Exercise 8.6.1

Question

\begin{enumerate}[label=(\alph*)] 
\item Fix $r \in \mathbf{Q}$. Show that the set $C_r = \{t \in \mathbf{Q} : t < r\}$ is a cut.
The temptation to think of all cuts as being of this form should be avoided. Which of the following subsets of $\mathbf{Q}$ are cuts?
\item $S = \{t \in \mathbf{Q}: t \leq 2\}$
\item $T = \{t \in \mathbf{Q} : t^2 < 2 \text{ or } t < 0\}$
\item $U = \{t \in \mathbf{Q} : t^2 \leq 2 \text{ or } t < 0\}$
 \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] \item $C_r$ contains $r - 1$ and does not contain $r$, so (c1) is satisfied. If $p \in C_r$ and $q < p$, then $q < p < r$ and thus $q \in C_r$, so (c2) is satisfied. Also, $p < \frac{p + r}{2} < r$ so $\frac{p+r}{2} \in C_r$ and (c3) is satisfied. \item Not a cut, $S$ has the maximum $p = 2$. There are no elements in $S$ that can be greater than $2$. \item Is a cut. $0 \in T$ and $2 \notin T$ so (c1) is satisfied. Let $r \in T$ and $q < r$. If $q < 0$ then $q \in T$ trivially. Otherwise, $r > q \geq 0$ implies $2 > r^2 > q^2$ and therefore $q \in T$, showing (c2) is satisfied. Finally, to show (c3), let $r \in T$ with $r \geq 1$. (If $r < 1$ then we can trivially identify $r < 1 \in T$ to confirm (c3).) Let $a = 2- r^2$, and note $1 \geq a > 0$. Consider the rational number $$q = \left(r + \frac{a}{4r}\right)^2 = r^2 + \frac{a}{2} + \frac{a^2}{4r} > r$$ It is easy to show $\frac{a^2}{4r} < \frac{a}{2}$, implying $q < r^2 + a = 2$ and thus $q \in 2$, and thus $r$ is not a maximum and (c3) is true. \item Is a cut. The only difference from part (c) is that we cannot immediately claim by definition that $a > 0$; instead the definition of $U$ only implies $a \geq 0$. However, Section 1.1. provides a proof that $a \neq 0$; therefore we can maintain $a > 0$ and reuse the rest of the logic. \end{enumerate}

Exercise 8.6.1

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Exercise 8.6.1

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