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 $A$ .