Exercise 8.6.5

Question

\begin{enumerate}[label=(\alph*)] 
\item Show that (c1) and (c3) also hold for $A+B$. Conclude that $A+B$ is a cut.
\item Check that addition in $\mathbf{R}$ is commutative (f1) and associative (f2).
\item Show that property (o4) holds.
\item Show that the cut
$$O = \{p \in \mathbf{Q} : p < 0\}$$
successfully plays the role of the additive identity (f3). (Showing $A+O = A$ amounts to proving that these two sets are the same. The standard way to prove such a thing is to show two inclusions: $A + O \subseteq A$ and $A \subseteq A + O$.)
 \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
\item For (c1), we can find $a \in A$, $a' \notin A$, $b \in B$, and $b' \notin B$. Then $a + b \in A + B$ so $A + B \neq \emptyset$. Also, since $a < a'$ and $b < b'$, $a + b \neq a' + b'$ and therefore $a' + b' \notin A + B$.

For (c3), let arbitrary $a + b \in A + B$ with $a \in A$ and $b \in B$, and let $a < a' \in A$ and $b < b' \in B$. Then $a + b < a' + b' \in A + B$.
\item Let arbitrary $a + b \in A + B$; then $b + a \in B + A$; hence (f1) holds for addition. Let arbitrary $(a + b) + c \in (A+B)+C $, then $a + (b+c) \in A + (B+C)$; hence (f2) holds for addition.
\item Let $Y \subseteq Z$, and let $x + y \in X + Y$ with $x \in X$ and $y \in Y$. Then $y \in Z$ so $x + y \in X + Z$, implying $X + Y \subseteq X + Z$.
\item Let $a + o \in A + O$ where $a \in A$ and $o \in O$. Then $a + o < a$ implying $a + o \in A$ so $A + O \subseteq A$. Now let arbitrary $a \in A$, and find $a < a' \in A$. Define $s = a - a' < 0$, so $s \in O$. Then $a = a' + s \in A + O$, so $A \subseteq A + O$.
 \end{enumerate}

Exercise 8.6.5

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Exercise 8.6.5

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