Exercise 4.8

Proof. Suppose that $A$ is an arbitrary nonempty set of real number that is bounded below by $a$. Now let $B=\left\{-x\mid x\in A\right\}$ and $b=-a$. First, we claim that $b$ is an upper bound of $B$. So consider any $y\in B$ so that $y=-x$ for some $x\in A$. Then $a\le x$ since $a$ a lower bound of $A$. It then follows from Exercise 4.2 part (d) that $y=-x\le -a=b$. Since $y\in B$ was arbitrary, this shows that $b$ is an upper bound of $B$.

Since $B$ is clearly nonempty (since $A$ is), we have that $B$ has a least upper bound $d=\sup <!--\; FUNCTION\; APPLICATION\; -->B$ since the reals have the least upper bound property. We claim that $c=-d$ is the greatest lower bound of $A$. So first consider any $x\in A$ so that $y=-x\in B$. Then we have $y\le d$ since $d=\sup <!--\; FUNCTION\; APPLICATION\; -->B$. Hence $c=-d\le -y=x$ again by Exercise 4.2 part (d). Since $x\in A$ was arbitrary, this shows that $c$ is in fact a lower bound of $A$.

Now suppose that $x$ is any lower bound of $A$. Then, by the same argument as above for $b=-a$, we have that $y=-x$ is an upper bound of $B$. It then follows that $d\le y$ since $d$ is the least upper bound of $B$. Then, again by Exercise 4.2 part (d), we have $x=-(-x)=-y\le -d=c$, which shows that $c$ is in fact the greatest lower bound since $x$ was arbitrary. This completes the proof. □

Proof. First, let $A=\left\{1∕n\mid n\in {\mathbb{Z}}_{\text{+}}\right\}$ so that we must show that $\inf <!--\; FUNCTION\; APPLICATION\; -->A=0$. For any $x\in A$ we have that $x=1∕n$ for some $n\in {\mathbb{Z}}_{\text{+}}$. Then $n>0$ so that $x=1∕n>0$ also by Exercise 4.2 part (i). Hence $0\le x$ is true, which shows that $0$ is a lower bound of $A$ since $x$ was arbitrary.

Now consider any $x>0$ so that also $1∕x>0$ by Exercise 4.2 part (i). Then, by the Archimedean ordering property there is an $n\in {\mathbb{Z}}_{\text{+}}$ such that $n>1∕x>0$ (since otherwise $1∕x$ would be an upper bound of ${\mathbb{Z}}_{\text{+}}$). It then follows from Exercise 4.2 part (j) that $1∕n<1∕(1∕x)=x$. Since clearly $1∕n\in A$ we have that $x$ is not a lower bound of $A$. Since $x>0$ was arbitrary, this shows that $0$ is the greatest lower bound of $A$ since, by the contrapositive, $x$ being a lower bound of $A$ implies that $x\le 0$. □

Proof. Consider any real $a$ where $0<a<1$. First we show that the set $\left\{1∕a^n\mid n\in {\mathbb{Z}}_{\text{+}}\right\}$ has no upper bound. To this end define $h=(1-a)∕a=1∕a-1$ so that $1+h=1+(1∕a-1)=1∕a$. Clearly we have

so that $1+h>1>0$ and

for any $n\in {\mathbb{Z}}_{\text{+}}$ since $n>0$.

We show by induction that ${(1+h)}^n\ge 1+\mathit{nh}$ for all $n\in {\mathbb{Z}}_{\text{+}}$. For $n=1$ we clearly have ${(1+h)}^n={(1+h)}^1=1+h\ge 1+h=1+1\cdot h=1+\mathit{nh}$. Now, supposing that ${(1+h)}^n\ge 1+\mathit{nh}$,we have

which completes the induction. So consider any real $x$. Then, since we know that ${\mathbb{Z}}_{\text{+}}$ has no upper bound, there is an $n\in {\mathbb{Z}}_{\text{+}}$ where $n>x∕h$ (noting that $h>0$) so that

Then we have $1∕a^n={(1∕a)}^n={(1+h)}^n\ge 1+\mathit{nh}>x$, which shows that the set $\left\{1∕a^n\mid n\in {\mathbb{Z}}_{\text{+}}\right\}$ is unbounded above since $x$ was arbitrary.

Now we show the main result. Let $A=\left\{a^n\mid n\in {\mathbb{Z}}_{\text{+}}\right\}$ so that we must show that $\inf <!--\; FUNCTION\; APPLICATION\; -->A=0$. First we show by induction that 0 is a lower bound of $A$. For $n=1$ we clearly have $a^n=a^1=a\ge 0$. Then, if $a^n\ge 0$, we have $a^{n+1}=a^n\cdot a\ge 0\cdot a=0$ since $a>0$. This completes the induction so that clearly $0$ is indeed a lower bound of $A$.

Now consider any real $x>0$ so that $1∕x>0$ also. Then, by what was shown above, we know that there is an $n\in {\mathbb{Z}}_{\text{+}}$ such that $1∕a^n>1∕x>0$. We then have $a^n=1∕(1∕a^n)<1∕(1∕x)=x$ by Exercise 4.2 part (j). This shows that $x$ is not a lower bound of $A$ since obviously $a^n\in A$. It then follows that $0$ is the greatest lower bound of $A$ since $x>0$ was arbitrary, because, by the contrapositive, $x$ being a lower bound of $A$ implies that $x\le 0$. Hence $0=\inf <!--\; FUNCTION\; APPLICATION\; -->A$ as desired. □