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Properties of Continuous Functions: Compact Sets, Bounded Sets, and Intermediate Value Theorem

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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Properties of Continuous Functions: Compact Sets, Bounded Sets, and Intermediate Value Theorem

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  1. Chapter 5 Limits and Continuity

  2. Section5.3 Properties of Continuous Functions

  3. 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.

  4. 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)

  5. 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:

  6. 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

  7. 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.

  8. (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). 

  9. (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

  10. 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, CS 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 

  11. 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, CS 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.  

  12. 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, CS and each side of S touches C. C  

  13. 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, CS and each side of S touches C. C  

  14. 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, CS and each side of S touches C. C  

  15. 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, CS and each side of S touches C. Now the circumscribing rectangle has become a square, with <  <  . C   

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