24 1. The Real Numbers
Theorem 1.4.4. If R is defined using Dedekind cuts of Q, as above, then every
non-empty subset of R which is bounded above has a least upper bound.
Proof. Let A be a bounded subset of R and let m be any upper bound for A. For
each x A, let Lx be the corresponding cut in Q. Then x m for all x A means
that Lx Lm for all x A. We set
L =
x∈A
Lx.
Then L is a proper subset of Q because L Lm. If r L and s r, then r Lx
for some x A and this implies s Lx and, hence, s L. If L had a largest
element t, then t would belong to Lx for some x, and it would have to be a largest
element for Lx a contradiction. Thus, L has no largest element. We have now
proved that L satisfies (a), (b), and (c) of Definition 1.4.1 and, hence, that L is a
Dedekind cut.
If y is the real number corresponding to L, that is, if L =
Ly, then, for all
x A, Lx Ly, and this means x y. Thus, y is an upper bound for A. Also,
Ly Lm means that y m. Since m was an arbitrary upper bound for A, this
implies that y is the least upper bound for A. This completes the proof.
This completes our outline of the construction of the real number system begin-
ning with Peano’s axioms for the natural numbers. The final result is the following
theorem, which we will state without further proof. It will be the starting point for
our development of calculus.
Theorem 1.4.5. The real number system R is a complete ordered field.
Example 1.4.6. Find all upper bounds and the least upper bound for the following
sets:
A = (−1,2) = {x R : −1 x 2};
B = (0,3] = {x R : 0 x 3}.
Solution: The set of all upper bounds for the set A is {x R : x 2}. The
smallest element of this set (the least upper bound of A) is 2. Note that 2 is not
actually in the set A.
The set of all upper bounds for B is the set {x R : x 3}. The smallest
element of this set is 3 and so it is the least upper bound of B. Note that, in this
case, the least upper bound is an element of the set B.
If the least upper bound of a set A does belong to A, then it is called the
maximum of A. Note that a non-empty set which is bounded above always has
a least upper bound, by axiom C. However, the preceding example shows that it
need not have a maximum.
The Archimedean Property. An ordered field always contains a copy of the
natural numbers and, hence, a copy of the integers (Exercise 1.4.5). Thus, the
following definition makes sense.
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