We have already seen two equivalent forms of the completeness axiom for the reals: the montone convergence theorem and the statement that every Cauchy sequence has a limit. The second of these is useful as it doesn't mention the order relation and so applies to the complex numbers for instance. (It turns out that the set of complex numbers is also complete in the Cauchy sense.) This web page describes a third approach due to Dedekind. This uses the order relation of but applies to arbitrary subsets of rather than sequences. There are a number of places (including results about the radius of convergence for power series, and several results in more advanced analysis) where Dedekind's approach is particularly helpful.
We start with a straightforward definition similar to many others in this course. Read the definitions carefully, and note the use of and here rather than and .
We say that is bounded above if there is such that . The number is called an upper bound for .
We say that is bounded below if there is such that . The number is called an lower bound for .
If we are interested in the
best upper bound
best lower bound we need to consider the following.
We say that is a least upper bound for if is an upper bound for and no is also an upper bound. In other words if and . The least upper bound of a set may not exist, but if it does it is unique, because if we have two distince upper bounds then one of these must be larger and so it cannot be a least upper bound. When it exists, the least upper bound of a set is called the supremum of and denoted .
We say that is a greatest lower bound for if is a lower bound for and no is also a lower bound. In other words if and . The greatest lower bound of a set may not exist, but if it does it is unique, and is called the infimum of and denoted .
Supremums and infimums are a bit like maximums and minimums for infinite sets .
The problem with maximums and minimums is that they are only guaranteed to exist for finite sets . For example, doesn't have a maximum element, though it does have a minimum element, . On the other hand has both a maximum (2) and a minumum (0).
In fact, for a nonempty set , the maximum element of exists if and only if exists and is an element of . Similarly, the minimum element of exists if and only if exists and is an element of .
Let . Then has no largest element, i.e., doesn't exist, but since is an upper bound and any is less than some by the Archimedean property. Note that is not an element of .
Let . Then has no smallest element, i.e., doesn't exist, but since is a lower bound and any is greater than some by the Archimedean property. Note that is not an element of .
Let . In this case does have largest and smallest elements, and and .
Let be the empty set. Then by convention every is both an upper bound and a lower bound for . So does not have least upper bound or greatest lower bound.
Let . Then does not have any upper bound, by the Archimedean property. But does have a lower bound, such as . The greatest lower bound is determined by your convention on the set of natural numbers. If you prefer then . Otherwise you will have .
Let . Then does not have any upper bound nor any lower bound, by the Archimedean property again.
Completeness of reals, supremum form.
Let be non-empty and bounded above. Then there is a least upper bound of , .
This is proved by a variation of a proof of an earlier result that every real number has a monotonic sequence of rationals converging to it.
We start by defining a sequence using induction on . Start with any with some such that . This just uses the assumption that is nonempty. Now inductively assume that is defined and for some , i.e., is not an upper bound for . If is an upper bound for then let . Otherwise, let where and is chosen to be the least natural number such that for all . Such a exists since is bounded above by some and we need only take so that , using the Archimedean Property. So since there is some such there must be a least such number, .
By construction, is a nondecreasing sequence of rationals and bounded above by , an upper bound for . It follows that converges to some . We shall show that this is the least upper bound of .
Suppose is not an upper bound of .
Then there is such that . But this gives a contradiction, for if is such that we consider the th stage of the construction, where we chose with greatest so that there is some with . But so contradicting this choice of .
Thus is after all an upper bound. To see it is the least upper bound, we again suppose otherwise.
Suppose is not the least upper bound of .
Then there is some such that every has . This again is impossible. Since , there must be some with . (To see this, put and with .) But by construction of there is always some with . Therefore and is not after all an upper bound for .
This completes the proof.
Completeness of reals, infimum form.
Let be non-empty and bounded below. Then there is a greatest lower bound of , .
Let be a lower bound for and let . Then is an upper bound for and hence by the previous result has a least upper bound, . It follows easily that is a greatest lower bound for , for if then as so , and if then so is not an upper bound for , so there is with hence and clearly .
These results are equivalent to the monotone convergence theorem. To see this, suppose is a bounded nondecreasing sequence. Then let . By the fact that the sequence is bounded and by completeness theorem above, exists, and it is a nice exercise to show that as .
You have seen another form of the completeness axiom for the reals, which is useful in that it doesn't involve any sequences and may be applied to subsets of .