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This chapter discusses the properties of continuous functions, including the behavior of their images on compact and bounded sets, as well as the intermediate value theorem.
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Chapter 5 Limits and Continuity
Section5.3 Properties of Continuous Functions
The continuous image of a closed set may not be closed. Note: A function f : D is said to be boundedif its range f (D) is a bounded subset of . The continuous image of a bounded set may not be bounded. So, a continuous function may not be bounded, even if its domain is bounded. But if a set is both closed and bounded, then its image under a continuous function will be both closed and bounded. The Heine-BorelTheorem 3.5.5 says S is compact iffS is closed and bounded. So, the continuous image of a compact set is compact. Recall from Definition 3.5.1 A set S is compact if every open cover of S contains a finite subcover.
Theorem 5.3.2 Let D be a compact subset of and suppose f : D is continuous. Then f (D) is compact. Idea of proof: Let G= {G} be an open cover of f (D). Since f is continuous, the pre-image of each set in G is open. These pre-images form an open cover of D. Since D is compact, there is a finite subcollection that covers D. The corresponding sets in the range form a finite subcovering of f (D). Hence f (D) is compact. Open cover of f (D) Open cover of D f D f (D)
Corollary 5.3.3 Let D be a compact subset of and suppose f : D is continuous. Then f assumes minimum and maximum values on D. That is, there exist points x1 and x2 in D such that f (x1) f (x) f (x2) for all x D. Proof: We know from Theorem 5.3.2 that f (D) is compact. Lemma 3.5.4 tells us that f (D) has both a minimum, say y1, and a maximum, say y2. Since y1 and y2 are inf (D), there exist x1 and x2 in D such that f (x1) = y1and f (x2) = y2. It follows that f (x1) f (x) f (x2) for all x D. If we require the domain of the function to be a compact interval, then the following holds:
Lemma 5.3.5 Let f : [a,b] be continuous and suppose f (a)<0<f (b). Then there exists a point c in (a,b) such that f (c) = 0. Proof: Let S = {x [a, b] : f (x) 0}. Since a S, S is nonempty. Clearly, S is bounded above by b, and we may let c = sup S. We claimf(c) = 0. Then there exists a neighborhood U of c such that f (x) < 0 for all x U [a, b]. Suppose f (c) < 0. And U contains a point p such that c<p<b. y = f(x) But f (p) < 0, since p U, so p S. This contradicts c being an upper bound for S. U S c p a b
Lemma 5.3.5 Let f : [a,b] be continuous and suppose f (a)<0<f (b). Then there exists a point c in (a,b) such that f (c) = 0. Proof: Let S = {x [a, b] : f (x) 0}. Since a S, S is nonempty. Clearly, S is bounded above by b, and we may let c = sup S. We claimf(c) = 0. Now suppose f (c) > 0. Then there exists a neighborhood U of c such that f (x) > 0 for all x U [a, b]. And U contains a point p such that a<p<c. y = f(x) Since f(x) > 0 for all x in U, S [p,c] = . This implies that p is an upper bound for S, and contradicts c being the least upper bound for S. U S p c a b We conclude that f (c) = 0. The Intermediate Value Theorem extends this to a more general setting.
(Intermediate Value Theorem) Suppose f : [a,b] is continuous. Then f has the intermediate value property on [a,b]. That is, if k is any value between f(a) and f(b) , i.e. f(a)<k<f(b) or f(b) < k <f(a), then there exists a point c in (a,b) such that f(c) = k. Theorem 5.3.6 Proof: Suppose f (a) < k <f (b). Apply Lemma 5.3.5 to the continuous function g : [a, b] given by g(x) = f (x) – k. Then g(a) = f (a) – k < 0 and g(b) = f (b) – k > 0. Thus there exists c (a, b) such that g(c) = 0. But then f (c) – k = 0 and f (c) = k. When f (b) < k <f (a), a similar argument applies to the function g(x)=k – f (x).
(Intermediate Value Theorem) Suppose f : [a,b] is continuous. Then f has the intermediate value property on [a,b]. That is, if k is any value between f(a) and f(b) , i.e. f(a)<k<f(b) or f(b) < k <f(a), then there exists a point c in (a,b) such that f(c) = k. Theorem 5.3.6 The idea behind the intermediate value theorem is simple when we graph the function. y = f(x) If k is any height between f(a) and f(b), f (b) there must be some point c in (a,b) such that f(c) = k. y = k That is, if the graph of f is below y = k at a and above y = k at b, then for f to be continuous on [a,b], it must cross y = k somewhere in between. f (a) The graph of a continuous function can have no “jumps.” c a b
Example 5.3.8 (an application in geometry) Let C be a bounded closed subset of the plane. We claim there exists a square S that circumscribes C. That is, CS and each side of S touches C. Given any [0, 2], let r be a ray from the origin having angle with the positive x axis. This ray determines a unique circumscribing rectangle whose sides are parallel and perpendicular to r. LetA() be the length of the sides parallel to r. A( ) LetB() be the length of the sides perpendicular to r. B( ) Now define f : [0, 2] by f () = A() – B(). C If for some the circumscribing rectangle is not a square, then A() B(), and we suppose f () = A() – B() > 0. r
Example 5.3.8 (an application in geometry) Let C be a bounded closed subset of the plane. We claim there exists a square S that circumscribes C. That is, CS and each side of S touches C. Given any [0, 2], let r be a ray from the origin having angle with the positive x axis. We have f () = A() – B() > 0. Now let = + /2. Then the circumscribing rectangle is unchanged except the labeling of its sides is reversed. B( ) A() B( ) So f ( ) = A( ) – B( ) < 0. A( ) Assuming that f is a continuous function, C the Intermediate Value Theorem implies there exists some angle with < < such that f () = 0. r But then A() = B() and the circumscribing rectangle for this angle is a square.
Example 5.3.8 (an application in geometry) Let C be a bounded closed subset of the plane. We claim there exists a square S that circumscribes C. That is, CS and each side of S touches C. C
Example 5.3.8 (an application in geometry) Let C be a bounded closed subset of the plane. We claim there exists a square S that circumscribes C. That is, CS and each side of S touches C. C
Example 5.3.8 (an application in geometry) Let C be a bounded closed subset of the plane. We claim there exists a square S that circumscribes C. That is, CS and each side of S touches C. C
Example 5.3.8 (an application in geometry) Let C be a bounded closed subset of the plane. We claim there exists a square S that circumscribes C. That is, CS and each side of S touches C. Now the circumscribing rectangle has become a square, with < < . C