316 PROPERTIES OF NUMBER SYSTEMS Chapter 9
parts of Article 4.3 at this stage.) Furthermore, this theorem is the basis
of important properties of the real number system, such as the existence
of a real nth root of any positive real number a, where n E N (see Exer-
cise 9). The proof of the intermediate value theorem, however, may not be
familiar to you. As we will soon see, the "IVT" is a consequence of the
completeness of R. In preparation for the proof, we state and prove the
following lemma.
LEMMA
Suppose the function y = f(x) is continuous on an interval I containing the real
number x,, and that c and dare real numbers such that c < f(x,) < d. Then there
exists a positive real number 6 such that c < f(x) < d for all x E (x, - 6,
x, + 6) n I.
Proof [Recall Exercise 16(b), Article 6.1.1 The proof uses the technique
of specialization as well as the definition of continuity at a point. We
separate the proof into two cases, according as x, either is or is not an
endpoint of the interval I. Applying the definition of continuity at a
point to the latter case, we have that, to any E > 0, there corresponds
6 > 0 such that f(x,) - E < f(x) < f(x0) + E when x E (x, - 6, X, + a),
where 6 may be chosen so small that the interval (x, - 6, xo + 6) is a
subset of I. In particular, letting E equal the smaller of d - f(x,) and
f (x,) - c, we may assert that a positive number 6 exists such that f(x,) -
(f (xo) - 4 < f (~0) - < f (4 < f (~0) + E < f (~0) + (d - f (~0)) for any
x between x, - 6 and x, + 6. From this, we conclude that c < f(x) < d
whenever x E (x, - 6, x0 + a), as desired.
The cases in which x, is either a left- or right-hand endpoint of I are
left to the reader in Exercise 10.
Proof of theorem Define a subset S of [a, b] by the rule S = (x E
[a, b] 1 f (x) < yo). Note that S is nonempty, since a E S, and S is bounded
above in R, by b. Hence, by the completeness property of R, S has a
least upper bound in R, call it x,. Note that x, E [a, b] (Why?). We
claim that f(x,) = yo; our approach is to prove that each of the other
two possibilities f(xo) < yo and f(x,) > yo leads to a contradiction.
Now if f(x,) < yo, then necessarily x, < b and we may conclude from
the lemma that there exists 6 > 0 such that x, + 6 < b and f(x) < yo for
any x E [x,, X, + 6). In particular, there exists x' E (x,, b] such that
f(x') < yo. Since x' E S (Why?) and x' > x,, we have a contradiction of
the fact that x, is an upper bound of S.
On the other hand, if f(x,) > yo, then x, cannot equal a and, again
using the lemma, we note that a positive 6 exists with a < x, - 6 such
that f (x) > yo for any x E (x, - 6, x,]. In particular, there is no element
of S greater than x, - 6, a violation of condition (b) of Definition 3 and
thus a contradiction of the fact that x, is the least upper bound of S.
Hence f(x,) = yo, as desired. Since f (a) < yo < f(b), we have x, # a
and xo # b, so that x, E (a, b), as claimed.