9.3 COMPLETENESS IN AN ORDERED FIELD 317The abstract definition of "interval" given earlier in the text (Definition
3, Article 1.1) makes possible a brief and rather nice proof that the image
of an interval under a continuous function is again an interval. This con-
sequence of the intermediate value theorem also provides us with a fairly
typical example of the use of the concept of image, introduced in Article
8.2, in formulating a theorem.
COROLLARY
Suppose the function y= f(x) is continuous on an interval I. Then f(l) is an
interval.Proof If f is constant on I, then f(I) is a singleton set, which is an
interval. So assume f is not constant on I, and let y,, y, E f(I) with
y, < y,. Let yo be given such that y, < yo < y,; we must prove yo E
f (I); that is, yo = f (x,) for some x, E I. Now since y,, y, E f (I), there
must exist x,, x, E I such that y, = f (x,) and y, = f (x,). Since f (x,) <
yo < f(x,) and f is continuous on the closed interval determined by x,
and x,, there must exist, by the intermediate value theorem, x, strictly
between x, and x, such that yo = f (x,). Since x,, x, E I, since x, is
between x, and x,, and since I is an interval, we have x, E I, as required.
We indicated, just prior to Definition 4 of Article 1.1, that the intervals
on the real line are precisely the subsets of R having one of nine possible
forms (e.g., [a, b], (- co, b), etc.). In Exercise 3(a), Article 5.2 you were asked
to prove that any set having one of these forms satisfies the definition of
interval. We propose now to prove the converse; that is, any subset of R
satisfying the definition of interval must have one of these forms.
THEOREM 4
If a subset I of R is an interval, then I has one of the nine forms listed in
Definition 3, Article 1.1; that is, I has one of the nine forms [a, b], (a, b), [a, b),
(a, b], (-m, b), (-m, bl, (a, co), [a, a), or (---a a)).
Proof Assume I c R and I is an interval. Consider the four possibilities, I
is bounded, I is bounded above only, I is bounded below only, and I is
bounded neither above nor below. Clearly one and only one of these
four possibilities is true for I. Suppose, for instance, that I is bounded
above, but not below. By the completeness of R, we may let b = lub I
and take note of the fact that, necessarily, either b E I or b 4 I.
Our claim is that I = ( - oo, b ] in the first case and I = ( - oo, b) in the sec-
ond. Suppose first that b E I. To show that I = (- co, b], we use a mu-
tual inclusion approach and start by letting x E I. Since b is an upper
bound for I, then x 5 b so that x E (-a, b]. Thus I E (- co, b]. Con-
versely, suppose x E (- oo, b]. If x = b, then x E I since we've assumed
b E I. If x < b, then since b = lub I, we may invoke condition (b) of the
definition of least upper bound to produce c E I such that x < c < b.