Let $$S$$ be any system or aggregate of real numbers $$s$$. If there is a number $$K$$ such that $$s \leq K$$ for every $$s$$ of $$S$$, we say that $$S$$ is bounded above. If there is a number $$k$$ such that $$s \geq k$$ 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, $$M \leq K$$, and when $$S$$ is bounded below, $$m \geq k$$. When $$S$$ is bounded, $$k \leq m \leq M \leq K$$.

1. An infinite aggregate of numbers does not necessarily possess a least member. The set consisting of the numbers $1,\ \frac{1}{2},\ \frac{1}{3},\ \dots,\ \frac {1}{n},\ \dots,$ for example, has no least member.↩︎