Chapter Limits and Continuity Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 5.3, Slide Section 5.3 Properties of Continuous Functions Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 5.3, Slide The continuous image of a closed set may not be closed The continuous image of a bounded set may not be bounded Note: A function f : D → a bounded subset of is said to be bounded if its range f (D) is 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-Borel Theorem 3.5.5 says S is compact iff S 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 Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 5.3, Slide 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 = {Gα} 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 D D Copyright © 2013, 2005, 2001 Pearson Education, Inc Open cover of f (D) f f (D) Section 5.3, Slide 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 in f (D), there exist x1 and x2 in D such that f (x1) = y1 and 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: Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 5.3, Slide Lemma 5.3.5 Let f : [a, b] → be continuous and suppose f (a) < < f (b) Then there exists a point c in (a, b) such that f (c) = 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 claim f (c) = Then there exists a neighborhood U of c such that f (x) < for all Suppose f (c) < x ∈ U ∩ [a, b] 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 S a U • c • p b Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 5.3, Slide Lemma 5.3.5 Let f : [a, b] → be continuous and suppose f (a) < < f (b) Then there exists a point c in (a, b) such that f (c) = 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 Now suppose f (c) > We claim f (c) = Then there exists a neighborhood U of c such that f (x) > for all x ∈ U ∩ [a, b] And U contains a point p such that a < p < c y = f (x) Since f ( x) > for all x in U, S ∩ [p, c] = ∅ This implies that p is an upper bound for S, and S a U • • p c contradicts c being the least upper bound for S b We conclude that f (c) = ♦ The Intermediate Value Theorem extends this to a more general setting Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 5.3, Slide (Intermediate Value Theorem) Theorem 5.3.6 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 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 < and g(b) = f (b) – k > Thus there exists c ∈ (a, b) such that g(c) = But then f (c) – k = and f (c) = k When f (b) < k < f (a), a similar argument applies to the function g(x) = k – f (x) ♦ Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 5.3, Slide (Intermediate Value Theorem) Theorem 5.3.6 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 The idea behind the intermediate value theorem is simple when we graph the function y = f (x) f (b) If k is any height between f (a) and 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 f (a) somewhere in between The graph of a continuous function can a c Copyright © 2013, 2005, 2001 Pearson Education, Inc b have no “jumps.” Section 5.3, Slide 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 α ∈ [0, 2π], let r be a ray from the origin having angle α with the positive x axis This ray determines a unique circumscribing rectangle whose sides are parallel and perpendicular to r Let A(α ) be the length of the sides parallel to r ) A(α ) B(α Let B(α ) be the length of the sides perpendicular to r Now define f : [0, 2π] → C by f (α ) = A(α ) – B(α ) If for some α the circumscribing rectangle is not a square, then A(α ) ≠ B(α ), and we r suppose f (α ) = A(α ) – B(α ) > α Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 5.3, Slide 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 α ∈ [0, 2π], let r be a ray from the origin having angle α with the positive x axis We have f (α ) = A(α ) – B(α ) > B Now let β = α + π /2 Then the circumscribing rectangle (β ) is unchanged except the labeling ) A(α of its sides is reversed ) B(α A(β ) So f (β ) = A(β ) – B(β ) < Assuming that f is a continuous function, C the Intermediate Value Theorem implies there exists some angle θ with α < θ < β such that f (θ ) = r β α Copyright © 2013, 2005, 2001 Pearson Education, Inc But then A(θ ) = B(θ ) and the circumscribing rectangle for this angle is a square Section 5.3, Slide 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 C β α Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 5.3, Slide 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 β α Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 5.3, Slide 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 β α Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 5.3, Slide 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 Now the circumscribing rectangle has become a square, with α < θ < β C θ β α Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 5.3, Slide 15 ... is, C ⊆ S and each side of S touches C Given any α ∈ [0, 2π], let r be a ray from the origin having angle α with the positive x axis This ray determines a unique circumscribing rectangle whose... 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 α ∈ [0, 2π], let r be a ray from the origin having angle α with the positive... are parallel and perpendicular to r Let A(α ) be the length of the sides parallel to r ) A(α ) B(α Let B(α ) be the length of the sides perpendicular to r Now define f : [0, 2π] → C by f (α ) =