Exercise 1.20

Question

Exercise 20:
  Suppose that we omitted property (III) from the definition of a cut. Keep the
  same definitions of order and addition. Show that the resulting set has the
  least-upper-bound property, that addition satisfies axioms (A1) to (A4)
  (with a slightly different zero element!) but that (A5) fails.

ghostofgarborg · Accepted Answer

( extit{Jack Gallagher}) \ Proof is omitted whenever the proof in the book does not use property (III). In this case, a cut is defined as any inhabited set $\alpha \subset Q$ such that \begin{easylist} & $\alpha eq Q$ & If $p \in \alpha$, $q \in Q$, and $q < p$, then $q \in \alpha$ \end{easylist} I'll adopt the same convention as the book, labelling rationals with Roman letters and cuts with Greek. The proof that $R'$ has the least-upper-bound property is the same as that given in the book. The proofs of axioms (A1-3) are again the same. For (A4), we define $$ 0_R = \left\{ n\ |\ n \in Q,\ n \leq 0 ight\} $$ \begin{proof}[(A4)] For any $a \in \alpha$ and $s \in 0_r$, we have either that $$ a + s = a ext{ or } a + s < a $$ In either case we have that $a + s \in \alpha$, either by equality or by (II). Thus $\alpha + 0_{R'} \subset \alpha$. Similarly, if $b \in \alpha + 0_{R'}$, we have that $b = a + s$ for some $a \in \alpha$ and $s \in 0_{R'}$, and the same analysis applies. \end{proof} Finally, to negate (A5), we shall first show that any addition with an open cut will produce another open cut. \begin{proof} Consider two cuts $\alpha \in R$, $\beta \in R'$. For any $r \in \alpha + \beta$, we have that $r = a + b$ for some $a \in \alpha$, $b \in \beta$. But, because $\alpha$ is open, we can pick some $a' \in \alpha$ such that $a' > a$, and therefore we have $a' + b > a + b \in \alpha + \beta$. \end{proof} \begin{proof}[(A5) fails!] Suppose we had a valid negation operation. Then we would have $$ 0^* - 0^* = 0_{R'} $$ But this leads to a contradiction, as with the given definition of addition we cannot add any number to the open $0^*$ to produce the closed $0_{R'}$. \end{proof}

Amit Awasthi · Answer

\begin{definition}
      The members of our set $X$ will be certain subsets of $\mathbb{Q}$, called \emph{cuts}. Our version of the cut is any set $\alpha \subset \mathbb{Q}$ with the following two properties.
      \begin{enumerate}
        \item[(I)] $\alpha$ is not empty, and $\alpha 
e \mathbb{Q}$.
        \item[(II)] If $p \in \alpha$, $q\in \mathbb{Q}$, and $q < p$, then $q \in \alpha$.
      \end{enumerate}
      Note $p,q,r,\ldots \in \mathbb{Q}$ and $\alpha,\beta,\gamma,\ldots$ denote these cuts.
    \end{definition}
    \begin{definition}
      ``$\alpha < \beta$'' means $\alpha \subsetneq \beta$ ($\alpha$ is a proper subset of $\beta$).
    \end{definition}
    \begin{lemma}
      $X$ is an ordered set, with the order as defined above.
    \end{lemma}
    \begin{proof}[Proof of Lemma]
      By Definition 1.6, we must prove that the order as defined above meets Definition 1.5. Definition 1.5(ii) holds since $\alpha < \beta$ and $\beta < \gamma$ implies $\alpha < \beta$ since $\alpha \subsetneq \beta$ and $\beta \subsetneq \gamma$ implies $\alpha \subsetneq \gamma$. By (II) and the properties of sets, at most one of the three relations $\alpha < \beta,\ \alpha = \beta,\ \beta < \alpha$ is true for any pair $\alpha,\beta$. To show that at least one holds, assume that the first two fail. Then $\alpha 
ot\subset \beta$. Hence there is a $p \in \alpha$ where $p 
otin \beta$. If $q \in \beta$, $q < p$ by (II) since $p 
otin \beta$, and so $q \in \alpha$. Thus, $\beta \subset \alpha$. Since $\beta 
e \alpha$, $\beta < \alpha$. Thus Definition 1.5(i) holds.
    \end{proof}
    \begin{proof}[Proof of least-upper-bound]
      Let $A \subset X$ where $A 
e \emptyset$, and let $\beta \in X$ such that $\alpha < \beta$ for all $\alpha \in A$. Let
      \begin{equation*}
        \gamma = \bigcup_{\alpha \in A} \alpha.
      \end{equation*}
      First prove $\gamma \in X$. Since $A 
e \emptyset$, there is an $\alpha_0 \in A$ such that $\alpha_0 
e \emptyset$. $\alpha_0 \subset \gamma$, so $\gamma 
e \emptyset$. $\gamma \subset \beta$ since $\alpha \subset \beta$ for all $\alpha \in A$, so $\gamma 
e \mathbb{Q}$ and (I) is met. For some $p \in \gamma$, $p \in \alpha_1$ for some $\alpha_1 \in A$. If $q < p$, then $q \in \alpha_1$, so $q \in \gamma$ and (II) is met, and so $\gamma \in X$. Now prove $\gamma = \sup A$. By construction, $\alpha \le \gamma$ for all $\alpha \in A$. Suppose $\delta < \gamma$ is an upper bound of $A$. Then there is an $s \in \gamma$ such that $s 
otin \delta$. Since $s \in \gamma$, $s \in \alpha_2$ for some $\alpha_2 \in \gamma$. Thus, $\delta < \alpha_2$, which is a contradiction.
    \end{proof}
    \begin{definition}
      If $\alpha,\beta \in X$, ``$\alpha+\beta$'' denotes the set of all $r + s$ where $r \in \alpha$, $s \in \beta$.
    \end{definition}
    \begin{remark}
      Omitting property (III) from the definition of a cut on page 17, which is that if $p \in \alpha$, then $p < r$ for some $r \in \alpha$, allows $\alpha$ to contain a largest member $r$ such that for all $p \in \alpha$, $p \le r$, since $r < p$ would contradict (II), and $r 
e p$ would contradict (I).
    \end{remark}
    \begin{definition}
      $0^*$ is the set of all negative rational numbers and $0$, justified by the remark above. $0^*$ is a cut since $0^* 
e \emptyset$ and $0^* 
e \mathbb{Q}$, satisfying (I), and if $p \in 0^*$, $q \in \mathbb{Q}$, and $p < q$, then $q \in 0^*$, satisfying (II).
    \end{definition}
    \begin{claim}
      Addition as defined above satisfies axioms \emph{(A1)} to \emph{(A4)}, but \emph{(A5)} fails, with $0^*$ as defined above serving as our identity.
    \end{claim}
    \begin{proof}[Proof of \emph{(A1)}]
      We have to prove $\alpha + \beta \subset X$. Since $\alpha,\beta 
e \emptyset$, $\alpha + \beta 
e \emptyset$. Take $r' 
otin \alpha$, $s' 
otin \beta$. Then, $r' + s' > r + s$ for all $r \in \alpha$, $s \in \beta$ by (II). Thus, $r' + s' 
otin \alpha + \beta$, and (I) is met. If we take $p \in \alpha + \beta$, $p = r + s$ for some $r \in \alpha$, $s \in \beta$ by the definition of addition above. If $q < p$, then $q - s < r$, so $q-s \in \alpha$ by (II), and $q = (q-s) + s \in \alpha + \beta$, and (II) is met.
    \end{proof}
    \begin{proof}[Proof of \emph{(A2)}]
      We have to prove $\alpha + \beta = \beta + \alpha$. By the definition of addition above, $\alpha + \beta$ is the set of all $r + s$ where $r \in \alpha$, $s \in \beta$. Likewise, $\beta + \alpha$ is the set of all $s + r$. Since $r + s = s + r$ for all $r,s \in \mathbb{Q}$ by Remark $1.13(b)$, $\alpha + \beta = \beta + \alpha$.
    \end{proof}
    \begin{proof}[Proof of \emph{(A3)}]
      We have to prove $(\alpha + \beta) + \gamma = \alpha + (\beta + \gamma)$. By the defintion of addition above, $(\alpha + \beta) + \gamma$ is the set of all $(r+s)+t$ where $r \in \alpha$, $s \in \beta$, $t \in \gamma$. Likewise, $\alpha + (\beta + \gamma)$ is the set of all $r+(s+t)$. Since $(r+s)+t = r+(s+t)$ for all $r,s,t \in \mathbb{Q}$ by Remark $1.13(b)$, $(\alpha + \beta) + \gamma = \alpha + (\beta + \gamma)$.
    \end{proof}
    \begin{proof}[Proof of \emph{(A4)}]
      We have to prove $0^* + \alpha = \alpha$. If $r \in \alpha$ and $s \in 0^*$, then $r + s \le r$, hence $r + s \in \alpha$ by our remark above. Thus $\alpha + 0^* \subset \alpha$. Now pick $p \in \alpha$, and pick $r \in \alpha$, $r \ge p$. Then $p - r \in 0^*$ by definition of $0^*$, and $p = r + (p-r) \in \alpha + 0^*$. Thus $\alpha \subset \alpha + 0^*$. By Definition 1.3, this implies $0^* + \alpha = \alpha$.
    \end{proof}
    \begin{proof}[Proof that \emph{(A5)} fails]
      We have to prove that for some $\alpha \in X$, there is no element $\beta \in X$ such that $\alpha + \beta = 0^*$. Choose an $\alpha$ such that it does not contain a largest member, i.e., if $p \in \alpha$, $p < r$ for some $r \in \alpha$. Suppose, to get a contradiction, that $\alpha + \beta = 0^*$. Then for some $s \in \alpha$, there exists an $t \in \beta$ such that $s + t = 0$, by the definition of addition above. By assumption for $\alpha$, we know $0 = s + t < r + t \in \alpha + \beta$. Since $0 < r+t$, $r+t 
otin \alpha + \beta$, which is a contradiction since $r + t 
otin 0^*$.
    \end{proof}

Exercise 1.20

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Exercise 1.20

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