1.3 The real numbers  (Page 3/3)

If $S$ is a subset of an ordered field $F,$ then an element $x\in F$ is called an upper bound for $S$ if $x\ge y$ for every $y\in S.$ An element $z$ is called a lower bound for $S$ if $z\le y$ for every $y\in S.$

A subset $S$ of an ordered field $F$ is called bounded above if it has an upper bound; it is called bounded below if it has a lower bound; and it is called bounded if it has both an upper bound and a lower bound.

An element $M$ is called the least upper bound or supremum of a set $S$ if it is an upper bound for $S$ and if $M\le x$ for every other upper bound $x$ of $S.$ That is, $M$ is less than or equal to any other upper bound of $S.$

Similarly, an element $m$ is called the greatest lower bound or infimum of $S$ if it is a lower bound for $S$ and if $z\le m$ for every other lower bound $z$ of $S.$ That is, $m$ is greater than or equal to any other lower bound of $S.$

Clearly, the supremum and infimum of a set $S$ are unique. For instance, if $M$ and $M^{\text{'}}$ are both least upper bounds of a set $S,$ then they are both upper bounds of $S.$ We would then have $M\le M^{\text{'}}$ and $M^{\text{'}}\le M.$ Therefore, by part (h) of [link] , $M=M^{\text{'}}.$

It is important to keep in mind that an upper bound of a set $S$ need not be an element of $S,$ and in particular, the least upper bound of $S$ may or may not actually belong to $S.$

If $M$ is the supremum of a set $S,$ we denote $M$ by $supS.$ If $m$ is the infimum of a set $S,$ we denote it by $infS.$

REMARK. The preceding exercise shows that peculiar things about upper and lower bounds happen when $S$ is the empty set. One point is that just because a set has an upper bounddoes not mean it has to have a least upper bound. That is, no matter whichupper bound we choose, there is always another one that is strictly smaller. This is a very subtle point, andit is in fact quite difficult to give a simple concrete example of this phenomenon. See the remark following [link] . However, part (d) of [link] contains the seed of an example.

An ordered field $F$ is called complete if every nonempty subset $S$ of $F$ that has an upper bound has a least upper bound.

REMARK. Although $Q$ is an ordered field, we will see that it is not a complete ordered field. In fact, the answer to part (d) of [link] is no. The set described there, though bounded above,has no least upper bound. In fact, it was one of nineteenth century mathematicians' major achievements to prove the following theorem.

There exists a complete ordered field.

We leave the proof of this theorem to the appendix.

Perhaps the most reassuring result along these lines is the following companion theorem, whose proof we also leave to the appendix.

If $F_1$ and $F_2$ are two complete ordered fields, then they are isomorphic.

Taken together, the content of the two preceding theorems is that, up to isomorphism, there exists one and only one complete ordered field. For no other reason that that, this special field should be an important object in mathematics. Our definition of the real numbers is then the following:

By the set $R$ of real numbers we mean the (unique) complete ordered field.