Exercise 1.3.10

Question

[Cut Property]

The Cut Property of the real numbers is the following:

If $A$ and $B$ are nonempty, disjoint sets with $A \cup B=\mathbf{R}$ and $a<b$ for all $a \in A$ and $b \in B$, then there exists $c \in \mathbf{R}$ such that $x \leq c$ whenever $x \in A$ and $x \geq c$ whenever $x \in B$.

\begin{enumerate}[label=(\alph*)] 
  \item Use the Axiom of Completeness to prove the Cut Property.
  \item Show that the implication goes the other way; that is, assume $\mathbf{R}$ possesses the Cut Property and let $E$ be a nonempty set that is bounded above. Prove $\sup E$ exists.
  \item The punchline of parts (a) and (b) is that the Cut Property could be used in place of the Axiom of Completeness as the fundamental axiom that distinguishes the real numbers from the rational numbers. To drive this point home, give a concrete example showing that the Cut Property is not a valid statement when $\mathbf{R}$ is replaced by $\mathbf{Q}$.
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] \item If $c = \sup A = \inf B$ then $a \le c \le b$ is obvious. So we must only prove $\sup A = \inf B$. If $\sup A < \inf B$ then consider $c=\frac{\sup A + \inf B}{2}$. $c > \sup A$ and therefore $c otin A$; similarly $c < \inf B$ and therefore $c otin B$, implying $A \cup B e \mathbf{R}$. If $\sup A > \inf B$ then we can find $a$ such that $a > b$ by subtracting $\epsilon > 0$ and using the least upper/lower bound facts, similarly to Lemma 1.3.8. Thus $\sup A$ must equal $\inf B$ since we have shown both alternatives are impossible. \item Let $B = \{x \mid e < x,\forall e\in E\}$ and let $A = B^c$. Clearly $a < b$ so the cut property applies. We have $a \le c \le b$ and must show the two conditions for $c = \sup E$ \begin{enumerate}[label=( oman*)] \item Since $E \subseteq A$, $a \le c$ implies $e \le c$ thus $c$ is an upper bound. \item $c \le b$ implies $c$ is the smallest upper bound. \end{enumerate} Note: Using (a) here would be wrong, it assumes the axiom of completeness so we would be making a circular argument. \item $A = \{r \in \mathbf{Q} \mid r^2 < 2 ext{ or } r < 0\}$, $B = A^c$ does not satisfy the cut property in $\mathbf{Q}$ since $\sqrt{2} otin \mathbf{Q}$. (Alternatively, $A = \{r \in \mathbf{Q} \mid r^3 < 2\}$ can be used to avoid needing to single out negative numbers in defining $A$, but requires a proof that $\sqrt[3]{2}$ is irrational - left to the reader.) \end{enumerate}

Exercise 1.3.10

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Exercise 1.3.10

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