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Chapter 5 Functions on Metric Spaces and Continuity

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Chapter 5 Functions on Metric Spaces and Continuity Chapter 5 Functions on Metric Spaces and Continuity When we studied real-valued functions of a real variable in calculus, the techniques and theory built on properties of continuity, differentiability, and integrability. All of these concepts are de¿ned using the precise idea of a limit. In this chapter, we want to look at functions on metric spaces. In particular, we want to see how mapping metric spaces to metric spaces relates to properties of subsets of the metric spaces. 5.1 Limits of Functions Recall the de¿nitions of limit and continuity of real-valued functions of a real vari- able. De¿nition 5.1.1 Suppose that f is a real-valued function of a real variable, p + U, and there is an interval I containing p which, except possibly for p is in the domain of f . Then the limit of f as x approaches p is L if and only if 1 0 " 2= == 0 F 1x0 x p =" f x L . In this case, we write lim f x L which is read as “the limit of f of x as x xp approaches p is equal to L.” De¿nition 5.1.2 Suppose that f is a real-valued function of a real variable and p + dom f .Thenf iscontinuous at p if and only if lim f x f p. xp 183 184 CHAPTER 5. FUNCTIONS ON METRIC SPACES AND CONTINUITY These are more or less the way limit of a function and continuity of a function at a point were de¿ned at the time of your ¿rst encounter with them. With our new terminology, we can relax some of what goes into the de¿nition of limit. Instead of going for an interval (with possibly a point missing), we can specify that the point p be a limit point of the domain of f and then insert that we are only looking at the real numbers that are both in the domain of the function and in the open interval. This leads us to the following variation. De¿nition 5.1.3 Suppose that f is a real-valued function of a real variable, dom f A, and p + A) (i.e., p is a limit point of the domain of f ). Then the limit of f as x approaches p is L if and only if 1 0 " 2= =d 0 e 1xx + A F 0 x p =" f x L b c Example 5.1.4 Use the de¿nition to prove that lim 2x2 4x 1 31. x3 Before we offer a proof, we’ll illustrate some “expanded ”scratch work that leads to the information needed in order to offer a proof. We want to show that, ncorrespondingb toc eachn 0 we can ¿nd a = 0 such that 0 x 3 =" n 2x2 4x 1 31n . The easiest way to do this is to come up with a = that is a function of . Note that nr s n n n n n n n n 2x2 4x 1 31n n2x2 4x 30n 2 x 3 x 5 . The x 3 is good news because it is ours to make as small as we choose. But if we restrict x 3 there is a corresponding restriction on x 5 to take care of this part we will put a cap on = which will lead to simpler expressions. Suppose that we place a 1st restriction on = of requiring that = n 1.If= n 1, then 0 x 3 = n 1 " x 5 x 3 8 n x 3 8 9.Now nr s n n n n 2x2 4x 1 31n 2 x 3 x 5 2 = 9 n Q R whenever = n . To get both bounds to be in effect we will take = max 1 . 18 18 This concludes that “expanded ”scratch work. Q R Proof. For 0,let= max 1 .Then 18 0 x 3 =n 1 " x 5 x 3 8 n x 3 8 9 5.1. LIMITS OF FUNCTIONS 185 and nr s n n n n 2x2 4x 1 31n 2 x 3 x 5 2 = 9 n 18 . 18 Since 0 was arbitrary, we conclude that, for every 0,thereexistsa= Q R nb c n min 1 0, such that 0 x 3 =" n 2x2 4x 1 31n i.e., b 18 c lim 2x2 4x 1 31. x3 b c Excursion 5.1.5 Use the de¿nition to prove that lim x2 5x 6. x1 Space for scratch work. A Proof. ***For this one, the = that you de¿ne will depend on the nature of the ¿rst restriction that you placed on = in order to obtain a nice upper bound on x 6 if you chose Q R = n 1 as your ¿rst restriction, then = min 1 would have been what worked 8 in the proof that was offered.*** You want to be careful not to blindly take = n 1asthe¿rst restriction. For 1 example, if you are looking at the greatest integer function as x , you would 2 1 need to make sure that = never exceeded in order to stay away from the nearest 2 1 “jumps” if you have a rational function for which is a zero of the denominator 2 1 1 and you are looking at the limit as x , then you couldn’t let = be as great at 4 4 186 CHAPTER 5. FUNCTIONS ON METRIC SPACES AND CONTINUITY 1 so you might try taking = n as a ¿rst restriction. Our next example takes such a 6 consideration into account. x2 3 Example 5.1.6 Use the de¿nition to prove that lim 4. x12x 1 Space for scratch work. | } 1 2 1 Proof. For 0,let= min .From0 x 1 =n , we have 4 25 4 that 1 25 x 7 x 1 6 x 1 6 6 4 4 and n n n n n n t u n n n n n n n 1n n 1n n 1n 1 1 2x 1 2 nx n 2 nx 1 n o 2 n x 1 n 2 . 2 2 2 4 2 Furthermore, t u nt u n n n = 25 n 2 n n 2 n n x 3 n nx 8x 7n x 1 x 7 4 n 4n n n 2x 1 2x 1 2x 1 1 2 25 = 25 2 n . 2 2 25 | } 1 2 Since 0 was arbitrary, for every 0 there exists a = min nt u n 4 25 n 2 n n x 3 n 0 such that 0 x 1 =implies that n 4n that is, 2x 1 x2 3 lim 4. x1 2x 1 5.1. LIMITS OF FUNCTIONS 187 In Euclidean U space, the metric is realized as the absolute value of the differ- ence. Letting d denote this metric allows us to restate the de¿nition of lim f x xp L as 1 0 " 2= = 0 d e 1x x + A F 0 d x p =" d f x L . Of course, at this point we haven’t gained much this form doesn’t look particu- larly better than the one with which we started. On the other hand, it gets us nearer to where we want to go which is to the limit of a function that is from one metric space to another–neither of which is U1.Asa¿rst step, let’s look at the de¿nition when the function is from an arbitrary metric space into U1.Againweletd denote the Euclidean 1metric. De¿nition 5.1.7 Suppose that A is a subset of a metric space S dS and that f is a function with domain A and range contained in U1 i.e., f : A U1. then “f tends to L as x tends to p through points of A” if and only if (i) p is a limit point of A, and (ii) 1 02= = 01xx + A F 0 dS x p= " d f x L . In this case, we write f x Lasx p for x + A, or f x Las x p, or A lim f x L xp x+A Example 5.1.8 Let f : F U be given by f z Re z.Provethat lim f z 3 z3i z+F Space for scratch work. 188 CHAPTER 5. FUNCTIONS ON METRIC SPACES AND CONTINUITY For this one, we will make use of the fact that for any complex number ? , Re ? n ? . Proof. For 0,let= . Then 0 z 3 i = implies that f z 3 Re z 3 Re z 3 i n z 3 i . Since 0 was arbitrary, we conclude that lim f z 3. z3i z+F Remark 5.1.9 Notice that, in the de¿nition of lim f x L, there is neither a xp x+A requirement that f be de¿ned at p nor an expectation that p be an element of A. Also, while it isn’t indicated, the = 0 that is sought may be dependent on p. Finally we want to make the transition to functions from one arbitrary metric space to another. De¿nition 5.1.10 Suppose that A is a subset of a metric space S dS and that f is a function with domain A and range contained in a metric space X dX i.e., f : A X. Then “ f tends to L as x tends to p through points of A” if and only if (i) p is a limit point of A, and (ii) 1 02= = 0 1x x + A F 0 dS x p = " dX f x L .
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