Exercise 1.3.4

Question

Let $A_{1}, A_{2}, A_{3}, \ldots$ be a collection of nonempty sets, each of which is bounded above.
  \begin{enumerate}[label=(\alph*)] 
  \item Find a formula for $\sup \left(A_{1} \cup A_{2}\right)$. Extend this to $\sup \left(\bigcup_{k=1}^{n} A_{k}\right)$.
  \item Consider $\sup \left(\bigcup_{k=1}^{\infty} A_{k}\right)$. Does the formula in (a) extend to the infinite case?
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
  \item $\sup \left(\bigcup_{k=1}^n A_k\right) = \sup \left\{\sup A_k \mid k=1,\dots,n\right\}$
  \item In general no, since $\bigcup_{k=1}^{\infty} A_k$ may be unbounded, for example with $A_n=[n,n+1]$.
   \end{enumerate}

Richard Ganaye · Answer

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\begin{proof} 
If a set $E$ is finite (and non empty), its least upper bound is the greatest element of $E$, 
$$\sup E = \max E.$$
\item[a)] We prove that
$$\sup (A_1 \cup A_2) = \max(\sup(A_1),\sup(A_2)).$$
Write $c_1 = \sup A_1$ and $c_2 = \sup(A_2)$. Without loss of generality, we can assume that $c_1 \leq c_2$.
\begin{enumerate}
\item[$\bullet$] Since $c_1$ is an upper bound of $A_1$, for every $a \in A_1$, $a\leq c_1$ (a fortiori $a \leq c_2$), and similarly for every $a \in A_2$, $a\leq c_2$. This shows that $c_2$ is an upper bound for $A_1 \cup A_2$.

\item[$\bullet$] Let $m$ be any upper bound of $A_1 \cup A_2$. Then $m$ is an upper bound of $A_2$, therefore $m \leq c_2$. This shows that $c_2$ is the least upper bound of $A_1 \cup A_2$.
\end{enumerate}
Therefore $c_2 = \max(c_1,c_2) = \sup(A_1 \cup A_2)$.

Same reasoning if $c_1 \geq c_2$.

We prove the generalization by induction. Assume that
$$\sup\left(\bigcup_{i=1}^n A_iight) = \max(\sup(A_i), 1\leq i \leq n).
$$
Then, using part (a), and the induction hypothesis,
\begin{align*}
\sup\left(\bigcup_{i=1}^{n+1} A_iight) &=\sup\left[\left(\bigcup_{i=1}^n A_iight) \cup A_{n+1}ight]\
&=\max\left(\sup\left(\bigcup_{i=1}^n A_iight), \sup(A_{n+1})ight)\
&=\max(\max(\sup(A_i), 1\leq i \leq n),\sup(A_{n+1})\
&= \max(\sup(A_i), 1\leq i \leq n+1).
\end{align*}
The induction is done, so, for all $n \in \N$,
$$\sup\left(\bigcup_{i=1}^n A_iight) = \max(\sup(A_i), 1\leq i \leq n).
$$
\item[b)] The same formula is not valid for $\sup\left(\bigcup_{i=1}^\infty A_iight) $, because the set $\{\sup(A_i) \mid i \in \N\}$ has not necessarily a maximum, or a least upper bound. The counterexample $A_i = [1,i]$ is such that
$$\{\sup A_i \mid i \in \N\} = \N$$
is not upper bounded.
\end{proof}

Exercise 1.3.4

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Exercise 1.3.4

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