136 Chapter 4. Functional Limits and Continuity (b) Let g be a continuous function on the open interval (a, b). Prove that g is uniformly continuous on (a, b) if and only if it is possible to define values g(a) and g(b) at the endpoints so that the extended function g is continuous on [a, b]. (In the forward direction, first produce candidates forg(a) andg(b), and then show the extendedgis continuous.)
Exercise 4.4.14. Construct an alternate proof of Theorem 4.4.7 using the open cover characterization of compactness from the Heine–Borel Theorem (Theorem3.3.8(iii)).
4.5. The Intermediate Value Theorem 137
a c b
f(a) L f(b)
Figure 4.8: Intermediate Value Theorem.
Proof. Intending to use the characterization of connected sets in Theorem3.4.6, let f(E) = A∪B where A and B are disjoint and nonempty. Our goal is to produce a sequence contained in one of these sets that converges to a limit in the other.
Let
C={x∈E :f(x)∈A} and D={x∈E:f(x)∈B}.
The setsCandDare called the preimages ofAandB, respectively. Using the properties ofAandB, it is straightforward to check thatCandDare nonempty and disjoint and satisfy E =C∪D. Now, we are assumingE is a connected set, so by Theorem3.4.6, there exists a sequence (xn) contained in one ofCor Dwithx= limxncontained in the other. Finally, becausef is continuous atx, we getf(x) = limf(xn). Thus, it follows thatf(xn) is a convergent sequence contained in eitherAorBwhile the limitf(x) is an element of the other. With another nod to Theorem3.4.6, the proof is complete.
InR, a set is connected if and only if it is a (possibly unbounded) interval.
This fact, together with Theorem4.5.2, leads to a short proof of the Interme- diate Value Theorem (Exercise 4.5.1). We should point out that the proof of Theorem4.5.2does not make use of the equivalence between connected sets and intervals inRbut relies only on the general definitions. The previous comment that this is the most useful way to approach IVT stems from the fact that, although it is not discussed here, the definitions of continuity and connected- ness can be easily adapted to higher-dimensional settings. Theorem4.5.2, then, remains a valid conclusion in higher dimensions, whereas the Intermediate Value Theorem is essentially a one-dimensional result.
138 Chapter 4. Functional Limits and Continuity
Completeness
A typical way the Intermediate Value Theorem is applied is to prove the exis- tence of roots. Given f(x) =x2−2, for instance, we see that f(1) =−1 and f(2) = 2. Therefore, there exists a pointc∈(1,2) wheref(c) = 0.
In this case, we can easily computec=√
2, meaning that we really did not need IVT to show thatf has a root. We spent a good deal of time in Chapter1 proving that√
2 exists, which was only possible once we insisted on the Axiom of Completeness as part of our assumptions about the real numbers. The fact that the Intermediate Value Theorem has just asserted that√
2 exists suggests that another way to understand this result is in terms of the relationship between the continuity off and the completeness ofR.
The Axiom of Completeness (AoC) from the first chapter states that
“Nonempty sets that are bounded above have least upper bounds.” Later, we saw that the Nested Interval Property (NIP) is an equivalent way to assert that the real numbers have no “gaps.” Either of these characterizations of complete- ness can be used as the cornerstone for an alternate proof of Theorem4.5.1.
Proof. I. (First approach using AoC.) To simplify matters a bit, let’s consider the special case wheref is a continuous function satisfyingf(a)<0< f(b) and show that f(c) = 0 for somec∈(a, b). First let
K={x∈[a, b] :f(x)≤0}.
• a
K
b f(a)
f(b)
✻
c=supK
❅❅
■ ✻ ✒
Notice thatK is bounded above byb, anda∈K so K is not empty. Thus we may appeal to the Axiom of Completeness to assert that c= supK exists.
There are three cases to consider:
f(c)>0, f(c)<0, andf(c) = 0.
The fact thatc is the least upper bound ofK can be used to rule out the first two cases, resulting in the desired conclusion that f(c) = 0. The details are requested in Exercise4.5.5(a).
II. (Second approach using NIP.) Again, consider the special case where L= 0 and f(a)<0< f(b). LetI0= [a, b], and consider the midpoint
z= (a+b)/2.
4.5. The Intermediate Value Theorem 139 Iff(z)≥0, then seta1=aandb1=z. Iff(z)<0, then seta1=z andb1=b.
In either case, the intervalI1 = [a1, b1] has the property that f is negative at the left endpoint and nonnegative at the right.
•
a z b
f(z)>0
I0
I1
I2
This procedure can be inductively repeated, setting the stage for an applica- tion of the Nested Interval Property. The remainder of the argument is left as Exercise4.5.5(b).
The Intermediate Value Property
Does the Intermediate Value Theorem have a converse?
Definition 4.5.3. A function f has the intermediate value property on an interval [a, b] if for all x < y in [a, b] and all L between f(x) and f(y), it is always possible to find a pointc∈(x, y) wheref(c) =L.
Another way to summarize the Intermediate Value Theorem is to say that every continuous function on [a, b] has the intermediate value property. There is an understandable temptation to suspect that any function that has the in- termediate value property must necessarily be continuous, but that is not the case. We have seen that
g(x) =
sin(1/x) ifx= 0
0 ifx= 0
is not continuous at zero (Example 4.2.6), but it does have the intermediate value property on [0,1].
The intermediate value propertydoes imply continuity if we insist that our function is monotone (Exercise4.5.3).
Exercises
Exercise 4.5.1. Show how the Intermediate Value Theorem follows as a corol- lary to Theorem4.5.2.
Exercise 4.5.2. Provide an example of each of the following, or explain why the request is impossible
140 Chapter 4. Functional Limits and Continuity (a) A continuous function defined on an open interval with range equal to a
closed interval.
(b) A continuous function defined on a closed interval with range equal to an open interval.
(c) A continuous function defined on an open interval with range equal to an unbounded closed set different from R.
(d) A continuous function defined on all ofRwith range equal toQ.
Exercise 4.5.3. A function f is increasing onA iff(x)≤f(y) for allx < y in A. Show that if f is increasing on [a, b] and satisfies the intermediate value property (Definition 4.5.3), thenf is continuous on [a, b].
Exercise 4.5.4. Letg be continuous on an interval Aand letF be the set of points whereg fails to be one-to-one; that is,
F ={x∈A:f(x) =f(y) for somey=xandy∈A}. ShowF is either empty or uncountable.
Exercise 4.5.5. (a) Finish the proof of the Intermediate Value Theorem using the Axiom of Completeness started previously.
(b) Finish the proof of the Intermediate Value Theorem using the Nested Interval Property started previously.
Exercise 4.5.6. Letf : [0,1]→Rbe continuous withf(0) =f(1).
(a) Show that there must exist x, y ∈ [0,1] satisfying |x−y| = 1/2 and f(x) =f(y).
(b) Show that for eachn∈Nthere existxn, yn∈[0,1] with|xn−yn|= 1/n and f(xn) =f(yn).
(c) If h ∈ (0,1/2) is not of the form 1/n, there does not necessarily exist
|x−y|= hsatisfying f(x) = f(y). Provide an example that illustrates this using h= 2/5.
Exercise 4.5.7. Let f be a continuous function on the closed interval [0,1]
with range also contained in [0,1]. Prove that f must have a fixed point; that is, showf(x) =xfor at least one value ofx∈[0,1].
Exercise 4.5.8 (Inverse functions). If a functionf :A→Ris one-to-one, then we can define the inverse function f−1 on the range of f in the natural way: f−1(y) =xwherey=f(x).
Show that iff is continuous on an interval [a, b] and one-to-one, thenf−1 is also continuous.