2

SOME BASIC TOPOLOGICAL PROPERTIES OF R^P

2.1UNIONS AND INTERSECTIONS OF OPEN AND CLOSED SETS

Properties that can be expressed in terms of open and closed sets are called topological. In this chapter, we develop results of this type. First, we need to know if we can take unions and intersections of open sets and still remain within the class of open sets (a similar question arises for closed sets). Suppose, for example, that A₁, A₂, and A₃ are open (all sets are subsets of a fixed metric space Ω); is A₁ ∪ Α₂ ∪ A₃ open? If x belongs to , then x is in A_i for some i, and we can find an open ball B_r(x) contained in A_i (see Fig. 2.1.1). But then B_r(x) is a subset of , proving that A₁ ∪ A₂ ∪ A₃ is open. In fact, this argument works for an arbitrary collection of open sets, and we have our first result.

2.1.1THEOREM. An arbitrary union of open sets is open. Thus, if A_i is open for all i (where the A_i’s can form a finite collection, a countably infinite collection, or even an uncountably infinite collection), then is open.

Now let’s consider the same question for intersections of open sets. If x belongs to A₁ ∩ A₂ ∩ A₃, where A₁, A₂, and A₃ are open sets, we can find open balls B_r1(x), B_r2(x), and B_r3(x) with B_rj(x) ⊆ A_j for each j. If we simply take the smallest radius, that is, r = min_j r_j, then B_r(x) ⊆ A₁ ∩ Α₂ ∩ A₃. This gives us the second result (see Fig. 2.1.2).

Figure 2.1.1 An Arbitrary Union of Open Sets Is Open

2.1.2THEOREM. A finite intersection of open sets is open. That is, if n is a positive integer and A₁,…, A_n are open sets, then is open.

Why doesn’t the above argument work when there are infinitely many sets? Simply because the “smallest radius” r might be 0; for example, we might have r_j = 1/j, j = 1, 2.… Now if a particular argument fails, we cannot conclude that a result is invalid. However, in this case we can give an explicit example to show that Theorem 2.1.2 is false for infinite intersections. Let A_n be the open interval (–1/n, 1/n) in R. Then is the set of real numbers x such that – 1/n < x < 1/n for every positive integer n; thus , the set consisting of {0} alone, which is not open.

    Caution. The notation , although standard, may be confusing. It means the set of points x such that x belongs to A_n for every positive integer n. There is no set A_∞; in other words, infinity is not a positive integer.

Figure 2.1.2 A Finite Intersection of Open Sets Is Open

To obtain appropriate results for sets that are closed rather than open, we can use the De Morgan laws and Theorem 1.3.2.

2.1.3THEOREM

        (a) An arbitrary intersection of closed sets is closed. In other words, if B_i is closed for all i, then is closed.

        (b) A finite union of closed sets is closed. That is, if n is a positive integer and B₁,…, B_n are closed sets, then is closed.

        Proof

        (a)By the De Morgan laws, , which is open by Theorems 1.3.2 and 2.1.1. Thus, by Theorem 1.3.2, is closed.

        (b)Again by the De Morgan laws, , which is open by Theorems 1.3.2 and 2.1.2. By Theorem 1.3.2, is closed. ■

Problems for Section 2.1

    1.Give an example of an infinite union of closed sets that is not closed.

    2.Show that an open subset of a metric space can be expressed as a union of open balls.

            The following problems will help to strengthen the result of Problem 2 later, in the special case when the metric space is R.

    3.Let V be an open subset of R. If x ∈ V, let V_x be the union of all open intervals I such that x ∈ I and I ⊆ V. For example, if V = (0, 1) ∪ (2, 3) ∪ (4, 5) and x = 2.7, then V_x = (2, 3), the largest open interval containing 2.7 and contained in V.

            Later, we show that V_x is always an open interval, but let’s assume this result for now. If x ≠ y, show that V_x and V_y are either disjoint or identical.

    4.Continuing Problem 3, show that there are only countably many distinct V_x’s. (The counting technique used here is very basic. Form a set S by choosing an element from each distinct V_x. Since the distinct V_x are disjoint, all the elements of S are distinct. If S turns out to be countable, there can only be countably many distinct V_x.)

2.2COMPACTNESS

When the real numbers are defined by a set of axioms, one property of a topological nature is built right into the axioms, and the other topological properties are then derived. There is a wide choice of the particular property to assume. Our choice will be somewhat unorthodox and is motivated by our desire to keep proofs as simple as possible.¹ Eventually we consider the standard approach and examine its relationship to ours. Of course, if the reals are obtained by a constructive procedure, then we don’t have to assume anything, but at some point we have to choose how to construct the real numbers, and the choice is made so that all the properties we regard as desirable can be obtained. First, some terminology.

2.2.1Definition

The set B is said to be bounded if it can be enclosed by a ball, that is, if B ⊆ B_r(x) for some x ∈ Ω and r > 0.

We now state our assumption (due to Cantor).

2.2.2Nested Set Property

If B₁, B₂, … is a sequence of closed, bounded, nonempty subsets of R^P, and the sequence is nested, that is, B_n+1 ⊆ B_n for all n = 1, 2,…, then is not empty.

Property 2.2.2 probably looks obscure, but if we examine closed intervals in R the result should appear more reasonable. Suppose B_j = [a_j, b_j], j = 1, 2, …. The nesting requirement means that the B_j are shrinking as j increases, in other words, the a_j will increase (not necessarily strictly) and the b_j will decrease. It is reasonable to expect that a_j will approach a limit a, and b_j will converge to b. Since a_j ≤ b_j for all j, we have a ≤ b; consequently, .

Note that the Nested Set Property does not apply to open sets. For example, in R the sets A_n = (0, 1/n) form a nested sequence, but .

Whenever we use the words “increase” and “decrease,” as above, the strict connotation is not implied.

Now consider the following problem. If x is a point of the interval [0, 1], let G be the open interval (x – 1/4, x + 1/4). The sets G_x cover [0,1]; that is, [0, 1] . Do we actually need all the G_x in order to form a covering? In fact G_.2, G_.5, and G_.8 are sufficient, since any x in [0, 1] will belong to at least one of these three sets. In this way, we have reduced the given open covering to a finite subcovering. If a set has the property that such a reduction can always be carried out, it is called compact.

2.2.3Definition

Let K be a subset of the metric space Ω; K is compact if every open covering of K has a finite subcovering; that is, if , where the G_i are open sets, then K ⊆ G_i1 ∪ G_i2 ∪ … ∪ G_in for some i₁,…, i_n.

It will be easy to recognize compact sets in R^P; they are precisely the sets that are closed and bounded. First we show that a compact set in any metric space is always closed and bounded.

2.2.4THEOREM. If K is a compact subset of the metric space Ω, then K is closed and bounded.

    Proof. To show K closed, we prove that K^c is open. Assume x ∉ K, and let G_m = {y : d(x, y) > 1/m}, m = 1, 2, …. If y ∈ K, then x ≠ y; hence, d(x, y) > 1/m for some m; therefore y ∈ G_m (see Fig. 2.2.1). Thus, , and by compactness we have a finite subcovering. Now observe that the G_m form an increasing sequence of sets (G₁ ⊆ G₂ ⊆ G₃ ⊆ …); therefore, a finite union of some of the G_m, for example G₃ ∪ G₇ ∪ G₁₁, is equal to the set (G₁₁) with the highest index. Thus, K ⊆ G_s for some s, and it follows that B_1/S (x) ⊆ K^c. (If d(x, y) < 1/s, then y ∉ G_s so y ∉ K.) Therefore, K^c is open.

To show K is bounded, fix r > 0. If x ∈ K, then x ∈ B_r(x), so K is covered by the B_r(x); by compactness, we have a finite subcovering, say K ⊆ B_r(x₁) ∪ ··· ∪ B_r(x_n). But a finite union of open balls can be placed inside a single open ball, so K is bounded. ■

We now show that closed, bounded subsets of R^P are compact.

Figure 2.2.1 Proof That a Compact Set Is Closed

2.2.5 Heine–Borel Theorem

If K is a subset of R^P, then K is compact if and only if K is closed and bounded.

    Proof. In view of Theorem 2.2.4, we need only show that if K is closed and bounded, it must be compact. Assume , with all G_i open. If x ∈ K, then x ∈ G_i for some i, and since G_i is open there is an open ball B_r(x) ⊆ G_i. Now there is a rational number y in R^P (this means y = (y₁, …, y_p), where each y_i is rational) and a rational number s > 0 such that x ∈ B_s(y) and B_s(y) ⊆ B_r(x): see Fig. 2.2.2. Let’s write down the relevant information in a table with three (very long) columns:

The first column contains the points x in K ; the second and third columns contain specific choices of G_i and B_s(y). It might happen that a particular B_s(y) appears more than once on the list, with a different G_i; if so, toss out the new G_i and replace it by the original. Since there are only countably many sets B_s(y), there will be only countably many G_i, in other words, K will be covered by a countable union of the G_i. The fact that K is closed and bounded has not yet been used; we have shown that if K is an arbitrary subset of R^P an open covering of K has a countable subcovering.

Figure 2.2.2 Proof of the Heine-Borel Theorem

It remains to reduce the countable covering to a finite subcovering. Assume ; if H_n = G₁ ∪ ··· ∪ G_n, then the H_n are open and H₁ ⊆ H₂ ⊆ H₃ ⊆ ··· and . Let ; since and K are closed and K is bounded, the B_n are closed and bounded. Furthermore, B_n+1 ⊆ B_n (because H_n ⊆ H_n+1, and therefore ). If all B_n are nonempty, the Nested Set Property implies that there is a point . By definition of B_n we have x ∈ K, and for all n, ; that is, x ∉ H_n. This contradicts the fact that .

It follows that for some . Now if y ∈ K, then y must be in H_m; otherwise . Thus, , a finite subcovering. ■

The following examples may aid the intuition. If A = (0, 1), then A is bounded but not closed, and, therefore, is not compact. In fact, if G_n = (0, 1 – 1/n), n = 1, 2,…, then , but there is no finite subcovering. If ∈ > 0 and H₁ = (–∈, ∈), H₂ = (1 – ∈, 1 + ∈), then the G_n together with H₁ and H₂ cover the compact set [0, 1], but there is a finite subcovering. (We can use H₁, H₂ and one of the G_n if 1/n < ∈.)

The sets (–r, r), r real and greater than zero, cover R, but there is no finite subcovering. However, there is a countable subcovering: .

If Z is the set of integers, then Z is closed (Z^c is open) but unbounded, and thus not compact. Explicitly, if G_n = (n – a, n + 0), where 0 < a < 1/2, n = 1, 2,…, the G_n form an open covering of Z with no finite subcovering.

Problems for Section 2.2

    1.Which of the following subsets of R are compact?

        (a)(–∞, 3)

        (b)(–7, 3]

        (c)[–7, 3]

        (d)[–7, 3)

        (e)[3, ∞)

    2.The set C = {x ∈ R : x ≥ 1} is closed but unbounded and, hence, not compact. Given an explicit example of an open covering of C with no finite subcovering.

    3.Let D = {(x, y) : x² + y² < 1}, a bounded but not closed, and hence not compact, subset of R².

        (a)Give an example of an open covering of D with no finite subcovering.

        (b)Give an example of an open covering of D that does have a finite subcovering.

    4.Why doesn’t the example of Problem 3(b) contradict the noncompactness of D?

    5.Is the empty set compact?

2.3SOME APPLICATIONS OF COMPACTNESS

Compactness is closely related to convergence of subsequences. It may be useful at this point to give the formal definitions of sequences and subsequences. A sequence x₁, x₂, … in a metric space Ω is really a function from the positive integers to Ω, in other words, for each positive integer n we have a point x_n in Ω. To form a subsequence of this sequence, we assign to each positive integer i a point x_ni, subject to the requirement that the indices are strictly increasing; that is, i < j implies n_i < n_j. In other words, x₂, x₆, x₁₅, x₁₇,… is allowable but not x₃, x₉, x₈, x₁₄, … or x₃, x₉, x₉, x₁₄,.…

Sometimes the notation {x₁, x₂,…}, or simply {x_n}, is used both for the sequence and for its set of values, but the mathematical ideas are different. For example, say we have a constant sequence: x_n = 3 for all n. The sequence is a function that assigns to each positive integer n the number 3; the set of values of the sequence is the set {3} consisting of 3 alone. We’ll try not to cause any notational confusion. If we want to use the notation {x_n} for a sequence, we’ll say “the sequence {x_n}” to make it clear that we mean the sequence, not its set of values.

A sequence of points in a compact set always has a convergent subsequence.

2.3.1THEOREM. Let K be a compact subset of the metric space Ω. If x₁, x₂, … is a sequence of points in K, there is a subsequence x_n1, x_n2,… converging to a limit x in K.

    Proof. Fix a point a ∈ K and consider the following inductive process. Find (if possible) a positive integer n₁ such that x_n1 ∈ B₁(a). If this can be done, then find (if possible) a positive integer n₂ > n₁ such that x_n2 ∈ Β_1/2(a). In general, we try to find positive integers n₁ < n₂ < ··· such that x_nj ∈ B_1/j(a), j = 1, 2,… (see Fig. 2.3.1). If the process does not terminate, then x_nj → a and we are finished. If the process terminates at n_i then B_1/(i+1)(a) contains no point x_n, n > n_i. Thus if r_a = 1/(i + 1), then x_n ∈ B_ra(a) for only finitely many n. Carry out the procedure for each a ∈ K. If for some a the process does not terminate, we are finished, so assume that for each a ∈ K we get an open ball G_a = B_ra(a) such that x_n ∈ B_ra(a) for only finitely many n. Since a ∈ G_a, we have , and, by compactness, there is a finite subcovering, say . It follows that x_n ∈ K for only finitely many n, contradicting the assumption that the entire sequence lies in K. ■

If we specialize Theorem 2.3.1 to R^p, we obtain the following basic result.

2.3.2Bolzano–Weierstrass Theorem

Let x₁, x₂,… be a bounded sequence in R^p (in other words, the set of values {x₁, x₂,…} is a bounded subset of R^p). Then there is a subsequence x_n1 x_n2, ··· converging to a point x in R^p.

    Proof. We may assume that all the x_n belong to a fixed closed ball C_r(a) = {x : d(x, a) ≤ r }. By the Heine–Borel Theorem 2.2.5, C_r(a) is compact, and the result follows from Theorem 2.3.1. ■

Figure 2.3.1 Proof of Theorem 2.3.1

2.3.3COROLLARY. If A is a bounded infinite subset of R^p, then A has a limit point

    Proof. Let x₁, x₂, … be a sequence of distinct points in A; such a sequence exists because A is infinite. By Theorem 2.3.2 we have a convergent subsequence x_ni → x ∈ R^p. Now consider the open ball B_r(x); for large enough i, x_ni ∈ B_r(x). If x_ni = x, then (because the points of the sequence are distinct) x_ni+1 ≠ x. Thus, we can find a pointy ∈ A ∩ B_r(x) with y ≠ x. Therefore, x is a limit point of A. ■

If x₁, x₂,… is a convergent sequence in R^p (say x_n → x), the sequence must be bounded, for if r > 0 then x_n will lie in the open ball B_r(x) for all sufficiently large n, say for n ≥ N. This leaves only finitely many points x₁, …, x_{N s–1}, so if we make s sufficiently large we can enclose the entire sequence in the ball B_s(x). Thus, an unbounded sequence cannot converge. It is sometimes convenient to relax this requirement, especially when working in the set of real numbers R. We can do this by adding +∞ and –∞ to R to obtain the set of extended real numbers . The rules of order and arithmetic in are natural:

We do not define 0·∞, although in some areas of mathematics (notably measure theory), it is required that 0 · ∞ = 0.

When working in , we encounter the problem that any point in R is infinitely far from +∞ and –∞. It is possible to modify the Euclidean metric so that we get a legitimate metric on without changing the open sets of R, but this involves more topological equipment than we have available at present. Instead, let’s define convergence to ±∞ directly:

        x_n → ∞ means for each positive real number M, x_n > M for all sufficiently large n; that is, there is a positive integer N such that x_n > M for all n ≥ N.

        x_n → –∞ means for each negative real number M, x_n < M for all sufficiently large n.

We can now establish an important property of unbounded sequences.

2.3.4THEOREM. If x₁, x₂, … is an unbounded sequence in R, there is either a subsequence converging to +∞ or a subsequence converging to –∞. Thus, by Theorem 2.3.2, every sequence in R has a convergent subsequence if +∞ and –∞ are allowed as limits.

    Proof. Since the sequence is unbounded, not all x_n can lie in the interval (–1, 1); hence, |x_n1| ≥ 1 for some n₁. The sequence x_{n1 + 1}, x_{n1 + 2}, … is also unbounded, so not all of its points can lie in (–2, 2); consequently, |x_n2| ≥ 2 for some n₂ > n₁. Inductively, we find a sequence x_n1, x_n2,… such that |x_nk| ≥ k for all k. Now either x_n1 > 0 for infinitely many i, or x_ni < 0 for infinitely many i. In the first case we can toss out the negative terms to obtain a subsequence converging to +∞; in the second case we remove the positive terms to obtain a subsequence converging to –∞. ■

We conclude this section with a convergence result that will be used many times.

2.3.5THEOREM. If {x_n} is a sequence of real numbers that converges to the real number L, and x_n ≤ c for all n, then L ≤ c. (Similarly, if x_n → L and x_n ≥ c for all n, then L ≥ c.)

    Proof. If L > c, then, since x_n → L, x_n will be greater than c for all sufficiently large n, a contradiction. ■

Problems for Section 2.3

    1.Find the limits, if they exist, of each of the following sequences {x_n} in R.

        (a)

        (b)n³e^–n

        (c)arctan n

        (d)

        (e)

    2.Suppose x₁, x₂, … ∈ R^p, y ∈ R^p, and d(x_n, y) ≤ k for all n. If x_n → x, show that d(x, y) ≤ k.

    3.Given an example of a sequence in the noncompact set (0, 1) with no subsequence converging to a point of (0, 1).

    4.Consider the extended reals and recall the definition of open set: G is open if and only if, for each x ∈ G, there is an open ball B_r(x) ⊆ G. If we try to extend this idea to , difficulties arise when x = +∞ or –∞. However, if x = +∞ we can replace B_r(x) by a set of the form , and if x = –∞ we can use . With this adjustment, show that every open covering of has a finite subcovering, so is compact.

    5.Continuing Problem 4, note that we may use the same definition of a closed subset of : C is closed if and only if for every sequence x₁, x₂,… of points of C converging to a limit , we must have x ∈ C. Give an example of a set A ⊆ R such that A is closed in R but not in ; in other words, if we regard R as the universe, A is a closed set, but if we regard as the universe, A is not closed.

    6.Continuing Problem 5, is it possible for a set A ⊆ R to be closed in but not in R? Explain.

2.4LEAST UPPER BOUNDS AND COMPLETENESS

We now examine the problem of finding the maximum of a set of real numbers, a familiar idea from calculus. For example, if E is the interval [0,1], we observe immediately that 1 is the largest element of E. But what if 1 is removed, so that we have a new set Ε₁ = [0, 1)? In this case, E₁ has no largest element, but 1 is the “closest thing to it.” Mathematically, we have an example of a least upper bound, which we now define.

2.4.1Definitions

Let E be a nonempty subset of R. The real number x is said to be an upper bound of E if x ≥ y for every y ∈ E : similary, x is a lower bound of E if x ≤ y for every y ∈ E. We say that x is a least upper bound or supremum (abbreviated sup) of E if x is an upper bound that is less than or equal to all other upper bounds of E ; similarly, x is a greatest lower bound or infimum (abbreviated inf) of E if x is a lower bound that is greater than or equal to all other lower bounds of E.

Note that if x and y are least upper bounds of E, we must have x = y, for if (say) x < y, then y cannot be the smallest upper bound. Thus, we may refer without ambiguity to “the” least upper bound. The following existence theorem is basic.

2.4.2THEOREM. Let E be a nonempty subset of R. If E has an upper bound, then E has a least upper bound. Similarly, a nonempty subset of R that has a lower bound has a greatest lower bound.

    Proof. Let x₁ be an upper bound of E, and let y₁ ∈ E, so that y₁ ≤ x₁. Let m be the midpoint of the interval [y₁, x₁] (see Fig. 2.4.1). If m is an upper bound of E, take x₂ = m, y₂ = y₁ so that x₂ is an upper bound of E and y₂ ∈ E. If m is not an upper bound of E, then some element of E, say y₂, is greater than m. (Necessarily y₂ ≤ x₁ since x₁ is an upper bound of E.) Take x₂ = x₁, and again x₂ is an upper bound of E and y₂ ∈ E. Inductively, we obtain closed intervals

furthermore, the x_n decrease and the y_n increase, so the C_n form a nested sequence of closed, bounded, nonempty sets. By the Nested

Figure 2.4.1 Proof of Theorem 2.4.2

Set Property 2.2.2, contains at least one point x. Since y_n ≤ x ≤ x_n and

we have x_n → x and y_n → x. We claim that x is the least upper bound of E. If y ∈ E, then y ≤ x_n for all n; let n → ∞ to conclude by Theorem 2.3.5 that y ≤ x, so that x is an upper bound. If x′ is an upper bound, then y_n ≤ x′ for all n, and consequently x ≤ x′.

The existence of the greatest lower bound is established by a similar argument that uses lower bounds instead of upper bounds. For example, we let x₁ be a lower bound of E and y₁ ∈ E so that x₁ ≤ y₁. If m is a lower bound, take x₂ = m, y₂ = y₁; if m is not a lower bound, let y₂ ∈ E with y₂ < m, and take x₂ = x₁. Inductively, we obtain C_n = [x_n, y_n], where x_n is a lower bound of E and y_n ∈ E. Just as above, if , then x = inf E. ■

If E is a nonempty subset of R with no upper bound, the smallest element of that is greater than or equal to every member or E is +∞; thus, we may write sup E = +∞. Similarly if E has no lower bound, inf E = –∞. Thus, if +∞ and –∞ are allowed, every nonempty subset of R has a sup and an inf.

The following properties of sups and infs are often used.

2.4.3THEOREM. Let E be a nonempty subset of R. If x < sup E, then there is a point y ∈ E such that x < y; if x > inf E, there is a point y ∈ E such that x > y. Furthermore, there is a sequence of points in E converging to sup E, and a sequence of points in E converging to inf E. Thus, if E is closed, then sup E and inf E (if finite) actually belong to E.

    Proof. Intuitively, if the largest member of a set is greater than x, then some member of the set is greater than x. Formally, if x is smaller than the least upper bound of E, then x cannot be an upper bound. Therefore, x must be smaller than some y ∈ E. (In terms of quantifiers, x is an upper bound of E iff (∀y ∈ E)y ≤ x. Thus x is not an upper bound of E iff (∃y ∈ E)(y > x).) Similarly, if x > inf E, then x is not a lower bound; hence, x > y for some y ∈ E.

If sup E is finite, then sup E – 1/n < sup E. Hence, by what we have just proved, we can find x_n ∈ E with sup E – 1/n < x_n (of course x_n ≤ sup E). If sup E = +∞, then n < sup E, and thus we can find x_n ∈ E with x_n > n. In either case, x_n → sup E. To obtain a sequence converging to inf E, use the fact that inf E + 1/n > inf E if inf E is finite, and – n > inf E if inf E = –∞. ■

You may remember the following problem from calculus. You are trying to show that a certain infinite series Σ_n a_n is convergent. If s_n is the nth partial sum, then, for n > m,

If becomes very small for large n and m, we may conclude that s_n – s_m → 0 as n, m → ∞. The inference that s_n must approach a limit is based on a key property of R called completeness, which we now explore.

2.4.4Definitions and Comments

The sequence {x_n} in the metric space Ω is said to be a Cauchy sequence if d(x_n, xm) → 0 as n, m → ∞. In other words, given ∈ > 0 there is a positive integer N such that whenever n and m are greater than or equal to N we have d(x_n, x_m) < ∈. Every convergent sequence is Cauchy. for if x_n → x, then

Furthermore, every Cauchy sequence is bounded, for if x₀ is any point of Ω, ∈ > 0 is arbitrary, and d(x_n, x_m) < ∈ whenever n, m ≥ N, then

But d(x_N, x_n) < ∈ for n ≥ N, and for n < N,

Thus, for some positive constant c, d(x₀, x_n) ≤ c for all n, as desired.

A metric space Ω in which every Cauchy sequence converges to an element of Ω is called complete. The following result is basic.

2.4.5THEOREM. R^p is complete.

    Proof. Let {x_n} be a Cauchy sequence in R^p. Since {x_n} is bounded, the Bolzano-Weierstrass Theorem 2.3.2 gives us a convergent subsequence {x_nj}. If x_nj → x, then

Given ∈ > 0, ∃N such that d(x_n, x_m) < ∈/2 for n, m ≥ N. Choose j so large that n_j ≥ N and d(x_nj, x) < ∈/2. If n ≥ N, it follows that d(x_n, x) < ∈. Therefore x_n → x. ■

Analysis in R is simplified by the result that monotone sequences, that is, sequences that are either increasing (x_n ≤ x_n+1 for all n) or decreasing (x_n ≥ x_n+1) always converge, if we allow ±∞ as limits.

2.4.6THEOREM. If {x_n} is a monotone bounded sequence in R, then x_n converges to a limit x ∈ R. An unbounded increasing sequence converges to +∞, and an unbounded decreasing sequence converges to –∞.

    Proof. Assume x₁ ≤ x₂ ≤ … The decreasing case is handled similarly. Let x be the sup of the set of values {x₁, x₂, …}. In the bounded case, x is finite. If ∈ > 0, then x – ∈ < x, so by Theorem 2.4.3, x – ∈ < x_n0 (≤ x) for some n₀. By monotonicity, x – ∈ < x_n ≤ x for all n ≥ n₀, and consequently x_n → x. In the unbounded case, x = +∞, and again by Theorem 2.4.3 for any positive number M we can find x_n0 > M, so x_n > M for all n ≥ n₀. Thus, x_n → ∞. ■

Problems for Section 2.4

    1.In Theorem 2.4.2, obtain the existence of the greatest lower bound by an alternative approach, namely, by considering – E = {–x : x ∈ E}.

    2.If {x_n} is a Cauchy sequence with a subsequence converging to x, show that x_n → x.

    3.Let K be a compact subset of the metric space Ω. Show that K is complete; that is, any Cauchy sequence in K converges to a point of K.

    4.Give an example of a metric space that is not complete.

    5.If K is a subset of R^p such that every sequence of points in K has a subsequence converging to a point of K, show that K is closed and bounded, and hence compact.

    6.Define , n = 1, 2,…, with .

        (a)Show that {x_n} is a monotone bounded sequence and, hence, converges to a finite limit L.

        (b)Find L (approximately).

    7.Let V be an open subset of R. If x ∈ V, let V_x be the union of all open intervals I such that x ∈ I and I ⊆ V (see Section 2.1, Problems 3 and 4). Define

            and show that V_x = (a_x, b_x).

    8.Continuing Problem 7, show that every open subset of R can be expressed as a disjoint union of countably many open intervals.

    9.True or false: if E is a nonempty subset of R, then sup E is a limit point of E.

    10.True or false: if sup E and inf E belong to E then E is closed.

REVIEW PROBLEMS FOR CHAPTER 2

    1.Give an example of a closed set of real numbers that is not compact.

    2.Let E be a nonempty subset of R, and assume that E has a least upper bound x ∈ R. If x does not belong to E, show that x is a limit point of E.

    3.Recall (Section 1.5, Problem 8) that a point x is a boundary point of the set E if every open ball B_r(x) contains both a point of E and a point of E^c. If E is a nonempty, bounded subset of R, show that E has at least one boundary point.

    4.A sequence is defined inductively by

            If {x_n} is known to converge to a limit L, find L explicitly.

    5.Let x₁ > 1, x_n+1 = 2 – (1/x_n), n ≥ 1. Show that {x_n} converges, and find the limit.

    6.Let f and g be bounded functions from Ω to R; in other words, for some M > 0 we have |f (x)| ≤ M for all x ∈ Ω, and similarly for g. Show that

            where, for example, sup_{x ∈ Ω} f(x) is the least upper bound of the set of all f(x), x ∈ Ω. (If you have trouble, think intuitively about maximizing f(x) + g(x) as x ranges over Ω.)

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¹Despite our best efforts, the proofs in this section may still be difficult, especially the Heine–Borel theorem. If you have trouble with the details, don’t worry, it’s normal. Concentrate on the explicit examples and on understanding the definitions, and you should be able to cope with the applications later.