Chapter 7. Continuity

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1 Chapter 7 Continuity There are many processes and eects that depends on certain set of variables in such a way that a small change in these variables acts as small change in the process. Changes of this type are commonly called continuous. As an example we can consider the growth of a living organism or dilation of heated wire. It is common to perform a computation of value of a function in an approximate argument. Doing this we assume that the values of the function obtained by such a manner are close enough to the exact values. However, there are function that do not obey such a nice property. For example, one cannot approximate values of the Dirichlet function in an irrational number by its values in an (arbitrary close) rational number. Therefore we are naturally forced to distinguish between two classes of functions with respect to the above mentioned property of continuity. We shall introduce the notion of continuous function (mapping). The notions of "small change" and so on will be replaced by exact notion of a neighborhood. In the following chapters we will draw our attention mainly to the real-valued functions of one or more real variables, however, this chapter will consider also the mappings dened on a subset of R m with values in R k. Let us mention that the proofs of the theorems considering such the mappings are essentially not dierent from the case of real-valued functions. Besides this, we shall prove that the problem of continuity of such a mapping is equivalent with the problem of continuity of the set of k real-valued functions each of them depending on m real variables. In spite of this fact we recommend to the beginners to formulate all the notions and prove relevant theorems in the case of real-valued function of one real variable. On the other hand, the experienced readers could try to generalize some notions and/or results to the cases of the mappings dened on a subset of a topological space and with values in a Hausdor topological space. Through this chapter the symbol R s, for s N means, as usually, Hausdor topological 55

2 7.1 Continuity of a mapping at a point 56 space equipped with the metric topology generated by the Euclidean metric d s d s (x, y) =ÌsXj=1(x j y j ) 2 for x = (x 1, x 2 m..., x s ) R s, y = (y 1, y 2,..., y s ) R s. 7.1 Continuity of a mapping at a point Denition Let f : (M R m ) R k. We say that the mapping f is continuous at a point a M, if for any neighborhood O(f(a)) of the point f(a) exists such a neighborhood O(a) of the point a that for all x in M O(a) we have f(x) O(f(a)). We write down (we read: the limit of f in a equals f(a)) In logical symbolic we write down: lim f(x) = f(a). x a lim f(x) = f(a) := O(f(a)) O(a) x M O(a) : f(x) O(f(a)). x a The set M O(a) is the neighborhood of the point a in a relative topology T r on M (see section 5.2). Thus the set M R m can be understood in denition as an topological subspace (M, T ) of the topological space (R m, T dm ). Let us write down another and often used denition of continuity of a mapping f : (M R m ) R k in a point a: Denition We say that the mapping f : (M R m ) R k is continuous at a point a M if the following holds ɛ R + δ R + x M : (d m (x, a) < δ d k (f(x), f(a)) < ɛ). The equivalency of the denitions and follows from the fact that any neighborhood of a point in R s contains some open ball centered at this point. Simultaneously, this ball is a neighborhood of its own center (see section 5.2). In the case k = 1 we have the notion of continuity of a real-valued function of many real variables. If m = k = 1 we have simplest case: continuity of a real-valued function of one real variable. Let us write denition in explicit forms for these two important cases: k = 1 the mapping f : (M R m ) R is said to be continuous at a point a M if: ɛ R + δ R + x M : (d m (x, a) < δ f(x) f(a) < ɛ). 56

3 7.1 Continuity of a mapping at a point 57 k = m = 1 the mapping f : (M R) R is said to be continuous at a point a M if: ɛ R + δ R + x M : ( x a < δ f(x) f(a) < ɛ). The geometrical interpretation of the notion of continuity of a real-valued function of one real variable in a point is given in g y f(a) + ɛ f(x) f(a) f(a) ɛ O a δ x a + δ x Figure 7.1: Let us examine the notion of continuity in a point in more detail. Let us consider the situation when a is an isolated point of the set M R m, where the mapping f is dened. Then the mapping f is continuous at a. In fact, a is isolated point in M implies that there exists such a neighborhood O 0 (a) of a that O 0 (a) M = {a}. Therefore for any x M O 0 (a) we have f(x) = f(a) O(f(a)) for arbitrary neighborhood of the point f(a), this simply means that f is continuous at a. This example shows us that the essential part of the denition of continuity of a mapping in a point a lies in the nontrivial case when the point a is limit point of the denition domain of the mapping f. It also follows the above mentioned that if a mapping is not continuous at given point then this point is limit point belonging to the denition domain of considered mapping. Moreover, denition implies: the mapping f : (M R m ) R k is continuous at a point a M if and only if the restriction f O(a) M, where O(a) is a neighborhood of the point a, is continuous at a. 57

4 7.1 Continuity of a mapping at a point 58 THEOREM Let f : (M R m ) R k and let a M be limit point of the set M. The mapping f is continuous at a if and only if lim f(x) = f(a). (7.1) x a Proof. a) Let the mapping f be continuous at a point a. Then it holds: This implies that O(f(a)) O(a) x M O(a) : f(x) O(f(a)). (7.2) O(f(a)) O(a) x M Ô(a) : f(x) O(f(a)), (7.3) thus (7.1) holds. b) Let lim f(a) = f(a). x a Then (7.3) holds. Since f(a) O(f(a)) for any neighborhood O(f(a)) of the point f(a), then (7.2) holds also, this means f is continuous at a. Example Let k R. Show that the constant function f : x k, x R is continuous at any a R. Solution: Any real number a is limit point of the set R and lim f(x) = k = f(a). x a Therefore, with respect to theorem 7.1.1, this function is continuous at a. Example Show that the function a) b) is continuous at 1. f : x x + [x], x M = {0, 1} [2, 5] g : x x, x R + 0 is continuous at any non-negative number a. 58

5 7.1 Continuity of a mapping at a point 59 Solution: a) The continuity of the considered function in the point 1 follows from the fact that 1 is isolated point of the denition domain of the function. b) Any non-negative number a is a limit point of the set D(g) = R + 0 and lim g(x) = a = g(a) x a (as shown in example ). With respect to the theorem 7.1.1, the function g is continuous at a. Example Let for x R \ {1} f : x 2x 0 for x = 1. Show that f is not continuous at 1. Solution: It is obvious that 1 is limit point of the set R = D(f) and lim f(x) = 2 f(1) = 0. x 1 Thus, theorem implies that f is not continuous at the point 1. Example Show that the function sin is continuous at any real a. Solution: Let ɛ > 0 and 0 < δ ɛ. Then for any x such that x a < δ holds: + a a a x a sin(x) sin(a) =2 cos x 2 sinx 2 2sinx x a < δ ɛ. And this means that f is continuous at a. Example Show that the function = f : (x 1, x 2 ) È x 1 x 2, (x 1, x 2 ) R 2 is continuous at a = (0, 0). Solution: For any (x 1, x 2 ) R 2 the inequality holds. Therefore x 1 x 2 x2 1 + x È x 1 x Èx x 2 2 = 1 2 d 2 (x, a), x = (x 1, x 2 ). Let ɛ be any positive number and 0 < δ 2ɛ. Then for any x R 2 such that d 2 (x, a) < δ the dierence f(x) f(0) obeys the following inequality f(x) f(0) 1 2 d 2 (x, a) And this shows that f is continuous at (0, 0). 59 δ 2 ɛ.

6 7.1 Continuity of a mapping at a point 60 Example Show that each of the following functions: exp a, log a, a polynomial (in one as well as in many variables) and a rational function (in one as well as in many variables) is continuous at any point of its denition domain. Solution: The statement follows from the theorem and from the relevant examples discussed in previous chapter (chapter 6). Namely, see examples , , 6.3.9, , 6.7.5, to verify that each of listed functions is continuous at any point of its denition domain. THEOREM (On continuity of composition). Let g : (E R s ) R k and f : (M R m ) E and let a M. If f is continuous at a and g is continuous at f(a) then the composition g f is continuous at a. (This theorem is, in fact, reformulation of the implication 2 of theorem 6.9.4) THEOREM Let f : (M R m ) R is continuous at a M. Then the following two statements hold: 1. there exists such a neighborhood O 0 (a) of the point a that f is bounded in O 0 (a) M 2. if f(a) > 0 (< 0) then there exists such an neighborhood O 0 (a) of the point a that f is positive (negative) in O 0 (a) M THEOREM Let f and g be a pair of real-valued functions dened on a subset M of the Euclidean space R m. Then their continuity in a M implies continuity in a of each function of the following list: f, λf (where λ R), f + g, fg. If g(a) 0 then the function f/g is continuous at a also. The proofs of the theorems can be done using the proofs of the theorems on limits from the section 6.3. One has to realize that the punctured neighborhoods in R are to be replaced by the neighborhoods of the points in R m. As an example, we will show the proof of the statement 1 of the theorem Proof. (Statement 1 of the theorem 7.1.3) The assumption that f is continuous at a implies that for ɛ = 1 exists such an neighborhood O 0 (a) of the point a that This can be easily rewritten into the form x M O 0 (a) : f(x) f(a) < 1. f(a) 1 < f(x) < f(a) + 1. The system of the last two inequalities directly tells us that f is bounded in M O 0 (a). With help of the proof of the theorem one can easily prove the following theorem 60

7 7.1 Continuity of a mapping at a point 61 THEOREM The mapping f : (M R m ) R k is continuous at a M if and only if for any sequence {x n } n N M such that lim n x n = a we have lim f(x n) = f(a). n Example Show that function cos is continuous at any real number. Solution: Let us dene the translation in the real axis f : x π 2 + x, x R and consider the sine function g : x sin(x), x R. Then obviously (g f)(x) = sin π 2 + x =cos(x), x R. This means the identity g f = cos holds. Let a be arbitrary real number. The continuity of f in a (as already shown in example 6) and continuity of the sine function in entire real axis imply (with respect to the theorem 7.1.2) that their composition g f = cos is continuous at a, too. Example Let α R. Show that the power function x x α, x R + is continuous at any positive x 0. Solution: We will use example 6 in which we have already shown that the exponential function and the logarithmic function are continuous at their denition domains. Let us dene g : y exp(y), y R and f : x α ln(x), x R +. Then our power function can be expressed as the following composition of f and g: (g f)(x) = x a, x R +. Then the continuity of ln function in x 0 as well as the continuity of exp function in ln(x 0 ) imply that their composition g f is also continuous at x 0. Example Let a process go on in such a way that in any time t the speed of creation of given substance is proportional to the total rate (volume, quantity) of this substance in this time t. Find the dependence of the total rate of the quantity on time in such a process. Solution: Let q 0 be the total rate of given substance at the time t = 0. Let us divide the interval [0, t] into n-equal subintervals 0, t n,t n, 2t 1)t n,..., (n n, nt n. Let the number n be large enough to be able to consider the speed of creation of the substance in each of the subintervals is constant. With respect to our assumption this 61

8 7.1 Continuity of a mapping at a point 62 speed is proportional to the total rate of the substance. Let this constant be denoted as k, (k > 0). Then the total rates of the substance in time moments nœ t n, 2t n,..., nt n = t are given by t q 1 = q 0 + kq 0 n = q kt t q 2 = q 1 + kq 1 n = q kt t q n = q n 1 + kq n 1 n = q kt. nœn The setting of the problem implies that the process of the substance creation continuously depends on time t. In order to obtain an exact expression for the total rate of the substance at any given time within considered interval [0, t] one needs to perform the limit n. This will result in nœ2 q(t) = lim q n kt. nœn Now, let us compute the limit in question: Let and Then lim + n 1 kt. nœn g 1 : x (1 + x) kt x, x R +, f 1 : n kt n, n N. lim f 1(n) = 0 n (f 1 (n) 0, n N), and (g 1 f 1 )(n) = 1 + kt, n N. nœn The theorem implies that if g 1 has a limit at point 0 then n 1 lim + kt = lim g(x). nœn x 0 Actually, function g 1 has a limit in 0, let us show this. Let g = exp and f : x (kt ln(1+x) x for x R + kt for x = 0. 62

9 7.1 Continuity of a mapping at a point 63 Then (g f)(x) = g 1 (x) for all x R +. Since ln(1 + x) lim x 0 + x (example 4.6.5), we have lim f(x) = kt = f(0). x 0 Thus, the theorem tells us the function f is continuous at 0. Then, the continuity of the function g = exp in kt together with the theorem guarantee continuity of the composition g f in 0. The point 0 is limit point of the denition domain of g f and therefore theorem implies = 1 lim (g f)(x) = (g f)(0) = x 0 ekt, and lim g 1(x) = lim(g f)(x) = e kt = lim + kt. x 0 x 0 n 1 nœn So, we can conclude q(t) = q 0 e kt. (7.4) This shows that at given setup the total rate of the substance q in time t is given by the formula (7.4). The exponential dependence like (7.4) arises in many situations. Exponential growth (k > 0) or decay (k < 0) describes such important processes as radioactive decay, bacteria reproduction and so on. This shows great importance of the e-number in many applications of mathematics. Example Show that the function is continuous at any (a 1, a 2 ) R 2. Solution: Let us denote (x 1, x 2 ) cos(x x 2 2), (x 1, x 2 ) R 2 f : (x 1, x 2 ) x x 2 2, (x 1, x 2 ) R 2, and g = cos. Then their composition is: (g f) = cos(x x 2 2) for (x 1, x 2 ) R 2. The continuity of this composition in any (a 1, a 2 ) R 2 is guaranteed by the theorem because from the previous examples we know that both f and g are continuous at any value of their arguments. Example Let b = (b 1, b 2,..., b m ) R m. Show that the function x d m (x, b) (x j b j ) =ÌmXj=1 2, x = (x 1, x 2,..., x m ) R m 63

10 7.1 Continuity of a mapping at a point 64 is continuous at any point a = (a 1, a 2,..., a m ) R m. Solution: Let us denote f : x = (x 1, x 2,..., x m ) (x j b j ) mxj=1 2, x R m and g : y y, y R + 0. The function f is continuous at a as was shown in example 6 and lim f(x) = f(a). x a Further, lim g(y) =Èf(a) = g(f(a)). y f(a) and theorem tells us that g is continuous at the point f(a). And nally, theorem guarantees the continuity in a of the composition g f : R m R, (g f)(x) = d m (x, b) in a. Example Show that each of the functions: tan, cot, sinh a, cosh a, tanh a and coth a is continuous at any point of its denition domain. Solution: We will show that the function =[ kπ tan(x) : x sin(x) cos(x), x M k Z π 2 + kπ, π 2 + is continuous at any number x 0 from M. The functions sin cos are continuous at x 0 (this was shown in examples 4 and 7 respectively) and cos(x 0 ) 0, therefore theorem implies that their fraction - the function tan - is continuous at x 0. Now we show that the function cosh a : x ax + a x, x R 2 is continuous at any real x 0. This statement follows the following: the function exp a is continuous at x 0 (example 6) and for any x 0 R: exp a (x 0 ) > 0. In fact, this means that the functions 1 exp 2 a and exp a are continuous at x 0 and then also their linear combination cosh a = 1 2 exp a exp a is, due to theorem 7.1.4, continuous at x 0. In other cases, the continuity in the point of denition domain of given functions can be proven analogically. 64

11 7.1 Continuity of a mapping at a point 65 Let f : (M R m R k ). Let us denote by f i (x) the i-th component of the value f(x) for x M. Then f(x) = (f 1 (x), f 2 (x),..., f k (x)) for x M. All the functions f i are dened on M: f i : (M R m ) R, (i = 1, 2,..., k). Within this notation it is natural to understand the mapping f as k-tuple of functions dened over M. This is usually represented by f = (f 1, f 2,..., f k ). The following theorem will allow for reducing the problem of continuity of the mapping f in given point a M to the problem of continuity in a of the system of k real-valued functions dened over M. THEOREM Let f 1, f 2,..., f k be real-valued functions dened on M R m. Then the mapping f : M R k dened by f(x) = (f 1 (x), f 2 (x),..., f k (x)), x M, i.e. f = (f 1, f 2,..., f k ) is continuous at a M if and only if any of the functions f 1, f 2,..., f k is continuous at a. Proof. a) Let the mapping f be continuous at a. For all ɛ > 0 exists such δ > 0 that for x M such that d m (x, a) < δ the inequality d k (f(x), f(a)) < ɛ holds. Making use of the inequality: f(x) f(a) ÌkXj=1(f j (x) f j (a)) 2 = d k (f(x), f(a)) (i = 1, 2,..., k) we derive that the following statement is true: x M : (d m (x, a) < δ f i (x) f i (a) < ɛ), (i = 1, 2,..., k). This states exactly the continuity in the point a of each of the functions f 1, f 2,..., f k. b) Let us suppose the functions f 1, f 2,..., f k be continuous at a M. This means that the following statement is true: i {1, 2,..., k} ɛ > 0 δ i > 0 x M : (d m (x, a) < δ i ) f i (x) f i (a) < Now, let 0 < δ min{δ 1, δ 2,..., δ k }. Then for x M such that d m (x, a) < δ we have d k (f(x), f(a)) (f j (x) f j (a)) =ÌkXj=1 2 <sk ɛ2 k = ɛ. And the last inequality shows that the mapping f is continuous at the point a. 65 ɛ k.

12 7.1 Continuity of a mapping at a point 66 This theorem together with theorem imply the following Corollary Let f 1, f 2,..., f k, g 1, g 2,..., g k be real-valued functions dened on M R m and let they be continuous at a point a M. Then the mappings: λf (λ R) and f + g dened in the following way: (λf)(x) = (λf 1 (x), λf 2 (x),..., λf k (x)), x M, (f + g)(x) = (f 1 (x) + g 1 (x), f 2 (x) + g 2 (x),..., f k (x) + g k (x)) are continuous at a. Moreover, the function (f, g) : M R dened as (f, g)(x) = f j (x)g j (x), kxj=1 x M is also continuous at a. Example Show that the mapping f : R R 2 given by f(t) = (cos(t), sin(t)), t R is continuous at any real t 0. Solution: We can write the mapping in question as a pair f = (cos, sin) where the components of this mapping, i.e. the functions cos and sin are continuous at any real number. Then theorem implies that f is continuous at t 0. Example Show that the mapping is continuous at (0, 0). Solution: Let us denote: f : (x 1, x 2 ) (È x 1 x 2, cos(x x 2 2),Èx x 2 2), (x 1, x 2 ) R 2 f 1 : (x 1, x 2 ) È x 1 x 2, (x 1, x 2 ) R 2, f 2 : (x 1, x 2 ) cos(x x 2 2), (x 1, x 2 ) R 2, f 3 : (x 1, x 2 ) Èx x 2 2, (x 1, x 2 ) R 2. We have already shown (see examples 5, 10 and 11) that all the three functions f 1, f 2, f 3 are continuous at (0, 0), thus, with respect to theorem 7.1.6, the mapping f is continuous at (0, 0). 66

13 7.2 Criteria of continuity of a mapping at a point. Points of discontinuity 67 Problems 1. Show that the following functions: x x, x R; x ln(1 + x), x ( 1, 0] are continuous at the point 0. Further, show that the function: x sgn(x), x R is not continuous at the point Show that the function is continuous at Show that the functions x (sin(x) x for x R \ {0} 1 for x = 0 x x cos(x) x sin 2 (x), x R + 0 ; are continuous at any point of their denition domains. 4. Show that the function x e 2x cos x π 3, x R (x 1, x 2 ) È x 1 x 2 cos(x x 2 2) x 2 sin(x 1 x 2 ), (x 1, x 2 ) R 2 ic continuous at any point of R Show that the mapping is continuous at any positive real number. 6. Show that the mapping t (e t, e t cos(t), ln(t)), t R + (x 1, x 2 ) (x 1 x 2, e x 1x 2, sin(èx x 2 2), 1), (x 1, x 2 ) R 2 is continuous at any point of R Criteria of continuity of a mapping at a point. Points of discontinuity Let = S M R m and let f : M R k. If the mapping is continuous at a point a S, i.e. if f S : x f(x), x S O(f(a)) O(a) x S O(a) : f(x) O(f(a)) 67

14 7.2 Criteria of continuity of a mapping at a point. Points of discontinuity 68 holds, or equivalently, if the statement: ɛ R + δ R + x S : (d m (x, a) < δ d k (f(x), f(a)) < ɛ) is true, then we will note down this by lim f(x) = f(a). S x a In the case m = k = 1, i.e. in the case of real-valued function of one real variable, we say that f : (M R) R is continuous at a point a M from the right (left) if the function f M + a (M M a ), where is continuous at the point a, i.e. if M + a = {x M; x a} (M a = {x M; x a}) ɛ R + δ R + x M + a : ( x a < δ f(x) f(a) < ɛ) ( ɛ R + δ R + x M a holds. This statement can be rewritten into the form : ( x a < δ f(x) f(a) < ɛ)) ɛ R + δ R + x M : (a x < a + δ f(x) f(a) < ɛ) ɛ R + δ R + x M : (a δ < x a f(x) f(a) < ɛ) and write down accordingly with the notation in the section 6.9 as lim f(x) = f(a) ( x a + lim f(x) = f(a)). x a Now we will introduce conditions that will allow for deciding whether a mapping is or is not continuous at a given point. THEOREM (Necessary condition of continuity in a point). Let = S M R m. If the mapping f : M R k is continuous at a point a S. then the restriction of this mapping f S is continuous at a, i.e. lim f(x) = f(a) lim f(x) = f(a). x a S x a THEOREM (Necessary and sucient condition of continuity in a point). Let f : (M R m ) R k and let S 1 and S 2 be such subsets in M that S 1 S 2 = M. Let a S 1 as well as a S 2. The mapping f is continuous at a if and only if both of its restrictions f S1 and f S2 are continuous at a. 68

15 7.2 Criteria of continuity of a mapping at a point. Points of discontinuity 69 Corollary Let f : (M R) R and let a M. The function f is continuous at a if and only if f is continuous at a from the left as well as from the right. The theorems and the corollary can be proven in a way analogical to that one used to prove the theorems from the section 6.4 (on the other hand, they follows from the theorems and ). Example Show that the function for x {0, 1, 2} f : x x 2 + ln(x) for x [1, 9) is continuous at 1. Solution: Let M := D(f). Then M1 = {1, 0, 1, 2}, M 1 + = [1, 9). The equalities: and lim f(x) = x 1 + lim f(x) = lim 2 = 2 = f(1) x 1 M 1 x 1 lim (1 + x 2 + ln(x)) = 2 = f(1) M + 1 x 1 imply (with respect to theorem 7.2.2) that f is continuous at 1. Example Show that the function f : (x 1, x 2 ) x 1 for (x 1, x 2 ) {(y 1, y 2 ) R 2 ; y 2 = y 1 > 0} x 1 + x 2 for (x 1, x 2 ) {(y 1, y 2 R 2 ; y 1 0, y 2 0)} is continuous at (0, 0). Solution: Let S 1 = {(x 1, x 2 ) R 2 ; x 2 = x 1 > 0} and S 2 = {(x 1, x 2 ) R 2 ; x 1 0, x 2 0)}. Let f 1 : (x 1, x 2 ) x 1, (x 1, x 2 ) R 2, f 2 : (x 1, x 2 ) x 1 + x 2, (x 1, x 2 ) R 2. The functions f 1, f 2 are continuous at (0, 0) and therefore their restrictions f 1 S1 and f 2 S2 are continuous at (0, 0), too. This means that the functions f S1 and f S2 are continuous at (0, 0) because f S1 = f 1 S1 and f S2 = f 2 S2. Obviously, S 1 S 2 = D(f). Therefore theorem guarantees continuity of f in the point (0, 0). Points of discontinuity Let f : (M R m ) R k. A point a M, in which the mapping f is not continuous, is called point of discontinuity of the mapping f. If a mapping f : (M R m ) R k is not continuous at a M then O(f(a)) O(a) x M O(a) : f(x) / O(f(a)) holds. Furthermore, the point of discontinuity is necessarily the limit point of the set M. So, we can speak about the limit of the mapping f in the point a. With respect to this and 69

16 7.2 Criteria of continuity of a mapping at a point. Points of discontinuity 70 theorem we obtain: A mapping f : (M R m ) R k is not continuous at a point a M if and only if the point a is limit point of the set M and the mapping f has no limit in the point a or if f has a limit in the point a then this limit is dierent from f(a). Example Show that the function f : (x 1, x 2 ) 1 for (x 1, x 2 ) S 1 = {(x 1, x 2 ) R 2 ; x x 2 2 > 1} 1 x 2 1 x 2 2 for (x 1, x 2 ) R 2 \ S 1 is not continuous at any point of the circle x x 2 2 = 1. Solution: Let S 2 = R 2 \ S 1 and let a = (a 1, a 2 ) be a point of the circle x x 2 2 = 1. The point a is limit point of both sets S 1 and S 2 and therefore we can compute the limits: and lim f(x 1, x 2 ) = 1, S 1 (x 1,x 2 ) (a 1,a 2 ) lim f(x 1, x 2 ) = lim (1 S 2 (x 1,x 2 ) (a 1,a 2 ) S 2 (x 1,x 2 ) (a 1,a 2 ) x2 1 x 2 2) = 0. We see that f has no limit at a and we conclude f is not continuous at a. In the case of real-valued functions we distinguish between two kinds of points of discontinuity. Let f : (M R) mbbr. A point a M is called point of discontinuity of the rst kind of the function f if : a) point a is limit point of the sets M a + and Ma and the function f has a limit from the left in a as well as it has a limit from the right in a, at least one of these limits is dierent from f(a), or b) the point a is limit point of the set M + a (M a ) and the function f has a limit from the right (left) in a and this limit diers from f(a). Point of discontinuity of a real-valued function of one real variable that is not of the rst kind is called point of discontinuity of the second kind. Example Let for x = g : x 1 m [0, 1] Q, where m is in the simple form n n n 0 for x [0, 1] \ Q. This function is commonly called Riemann function. Show that this function is continuous at any irrational number in the interval [0, 1] and that it has point of discontinuity of the rst kind in any rational number in the interval [0, 1]. 70

17 7.2 Criteria of continuity of a mapping at a point. Points of discontinuity 71 Solution: Let a be any number in [0, 1]. Let ɛ be arbitrary positive number. There is only a nite number of fractions m in simple form such that n 1/ɛ. This implies that there is n a truncated δ-neighborhood of a that contains rational numbers (in simple form) with the denominator greater than 1/epsilon. Thus, for x [0, 1], 0 < x a < δ we have g(x) < ɛ. This means that in any point a from [0, 1] the Riemann function g has a limit and this limit is given by lim g(x) = 0. x a Now, if a is an irrational number then g(a) = 0, thus g is continuous at a. On the other hand, if a is a rational number, then g(a) 0 and the Riemann function is not continuous at a. Existence of the limit in a means that this point of discontinuity is of the rst kind. Example Show that the so-called Dirichlet function: for x Q χ : x 1 0 for x R \ Q has point of discontinuity of the second kind in any real number. Solution: Example tells us that the χ-function has no limit in a real number and therefore it has in any given real number the point of discontinuity of the second kind. Example Let for x R f : x 8><>:sin(1/x) + 0 for x = 0 1 for x ( 5, 0) Show that the point 0 is the point of discontinuity of the second kind of the function f. Solution: The function f evidently has no limit from the right in the point 0, therefore the point 0 is the point of discontinuity of the second kind if f. Let us note that f has a limit from the left in 0 - it is simply 1. Example Let f : [0, 1] R be dened by: f(x) = 1 x for x (0, 1] 1 for x = 0. Show that 0 is the point of discontinuity of the second kind of this function. Solution: Obviously, there is no limit of the function f in the point 0, and this implies that the point 0 is the point of discontinuity of the second kind of the function f. Problems 1. Show that the function x ( 1) [x], x R is continuous at any point of the set R \ Z and that it has point of discontinuity of the rst kind in any integer. 71.

18 7.2 Criteria of continuity of a mapping at a point. Points of discontinuity Show that the function given by: x x x for x Q for x R \ Q is continuous at the point 0 and that it has point of discontinuity of the second kind in any other number. 3. Prove that the functions: and x sgn(x), x R x (sin(x) x for x R \ {0} 33 for x = 0 are continuous at any real number except 0 and that they have point of discontinuity of the rst kind in Prove that the functions: and 1 x 2, (x x (x 1, x 2 ) (x 2 1, x 2 ) R 2 \ {(0, 0)} 1 +x2 2 0, (x 1, x 2 ) = (0, 0) (x 1, x 2 ) 8<:x2, (x 1, x 2 ) R 2 \ {(0, 0)} 0, (x 1, x 2 ) = (0, 0) 1 x 2 x 4 1 +x4 2 are discontinuous at the point (0, 0) and continuous at any other point of the Euclidean plane. 5. Show that the function (x 1, x 2 ) 1, (x 1, x 2 ) S = {(y 1, y 2 ) R 2 ; y 1 y 2 = 0} 0, (x 1, x 2 ) R 2 \ S is discontinuous at any point of the set S and is continuous at any point of the set R 2 \ S. 6. Prove that the mapping is not continuous at the point Let and x (x + 1,sgn(x)), x R 1 x 2, (x x f 1 : (x 1, x 2 ) (x 2 1, x 2 ) R 2 \ {(0, 0)} 1 +x2 2 0, (x 1, x 2 ) = (0, 0) f 2 : (x 1, x 2 ) x 1 x 2, (x 1, x 2 ) R 2 and let f : R 2 R 2 be given by f = (f 1, f 2 ). Prove that the mapping f is not continuous at the point (0, 0). 72

19 7.3 Continuity of a mapping on a set Continuity of a mapping on a set Denition We say that the mapping f : (M R m ) R k is continuous on M if f is continuous at any point of M. The set of all continuous mappings from M to R k is denoted by C(M; R k ) or briey by C(M). The examples 7.1.4, 7.1.6, 7.1.7, 7.1.8, show us that the functions sin, cos, exp a, log a, tan, cot, sinh a, cosh a, tanh a, coth a, a power function, a polynomial and a rational function of one or many real variables are continuous on their denition domains. Moreover, example and theorem imply that the functions (x 1, x 2,..., x m ) x i, (x 1, x 2,..., x m ) R, (i = 1, 2,..., m) are continuous on R m. The function x d m (x, b), x R m, where b is a point of R m, is continuous on R m as shown in example The following theorem is a direct consequence of the theorem 7.1.6: THEOREM Let f 1, f 2,..., f k be real-valued functions dened on a set M R m. The mapping f : (M R m ) R k given by f = (f 1, f 2,..., f k ) is continuous on M if and only if each of the functions f 1, f 2,..., f k is continuous on M. This theorem and example imply that the mapping f : R R 2, f = (cos, sin) is continuous on R, i.e. f C(R, R 2 ). Example Let a = (a 1, a 2,..., a m ), b = (b 1, b 2,..., b m ) be two points of R m. Let the mapping ϕ : [0, 1] R m be dened by the prescription: ϕ(t) = bt + a(1 t) for t [0, 1], i.e. ϕ = (ϕ 1, ϕ 2,..., ϕ m ) where ϕ i : t b i t + a i (1 t), t [0, 1] (i = 1, 2,..., m). Show that the mapping ϕ C([0, 1]; R m ). Solution: The functions ϕ 1, ϕ 2,..., ϕ m are continuous on [0, 1], consequently (with respect to theorem 7.3.1) the mapping ϕ is continuous on [0, 1], i.e. ϕ C([0, 1]; R m ). Example Let a ij, i = 1, 2,..., k; j = 1, 2,..., m be given real jž numbers. Prove that the linear mapping l : R m R k dened by l(x 1, x 2,..., x m ) a 1j x j, a kj x = mxj=1 73 a 2j x j,..., mxj=1 mxj=1

20 7.3 Continuity of a mapping on a set 74 is continuous on R m. Solution: The polynomials in m-variables: (x 1, x 2,..., x m ) a 1j x j, (i = 1, 2,..., k) mxj=1 are continuous on R m, thus, with respect to theorem 7.3.1, the mapping l is continuous on R m, i.e. l C(R m ; R k ). Problems 1. Show that the function x ln(x x 2 ), x R is continuous on R. 2. Prove that the function x x cos(x) + e x sin(x) + 3x 1, x R is continuous on R. 3. Show that the function is continuous on R Prove that the function (x 1, x 2, x 3 ) e x 2 1 +x 2 2 +x2 3 + x3 sin(x 1 x 2 ), (x 1, x 2, x 3 ) R 3 (x 1, x 2, x 3 ) ( x 1 x 2 x 3 x 2 1 +x2 2 +x2 3, (x 1, x 2, x 3 ) R 3 \ {(0, 0, 0)} 0, (x 1, x 2, x 3 ) = (0, 0, 0) is continuous on R Let M R m and f, g C(M; R). Show that then the following functions: x max{f(x), g(x)}; x min{f(x), g(x)} are also from C(M; R). 6. Let f C(R; R). Show that the mapping t (t, f(t)), t R is continuous on R. 7. Show that any mapping Z Z R 5 is continuous on Z Z. 8. Show that the mapping (x 1, x 2 ) (x 2 1 x 2 2,È x 1 x 2, cos(x x 2 2), 3), (x 1, x 2 ) R 2 belongs to C(R 2, R 4 ). 9. Let M R m and let (B(M), d B ) be a metric space dened in example Let BC(M; R = B(M) C(M; R). Prove that (BC(M; R), d B ) is complete metric space. (Hint: use of example ) 74

21 7.4 Uniformly continuous mappings Uniformly continuous mappings Let a mapping f : (M R m ) R k be continuous on M. This means f is continuous at each point of M: a M ɛ > 0 δ > 0 x M : (d m (x, a) < δ d k (f(x), f(a)) < ɛ). In this denition the number δ may depend upon choice of the point a except its dependence on ɛ. (δ may vary from point to point without change in ɛ). However, there are such mappings for which δ does not depend on a M. These mappings form an important subset of all continuous mappings and are called: uniformly continuous. Let us write down the denition: Denition We say that a mapping f : (M R m ) R k is uniformly continuous on M if ɛ > 0 δ > 0 x M, y M : (d m (x, y) < δ d k (f(x), f(y)) < ɛ). holds. In the special case m = k = 1 we obtain: A real-valued function of one real variable f : (M R) R is uniformly continuous on M if ɛ > 0 δ > 0 x M, y M : ( x y < δ f(x) f(y) < ɛ). THEOREM If a mapping f : (M R m ) R k is uniformly continuous on M then it is continuous on M. Proof. Let ɛ be positive and δ > 0 be such that for any pair x, y M obeying d m (x, y) < δ we have: d k (f(x), f(y)) < ɛ. Let a be a point of the set M. Then for all x M such that d M (x, a)δ we have d k (f(x), f(a)) < ɛ and this means continuity of f in a. As a was an arbitrary point of M we have proved that f is continuous on M. If a mapping f : (M R m ) R k is continuous, it has not to be uniformly continuous. This is shown in the following example. Example Let x 1, x (0, 1). x Show that this continuous function on (0, 1) is not uniformly continuous on (0, 1). Solution: Let ɛ > 0. Let δ be arbitrary positive number. Let us consider x 1 (0, 1) (0, δ) and dene x 2 = 75 x (2 + ɛ)x 1.

22 7.4 Uniformly continuous mappings 76 Then x 2 (0, δ), x 2 < x 1 < δ and x 1 x 2 < δ. However, 1 x 1 1 x 2=2 + ɛ > ɛ. The last inequality, since δ was considered as arbitrary positive number, veries that our function is not uniformly continuous on (0, 1). Example Let a, b be real numbers such that a < b. Prove then that the function f : x x 2, x [a, b] is uniformly continuous on [a, b]. Solution: Let ɛ > 0. Let x, y [a, b] be such that x y < δ with some positive δ. Then: f(x) f(y) = x 2 y 2 = x y x + y δ x + y 2δ( a + b ). So, if we choose (for given ɛ > 0) as: δ < ɛ 2( a + b ) we obtain: f(x) f(y) < ɛ. We have shown f is uniformly continuous on [a, b]. Example Show that the function f : x x 2, x [0, ) is not uniformly continuous on [0, ). Solution: Let us x ɛ > 0. Let δ be arbitrary positive number. Then: for x = n and y = n + 3ɛ, with n > 9ɛ 2 /δ 2, we obtain At the same time x y = n + 3ɛ n = f(x) f(y) = 3ɛ > ɛ. 3ɛ n + n + 3ɛ < This implies that f is not uniformly continuous on [0, ). Example Prove that the function f : (x, y) 1 + x + y, (x, y) R 2 is uniformly continuous on R 2. Solution: Let ɛ be arbitrary positive number. Let A = (x, y) R 2, B = (x, y ) R 2. 3ɛ 2 n < δ 2 < δ. 76

23 7.4 Uniformly continuous mappings 77 Then f(a) f(b) = 1 + x + y 1 x y x x + y y 2d 2 (A, B). So we see that if we choose 0 < δ ɛ/ 2, we obtain, for any pair A, B such that d 2 (A, B) < δ, that f(a) f(b) < ɛ. And this demonstrates that the linear function f is uniformly continuous on R 2. THEOREM Let f 1, f 2,..., f k be real-valued functions dened on M R m. The mapping f : M R k dened by f(x) = (f 1 (x), f 2 (x),..., f k (x)), x M is uniformly continuous on M if and only if each of the functions f 1, f 2,..., f k is uniformly continuous on M. Proof. This theorem can be easily proved in the same way as theorem Theorem implies that uniform continuity of the mapping f = (f 1, f 2,..., f k ) : (M R m ) R k is equivalent with the uniform continuity on the set M of k-tuple of real-valued functions f 1, f 2,..., f k. Problems 1. Prove that the functions x x, x R + 0 ; x x x 2, x R are uniformly continuous on their denition domains. 2. Find the functions f : (M R) R, g : (M R) R such that: they are uniformly continuous on M and their product fg is not uniformly continuous on M. 3. Prove that the functions x sin1 x, x (0, 1); x cos(x 2 ), x R + are not uniformly continuous on their denition domains. 4. Show that the function is uniformly continuous on [0, 1] ( 1, 2). (x 1, x 2 ) x 1 x 2, (x 1, x 2 ) [0, 1] ( 1, 2) 77

24 7.5 Properties of uniformly continuous mappings on compact sets78 5. Prove that the function is not uniformly continuous on R Show that the mappings (x 1, x 2 ) x 1 x 2, (x 1, x 2 ) R 2 t (cos(t), sin(t)), t R; t (1, t, t 2 ), t [0, 1] are uniformly continuous on their denition domains. 7. Prove that the mapping (x 1, x 2 ) ( x 1, x 1 + x 2 ), (x 1, x 2 ) R R + 0 is uniformly continuous on (x 1, x 2 ) R R Prove that the mapping is not uniformly continuous on R Show that the mapping (x 1, x 2 ) (1, x 1 x 2 ), (x 1, x 2 ) R 2 (x 1, x 2, x 3 ) (x 1, x 2 2, x 3 3), (x 1, x 2, x 3 ) Z Z Z is uniformly continuous on Z Z Z. 10. Show that the linear map l : R m R k dened in example is uniformly continuous on R m. Answers 2 For example the functions f = g : x x, x R Properties of uniformly continuous mappings on compact sets In general it is not necessary that a mapping f : (M R m ) R k continuous on M must be bounded on M, must attain its maximum value or minimum value, and must be uniformly continuous on M. For example, the function considered in example 7.4.1: f : x 1, x (0, 1) x 78

25 7.5 Properties of uniformly continuous mappings on compact sets79 is continuous on (0, 1), however, it is unbounded, does not reach maximal neither minimal value and is not uniformly continuous on (0, 1). We will show that if M R m is a compact set, then a function f C(M; R) is necessarily bounded, uniformly continuous on M and also reaches its maximal and minimal value on M. THEOREM Let M R m be a compact set and let a mapping f : M R k be continuous on M. Then the range of the mapping f, i.e. the set f(m), is a compact subset of R k. Proof. Let {S a } a I, where I is an index set, be an open covering of f(m). Then: and τ M α τ I : f(τ) S ατ [ f(m) S ατ S α. α τ M [α I f is continuous on M, so it is continuous at τ, thus for S ατ exists such a neighborhood O(τ) of the point τ that for all x M O(τ) we have f(x) S ατ, i.e. f(m O(τ)) S ατ. We can assume that O(τ) is an open ball centered at τ. It is self-evident that [ M O(τ). τ M Compactness of M allow for choosing a nite sub-covering: M O(τ i ). n[i=1 Now, M = M O(τ i ). n[i=1 However, f(m O(τ i )) S ατi, for i = 1, 2,..., n. Thus f(m) S ατi S α. n[i=1 [α I {S ατi } i {1,2,...,n} is a nite sub-covering of the covering {S α } α I. This means that f(m) is a compact set. The following theorem is a consequence of the previous one and theorem THEOREM Let M R m be a compact set and let a mapping f : continuous on M. Then f(m) is a closed and bounded set in R k. The special case of this theorem is the following one: 79 M R k be

26 7.5 Properties of uniformly continuous mappings on compact sets80 THEOREM (Boundedness theorem (1. Weierstrass theorem)). Let M R m be compact and let a function f : M R be continuous. Then f is bounded on M. With respect to the theorems and 7.5.3, we easily obtain the following corollary Corollary Let M R be closed and bounded set (e.g. an interval [α, β]) and let f : M R be a continuous function on M. Then f is bounded on M. Furthermore, theorem (for k = 1) together with theorem imply the following: THEOREM (Extreme value theorem (2. Weierstrass theorem)). Let M R m be a compact set and let a function f : M R be continuous on M. Then f attains on M its maximum value and its minimum value, i.e. there are points c 1, c 2 M such that f(c 1 ) = min x M f(x), f(c 2) = max x M f(x). As a consequence of this theorem and theorem we obtain: Corollary Let f : ([α, β] R) R be continuous function on an interval [α, β]. Then there are numbers c 1, c 2 [α, β] such that f(c 1 ) = min x [α,β] f(x), f(c 2) = max x [α,β] f(x). A function continuous on a compact set can attain its maximal value as well as minimal value more then once. It can attain these values innitely many times as shown on gure 7.2, where a function continuous on the interval [α, β] attains its minimal value in the entire interval [c 1, c 2]. Example Let M R m be a compact set and let b R m \ M. Prove that there is a point c in M minimally distanced from b, (see Fig. 7.3), i.e. c M : d m (c, b) = min{d m (x, b); x M}. (Let us note that the number d m (c, b) is naturally called distance of the set M and the point b. Solution: Example and theorem imply that the function x d m (x, b), x M is continuous on M. Compactness of M then tells us (with respect to the Extreme value theorem) that there is c in M in which our function attains its minimum value, i.e. d m (c, b) = min x M d m(x, b). 80

27 7.5 Properties of uniformly continuous mappings on compact sets81 y f(c 2) α = c 1 O c2 c 1 c 1 c 2 β x f(c 1) Figure 7.2: THEOREM (Heine-Cantor theorem). Let M R m be a compact set and let a function f : M R be continuous on M. Then f is uniformly continuous on M. Proof. Let ɛ be arbitrary positive number. Due to continuity of f in any point y M there is a positive δ y such that: Let d m (x, y) < δ y f(x) f(y) < ɛ 2. O(y; δ y ) = x R m ; d m (x, y) < δ y 2«. The system {O(y; δ y )} y M is an open covering of M. Due to compactness of M we can choose a nite sub-covering, i.e. there is a nite sequence y 1, y 2,..., y n M such that M O(y i, δ yi ). n[i=1 Let δ = 1 2 min{δ y 1, δ y2,..., δ yn }. Let x 1, x 2 M be such that d m (x 1, x 2 ) < δ. The nite system of open sets {O(y i ; δ yi )} n i=1 covers the set M. Therefore, there is an index j {1, 2,... n} such that x 1 O(y j, δ yj ) and thus d m (x 1, y j ) < δ yj /2. Making use of the triangle inequality we obtain d m (x 2, y j ) d m (x 2, x 1 ) + d m (x 1, y j ) < δ + δ y j 2 δ y j 2 + δ y j 2 = δ y j. 81

28 7.5 Properties of uniformly continuous mappings on compact sets82 c b Figure 7.3: Furthermore, f(x 1 ) f(y j ) < ɛ and f(x 2 ) f(y j ) < ɛ 2 2 and nally we can estimate the dierence between the values of f at x 1 and x 2, respectively f(x 1 ) f(x 2 ) f(x 1 ) f(y j ) + f(y j ) f(x 2 ) < ɛ 2 + ɛ 2 = ɛ. This shows that f is uniformly continuous on M. Corollary Let M R be closed and bounded set (M = [α, β] for example) and let f : M R be continuous on M. Then f is uniformly continuous on M. Corollary Let M R m be a compact set and let a mapping f = (f 1, f 2,..., f k ) : M R k be continuous on M. Then f is uniformly continuous on M. Example Let α, β be a pair of real numbers such that α < β. Let f : (α, β) R be monotonous, bounded and continuous on (α, β). Show that such a function f is uniformly continuous on (α, β). Solution: There exists the limits lim f(x) = A, x α lim f(x) = B, x β since f is monotonous and bounded. Let us consider the following function dened on [α, β]: 8><>:f(x), g : x 82 x (α, β) A, x = α B, x = β.

29 7.6 Properties of continuous mappings on connected sets 83 Due to f C(α, β) and lim g(x) = g(α), lim x α g(x) = g(β) x β we see that g is continuous on [α, β] and therefore (corollary 7.5.3) g is uniformly continuous on [α, β]. Since f = g (α,β), f is uniformly continuous on (α, β). Problems 1. Construct a continuous and bounded function f dened on R such that f does not attain neither maximum value nor minimum value. 2. Let f : ([a, b] R) R be continuous on [a, b]. Prove that the functions: are continuous on [a, b]. x min f(t), x [a, b]; x max f(t), x [a, b] a t x a t x 3. Let α, β be a given pair of real numbers such that α < β. Let f : (α, β) R be continuous on (α, β) and f have limits in the end-points α, β, i.e. let exist real numbers A, B such that lim f(x) = A, lim x α Then f is uniformly continuous on (α, β). Prove it! f(x) = B. x β 4. Let M R m be a closed set and let b R m \ M. Show that there exists such a point c M that d m (c, b) = min d m(x, b). x M (Hint: Let a M and A = {x M; d m (x, b) d m (a, b)}. Then A is compact set and d m (c, b) = min x A d m (x, b).) 7.6 Properties of continuous mappings on connected sets Lemma Let f : (M R m ) R k be a continuous on M. Let A R k be closed in R k and let S M be closed in R m. Then the set S A = {t S; f(t) A} is closed in R m. Proof. Let a R m be an arbitrary limit point of the set S A. Then a is limit point of the set S, too. Since S is closed, a S and a belongs also to M. The mapping f is continuous at a. Theorem implies lim f(x) = f(a). x a 83

30 7.6 Properties of continuous mappings on connected sets 84 This means that (with respect to theorem 6.4.1) lim f(x) = f(a). S A (1) x a The following two cases are possible: a) f(a) is a limit point of the set A b) f(a) is not a limit point of the set A. In the case a) the fact A is closed implies f(a) A and therefore a S A. In the case b) we obtain there exists a neighborhood O 0 (a) of a such that O 0 (f(a)) A =. Then (1) implies: for O 0 (f(a)) exists such a neighborhood of the point a that that for all x O(f(a)) S A we have f(x) O 0 (f(a)). Among the values f(x), x O(f(a)) S A only f(a) belongs to O(f(a)), thus f(x) = f(a) for all x O(f(a)) S A. This implies that a S A. The intervals [c, + ) as well as (, c], c R are closed sets in R. This fact together with lemma imply the following: Corollary Let f : (M R m ) R be a continuous function on M and let S M be a closed set in R m. Then for all c R the sets: {x S : f(x) c}, and {x S : f(x) c} closed in R m. Denition A closed set M R m is called connected if there are no non-empty closed sets A, B in R m such that M = A B and A B =. As a simple example of a connected set we have any one-point set in R m. Lemma Any closed interval [α, β] R is a connected set. Proof. Let us suppose the interval in question [α, β] is not a connected set. This mens there are two closed non-empty number sets A, B such that [α, β] = A B and A B =. Let a A and b B and let us choose them in such a way that a < b. Let σ = sup{a [a, b]}. It is self-obvious that a σ b. We will show that σ / A B. If σ A then σ < b and (σ, b] B. This implies that σ is a limit point of the set B, and therefore σ B - but this is 84

31 7.6 Properties of continuous mappings on connected sets 85 impossible because the sets A and B are disjoint. If σ B then σ / A and with respect to example σ must be a limit point of the set A. This, however, implies a contradiction because A was supposed to be closed. So, we have shown that σ / A B, i.e. σ / [α, β]. Thus, the number σ obeys a < σ < b where a, b is a pair of numbers belonging to the interval [α, β]. This means that [α, β] cannot be an interval and this is an contradiction. As a result we have the interval [α, β] is a connected set. Lemma Let φ : ([α, β] R) R m be continuous on [α, β]. Then the values of φ, i.e. the set φ([α, β]) is a closed and connected set. Proof. Theorem implies that φ([α, β]) is a closed set in R m. We will show it is also connected. Let us suppose that φ([α, β]) is not a connected set. Then there are two nonempty closed disjoint sets A, B in R m such that their union is equal exactly to φ([α, β]). With respect to lemma the sets S A and S B are closed in R and obviously S A as well as S B, and [α, β] = S A S B, S A S B =. This means the interval [α, β] is not a connected set and this is a contradiction with the statement of lemma Thus φ([α, β]) is a connected set. Now we will extend the notion of connectedness to the sets that are not closed. Denition We say that a set M R m is connected if for any pair of points a, b M exists a closed and connected set S M such that a, b S. It is obvious that M R m - a closed connected set with respect to denition is also a connected set with respect to denition (it suces to consider S = M). Any interval I R is a connected set. In fact, for any two points a, b from I is the set {x I : a x b} closed and connected and containing both a and b (lemma 7.6.2) and this means (with respect to denition 7.6.2) that the interval I is a connected set. Example Show that a connected set M R is either an one-point set or an interval. Solution: If M is an one-point set then it is connected. Let M R be connected and let it contain at least two dierent points. We will show that then M must be an interval by indirect proof. Let us suppose that M is not an interval. Then there are two points a, b, a < b in M such that there exists a point c: a < c < b such that c / M. Since M is connected there exists a closed and connected set S M such that it contains both points a and b. With respect to corollary (for m = 1, f = i d ) are the sets S c = {x S; x c}, S + c = {x S; x c} 85

32 7.6 Properties of continuous mappings on connected sets 86 closed in R, furthermore S c, S + c (a S c, b S + c ), and S = S c S + c, S c S + c =. However, this is a contradiction with the connectedness of S. This shows that a connected set M R (if not an one-point set) is an interval. Denition Let a mapping φ : ([α, β] R) R m be continuous on [α, β] and let φ(α) = a and φ(β) = b. Then the set φ([α, β]) is called the path from a to b (or equivalently, the path connecting a and b). Denition A non-empty set M R m is called path-connected if any two points of M can be connected by a path that belongs to M. THEOREM If a set M R m is path-connected then it is connected. Proof. Let a and b are arbitrary points from M. Then there is a path connecting these two points, i.e. there is a continuous (on [α, β]) mapping φ : ([α, β] R) R m such that φ(α) = a, φ(β) = b and φ([α, β]) M. We know (lemma 7.6.3) that φ([α, β]) is closed and connected and this shows that for any two points a, b M exists a closed and connected set φ([α, β]) that contains the points a and b. This means that M is connected. Example Let c R m and r > 0. Show that the closed ball and the open ball B(c, r) = {x R m, d m (c, x) r} K(c, r) = {x R m, d m (c, x) < r} are connected sets. Solution: We will show that our closed ball B(c, r) is a connected set. Let c = (c 1,..., c m ) and let a = (a 1...., a m ) and b = (b 1,..., b m ) be any two points in B(c, r). Let us dene a mapping φ as follows: φ : [0, 1] R m, This means we have φ = (φ 1,..., φ m ) with φ(t) = bt + (1 t)a. φ j : t b j t + (1 t)a j, t [0, 1] (j = 1, 2,..., m). 86

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