Exercise 4.9

Main Problem.

(a)

Proof. Suppose that $A$ is a nonempty subset of $\mathbb{Z}$ and that it is bounded above by $\alpha \in ℝ$. Since $A\ne \varnothing$, there is an $a\in A$, so define $A^{\prime }=\left\{n-a+1\mid n\in A\right\}$. First we claim that ${\alpha }^{\prime }=\alpha -a+1$ is an upper bound of $A^{\prime }$ So consider any $n^{\prime }\in A^{\prime }$ so that $n^{\prime }=n-a+1$ for some $n\in A$. Since $\alpha$ is an upper bound of $A$ we have

$$\begin{array}{llll}\hfill n& \le \alpha & \hfill & \\ \hfill n-a& \le \alpha -a& \hfill & \\ \hfill n-a+1& \le \alpha -a+1& \hfill & \\ \hfill n^{\prime }& \le {\alpha }^{\prime },& \hfill & \end{array}$$

which shows that ${\alpha }^{\prime }$ is an upper bound of $A^{\prime }$ since $n^{\prime }$ was an arbitrary element. We also have that there is an $N^{\prime }\in {\mathbb{Z}}_{\text{+}}$ such that ${\alpha }^{\prime }<N^{\prime }$ since ${\mathbb{Z}}_{\text{+}}$ has no upper bound.

Now let $B^{\prime }=A^{\prime }\cap {\mathbb{Z}}_{\text{+}}$. Then, for any $n^{\prime }\in B^{\prime }$, we have that $n^{\prime }\in A^{\prime }$ so that $n^{\prime }\le {\alpha }^{\prime }<N^{\prime }$. Since also clearly $n^{\prime }\in {\mathbb{Z}}_{\text{+}}$, we have that $n^{\prime }\in S_{N^{\prime }}=\left\{k\in {\mathbb{Z}}_{\text{+}}\mid k<N^{\prime }\right\}=\mathit{\mathbb{Z}fin}{N}^{\prime }-1$. Hence $B^{\prime }\subset S_{N^{\prime }}$ since $n^{\prime }$ was arbitrary. We also have that $1\in A^{\prime }$ since $a\in A$ and $a-a+1=1$. Hence $1\in B^{\prime }$ since clearly also $1\in {\mathbb{Z}}_{\text{+}}$ since it is inductive. Thus $B^{\prime }$ is a nonempty subset of $S_{N^{\prime }}$ so that it has a largest element $b^{\prime }$ by Exercise 4.4 part (a).

Since $b^{\prime }\in B^{\prime }$, we have that $b^{\prime }\in A^{\prime }$ so that there is a $b\in A$ such that $b^{\prime }=b-a+1$. We claim that $b$ is the largest element of $A$. We already know that $b\in A$ so we need only show that it is also an upper bound of $A$. So consider any $n\in A$ so that clearly $n^{\prime }=n-a+1\in A^{\prime }$. Now, it follows from Exercise 4.5 part (d) that $n^{\prime }\in \mathbb{Z}$ since $n,a,1\in \mathbb{Z}$. Thus we have the following:

Case: $n^{\prime }\in {\mathbb{Z}}_{\text{+}}$. Then, clearly $n^{\prime }\in A^{\prime }\cap {\mathbb{Z}}_{\text{+}}=B^{\prime }$ so that $n^{\prime }\le b^{\prime }$ since $b^{\prime }$ is the largest element of $B^{\prime }$.

Case: $n^{\prime }\in {\mathbb{Z}}_{\text{−}}\cup \left\{0\right\}$. Then $n^{\prime }<1\le b^{\prime }$ since $1\in B^{\prime }$ and $b^{\prime }$ is the largest element of $B^{\prime }$.

Thus in either case $n^{\prime }\le b^{\prime }$ is true so that

$$\begin{array}{llll}\hfill n^{\prime }& \le b^{\prime }& \hfill & \\ \hfill n-a+1& \le b-a+1& \hfill & \\ \hfill n-a& \le b-a& \hfill & \\ \hfill n& \le b,& \hfill & \end{array}$$

which shows that $b$ is an upper bound and thus the largest element of $A$ since $n$ was arbitrary. □

(b)

Proof. Suppose an $x\in ℝ$ where $x\notin \mathbb{Z}$ and let $A=\left\{n\in \mathbb{Z}\mid n<x\right\}$. It follows from Lemma 1 that there is an $m\in \mathbb{Z}$ where $m<x$ since $\mathbb{Z}$ has no lower bounds. Hence by definition $m\in A$ so that $A\ne \varnothing$. Clearly also $x$ is an upper bound of $A$ so that $A$ is a nonempty subset of $\mathbb{Z}$ that is bounded above. It then follows from part (a) that $A$ has a largest element $n$, where clearly $n<x$ since $n\in A$.

Now, suppose for the moment that $n+1\le x$. Then, since $\mathbb{Z}$ is inductive (again by Lemma 1 ) and $n\in \mathbb{Z}$, we have that $n+1\in \mathbb{Z}$ as well. But $x\notin \mathbb{Z}$ so that it must be that $n+1\ne x$, and hence $n+1<x$. Then $n+1\in A$ so that $n+1\le n$ since $n$ is the largest element of $A$. However, this contradicts the obvious fact that $n+1>n$ so that it must be that $n+1\le x$ is not true. Hence $n+1>x$ and thus we have shown that $n<x<n+1$.

Lastly, suppose that there is an integer $m$ such that $m<x<m+1$. Then $m\in A$ so that $m\le n$ since $n$ is the largest element of $A$. Suppose for a moment that $m<n$. Then we would have $m<n<x<m+1$ so that $n$ is an integer between $m$ and $m+1$, which violates Corollary 1 . Thus is has to be that $m=n$ (since $m\le n$), which shows that $n$ is the unique integer such that $n<x<n+1$. □

(c)

Proof. Suppose that $x,y\in ℝ$ and $x-y>1$. If $x\in \mathbb{Z}$ then let $n=x-1$ so that clearly $n\in \mathbb{Z}$ by Exercise 4.5 part (d). First, we have

$$\begin{array}{llll}\hfill x-y& >1& \hfill & \\ \hfill x& >1+y& \hfill & \\ \hfill x-1& >y& \hfill & \\ \hfill n& >y.& \hfill & \end{array}$$

We also clearly have $n=x-1<x$ so that $y<n<x$.

On the other hand, if $x\notin \mathbb{Z}$, then we know from part (b) that there is a unique integer $n$ such that $n<x<n+1$. We also have that

$$\begin{array}{llll}\hfill x& <n+1& \hfill & \\ \hfill 1<x-y& <n+1-y& \hfill & \\ \hfill 0& <n-y& \hfill & \\ \hfill y& <n& \hfill & \end{array}$$

so that again $y<n<x$.

Hence in both cases we have found an integer $n$ such that $y<n<x$, which proves the result. □

(d)

Proof. Suppose that $x,y\in ℝ$ where $y<x$. Then $0<x-y$ so that $1∕(x-y)$ exists.. Since ${\mathbb{Z}}_{\text{+}}$ is unbounded above there is a $b\in {\mathbb{Z}}_{\text{+}}$ where $b>1∕(x-y)$. Hence

$$\begin{array}{llllll}\hfill b& >\frac{1}{x-y}& \hfill & & \hfill & \\ \hfill b(x-y)& >1& \hfill \text{(since }x-y>0\text{)}& & \hfill & & \hfill \\ \hfill \mathit{bx}-\mathit{by}& >1.& \hfill & & \hfill & \end{array}$$

It then follows from part (c) that there is an integer $a$ such that $\mathit{by}<a<\mathit{bx}$. We then have that $y<a∕b<x$ since $b>0$ (since $b\in {\mathbb{Z}}_{\text{+}}$). This shows the result since clearly $a∕b$ is rational because $a,b\in \mathbb{Z}$. □