Math 117: Topology of the Real Numbers

Math 117: Topology of the Real Numbers John Douglas Moore November 10, 2008 The goal of these notes is to highlight the most important topics presented in Chapter 3 of the text [1] and to provide a few additional topics on metric spaces, in the hopes of providing an easier transition to more advanced books on real analysis, such as [2]. 1 Open and closed sets If a and b are real numbers with a < b, we let (a, b) = {x R : a < x < b}, [a, b] = {x R : a x b}. Moreover, if x R and ɛ > 0, we let N(x; ɛ) = {y R : x y < ɛ}, N (x; ɛ) = {y R : 0 < x y < ɛ}. Definition. A set S R is open if whenever x S, there exists ɛ > such that N(x; ɛ) S. A set S R is closed if R S is open. Thus, for example, (a, b) is open while [a, b] is closed. Definition. Suppose S R. Then the interior of S is The boundary of S is int(s) = {x S : N(x; ɛ) S for some ɛ > 0 }. bd(s) = R (int(s) int(r S)). Clearly, S is open if and only if S = int(s). Note also that R is the disjoint union of int(s), int(r S) and bd(s). Proposition 1.1. S is closed if and only if bd(s) S. Proof: : Suppose that S is closed. Then R S is open so int(r S) = R S. Thus S = int(s) bd(s), and bd(s) S. : Suppose that bd(s) S. Then int(s)bd(s) S, so R S (R S). Hence int(r S) = R S and S is closed. 1

Proposition 1.2. The empty set and the whole space R are open. The union of an arbitrary collection of open sets is open. The intersection of a finite collection of open sets is open. Proof: We leave it to the reader to check that the empty set and the whole space R are open. Suppose that {U α : α A} is a collection of open sets and that U = {U α : α A}. If x U, then x U α for some α A. Hence there is an ɛ > 0 such that N(x; ɛ) U α. But then N(x; ɛ) U, and this shows that U is open. Suppose that {U 1,..., U m } is a finite collection of open sets and that U = U 1 U m. If x U, then x U i for every i, 1 i m. Hence for each i, 1 i m, there exists ɛ i > 0 such that N(x; ɛ i ) U i. Let ɛ = min(ɛ 1,..., ɛ m ), and note that ɛ > 0 since the minimum of a finite number of positive numbers is positive. Then N(x; ɛ) U i for every i, 1 i m. Hence N(x; ɛ) U and U is open. Proposition 1.3. The empty set and the whole space R are closed. The intersection of an arbitrary collection of closed sets is closed. The union of a finite collection of closed sets is closed. Proof: Once again it is easy to see that the empty set and the whole space R are closed. Suppose that {C α : α A} is a collection of closed sets and that C = {C α : α A}. Then U α = R C α is open, and by the previous theorem U = {U α : α A} is open. It follows from the De Morgan laws that R U = R {U α : α A} = (R U α ) = {C α : α A} = C. Hence C is closed. Suppose that {C 1,..., C m } is a finite collection of open sets and that C = C 1 C m. Then U i = R C i is open and by the previous theorem, U = U 1 U m is also open. It follows from the De Morgan laws that R U = R (U 1 U m ) = (R U 1 ) (R U m ) = C 1 C m = C, 2

and hence C is closed. The preceding propositions allow one to check whether many subsets of R are open or closed. For example, consider the subset Z R. Since R Z is the union of the intervals (n, n + 1) for n Z, we see that R Z is open, and hence Z is closed. On the other hand we claim that the rationals Q R is neither open nor closed. Equivalently, neither Q nor R Q is open. If Q were open there would exist ɛ > 0 such that ( ɛ, ɛ) Q. But then there would exist n N such that ( 1/n, 1/n) Q. Since multiplication by n takes Q to itself, it would then follow that ( 1, 1) Q. This is impossible, because we have seen that ( 1, 1) is uncountable while Q is countable. To show that R Q is not open, we use: Proposition 1.4. Any nonempty open interval (a, b) contains a rational number r. Proof: In proving this, we can assume without loss of generality (WOLOG) that 2 < a < b. (Indeed, we could arrange this by replacing (a, b) by (a + n, b + n) for some n N.) By the Archimedean property of R, we know that N is not bounded above, so we can choose n N such that n > 1/(b a) Then 1/n < b a. We let S = {m N : m/n > a}, a set which is nonempty because of the Archimedean property once again. (There exists m N such that m > na.) Since N is well-ordered, S contains a least element m, which must be 2. Then m 1 n a m n 1 n a m n a + 1 n < b. Hence r = m/n is a rational number such that r (a, b). Suppose now that R Q were open. Then it would contain a nonempty open interval (a, b) in contradiction to the previous proposition. We conclude that R Q cannot be open. Definition. Suppose S R. Then a point x R is an accumulation point of S if for every ɛ > 0, N (x; ɛ) S. Proposition 1.5. If x R is an accumulation point of S R, then for every ɛ > 0, N (x; ɛ) S is infinite. Proof: If x is an accumulation point of S, there exists a point x 1 N (x; ɛ) S. Let ɛ 1 = x 1 x. Then there exists a point x 2 N (x; ɛ 1 ) S N (x; ɛ) S, and we let ɛ 2 = x 2 x. Continuing in this fashion, we get an infinite collection of distinct points {x 1, x 2, x 3,...} in N (x; ɛ) S. We let S denote the set of accumulation points of S. For example, if S = {1, 1/2, 1/3, 1/4,...}, the only accumulation point of S is the point 0. In this case, S = {0}. 3

Proposition 1.6. A subset S of R is closed if and only if S S. Proof: : Suppose that S is closed and x S. If x R S, then N(x; ɛ) R S for some ɛ > 0. But the x / S. : Suppose that S S. If x R S, then x / S and hence x / S, so N (x; ɛ) S = for some ɛ > 0. But then N(x; ɛ) R S for some ɛ > 0. Thus R S) is open and S is closed. The Cantor set. We start with C 0 = [0, 1]. We then let [ C 1 = 0, 1 ] [ ] 2 3 3, 1 = C ( 0 2 3 3 + C ) 0, 3 [ C 2 = 0, 1 ] [ 2 9 9, 1 ] [ 2 3 3, 7 ] [ ] 8 9 9, 1 = C ( 1 2 3 3 + C ) 1, 3 and so forth. Thus for each n N, we let C n = C n 1 3 ( 2 3 + C ) n 1. 3 Note that all of the sets C n are closed. The intersection C = {C n : n N} is also closed, and called the Cantor set. How long is the Cantor set? To answer this question, we carry out the calculation: Length of ([0, 1] C) = 1 3 + 2 9 + 4 27 + = 1 3 n=1 2 n 3 n = 1 ( ) 1 3 1 (2/3) = 1. Thus we see that the Cantor set has length (or measure) zero. The Cantor set also enjoys many other interesting properties. For example, it can be shown that every point of the Cantor set is an accumulation point; thus C = C. A proof of this fact can be found in [2]; see pages 36,37. This source also provides a proof that the Cantor set has uncountably many points. 2 Compactness Suppose that S is a subset of R. We say that a collection U = {U α : α A} is an open cover of S if 1. each U α is open, and 2. S {U α : α A}. 4

Definition. A subset K R is compact if whenever {U α : α A} is an open cover of K, then K U α1 U αm, for some finite subset {α 1,..., α m } A. In other words, K is compact if any open cover of K has a finite subcover. Here are some examples of compact and noncompact sets. First, any finite subset of R is compact. Next, the set N of natural numbers is not compact, because {(0, n) : n N} is an open cover of N, but it has no finite subcover. Indeed, if N (0, n 1 ) (0, n 2 ) (0, n m ) for some finite m, then we could let N = max(n 1,..., n m ). Since (0, n i ) (0, N), we would then have N (0, N) but this would contradict the fact that N is unbounded (which follows from the Archimedean property of the real numbers. Similarly, the interval (0, 1] = {x R : 0 < x 1} is not compact. Indeed if we let ( ) 1 U n = n, 2, then (0, 1] {U n : n N}. Note that n < m U n U m. Thus if it were the case that it would follow that (0, 1] U n1 U nm, (0, 1] U N, where N = max(n 1,..., n m ). But by the Archemedean Property of R, there exists an element m N such that 0 < 1/m < 1/N and hence it is not the case that (0, 1] ((1/N), 2). A subset K R is said to be bounded if K N(x; R) for some R > 0. Proposition 2.1. If a subset of R is compact, it is closed and bounded. Proof: Suppose that K is a compact subset of R. For n N, the set U n = ( n, n) is open. Moreover, {Un : n N} = R. Thus U = {U n : n N} is an open cover of K. Since K is compact, it must have a finite subcover {U n1,..., U nm }, so K U n1 U nk. Let N = max(n 1,..., n m ). Then K ( N, N) and hence K is bounded. Suppose now that K is compact and x bd(k) K. Let ( U n =, x 1 ) (x + 1n ) n,. Then K {U n : n N} = R {x}. 5

Thus U = {U n : n N} is an open cover of K. Since K is compact, it must have a finite subcover Let N = max(n 1,..., n m ). Then {U n1,..., U nm }, so K U n1 U nk. K (, x 1 N ) (x + 1N ), and hence (x ɛ, x + ɛ) R K, for ɛ 1 N. This contradicts the assumption that x bd(k). Hence bd(k) K, and K is closed. Proposition 2.2. Any nonempty finite closed interval [a, b] is compact. The argument (following [2]) is based upon: Nesting Lemma. Suppose that {I n : n N} is a decreasing nested sequence of nonempty closed intervals in R, Then {I n : n N} is nonempty. I 1 I 2 I n I n+1. Proof of Lemma: Suppose that I n = [a n, b n ] and let S = {a n : n N}. S is nonempty and since a n b 1, it is bounded above. Thus S has a least upper bound x. Since x is an upper bound, a n x for all n. Since x is a least upper bound and a n a n+k b n+k b n, it follows that x b n, for all n. Thus x [a n, b n ], for every n, and x lies in the intersection. Proof of Proposition 2.2: We assume that U = {U α : α A} is an open cover of [a, b] with no finite subcover. We can divide the interval I 1 = [a, b] into two equal intervals [ [a, b] = a, a + b ] [ ] a + b 2 2, b. At least one of the two subintervals cannot be covered by a finite subcollection of U; otherwise the original interval could be so covered. Let I 2 = [a 2, b 2 ] be one of the two subintervals, chosen so that it cannot be covered by a finite subcollection of U. We divide I 2 into two intervals of equal length I 2 = [a 2, b 2 ] = [ a 2, a ] 2 + b 2 2 [ a2 + b 2 2, b 2 ]. 6

Once again, at least one of the two subintervals, say I 3 cannot be covered by a finite subcollection of U. We continue in a similar fashion, thereby obtaining a decreasing nested sequence of nonempty intervals I 1 I 2 I n I n+1 with Length(I n+1 ) = 1 2 Length(I n) such that no I n is covered by a finite subcover of U. By the Lemma, {I n : n N} contains some element x. Let U α be an element of U which contains x. Since U α is open there exists ɛ > 0 such that (x ɛ, x + ɛ) U α. But then I n (x ɛ, x + ɛ) U α for n sufficiently large. This contradicts the fact that I n was not covered by a finite subcover of U = {U α : α A}. Proposition 2.3. If S is a closed subset of a compact set K, then S is also compact. Proof: Let U = {U α : α A} be an open cover of S. Then {U α : α A}{R S} is an open cover of K. It must have a finite subcover But then {U α1,..., U αk, R S}. {U α1,..., U αk } is a finite subset of U which covers S. Hence S is compact. Putting the last three propositions together gives us the key theorem regarding compactness: Heine-Borel Theorem. A subset of R is compact if and only if it is closed and bounded. Proof: The implication follows from Proposition 2.1. Conversely, if K a closed bounded subset of R, then K is a closed subset of [ N, N] for some N N. Proposition 2.2 implies that [ N, N] is compact, and Proposition 2.3 implies that K is compact. As a corollary, we get: Bolzano-Weierstrass Theorem. Any bounded infinite subset of R must have an accumulation point. Proof: Suppose that S is an infinite subset of R which has no accumulation point. Then since S contains all its accumulation points, it must be closed. Since it also bounded, it must also be compact. For each x S there is an open neighborhood N(x) of x such that N(x) S = {x}. Thus we have an open cover {N(x) : x S} of S. Since S is compact, this cover must have a finite subcover, S N(x 1 ) N(x 2 ) N(x m ), 7

for some m N. But then we would have S = {x 1,..., x m }, a finite set, a contradiction. 3 The dot product If x = (x 1...., x n ) and y = (y 1...., y n ) are elements of R n, we define their dot product by x y = x 1 y 1 + + x n y n. The dot product satisfies several key axioms: 1. it is symmetric: x y = y x; 2. it is bilinear: (ax + x ) y = a(x y) + x y; 3. and it is positive-definite: x x 0 and x x = 0 if and only if x = 0. We define the length of x by x = x x. Note that the length of x is always 0. Cauchy-Schwarz Theorem. The dot product satisfies Sketch of proof: Expand the inequality and simplify to obtain which simplifies to or 1 x y 1. (1) x y 0 (x(y y) y(x y)) (x(y y) y(x y)) 0 (x x)(y y) 2 2(x y) 2 (y y) + (y y)(x y) 2, (x y) 2 (y y) (x x)(y y) 2, (x y) 2 x 2 y 2. Taking the square root yields the Cauchy-Schwarz inequality(1). The importance of the Cauchy-Schwarz inequality is that it allows us to define angles between vectors x and y in R n. Given an number t [ 1, 1], there is a unique angle θ such that θ [0, π] and cos θ = t. Thus we can define the angle between two nonzero vectors x and y in R n by requiring that θ [0, π] and cos θ = x y x y. 8

Thus we can say that two vectors vectors x and y in R n are perpendicular or orthogonal if x y = 0. This provides much of the intuition for dealing with vectors in R n. Corollary of Cauchy-Schwarz Theorem. If u, v R n, then u + v u + v. Proof: It suffices to check that or u + v 2 ( u + v ) 2 u 2 + 2u v + v 2 u 2 + 2 u v + v 2. But this follows immediately from the Cauchy-Schwarz inequality. 4 Topology of metric spaces Definition 4.1. A metric space is a set X together with a function d : X X R such that 1. d(x, y) 0 and d(x, y) = 0 x = y, 2. d(x, y) = d(y, x), and 3. d(x, z) d(x, y) + d(y, z). The last of these conditions is known as the triangle inequality. Example 1. The most basic example is X = R with d : R R R defined by d(x, y) = x y. Example 2. The example needed for multivariable calculus is X = R n with d : R n R n R defined by d(x, y) = x y, the length of x y being defined as in the preceding section. We call this metric the standard metric on R n. In verifying that this really is a metric space, the only difficulty is checking the triangle inequality. But this follows from the Corollary in the preceding section when u = x y and v = y z. Of course, this example includes the previous one as a special case. Example 3. Let S be a subset of R n. If x and y are elements of S, we let d(x, y) be the distance from x to y in R n. This provides numerous additional examples of metric spaces. 9

Further examples can be found in the text [1], 15. If (X, d) is a metric space and ɛ > 0 is given the ɛ-neighborhood N(x; ɛ) of a point x X is N(x; ɛ) = {y X : d(x, y) < ɛ}, while the deleted ɛ-neighborhood N(x; ɛ) of x is N (x; ɛ) = {y X : d(x, y) < ɛ, y x}. Definition 4.2. A subset U of X is said to be open if x U N(x; ɛ) U, for some ɛ > 0. A subset C of X is closed if its complement X C is open. A subset K X is bounded if K N(x; R) for some R > 0. For example, the set N(x; R) = {y X : d(x, y) < R} is open whenever R > 0. To see this it suffices to show that But y N(x; R) N(y; R d(x, y)) N(x; R). z N(y; R d(x, y)) d(y, z) < R d(x, y) It follows that the set d(x, z) < d(x, y) + d(y, z) < R d(x, z) < R z N(x : R). X N(x; R) = {y X : d(x, y) R} is closed. It is useful to become familiar with which sets in R 2 are open and closed. Thus the open ball of radius R about any given point (x 0, y 0 ) R 2, N((x 0, y 0 ); R) = {(x, y) R 2 : (x x 0 ) 2 + (y y 0 ) 2 < R 2 }, is an open subset of R 2 by the argument given in the preceding paragraph. However, it is not a closed subset of R 2 because the point (1, 0) lies in R 2 N((x 0, y 0 ); R), but there is no open neighborhood about (1, 0)) which is entirely contained in R 2 N((x 0, y 0 ); R). Similarly, one can show that the closed ball B((x 0, y 0 ); R) = {(x, y) R 2 : (x x 0 ) 2 + (y y 0 ) 2 R 2 } is closed but not open. 10

If (a, b) and (c, d) are nonempty open intervals, the Cartesian product (a, b) (c, d) is open but not closed. If [a, b] and [c, d] are nonempty closed intervals, [a, b] [c, d] is closed but not open. The metric d of a metric space (X, d) can be restricted to any subset S X, making S into a metric space in its own right. This gives rise to the notion of an open subset of S. (Note that for x S, the ɛ-neighborhoods of x in S are just the intersections N(x; ɛ) S, where N(x; ɛ) is an ɛ-neighborhood of x in X.) Proposition 4.1. Let (X, d) be a metric space and let S be a subset of X. Then a subset U of S is open if and only if U = V S, where V is open in X. Proof: Suppose that V is an open subset of X and U = V S. If x V S, then since V is open in X, there exists an ɛ > 0 such that N(x; ɛ) V. But then N(x; ɛ) S V S = U. So U is open in S. Conversely, if U is an open subset of S, then for x U S, there exists an ɛ x > 0 such that N(x; ɛ x ) S U. We let V = {N(x; ɛ x ) : x U S}. Then V is an open subset of X and V S = U. Much of the theory of open and closed sets can be generalized to metric spaces. Thus, for example, we have: Proposition 4.2. If (X, d) is a metric space, the empty set and the whole space X are open. The union of an arbitrary collection of open sets is open. The intersection of a finite collection of open sets is open. The proof is virtually the same as the proof given in 1. Definition. We say that a subset K of a metric space (X, d) is compact if every open cover of K has a finite subcover. 5 The Heine-Borel Theorem for R n The Heine-Borel Theorem for R we have proven above is actually a special case of: Heine-Borel Theorem for R n. A subset of R n with the standard metric is compact if and only if it is closed and bounded. The proof we gave before actually extends from R to R n. First we need to show that a compact subset K of R n is bounded. If not we can construct a collection of open sets, U = {U n : n N}, U n = N(0; n), which covers K and has no finite subcover. 11

Next we must show that a compact subset K of R n is closed. If K is not closed there exists x R n K which lies in K. In this case, we can construct a collection of open sets, U = {U n : n Z}, U n = N(x; 2 n ), which covers K but has no finite subcover. Finally, we must show that if K is a closed bounded subset of R n, K is compact. For this we need a modification of our earlier nesting lemma. We say that I R n is a closed n-cube if I = [a 1, b 1 ] [a n, b n ], with b 1 a 1 = = b n a n. We say that the length of the closed n-cube is b 1 a 1. Nesting Lemma. Suppose that {I k : k N} is a decreasing nested sequence of nonempty closed n-cubes in R n, Then {I n : n N} is nonempty. Proof of Lemma: Suppose that Then for each j, 1 j n, I 1 I 2 I n I n+1. I k = [a k,1, b k,1 ] [a k,n, b k,n ]. [a 1,j, b 1,j ] [a 1,j, b 1,j ] [a 1,j, b 1,j ] is a nested sequence of intervals. By the earlier nesting lemma, there exists a point x j {[a k,j, b k,j ] : k N}. Then the point (x 1,..., x n ) lies in {I n : n N}. Once we have this Nesting Lemma, we can prove that for any M > 0, the closed unit n-cube I 1 = [ M, M] [ M, M] of length 2M is compact by a modification of the argument we gave before. Suppose that U = {U α : α A} is an open cover of I 1 with no finite subcover. We can divide ti 1 = [a, b] by the coordinate hyperplanes into 2 n closed cubes of length M. At least one of these cubes I 2 cannot be covered by a finite subcollection of U; otherwise the original interval could be so covered. We then divide I 2 into 2 n closed cubes of length M/2. Once again, at least one of these cubes, say I 3 cannot be covered by a finite subcollection of U. We continue in a similar fashion, thereby obtaining a decreasing nested sequence I 1 I 2 I n I n+1 with Length(I n+1 ) = 1 2 Length(I n) 12

such that no I n is covered by a finite subcover of U. By the new Nesting Lemma, {I n : n N} contains some element (x 1,..., x n ). Let U α be an element of U which contains x. Since U α is open there exists ɛ > 0 such that But then (x 1 ɛ, x 1 + ɛ) (x n ɛ, x n + ɛ) U α. I n (x 1 ɛ, x 1 + ɛ) (x n ɛ, x n + ɛ) U α for n sufficiently large. This contradicts the fact that I n was not covered by a finite subcover of U = {U α : α A}. Hence the n-cube I 1 = [ M, M] [ M, M] is compact. Finally, if K is any closed bounded subset of R n, then K is a closed subset of [ M, M] [ M, M] for some M > 0. The proof of Proposition 2.3 now shows that K is compact. Bolzano-Weierstrass Theorem for R n. Any bounded infinite subset of R n must have an accumulation point. This follows from the Compactness Criterion, once we have the Heine-Borel Theorem for R n. References [1] Steven R. Lay, Analysis: with an introduction to proof, Pearson Prentice Hall, Upper Saddle Riven, NJ, 2005. [2] W. Rudin, Principles of mathematical analysis, Third edition, McGraw- Hill, New York, 1976. 13