The next big theorem is called the intermediate value theorem and the following picture
illustrates its conclusion. It gives the existence of a certain point.
PICT
You see in the picture there is a horizontal line, y = c and a continuous function which
starts off less than c at the point a and ends up greater than c at point b. The intermediate
value theorem says there is some point between a and b shown in the picture as
z such that the value of the function at this point equals c. It may seem this is
obvious but without completeness the conclusion of the theorem cannot be drawn.
Nevertheless, the above picture makes this theorem very easy to believe. Here is a useful
lemma.
Lemma 6.4.1If f is continuous on the closed interval
[a,b]
and for some x ∈
[a,b]
, f
(x)
≠c, then there exists δ > 0 such that for y ∈
[a,b]
∩
(x − δ,x+ δ)
, the signof f
(y)
−c is constant. That is, it is either always positive or always negative dependingon the sign of f
(x)
− c.
Proof: Let ε =
|f (x)− c|
> 0 by assumption. Then let δ correspond to this ε in the
definition of continuity.
Case 1: f
(x)
−c > 0. Then for y ∈
[a,b]
∩
(x− δ,x+ δ)
, f
(x )
−f
(y)
< f
(x)
−c and so
0 < f
(y)
− c.
Case 2: f
(x)
−c < 0. Then for y ∈
[a,b]
∩
(x− δ,x+ δ)
, f
(y)
−f
(x)
< c−f
(x)
and so
0 > f
(y)
− c. ■
Next here is a proof of the intermediate value theorem.
Theorem 6.4.2Suppose f :
[a,b]
→ ℝ is continuousand suppose f
(a)
< c <f
(b)
. Then there exists x ∈
(a,b)
such that f
(x )
= c.
Proof: Let d =
a+2b
and consider the intervals
[a,d]
and
[d,b]
. If f
(d)
≥ c,
then on
[a,d]
, the function is ≤ c at one end point and ≥ c at the other. On the
other hand, if f
(d)
≤ c, then on
[d,b]
, f ≥ 0 at one end point and ≤ 0 at the
other. Pick the interval on which f has values which are at least as large as c and
values no larger than c. Now consider that interval, divide it in half as was done for
the original interval and argue that on one of these smaller intervals, the function
has values at least as large as c and values no larger than c. Continue in this way.
Next apply the nested interval lemma to get x in all these intervals. In the n^{th}
interval, let x_{n},y_{n} be points of this interval such that f
(xn)
≤ c,f
(yn)
≥ c. Now
|xn − x|
≤
(b− a)
2^{−n} and
|yn − x|
≤
(b − a)
2^{−n} and so x_{n}→ x and y_{n}→ x.
Therefore,
f (x)− c = lim (f (xn) − c) ≤ 0
n→∞
while
f (x) − c = nli→m∞ (f (yn)− c) ≥ 0.
Consequently f
(x)
= c and this proves the theorem. The last step follows from Theorem
6.1.1. ■
This proves the theorem in the case where f is strictly decreasing. The case where f is
increasing is similar. ■
Proof 2: Note that since f is either strictly increasing or strictly decreasing, it maps an
open interval to an open interval. Now let U be an open set. Thus U = ∪_{x∈U}I_{x} where I_{x} is an
open interval. Then
( − 1)−1 ( −1)−1 ( −1)−1
f (U ) = f (∪x∈UIx) = ∪x ∈U f (Ix)
= ∪x∈U (f−1)−1(Ix ∩ (a,b)) = ∪x∈Uf (Ix ∩ (a,b))
which is a union of open intervals and is therefore, open. ■