Exercise 1.3.9

Question

\begin{enumerate}[label=(\alph*)] 
  \item If $\sup A<\sup B$, show that there exists an element $b \in B$ that is an upper bound for $A$.
  \item Give an example to show that this is not always the case if we only assume $\sup A \leq \sup B$
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] \item By Lemma 1.3.8 we know there exists a $b$ such that $(\sup B) - \epsilon < b$ for any $\epsilon > 0$, We set $\epsilon$ to be small enough that $\sup A < (\sup B)-\epsilon$ meaning $\sup A < b$ for some $b$, and thus $b$ is an upper bound on $A$. \item $A = \{x \mid x \le 1\}$, $B = \{x \mid x < 1\}$ no $b \in B$ is an upper bound since $1 \in A$ and $1 > b$. \end{enumerate}

Exercise 1.3.9

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Exercise 1.3.9

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