Criticize the following “proof” that \(\overline{\bigcup A_\alpha} \subset \bigcup \overline{A}_\alpha\): if \(\ts{A_\alpha}\) is a collection of sets in \(X\) and if \(x \in \overline{\bigcup A_\alpha}\), then every neighborhood \(U\) of \(x\) intersects \(\bigcup A_\alpha\). Thus \(U\) must intersect some \(A_\alpha\), so that \(x\) must belong to the closure of some \(A_\alpha\). Therefore, \(x \in \bigcup \overline{A}_\alpha\).
Solution
1. The step “so that \(x\) must belong to the closure of some \(A_\alpha\)” does not follow.
The previous step gives, for each neighborhood \(U\) of \(x\), an index \(\alpha(U)\) with \(U\cap A_{\alpha(U)}\ne\varnothing\). By the definition of closure, \(x\in\overline{A_\beta}\) requires a single index \(\beta\) such that every neighborhood of \(x\) meets \(A_\beta\). The index \(\alpha(U)\) may change with \(U\), so no such \(\beta\) is produced.
2. The inclusion \(\overline{\bigcup A_\alpha}\subseteq\bigcup\overline{A_\alpha}\) is false in general.
In \(\RR\) let \(A_n=\{1/n\}\) for \(n\ge1\). Then \(\overline{\bigcup_nA_n}=\{0\}\cup\{1/n:n\ge1\}\), while \(\bigcup_n\overline{A_n}=\{1/n:n\ge1\}\) does not contain \(0\).
3. Q.E.D.
Step 1 locates the invalid inference, and step 2 shows that the conclusion itself fails.