Exercise 8.6.8

Question

Let $\mathcal{A} \subseteq \mathbf{R}$ be nonempty and bounded above, and let $S$ be the \emph{union} of all $A \in \mathcal{A}$.
\begin{enumerate}[label=(\alph*)] 
\item First, prove that $S \in \mathbf{R}$ by showing that it is a cut.
\item Now, show that $S$ is the least upper bound for $\mathcal{A}$.
 \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
\item For (c1):
Let $A \in \mathcal{A}$, and let $a \in A$. Then $a \in S$, so $S \neq \emptyset$. Now let $B$ be a bound on $\mathcal{A}$, and let $b' \notin B$. Then for any $s \in S$, we have $s \in A \in \mathcal{A}$ for some $A$. Thus since $A \leq B$, $s \in B$ and so $b' > s$, so $b' \notin S$ and $S \neq \mathbf{Q}$.

For (c2) and (c3):
Consider arbitrary $s \in S$, with $s \in A \in \mathcal{A}$. Then for any $q < s$ we must have $q \in A \implies q \in S$, so (c2) is satisfied. Also, we can find $s < a \in A$, with $a \in A$, so (c3) is satisfied.

\item Let $B$ be an upper bound for $\mathcal{A}$. Now for any $s \in S$, with $s \in A \in \mathcal{A}$, we have $A \subseteq B$ so $s \in B$. Therefore $S \subseteq B$ and $S$ is the least upper bound for $\mathcal{A}$.
 \end{enumerate}

Exercise 8.6.8

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Exercise 8.6.8

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