80. The bounds of a bounded aggregate

Let $S$ be any system or aggregate of real numbers $s$. If there is a number $K$ such that $s\le K$ for every $s$ of $S$, we say that $S$ is bounded above. If there is a number $k$ such that $s\ge k$ for every $s$, we say that $S$ is bounded below. If $S$ is both bounded above and bounded below, we say simply that $S$ is bounded.

Suppose first that $S$ is bounded above (but not necessarily below). There will be an infinity of numbers which possess the property possessed by $K$; any number greater than $K$, for example, possesses it. We shall prove that among these numbers there is a least,¹ which we shall call $M$. This number $M$ is not exceeded by any member of $S$, but every number less than $M$ is exceeded by at least one member of $S$.

We divide the real numbers $\xi$ into two classes $L$ and $R$, putting $\xi$ into $L$ or $R$ according as it is or is not exceeded by members of $S$. Then every $\xi$ belongs to one and one only of the classes $L$ and $R$. Each class exists; for any number less than any member of $S$ belongs to $L$, while $K$ belongs to $R$. Finally, any member of $L$ is less than some member of $S$, and therefore less than any member of $R$. Thus the three conditions of Dedekind’s Theorem (§ 17) are satisfied, and there is a number $M$ dividing the classes.

The number $M$ is the number whose existence we had to prove. In the first place, $M$ cannot be exceeded by any member of $S$. For if there were such a member $s$ of $S$, we could write $s=M+\eta$, where $\eta$ is positive. The number $M+\frac{1}{2}\eta$ would then belong to $L$, because it is less than $s$, and to $R$, because it is greater than $M$; and this is impossible. On the other hand, any number less than $M$ belongs to $L$, and is therefore exceeded by at least one member of $S$. Thus $M$ has all the properties required.

This number $M$ we call the upper bound of $S$, and we may enunciate the following theorem.

Any aggregate $S$ which is bounded above has an upper bound $M$. No member of $S$ exceeds $M$; but any number less than $M$ is exceeded by at least one member of $S$.

In exactly the same way we can prove the corresponding theorem for an aggregate bounded below (but not necessarily above).

Any aggregate $S$ which is bounded below has a lower bound $m$. No member of $S$ is less than $m$; but there is at least one member of $S$ which is less than any number greater than $m$.

It will be observed that, when $S$ is bounded above, $M\le K$, and when $S$ is bounded below, $m\ge k$. When $S$ is bounded, $k\le m\le M\le K$.