Themes > Science > Mathematics > Calculus > Real Number > Least upper bounds

Suppose that Y is an ordered field, and S is a nonempty subset of Y, and b is a member of Y. We say that b is an upper bound for the set S if we have s < b satisfied for every s in S.

If the set S has an upper bound, then in general it has many upper bounds. Say B is the set of upper bounds of S, and B is nonempty. Does B have a lowest member? If it does, that member is called the least upper bound of the set S.

The word "complete" has different meanings in different branches of mathematics. Generally, an object is called "complete" if there are no "holes" in it -- i.e., if nothing that seemingly "ought to" be there is missing -- but this vague description has different meanings for different kinds of mathematical objects. Here, we will only consider the meaning of completeness for ordered fields.

An ordered field Y is said to be complete, or Dedekind complete, if it has this property:

 Whenever S is a nonempty subset of Y, and S has at least one upper bound, then S has a least upper bound.

Dedekind completeness turns out to be crucial in analysis, because it enables us to take limits.

Some ordered fields are Dedekind complete, and some aren't. Here are two quick examples of ordered fields that aren't complete:

• The set of rational numbers is incomplete. To see this, let S be the set of all rational numbers r that satisfy r2 < 5. Then S has many upper bounds -- for instance, 3 is an upper bound, and 2.24 is another upper bound, and 2.23607 is another upper bound. We can keep finding more of these numbers -- whatever rational number we propose for an upper bound for S, it is possible to find another rational number that is still a little lower and that is also an upper bound for S. You can probably see why already: These numbers are converging to

Ö5 = 2.23606797749978969640917366873128...

But that number is not rational. Any rational upper bound for S would have to be slightly higher than Ö5, and between that rational number and Ö5 we can always find still another rational number. In the field of rational numbers, the set S does not have a least upper bound.

• If Y is a non-Archimedean field -- i.e., an ordered field that has infinitesimals -- then Y is incomplete. One way to see this is to let S be the set of all infinitesimals. Since some of the infinitesimals are positive, any upper bound for S must be greater than 0. Note that 1 is an upper bound for S, and 1/2 is another upper bound for S, and 1/3 is another upper bound for S, and so on. Suppose (for contradiction) that b were the least upper bound for S. Then b must be positive, and must be less than or equal to all of the numbers 1, 1/2, 1/3, etc. -- thus b must be a positive infinitesimal. Then 2b is also an infinitesimal, so 2b is a member of S. Since b is an upper bound for S, that tells us 2b < b. But b < 2b since b is positive. This is a contradiction.
(Note that, conversely, any complete ordered field must be Archimedean.)

Information provided by Eric Schechter