Consequences of Continuity - PDF Free Download

Consequences of Continuity James K. Peterson Department of Biological Sciences and Department of Mathematical Sciences Clemson University October 4, 2017

Outline 1 Domains of Continuous Functions 2 The Intermediate Value Theorem 3 Consequences of a compact domain for a continuous function 4 Continuity in terms of inverse images of open sets

Domains of Continuous Functions In our discussions of continuity, we always assume that f is defined locally at the point p which means there is a radius r so that if x B r (p), f (x) is defined.

Domains of Continuous Functions In our discussions of continuity, we always assume that f is defined locally at the point p which means there is a radius r so that if x B r (p), f (x) is defined. If at each point p in dom(f ), f is locally defined, this says each such p is an interior point of dom(f ) and so dom(f ) is an open set. Because of the requirement of f being locally defined at each p, we see dom(f ) can not have isolated boundary points. Hence, any boundary points dom(f ) has must be accumulation points. This means if p (dom(f )), there has to be a sequence (x n ) in dom(f ), each x n p, with x n p.

Domains of Continuous Functions Now subsets of R can be very complicated, so let s restrict our attention to intervals. An interval of R is a subset I of the following form I = (, a) infinite open interval I = {, a} I = (, a] infinite closed interval I = {, a} I = (a, ) infinite open interval I = {a,, } I = [a, ) infinite closed interval I = {a,, } I = (, ) infinite open and closed interval I = {, } I = (a, b), a < b finite open interval I = {a, b} I = [a, b), a < b finite half open interval I = {a, b} I = (a, b], a < b finite half open interval I = {a, b} I = [a, b], a < b finite closed interval I = {a, b} Continuity at the points in I is possible but not guaranteed. Examples: 1/x is continuous on (0, ) and the boundary point 0 can not be included. x sin(1/x) on (0, 1] can be extended to the boundary point 0 so that x sin(1/x) has a removeable discontinuity at 0. It is possible to talk about boundary points at ± but we won t do that here.

The Intermediate Value Theorem Theorem The Intermediate Value Theorem for [a, b] Let f : [a, b] R be continuous with f (a) f (b). Then if y is between f (a) and f (b), there is at least one c between a and b so that f (c) = y. Proof For concreteness, let s assume f (a) < f (b). We then have f (a) < y < f (b) and we want to find a c so that a < c < b and f (c) = y. Our argument here will be very abstract! So lot s of fun! Let A = {x [a, b] : f (x) y}. Since f (a) < y we know A is non empty. Further, since x = b is the largest value allowed, we see A is bounded. So by the completeness axiom, sup(a) exists and is finite. Let c = sup(a). Since b is an upper bound of A, we see by the definition of supremum, that c b.

The Intermediate Value Theorem Proof We now show f (c) = y. (Step 1: f (c) y): (Case (a) ): if c A, then by the definition of A, we must have f (c) y. (Case (b) ): Assume c A. Then applying the Supremum Tolerance Lemma using the sequence of tolerances ɛ = 1/n, we can construct a sequence (x n ) with c 1/n < x n c and x n A. Since we are in Case (b), none of these x n can be c. Since f is continuous at c, we then must have lim n f (x n ) = f (c). But since x n A, we also must have f (x n ) y for all n. In general, if a n a and b n b with a n b n, this implies a b. To see this, note for any ɛ > 0, there is an N 1 so that n > N 1 implies a ɛ/2 < a n < a + ɛ/2 and there is an N 2 so that n > N 2 implies b ɛ/2 < b n < b + ɛ/2. So for n > max{n 1, N 2 }, we have a ɛ/2 < a n b n < b + ɛ/2. But this tells us that a b < ɛ.

The Intermediate Value Theorem Proof Now if a > b, we could choose ɛ = (a b)/2 and then we would have a b < (a b)/2 implying (a b)/2 < 0 which contradicts our assumption. Hence, we must have a b. Applying this result here, f (x n ) y for all n thus implies f (c) y. So in either Case (a) or Case (b). f (c) y. (Step 2: f (c) = y): Assume f (c) < y. Since we know y < f (b), this says c < b. Choose ɛ = (1/2)(y f (c)) > 0. Since f is continuous at c, for this ɛ, there is a δ > 0 so that B δ (c) [a, b] and x c < δ f (x) f (c) < ɛ. Rewriting, we have c δ < x < c + δ f (c) ɛ < f (x) < f (c) + ɛ Since B δ (c) [a, b], we know (c, c + δ) (c, b]. Now pick a point z in (c, c + δ). Then, f (z) < f (c) + ɛ.

The Intermediate Value Theorem Proof But ɛ = (1/2)(y f (c)). Thus, f (z) < (y + f (c))/2. Now we assumed f (c) < y so the average (f (c) + y)/2 < y also. So we have f (z) < y. But this says z A with z > c which is the supremum of A. This is not possible and so our assumption that f (c) < y is wrong and we must have f (c) = y. Theorem The Intermediate Value Theorem for Intervals Let f : I R be continuous. Let a and b be any two points in I with a b and f (a) f (b). Then if y is between f (a) and f (b), there is at least one c between a and b so that f (c) = y.

The Intermediate Value Theorem Proof Apply the Intermediate Value Theorem for [a, b] I and the result follows. Theorem Let I be an interval and assume f is continuous on I. Then f (I ) is also an interval. Proof Let u, v be in f (I ) with u < v. Then there are s, t I so that f (s) = u and f (t) = v. Choose any number y so that u < y < v. (Case (a)): If s < t, since f is continuous on [s, t], by the Intermediate Value Theorem for finite closes intervals, there is a number c with s < c < t with f (c) = y. Thus, y f (I ).

The Intermediate Value Theorem Proof (Case (b)) If t < s, we apply the same sort of argument to again show y f (I ). Combining these arguments, we see for any u, v f (I ) with u < v, all u < y < v are also in f (I ). This shows f (I ) is an interval.

Consequences of a compact domain for a continuous function What if the domain of a continuous function f is compact? What are the consequences? Theorem If f : dom(f ) R with dom(f ) a compact set, then f (dom(f )) is a compact set also. Proof For convenience, let K = dom(f ). Then f (K) = {f (x) : x K}. Let (y n ) be any sequence in f (K). Then there is a sequence (x n ) K with y n = f (x n ) for all n. Since K is compact, there is a subsequence (x 1 n ) of (x n ) so that x 1 n p for some p K. Since f is continuous at p, we then have f (x 1 n ) f (p). ( Recall at a boundary point of K, we would interpret continuity at this point in terms of right or left continuity). Hence (y n ) has a subsequence (y 1 n ) = (f (x 1 n )) which converges to an element of f (K). Thus, f (K) is sequentially compact and topologically compact and closed and bounded.

Consequences of a compact domain for a continuous function Let s recall what we know about minima and maxima of sets. First if a set Ω is nonempty and bounded, it has a finite infimum and supremum.

Consequences of a compact domain for a continuous function Let s recall what we know about minima and maxima of sets. First if a set Ω is nonempty and bounded, it has a finite infimum and supremum. It is a straightforward argument to show there are sequences (x m n ) and (x M n ) so that x m n inf(ω) and x M n sup(ω). If the inf(ω) Ω, then we say the minimum of the set Ω is achieved by at least the point x m = inf(ω) and xn m x m. This is also called the absolute minimum of Ω. If the sup(ω) Ω, then we say the maximum of the set Ω is achieved by at least the point x M = sup(ω) and xn M x M. This is also called the absolute maximum of Ω. So Ω compact implies Ω always has an absolute minimum and maximum value.

Consequences of a compact domain for a continuous function Here are more consequences. If f : [a, b] R is continuous on [a, b], first note this means f is left continuous at b and right continuous at a. Note, to use full circles, we could redefine f like this: f (a), x < a ˆf (x) = f (x), x [a, b] f (b), x > b i.e. we extend f to the right and left based on the f (a) and f (b) value. Then f is continuous at a and b in the usual sense. We don t usually need to be so precise however.

Consequences of a compact domain for a continuous function Theorem Let f : [a, b] R be continuous. Then f ([a, b]) = [f (x m ), f (x M )] where x m, x M are in [a, b] and f (x m ) and f (x M ) are the absolute minimum and maximum of f on [a, b], respectively. Proof We already know f ([a, b]) is an interval and it is a compact set. and we have f (x m ) f (x) f (x M ) for all x [a, b]. Thus f ([a, b]) = [f (x m ), f (x M )].

Continuity in terms of inverse images of open sets There are even more ways to look at continuity. We know that f is continuous at p if ɛ > 0 δ > 0 x p < δ f (x) f (p) < ɛ, where δ is small enough to fit inside the circle B r (p) on which f is locally defined. Now suppose V was an open set in the range of f. This means every point in V can be written as y = f (x) for some point in the domain of f. Consider f 1 (V ), the inverse image of V under f. This is defined to be f 1 (V ) = {x dom(f ) : f (x) V } Is f 1 (V ) open? Let y 0 = f (x 0 ) V for some x 0 dom(f ). Since V is open, y 0 is an interior point of V and so there is an R > 0 so that B R (y 0 ) V. Then since f is continuous at x 0, choose ɛ = R and we see there is a δ > 0 so that if x x 0 < δ, where we choose δ < r, with f (x) f (x 0 ) < R. This says B δ (x 0 ) f 1 (V ) telling us x 0 is an interior point of f 1 (V ). Since x 0 is arbitrary, we see f 1 (V ) is open. Thus, the inverse image f 1 (V ) of an open set V is open.

Continuity in terms of inverse images of open sets We can use the idea of inverse images to rewrite continuity at a point p this way. ɛ > 0 δ > 0 x B δ (p) f (x) B ɛ (f (p)) or ɛ > 0 δ > 0 f (B δ (p)) B ɛ (f (p)) or ɛ > 0 δ > 0 B δ (p) f 1 (B ɛ (f (p))) Theorem f is continuous on the set D if and only if f 1 (V ) is an open set whenever V is open.

Continuity in terms of inverse images of open sets Proof ( ): We assume f is continuous on the set D. Let V in the range of f be an open set. By the arguments we have already shown you earlier, we know every point x 0 f 1 V is an interior point and so f 1 (V ) is an open set. ( ): We assume f 1 (V ) is an open set when V is open. Since V is open, given y 0 in V, there is a radius R so that B R (y 0 ) is contained in V. Choose any ɛ > 0. If ɛ R, argue this way: The set B R (y 0 ) is open and so by assumption, f 1 (B R (y 0 )) is open. Thus there is an x 0 f 1 (B R (y 0 )) so that y 0 = f (x 0 ). By assumption, x 0 must be an interior point of f 1 (B R (y 0 )). So there is a radius r so that B r (x 0 ) is contained in f 1 (B R (y 0 )). So x B r (x 0 ) f (x) B R (y 0 ). Said another way, x x 0 < r f (x) f (x 0 ) < R ɛ.

Continuity in terms of inverse images of open sets Proof On the other hand, if ɛ < R, we can argue almost the same way: instead of using the set B R (y 0 ), we just use the set B ɛ (y 0 ) which is still open. The rest of the argument goes through as expected! Comment Note here we are explicitly talking about continuity on the full domain not just at one point! This suggests how to generalize the idea of continuity to very general situations. A topology is a collection of subsets T of a set of objects X which is closed under arbitrary unions, closed under finite intersections and which contains both the empty set and the whole set X also. Our open sets in R as we have defined them are a collection like this. In this more general setting, we simply define the members of this set T to be what are called open sets even though they may not obey our usual definition of open.

Continuity in terms of inverse images of open sets Comment (Continued): Now the set X with the topology T is called a topological space and we denote it by the pair (X, T ). If (Y, S) is another topological space, based on what we have seen, an obvious way to generalize continuity is to say f : Ω X Y is continuous on Ω means f 1 (V ) is in T for all V S. Again, inverse images of open sets are open is a characterization of continuity. Note a set of objects can have many different topologies so continuity depends on the choice of topology really. For us we always use the topology of open sets in R which is nice and comfortable!

Continuity in terms of inverse images of open sets Homework 17 17.1 Prove lim x 1 14x + 25 exists. using an ɛ δ argument. 17.2 Prove lim x 2 (3x + 2)/(4x 2 + 8) exists using an ɛ δ argument. 17.3 f (x) = 3x 2 5 on [ 3, 6]. Find f ([ 3, 6]) which requires you to find the minimum and maximum of f on this domain. f (x) = 1/x 2 on (0, ). Find f (0, ). Note this is an interval. f (x) = 4x 4 + 35x 2 + 10 on (0, 4). Is f (0, 4) an interval? f (x) = 5x 7 23x 2 + 85 on ( 1, 6). Is f ( 1, 6) compact? This is meant to be a thought provoking question. In general, this is hard to answer. For example, if f (x) = sin(x), on ( π, π), we know the image f (( π, π)) is an interval. The domain is not compact, so we are NOT guaranteed that the image is compact. But here, the maximum occures at +1 and the minimum occurs at 1. So the range f (( π, π)) = [ 1, 1] which is compact even though the domain is not compact. So to answer this question, you need information about the minimum and maximum of this function. So graph it in MatLab, Sage etc and see what you find.