Exercise 8.6.6

Question

\begin{enumerate}[label=(\alph*)] 
\item Prove that $-A$ defines a cut.
\item What goes wrong if we set $-A = \{r \in \mathbf{Q} : -r \notin A\}$?
\item If $a \in A$ and $r \in -A$, show $a + r \in O$. This shoes $A + (-A) \subseteq O$. Now, finish the proof of property (f4) for addition in Definition 8.6.4.
 \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] \item Let $t \neq A$ and $r < -t$. Then $-r > t$ so $r \in -A$ and $-A \neq \emptyset$. Let $a \in A$. Then $\forall t \neq A$, $t > a$; therefore $-a \notin -A$ and $A \neq \mathbf{Q}$, proving (c1). To prove (c2), let arbitrary $r \in -A$ and let $q < r$. Then $-q > -r > t$ for some $t \notin A$; hence $q \in -A$. To prove (c3), let $r \in -A$, and let $t \notin A$ satisfy $t < -r$. Let $q = (t - r) / 2$, and note that $t < q < -r$, so $-q \in -A$. Moreover $-q > r$, so (c3) is proved. \item This would fail on the cut $A = \{t \in \mathbf{Q} : t < -2\}$, since then $-A = \{t \in \mathbf{Q} : t \leq 2\}$ which from Exercise 8.6.1 (b) is not a cut. \item Since $-r \in -A$, then we have some $t \notin A$ where $-r > t > a$. Then $0 > a + r$ implying $a + r \in O$. Now suppose $o \in O$; we would like to find $a \in A$ and $r \in -A$ satisfying $o \leq a + r$; this would imply $O \subseteq A + (-A)$ and thus $O = A + (-A)$. Since $o \in \mathbf{Q}$ and $o < 0$, let $o = -p/q$ where $p, q \in \mathbf{N}$. We can prove the following lemma: for any cut $A$ and $n \in \mathbf{N}$, we can find $z \in \mathbf{Z}$ where $$\frac{z}{n} \in A \text{\quad and \quad} \frac{z + 1}{n} \notin A$$ To do this, start with $a \in A$ and $a' \notin A$, and find $N, M \in \mathbf{Z}$ satisfying $$\frac{N}{n} < a \text{\quad and \quad} \frac{M}{n} > a'$$ Clearly $\frac{N}{n} \in A$ and $\frac{M}{n} \notin A$. If we count upwards from $i = N$, at some point $\frac{i}{n}$ is going to transition from being in $A$ to not being in $A$, so that by the time we reach $i = M$ we have $\frac{i}{n} \notin A$. This transition point implies the existence of $z$. Now, let $a = \frac{n}{2q} \in A$ with $\frac{n+1}{2q} \notin A$. Then $r = -\frac{n+2}{2q} \notin A$, and $$a + r = \frac{-2}{2q} = -\frac{1}{q} \geq \frac{-p}{q} = o$$ completing the proof. \end{enumerate}

Exercise 8.6.6

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Exercise 8.6.6

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