcup A_\alpha} \supset \bigcup \overline{A}_\alpha\); give an example where equality fails.
Solution
If \(A\subseteq B\), every closed set containing \(B\) also contains \(A\). Since \(\overline A\) is the intersection of all closed sets containing \(A\) and \(\overline B\) is one such closed set, \(\overline A\subseteq\overline B\).
By monotonicity, \[ \overline A\cup\overline B\subseteq\overline{A\cup B}. \] Conversely, \(\overline A\cup\overline B\) is closed and contains \(A\cup B\), so minimality of closure gives \[ \overline{A\cup B}\subseteq\overline A\cup\overline B. \] Hence equality holds.
For every \(\alpha\), \(A_\alpha\subseteq\bigcup_\beta A_\beta\), so \[ \overline{A_\alpha}\subseteq\overline{\bigcup_\beta A_\beta}. \] Taking the union over \(\alpha\) gives \[ \bigcup_\alpha\overline{A_\alpha}\subseteq\overline{\bigcup_\alpha A_\alpha}. \] Equality can fail. In \(\mathbb R\), let \(A_n=\{1/n\}\). Then \[ \bigcup_n\overline{A_n}=\{1/n:n\ge1\}, \] whereas \[ \overline{\bigcup_nA_n}=\{1/n:n\ge1\}\cup\{0\}. \]