Show that if \(A_i \subseteq X\), then \(\cl_X(\union_i A_i) = \union_i \cl_X(A_i)\).
Solution
1. The displayed assertion is true for finite unions but false for arbitrary indexed unions.
For two sets, \(\overline{A\cup B}=\overline A\cup\overline B\): the right side is closed and contains \(A\cup B\), while monotonicity of closure gives the reverse inclusion. Induction handles finite unions.
2. For an infinite counterexample in \(\mathbb R\), take \(A_n=\{1/n\}\) for \(n\ge1\).
Each \(A_n\) is closed, so \(\bigcup_n\overline{A_n}=\{1/n:n\ge1\}\). But \(0\) is a limit point of the union, hence \[0\in\overline{\bigcup_nA_n}\setminus\bigcup_n\overline{A_n}.\]
3. Thus the correct general relation is only\[\boxed{\bigcup_i\overline{A_i}\subseteq\overline{\bigcup_iA_i}},\]with equality guaranteed for finite index sets.
Each \(A_i\subseteq\bigcup_jA_j\), so monotonicity gives \(\overline{A_i}\subseteq\overline{\bigcup_jA_j}\); take the union over \(i\).