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
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:
(Note that, conversely, any
complete ordered field must be Archimedean.)
- 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
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.