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.↩︎

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