Exercise 1.5

Question

Let $\mathcal{A}$ be a nonempty collection of sets.
Determine the truth of each of the following statements and of their converses:
\begin{enumerate}[label=(\alph*)]
\item $x \in \bigcup_{A \in \mathcal{A}} A \Rightarrow x \in A$ for at least one $A \in \mathcal{A}$.
\item $x \in \bigcup_{A \in \mathcal{A}} A \Rightarrow x \in A$ for every $A \in \mathcal{A}$.
\item $x \in \bigcap_{A \in \mathcal{A}} A \Rightarrow x \in A$ for at least one $A \in \mathcal{A}$.
\item $x \in \bigcap_{A \in \mathcal{A}} A \Rightarrow x \in A$ for every $A \in \mathcal{A}$.
\end{enumerate}

Dan Whitman · Accepted Answer

(a) The statement on the right is the definition of the statement on the left so of course the implication and its converse are true.

(b) The implication is generally false.
\begin{proof}
As a counterexample, consider $\mathcal{A} = \left\{\left\{1ight\}, \left\{2ight\}ight\}$.
Then clearly $\bigcup_{A \in \mathcal{A}} A = \left\{1, 2ight\}$ so that $1 \in \bigcup_{A \in \mathcal{A}} A$, but $1$ is not in $A$ for every $A \in \mathcal{A}$ since $1 
otin \left\{2ight\}$.
\end{proof}
However, the converse \emph{is} true.
\begin{proof}
Suppose that $x \in A$ for every $A \in \mathcal{A}$.
Since $\mathcal{A}$ is nonempty there is an $A_0 \in \mathcal{A}$.
Then $x \in A_0$ since $A_0 \in \mathcal{A}$.
Hence by definition $x \in \bigcup_{A \in \mathcal{A}} A$ since $x \in A_0$ and $A_0 \in \mathcal{A}$.
\end{proof}

(c) The implication here is true.
\begin{proof}
Suppose that $x \in \bigcap_{A \in \mathcal{A}} A$ so that by definition $x \in A$ for every $A \in \mathcal{A}$.
Since $\mathcal{A}$ is nonempty there is an $A_0 \in \mathcal{A}$ so that in particular $x \in A_0$.
This shows the desired result since $A_0 \in \mathcal{A}$.
\end{proof}
The converse is not generally true.
\begin{proof}
As a counterexample consider $\mathcal{A} = \left\{\left\{1,2ight\}, \left\{2,3ight\}ight\}$.
Then $1 \in \left\{1,2ight\}$ and $\left\{1,2ight\} \in \mathcal{A}$, but $1 
otin \bigcap_{A \in \mathcal{A}} A$ since clearly $\bigcap_{A \in \mathcal{A}} A = \left\{2ight\}$.
\end{proof}

(d) The statement on the right is the definition of the statement on the left so of course the implication and its converse are true.

Exercise 1.5

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Exercise 1.5

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