Math 117: Continuity of Functions

Transcription

1 Math 117: Continuity of Functions John Douglas Moore November 21, 2008 We finally get to the topic of ɛ δ proofs, which in some sense is the goal of the course. It may appear somewhat laborious to use such proofs to prove facts about real-valued functions of one variable, facts that appear to be true from an intuitive perspective. But the techniques developed can be used for far more complicated problems, for which the resolution is by no means self-evident. The reader should focus first on learning the definitions of limits and continuity. Several propositions will illustrate how these definitions can be used in mathematical arguments. Then focus on the main theorems, the Intermediate Value Theorem and the Compactness Theorem. 1 Limits of functions Definition Let D R and let c be an accumulation point of D. A real number L is a limit of the function f : D R at c if for every ɛ > 0 there exists δ > 0 such that x D and 0 < x c < δ f(x) L < ɛ. (1) If it exists, the limit L is unique and we write lim x c f(x) = L. Example. Suppose that f : R R is the function defined by { 2x sin(1/x), if x 0, f(x) = 5, if x = 0, and take c = 0. Then given ɛ > 0 we can let δ = ɛ/2, and 0 < x 0 < δ f(x) 0 = 2 x sin(1/x) 2 x < 2δ = ɛ, so lim x 0 f(x) = 0. Note that the limit does not depend on the value of f at 0. Although the above definition is very useful for proving theorems, it should be remarked that for a given choice of ɛ > 0 it is not always easy to find a δ > 0 that works in (1). (The text [2] provides several examples.) 1

2 Proposition. Let D R and let c be an accumulation point of D. Then lim x c f(x) = L if and only if for every sequence (s n ) of elements of D such that s n s which converges to c, the sequence (f(s n )) converges to L. Proof: Suppose that lim x c f(x) = L and let ɛ > 0 be given. then there exists δ > 0 such that x D and 0 < x c < δ f(x) L < ɛ. If (s n ) is a sequence of elements of D such that s n s which converges to c, then there exists N N such that But then n > N 0 < s n s < δ. n > N 0 < f(s n ) L < ɛ, and (f(s n )) converges to L. Conversely, suppose that L is not a limit of f(x) at c. Then there exists an ɛ > 0 such that for every δ > 0, there is an element x D with 0 < x c < δ and f(x) L ɛ. Then since s is an accumulation point of D, for each n N, we can choose s n D {s} so that 0 < s n c < 1/n and f(s n ) L > ɛ. Then (s n ) is a sequence of elements of D such that s n s which converges to c, but (f(s n )) does not converge to L. Example. Suppose that f : R R is the function defined by { sin(1/x), if x 0, f(x) = 0, if x = 0, and let s n = 2/(nπ). Then s n 0 for n N and the sequence (s n ) converges to zero. Moreover, 1, if s = 1, 5, 9,..., 0, if s = 2, 6, 10,..., f(s n ) = sin(nπ/2) = 1, if s = 3, 7, 11,..., 0, if s = 4, 8, 12,.... We conclude that (f(s n ) does not converge. Thus it follows from the preceding proposition that the function f does not have a limit at c = 0. The previous proposition can be used to translate many properties of convergence of sequences to convergence of functions. Proposition. Suppose that c is an accumulation point of D R. Suppose that f : D R and g : D R are functions such that lim x c f(x) = L and lim x c g(x) = M. Then 1. lim x c (f(x) + g(x)) = L + M, 2

3 2. lim x c (f(x)g(x)) = LM, 3. if g(x) 0 for x D and M 0, then lim x c (f(x)/g(x)) = L/M. This follows from the corresponding proposition on limits of sequences. 2 Continuous functions We now come to what is arguably the most important definition in the course: Definition Let D R and let c be an element of D. A function f : D R is continuous at c if for every ɛ > 0 there exists δ > 0 such that x D and x c < δ f(x) f(c) < ɛ. We say that a function f : D R is continuous if it is continuous at every point c D. Example. We claim that the function f : R R defined by f(x) = 3x is continuous at every point c R. Indeed, suppose c R. If ɛ > 0 is given, we let δ = ɛ/3. Then x c < δ f(x) = f(c) = 3x 3c = 3 x c < 3δ = ɛ. In other words, and f is continuous at c. x c < δ f(x) = f(c) < ɛ, Example. We claim that the function f : R R defined by { 1, if x Q, f(x) = 0, if x R Q. is not continuous at any point c R. Indeed, let c R be a given point. Given any δ > 0, there are both rational and irrational points x (c δ, c+δ), that is there are points x, y (c δ, c+δ) such that f(x) = 1 and f(y) = 0. It follows that there is no δ > 0 such that f(x) f(c) < 1 2 for all x (c δ, c + δ). Proposition. Suppose that f : D R where D R and c D. Then f is continuous at c if and only if it satisfies the following condition: If (x n ) is any sequence in D which converges to c, then (f(x n )) converges to f(c). Proof: Suppose that f is continuous at c and that ɛ > 0 is given. Then there exists δ > 0 such that x D and x c < δ f(x) f(c) < ɛ. 3

4 Suppose now that (x n ) is any sequence in D which converges to c. Then there is N N such that n > N implies that x n c < δ. But then n > N f(x n ) f(c) < ɛ. Hence (f(x n )) converges to f(c). Suppose that f is not continuous at c. Then there exists an ɛ > 0 such that it is not the case that x D and x c < 1 n f(x) f(c) < ɛ. Choose x n D such that x n c < 1/n and f(x n ) f(c) > ɛ. Then (x n ) converges to c but (f(x n )) does not converge to f(c). The preceding proposition allows us to transport many results regarding limits to continuous functions. Thus we obtain: Proposition. If f : D R and g : D R are continuous functions so is the sum f + g and the product fg. Moreover, if g(x) 0 for xind, then (f/g) : D R is a continuous function. This proposition allows us to build up many new continuous functions from old. Thus starting with the fact that the constant function is continuous and the function f(x) = x is continuous, we can conclude that any polynomial is continuous, for example. 3 Intermediate Value Theorem Intermediate Value Theorem. Suppose that f : [a, b] R is continuous and f(a) < 0 < f(b). Then there exists c (a, b) such that f(c) = 0. Proof: Let S = {x [a, b] : f(x) 0}. Then S is nonempty and bounded above, so S has a supremum c by the completeness axiom of the real numbers. Moreover, since a S, c a while clearly c b. The idea behind the proof is to show that f(c) = 0, from which it follows that c (a, b). If f(c) < 0, we can suppose that f(c) = y, where y < 0. Let ɛ = y. Then there exists δ > 0 such that x [a, b] and x c < δ f(x) f(c) < ɛ f(x) < 0. It follows that there is a point y (c, b] N(c; δ) such that f(y) < 0 and c is not the supremum of S, a contradiction. Thus f(c) 0. Similarly, if f(c) > 0, we can suppose that f(c) = y, where y > 0. Let ɛ = y. Then there exists δ > 0 such that x [a, b] and x c < δ f(x) f(c) < ɛ f(x) > 0. 4

5 It follows that if x [a, c) N(c; δ), then f(x) > 0 and hence any point of [a, c) N(c; δ) is an upper bound of S and c is not the supremum of S, a contradiction once again. Thus f(c) 0. This theorem often allows us to construct the solutions to equations. Thus, for example, suppose that we know that the function f(x) = e x is continuous and let g(x) = 4x, also a continuous function. If we want to find a solution to the equation e x = 4x, it suffices to show that the continuous function g(x) f(x) = 4x e x has a zero. But g(0) f(0) = 0 1 = 1 < 0, g(1) f(1) = 4 e > 0, so it follows from the Intermediate Value Theorem that there exists c [0, 1] such that g(c) f(c) = 0. This c is a solution to e x = 4x. 4 Maximum values of continuous functions There is an important criterion for continuity which is expressed in terms of open sets. Suppose that D is a subset of R. A subset V of D is said to be open if V = W D, where W is open in R. Proposition. A function f : D R is continuous if and only if whenever U is an open subset of R, f 1 (U) is an open subset of D. Proof: Suppose that c f 1 (U). Then f(c) U and since U is open, N(f(c); ɛ) U for some ɛ > 0. Since f is continuous, there exists δ c > 0 such that x c < δ c f(x) f(c) < ɛ. In other words, f(n(c; δ c ) D) N(f(c); ɛ). Let W = {N(c, δ c ) : c f 1 (U)}, V = Ṽ D. Since W is the union of open sets, it is open, and hence V = f 1 (U) is an open subset of D. Conversely, if f satisfies the condtion that f 1 (U) is an open subset of D whenever U is an open subset of R, then given c D and ɛ > 0, the set N(f(c); ɛ) is open in R. Thus f 1 (N(f(c); ɛ)) is open in D and hence f 1 (N(f(c); ɛ)) = Ṽ, where Ṽ is open in R. Since c Ṽ, there exists δ > 0 such that N(c : δ) Ṽ. But then x D and x c < δ f(x) f(c) < ɛ, and hence f is continuous. The definition of compactness in terms of open covers and the open set criterion for continuity of functions allows us to provide an efficient proof of existence of maxima and minima in many cases. 5

6 Compactness Theorem. If K is a compact subset of R and f : K R is a continuous function, then f(k) is compact. Proof: Suppose that F = {U α : α A} is an open cover of f(k). For each U α, there is an open subset V α R such that V α K = f 1 (U α ). We then have K {V α : α A}, so that {V α : α A} is an open cover of K. Since K is compact, there exists a finite subcover {V α1,..., V αn }; in other words, But then K V α1 V αn. f(k) f(v α1 ) f(v αn ) U α1 U αn. So {U α1,..., U αn } is a finite subcover of F. Since F was an arbitrary open cover, we have shown that ANY open cover of f(k) has a finite subcover, so f(k) is indeed compact, as needed. Corollary. Suppose that K is a compact subset of R and f : K R is a continuous function. Then f assumes its maximum and minimum values on K. Proof: We prove only that f assumes its maximum value; the proof that it assumes its minimum is similar. By the previous theorem, f(k) is compact and therefore closed and bounded by the Heine-Borel Theorem. Let S = f(k) and let m be the least upper bound of S, which exists by the Completeness Axiom for real numbers. It suffices to show that m S. If m / S, then for any ɛ > 0 there exists and element s S such that m ɛ < s < m, because otherwise m ɛ would be an upper bound, contradicting the fact that m is the least upper bound. Thus m is an accumulation point of S. But since S is closed it contains all of its accumulation points. Therefore m S. Thus m is a maximum for S and f achieves its maximum value on K. 5 Uniform continuity Definition Let D R and let c be an element of D. A function f : D R is uniformly continuous if for every ɛ > 0 there exists δ > 0 such that x, y D and x y < δ f(x) f(y) < ɛ. The key point is that δ does not depend upon the points x or y within D. 6

7 For example, the function f : R R defined by f(x) = 2x is uniformly continuous, because given ɛ > 0, we can set δ = (/2)ɛ and x y < δ 2x 2y < 2δ = ɛ f(x) f(y) < ɛ. On the other hand, suppose f : (0, ) R by f(x) = 1/x. Then f(x) f(c) = 1 x 1 c = 1 xc (c x) f(x) f(c) = 1 xc c x. Suppose that we have chosen a δ > 0 such that x c < δ f(x) f(c) < ɛ, (2) and take x = c δ/2. Then 1 xc 1 c 2 f(x) f(c) 1 c 2 c x, so x c c2 f(x) f(c). Thus in order to achieve (2), we must take δ c 2 ɛ. As c 0, the δ > 0 we must choose 0, so there is no δ > 0 which works for all choices of c (0, ). In other words, f : (0, ) R is continuous, but not uniformly continuous in this case. The open cover definition of compactness makes the following theorem easy to prove: Uniform Continuity Theorem. If K is a compact subset of R and f : K R is a continuous function, then f is uniformly continuous. Proof (following [1]): The proof is based upon the ɛ/2 trick. Suppose that ɛ > 0 is given. Since f is continuous, for each x K there exists δ x > 0 such that For each x K, let Then y x < δ x 2 U x = f(y) f(x) < ɛ 2. ( x δ x 2, x + δ ) x. 2 U = {U x : x K} is an open cover of K. By compactness of K, this cover has a finite subcover, ( K U x1 U x2 U xk, U xi = x i δ x i 2, x i + δ ) x i. 2 Let δ = (1/2) min(δ x1,..., δ xk ). Suppose now that x, y D and x y < δ. Then for some i {1,..., k}, Thus x x i < δ x i 2, y x i < y x + x x i < δ x i 2 + δ x i 2 = δ x i. f(x) f(y) f(x) f(x i ) + f(x i ) f(x) < ɛ 2 + ɛ 2 = ɛ. This shows that f is indeed uniformly continuous. 7

8 6 Continuity of functions on metric spaces The preceding theory can be generalized with very little work to metric spaces. Definition Suppose that (X 1, d 1 ) and (X 2, d 2 ) are metric spaces. A function f : X 1 X 2 is continuous if for every c X 1 and every ɛ > 0 there exists a δ c > 0 such that We can then (3) as x X 1 and d 1 (x, c) < δ c d 2 (f(x), f(c)) < ɛ. (3) x N(x; δ c ) f(x) N(f(c); ɛ). (4) A function f : X 1 X 2 is uniformly continuous if for every ɛ > 0 there exists a δ > 0 such that x, y X 1 and d 1 (x, y) < δ d 2 (f(x), f(y)) < ɛ. Proposition. If (X 1, d 1 ) and (X 2, d 2 ) are metric spaces, then a function f : X 1 X 2 is continuous if and only if whenever U is an open subset of X 2, f 1 (U) is an open subset of X 1. The proof is very similar to the special case presented before. Suppose that c f 1 (U). Then f(c) U and since U is open, N(f(c); ɛ) U for some ɛ > 0. Since f is continuous, there exists δ c > 0 such that (4) holds; thus f(n(c; δ c )) N(f(c); ɛ). Let V = {N(c; δ c ) : c f 1 (U)}. Then V = f 1 (U), and since V is the union of open sets, it is open. Conversely, if f satisfies the condtion that f 1 (U) is an open subset of X 1 whenever U is an open subset of X 2, then given c X 1 and ɛ > 0, the set N(f(c); ɛ) is open in X 2. Thus f 1 (N(f(c); ɛ)) is open in X 2 and since c f 1 (N(f(c); ɛ)), there exists δ > 0 such that N(c : δ) f 1 (N(f(c); ɛ)). But then x N(c; δ) f(x) N(f(c); ɛ), and hence f is continuous. Main Theorem. If (X 1, d 1 ) and (X 2, d 2 ) are metric spaces, f : X 1 X 2 is a continuous function and K is a compact subset of X 1, then f(k) is compact. Once again, the proof is almost identical to the previous special case. Suppose that F = {U α : α A} is an open cover of f(k). For each U α, there is an open subset V α X 1 such that V α K = f 1 (U α ). We then have K {V α : α A}. 8

9 In other words, {V α : α A} is an open cover of K. Since K is compact, there exists a finite subcover {V α1,..., V αn }; in other words, But then K V α1 V αn. f(k) f(v α1 ) f(v αn ) U α1 U αn. So {U α1,..., U αn } is a finite subcover of F. Since F was an arbitrary open cover, we have shown that ANY open cover of f(k) has a finite subcover, so f(k) is indeed compact, as needed. Corollary. Suppose that (X, d) is a metric space and f : X R is a continuous function. If K is a compact subset of X, then f assumes its maximum and minimum values on K. Proof: Once again, we prove only that f assumes its maximum value. By the previous theorem, f(k) is compact and therefore closed and bounded by the Heine-Borel Theorem. Let S = f(k) and let m be the least upper bound of S, which exists by the completeness axiom for real numbers. It suffices to show that m S. If m / S, then for any ɛ > 0 there exists and element s S such that m ɛ < s < m, because otherwise m ɛ would be an upper bound, contradicting the fact that m is the least upper bound. Thus m is an accumulation point of S. But since S is closed it contains all of its accumulation points. Therefore m S. Thus m is a maximum for S and f achieves its maximum value on K. 7 Spaces of functions* Much of the preceding theory can be extended to metric spaces whose elements are functions. We now give an important example of such a metric space. Let and define X = C([a, b], R) = {f : [a, b] R : f is continuous }, d : C([a, b], R) C([a, b], R) R by d(f, g) = sup{ f(x) g(x) : x [a, b]}. The supremum exists by the corollary in the preceding section. It is straightforward to check that this distance function satisfies the axioms for a metric space. Using this example, we can apply many of the techniques that we have learned for dealing with the real numbers to spaces of functions. This leads to an important subject functional analysis that plays a key role in proving existence of solutions to differential equations. Definition A metric space (X, d) is complete if every Cauchy sequence in X converges to a point of X. 9

10 For example, R n is complete. The following theorem says that C([a, b], R) is also complete: Completeness Theorem. Every Cauchy sequence in C([a, b], R) converges in C([a, b], R). The idea behind the proof is simple. One imagines that (f n ) is a Cauchy sequence in C([a, b], R). Then if x [a, b], (f n (x)) is a Cauchy sequence in R and by the Cauchy Sequence Theorem, this sequence must converge to a limit, which we call f(x). This defines a function f : [a, b] R. One must then check that this function is continuous and that it is the limit of the sequence (f n ) within C([a, b], R). Suppose that ɛ > 0 is given. Then there exists N N such that m, n > N d(f m, f n ) < ɛ 2. If x [a, b], there exists N x N such that N x N and But then m > N x f m (x) f(x) < ɛ 2. n > N f n (x) f(x) f n (x) f m (x) + f m (x) f(x) < ɛ 2 + ɛ 2 = ɛ. Since this holds for every x [a, b], we see that (f n ) converges to f within C([a, b], R). To prove that f is continuous, we suppose that c [a, b] and ɛ > 0 is given. Then since (f n ) converges to f, there exists N N such that n > N f n (x) f(x) < ɛ, for all x [a, b]. 3 Since f n is continuous, there exists δ > 0 such that But then x c < δ f(x) f(c) x c < δ f n (x) f n (c) < ɛ 3. < f(x) f n (x) + f n (x) f n (c) + f n (c) f(c) < ɛ 3 + ɛ 3 + ɛ 3 = ɛ. This shows that f is continuous at c. Remark 1. If a sequence (f n ) converges to f in C([a, b], R), we say that it converges uniformly to f. Remark 2. The study of spaces of functions such as C([a, b], R) is called functional analysis. It is an important branch of modern mathematics. 10

11 To proceed further, we use some notions from the later parts of real analysis, derivatives and integrals, notions that you already studied in calculus, and that are discussed with more rigor in the Math 118 sequence. The metric space of functions gives a strategy for proving the existence of solutions to differential equations. Suppose that we want to solve the initial value problem d dx (y(x)) = f(x, y(x)), y(x 0) = y 0, (5) where f : R 2 R is a continuous function which satisfies the condition f(x, y 1 ) f(x, y 2 ) L y 1 y 2, L being a positive constant. (This is called a Lipschitz condition of f.) We can rewrite this differential equation as an integral equation y(x) = y 0 + x x 0 f(ξ, y(ξ))dξ. If y(x) C([x 0, x 0 + ɛ], R), where ɛ > 0 is small, we can define T (y)(x) C([x 0, x 0 + ɛ], R) by T (y)(x) = y 0 + for x [x 0, x 0 + ɛ], thus obtaining a function Note that T (y 1 )(x) T (y 2 )(x) = Thus x x T : C([x 0, x 0 + ɛ], R) C([x 0, x 0 + ɛ], R). x x 0 f(ξ, y 1 (ξ) y 2 (ξ))dξ x 0 f(ξ, y(ξ))dξ, x 0 L y 1 (ξ) y 2 (ξ) dξ Lɛ sup{ y 1 (x) y 2 (x) : x [x 0, x 0 + ɛ]}. d(t (y 1 ), T (y 2 )) ɛld(y 1, y 2 ), and it follows that T is a contraction mapping when ɛ > 0 is sufficiently small. Starting with an approximate solution y 0 (x), one can define a sequence of solutions y n (x) by y n+1 (x) = T (y n )(x). Using the Contraction Mapping Theorem (from the notes on infinite sequences), one can show that (y n (x)) is a Cauchy sequence in C([x 0, x 0 + ɛ], R). It follows from the Completeness Theorem that this Cauchy sequence converges to a limit y (x). It can then be shown that the limit function is a solution to the initialvalue problem (5). The key point is that the notion of metric space provides an avenue for extending many of the theorems used in the foundations of calculus to settings that 11

12 allow us to find solutions to differential equations, both ordinary and partial. This has turned out to be an extremely powerful approach to understanding the theory of differential equations, which is of course widely used in applications of mathematics. 8 Sequential compactness* For metric spaces, there is an alternative definition of compactness is often quite useful: Definition. A metric space (X, d) is said to be sequentially compact if every sequence (x n ) of points in X has a convergent subsequence. The following theorem is proven in Munkres [3], 28: Theorem. A metric space (X, d) is compact if and only if it is sequentially compact. Proof: First suppose that (X, d) is a compact metric space and let (x n ) be a sequence in X. If the set S = {x n : n N} is finite, (x n ) clearly has a convergent subsequence. If S were infinite and had no accumulation point, then U n = {x n } would be an open subset of X for each n N and the open cover {U n : n N} would have no finite subcover. This would contradict compactness. Suppose, on the other hand that (X, d) is sequentially compact and that U = {U α : α A} is an open cover of X. We claim there exists δ > 0 such that every subset of X of diameter < δ lies in some element of mathcalu. If this were not true, we could construct a subset C n X such that C n has diameter < 1/2 n and C n is not contained in any element of U. Choose a point x n C n for each n N. Then the sequence (x n ) would have a sequence converging to some x X. Since x is a member of the open set U α for some α A, we have N(x; 2δ) U α for some δ > 0. But then N(x n ; δ) U α when n is sufficiently large, a contradiction. We next claim that for every ɛ > 0, there exists a finite cover of X by balls N(x i ; ɛ). If not we could construct a sequence (x n ) such that x n / N(x 1 ; ɛ) N(x n 1 ; ɛ). Then this sequence would have no convergent subsequence. We now choose ɛ = δ/3. Our previous claim shows that X is covered by finitely many ɛ-balls each of which is contained in some element U α of U. We choose one such element U α for each ɛ-ball, thereby obtaining our finite subcover. Thus (X, d) is compact. Remark. Thus if we were to restrict attention to metric spaces, the sequential compactness definition would be perfectly acceptable. The reason the open cover definition of compactness is preferred is that it can be applied to topological spaces which are not metric spaces. See [3] for the general definition of topological space. 12

13 It would be nice if the Heine-Borel Theorem held for the infinite-dimensional metric space C([a, b], R), in other words, if closed bounded subsets of C([a, b], R) were compact. This is not true, but certain subspaces of C([a, b], R) are indeed compact. This is a fact which turns out to be quite useful in applications to differential equations. Definition. A collection K of functions in C([a, b], R) is said to be equicontinuous if for every ɛ > 0 there exists a δ > 0 such that x y < δ f(x) f(y) < ɛ, for all f K. The important point is that δ depends only on ɛ, not on f. We say that a subset K of C([a, b], R) is uniformly bounded if there is a constant M > 0 such that f(x) M for all x [a, b] and all f K. Arzela-Ascoli Theorem. let K be a closed bounded equicontinuous subset of C([a, b], R). Then K is sequentially compact. In other words, if (f n ) is a sequence of uniformly bounded equicontinuous functions in C([a, b], R), then (f n ) has a subsequence which converges to an element of C([a, b], R). Proof: The proof makes use of Cantor s diagonalization argument. We can assume without loss of generality that [a, b] = [0, 1]. Let (f n ) be a uniformly bounded equicontinuous family of functions. Choose a sequence (x n ) consisting of all the elements of [0, 1] Q. Choose subsequences (f n1 ) (f n2 ) (f n3 ) (f nk ) such that (f nk )(x k ) converges. The diagonal subsequence (f nn ) then converges at every point of [0, 1] Q. Let ɛ > 0 be given. For this choice of ɛ, there exists δ > 0 such that f, g K and x y < δ f(x) f(y) < ɛ 3. We cover [0, 1] by finitely many intervals N(y i, δ) where y i [0, 1] Q, for 1 i m. For each i there exists N i N such that m, n > N i f nn (y i ) f mm (y i ) < ɛ 3. We let N be the maximum of {N i : 1 i m}. Then if x [0, 1], there exists y i such that x y i < δ, and m, n > N f nn (x) f mm (x) < f nn (x) f nn (y i ) + f nn (y i ) f mm (y i ) + f mm (y i ) f mm (x) < ɛ 3 + ɛ 3 + ɛ 3 = ɛ. Hence the subsequence (f nn ) is a Cauchy sequence in C([a, b], R), and must converge by completeness of C([a, b], R). 13

14 References [1] Edward D. Gaughan, Introduction to analysis, Brooks-Cole, Pacific Grove, [2] S. Lay, Analysis with an introduction to proof, Fourth edition, Pearson Prentice Hall, Upper Saddle River, NJ, [3] J. Munkres, Topology, Second edition, Pearson Prentice Hall, Upper Saddle River, NJ, [4] W. Rudin, Principles of mathematical analysis, Third edition, McGraw- Hill, New York,