« Section 2.8: Problem 6 Solution

# Section 2.8: Problem 7 Solution

Working problems is a crucial part of learning mathematics. No one can learn... merely by poring over the definitions, theorems, and examples that are worked out in the text. One must work part of it out for oneself. To provide that opportunity is the purpose of the exercises.
James $^{*}\mathfrak{R}$ . Munkres
Let $A$ be a subset of $\mathbb{R}$ having no greatest member. Then as a subset of $^{∗}\mathbb{R}$ , $A$ will have upper bounds (with respect to the ordering $^{∗}<$ ) in $^{∗}\mathbb{R}$ . But show that $A$ does not have a least such bound.
We, first, argue that $b$ is an upper bound of $A$ in $^{∗}\mathbb{R}$ iff $b$ is positive and infinite. Indeed, if $b$ is an upper bound of $A$ in $^{∗}\mathbb{R}$ , then, since there is $a\in A$ such that $a>0$ , $b\ge a>0$ . Further, suppose that $b$ is finite. Then $b\simeq\mbox{st}b,$ and there is $a'\in A$ such that $a'>\mbox{st}b\simeq b$ , $a'-\mbox{st}b>0\simeq b-\mbox{st}b$ , implying (by the definition of infinitesimal numbers) $a'-\mbox{st}b>b-\mbox{st}b$ , and $a'>b$ . Therefore, $b$ must be infinite. Now, if $b$ is positive and infinite, then, by definition, $|b|=b\ge x$ for every $x\in\mathbb{R}$ , in particular, $x\in A$ .
Next, we show that if $b$ is an upper bound of $A$ , then there is a strictly lower upper bound of $A$ . This follows from the fact that if $b$ is positive and infinite, then $b-1$ is positive and infinite. Indeed, for every $x\in\mathbb{R}$ , $x>0$ , $b\ge x+1>1$ , so that $b-1\ge x>0$ .
We conclude that there is no least upper bound of $A$ .