Let \(\mathcal{A}\) be a nonempty collection of sets. Determine the truth of each of the following statements and of their converses:
\(x \in \bigcup_{A \in \mathcal{A}} A \Rightarrow x \in A\) for at least one \(A \in \mathcal{A}\) .
\(x \in \bigcup_{A \in \mathcal{A}} A \Rightarrow x \in A\) for every \(A \in \mathcal{A}\) .
\(x \in \bigcap_{A \in \mathcal{A}} A \Rightarrow x \in A\) for at least one \(A \in \mathcal{A}\) .
\(x \in \bigcap_{A \in \mathcal{A}} A \Rightarrow x \in A\) for every \(A \in \mathcal{A}\) .
Solution
The counterexamples take \(\mathcal A=\{\{1\},\{2\}\}\) and \(x=1\), so that \(\bigcup_{A\in\mathcal A}A=\{1,2\}\) and \(\bigcap_{A\in\mathcal A}A=\varnothing\).
1. Statement (a) and its converse are both true.
By the definition of the union, \(x\in\bigcup_{A\in\mathcal A}A\) if and only if \(x\in A\) for at least one \(A\in\mathcal A\).
2. Statement (b) is false, and its converse is true.
For the counterexample, \(x\in\bigcup_{A\in\mathcal A}A\) but \(x\notin\{2\}\). For the converse, if \(x\in A\) for every \(A\in\mathcal A\), then, since \(\mathcal A\) is nonempty, \(x\) lies in at least one member of \(\mathcal A\), hence in the union.
3. Statement (c) is true, and its converse is false.
If \(x\in\bigcap_{A\in\mathcal A}A\), then \(x\) lies in every member of \(\mathcal A\), hence, since \(\mathcal A\) is nonempty, in at least one. For the converse, the counterexample has \(x\in\{1\}\) but \(x\notin\bigcap_{A\in\mathcal A}A=\varnothing\).
4. Statement (d) and its converse are both true.
By the definition of the intersection, \(x\in\bigcap_{A\in\mathcal A}A\) if and only if \(x\in A\) for every \(A\in\mathcal A\).
5. Q.E.D.
Steps 1 through 4 decide each of the four statements and its converse.