Advanced Calculus of Several Variables (1973)
Part I. Euclidean Space and Linear Mappings
Chapter 8. ELEMENTARY TOPOLOGY OF Rn
In addition to its linear structure as a vector space, and the metric structure provided by the usual inner product, Euclidean n-space n possesses a topological structure (defined below). Among other things, this topological structure enables us to define and study a certain class of subsets of n, called compact sets, that play an important role in maximum-minimum problems. Once we have defined compact sets, the following two statements will be established.
(A)If D is a compact set in n, and f : D → m is continuous, then its image f(D) is a compact set in m (Theorem 8.7).
(B)If C is a compact set on the real line , then C contains a maximal element b, that is, a number such that for all .
It follows immediately from (A) and (B) that, if f : D → is a continuous real-valued function on the compact set , then f(x) attains an absolute maximum value at some point . For if b is the maximal element of the compact set , and a is a point of D such that f(a) = b, then it is clear that f(a) = b is the maximum value attained by f(x) on D. The existence of maximum (and, similarly, minimum) values for continuous functions on compact sets, together with the fact that compact sets turn out to be easily recognizable as such (Theorem 8.6), enable compact sets to play the same role in multivariable maximum-minimum problems as do closed intervals in single-variable ones.
By a topology (or topological structure) for the set S is meant a collection of subsets, called open subsets of S, such that satisfies the following three conditions:
(i)The empty set and the set S itself are open.
(ii)The union of any collection of open sets is an open set.
(iii)The intersection of a finite number of open sets is an open set.
The subset A of n is called open if and only if, given any point , there exists an open ball Br(a) (with r > 0) which is centered at a and is wholly contained in A. Put the other way around, A is open if there does not exist a point such that every open ball Br(a) contains points that are not in A. It is easily verified that, with this definition, the collection of all open subsets of n satisfies conditions (i)–(iii) above (Exercise 8.1).
Examples (a) An open interval is an open subset of , but a closed interval is not. (b) More generally, an open ball in n is an open subset of n (Exercise 8.3) but a closed ball is not (points on the boundary violate the definition). (c) If F is a finite set of points in n, then n − F is an open set. (d) Although is an open subset of itself, it is not an open subset of the plane 2.
The subset B of n is called closed if and only if its complement n − B is open. It is easily verified (Exercise 8.2) that conditions (i)–(iii) above imply that the collection of all closed subsets of n satisfies the following three analogous conditions:
(i′) and n are closed.
(ii′)The intersection of any collection of closed sets is a closed set.
(iii′)The union of a finite number of closed sets is a closed set.
Examples: (a) A closed interval is a closed subset of . (b) More generally, a closed ball in n is a closed subset of n (Exercise 8.3). (c) A finite set F of points is a closed set. (d) The real line is a closed subset of 2. (e) If Ais the set of points of the sequence , together with the limit point 0, then A is a closed set (why?)
The last example illustrates the following useful alternative characterization of closed sets.
Proposition 8.1 The subset A of n is closed if and only if it contains all of its limit points.
PROOFSuppose A is closed, and that a is a limit point of A. Since every open ball centered at a contains points of A, and n − A is open, a cannot be a point of n − A. Thus .
Conversely, suppose that A contains all of its limit points. If , then b is not a limit point of A, so there exists an open ball Br(b) which contains no points of A. Thus n − A is open, so A is closed.
If, given , we denote by the union of A and the set of all limit points of A, then Proposition 8.1 implies that A is closed if and only if A = .
The empty set and n itself are the only subsets of n that are both open and closed (this is not supposed to be obvious—see (Exercise 8.6). However there are many subsets of n that are neither open nor closed. For example, the set Q of all rational numbers is such a subset of .
The following theorem is often useful in verifying that a set is open or closed (as the case may be).
Theorem 8.2 The mapping f: n → m is continuous if and only if, given any open set , the inverse image f−1(U) is open in n. Also, f is continuous if and only if, given any closed set is closed in n.
PROOFThe inverse image f−1(U) is the set of points in n that map under f into U, that is,
We prove the “only if” part of the Theorem, and leave the converse as Exercise 8.4.
Suppose f is continuous. If is open, and , then there exists an open ball Since f is continuous, there exists an open ball Bδ(a) such that . Hence this shows that f−1(U) is open.
If is closed, then m − C is open, so f−1 (m − C) is open by what has just been proved. But f−1(m − C) = n − f−1(C), so it follows that f−1(C) is closed.
As an application of Theorem 8.2, let f : n → be the continuous mapping defined by f(x) = x − a, where is a fixed point. Then f−1((−r, r)) is the open ball Br(a), so it follows that this open ball is indeed an open set. Also f−1([0, r]) = r(a), so the closed ball is indeed closed. Finally,
so the (n − 1)-sphere of radius r, centered at a, is a closed set.
The subset A of n is said to be compact if and only if every infinite subset of A has a limit point which lies in A. This is equivalent to the statement that every sequence of points of A has a subsequence which converges to a point . (This means the same thing in n as on the real line: Given > 0, there exists N such that ) The equivalence of this statement and the definition is just a matter of language (Exercise 8.7).
Examples: (a) is not compact, because the set of all integers is an infinite subset of that has no limit point at all. Similarly, n is not compact. (b) The open interval (0, 1) is not compact, because the sequence is an infinite subset of (0, 1) whose limit point 0 is not in the interval. Similarly, open balls fail to be compact. (c) If the set F is finite, then it is automatically compact because it has no infinite subsets which could cause problems.
Closed intervals do not appear to share the problems (in regard to compactness) of open intervals. Indeed the Bolzano–Weierstrass theorem says precisely that every closed interval is compact (see the Appendix). We will see presently that every closed ball is compact. Note that a closed ball is both closed and bounded, meaning that it lies inside some ball Br(0) centered at the origin.
Lemma 8.3Every compact set is both closed and bounded.
PROOFSuppose that is compact. If a is a limit point of A then, for each integer n, there is a point an such that an − a < 1/n. Then the point a is the only limit point of the sequence . But, since A is compact, the infinite set must have a limit point in A. Therefore , so it follows from Proposition 8.1 that A is closed.
If A were not bounded then, for each positive integer n, there would exist a point with bn > n. But then would be an infinite subset of A having no limit point (Exercise 8.8), thereby contradicting the compactness of A.
Lemma 8.4A closed subset of a compact set is compact.
PROOFSuppose that A is closed, B is compact, and . If S is an infinite subset of A, then S has a limit point , because B is compact. But also, because b is a limit point of A, and A is closed. Thus every infinite subset of A has a limit point in A, so A is compact.
In the next theorem and its proof we use the following notation. Given and , write . If and , then the Cartesian product
is a subset of m + n.
Theorem 8.5If A is a compact subset of m and B is a compact subset of n, then A × B is a compact subset of m+n.
PROOFGiven a sequence of points of A × B, we want to show that it has a subsequence converging to a point of A × B. Since A is compact, the sequence has a subsequence which converges to a point . Since B is compact, the sequence has a subsequence which converges to a point . Then is a subsequence of the original sequence which converges to the point .
We are now ready for the criterion that will serve as our recognition test for compact sets.
Theorem 8.6A subset of n is compact if and only if it is both closed and bounded.
PROOFWe have already proved in Lemma 8.3 that every compact set is closed and bounded, so now suppose that A is a closed and bounded subset of n.
Figure 1.8
Choose r > 0 so large that . (See Fig. 1.8.) If I = [−r, r], then A is a closed subset of the product I × I × · · · × I (n factors), which is compact by repeated application of Theorem 8.5. It therefore follows from Lemma 8.4that A is compact.
For example, since spheres and closed balls are closed and bounded, it follows from the theorem that they are compact.
We now prove statement (A) at the beginning of the section.
Theorem 8.7If A is a compact subset of n, and f : A → m is continuous, then f(A) is a compact subset of m.
PROOFGiven an infinite set T of points of f(A), we want to prove that T contains a sequence of points which converges to a point of f(A). If S = f−1(T), then S is an infinite set of points of A. Since A is compact, S contains a sequence of points that converges to a point . Since f is continuous, is then a sequence of points of T that converges to the point (see Exercise 7.8). Therefore f(A) is compact.
Statement (B) will be absorbed into the proof of the maximum–minimum value theorem.
Theorem 8.8If D is a compact set in n, and f : D → is a continuous function, then f attains maximum and minimum values at points of D. That is, there exist points a and b of D such that for all .
PROOFWe deal only with the maximum value; the treatment of the minimum value is similar. By the previous theorem, f(D) is a compact subset of , and is therefore closed and bounded by (Theorem 8.6. Then f(D) has a least upper bound b, the least number such that for all (see the Appendix). We want to show that . Since b is the least upper bound for f(D), either or, for each positive integer n, there exists a point with b − 1/n < tn < b. But then the sequence of points of f(D) converges to b, so b is a limit point of f(D). Since f(D) is closed, it follows that as desired. If now is a point such that f(b) = b, it is clear that for all .
For example, we now know that every continuous function on a sphere or closed ball attains maximum and minimum values.
Frequently, in applied maximum–minimum problems, one wants to find a maximum or minimum value of a continuous function f : D → where D is not compact. Often Theorem 8.8 can still be utilized. For example, suppose we can find a compact subset C of D and a number c such that for all , whereas f attains values less than c at various points of C. Then it is clear that the minimum of f on C, which exists by Theorem 8.8, is also its minimum on all of D.
Two additional applications of compactness will be needed later. Theorem 8.9 below gives an important property of continuous functions defined on compact sets, while the Heine–Borel theorem deals with coverings of compact sets by open sets.
First recall the familiar definition of continuity: The mapping f : D → is continuous if, given and > 0, there exists δ > 0 such that
In general, δ will depend upon the point . If this is not the case, then f is called uniformly continuous on D. That is, f : D → is uniformly continuous if, given > 0, there exists δ > 0 such that
Not every continuous mapping is uniformly continuous. For example, the function f : (0, 1) → defined by f(x) = 1/x is continuous but not uniformly continuous on (0, 1) (why?).
Theorem 8.9If f : C → is continuous, and is compact, then f is uniformly continuous on C.
PROOFSuppose, to the contrary, that there exists a number > 0 such that, for every positive integer n, there exist two points xn and yn of C such that
Since C is compact we may assume, taking subsequences if necessary, that the sequences and both converge to the point . But then we have an easy contradiction to the continuity of f at a.
The Heine–Borel theorem states that, if C is a compact set in n and is a collection of open sets whose union contains C, then there is a finite sub-collection of these open sets whose union contains C. We will assume that the collection is countable (although it turns out that this assumption involves no loss of generality). So let be a sequence of open sets such that . If for each k > 1, then is an increasing sequence of sets—that is, for each —and it suffices to prove that for some integer k.
Theorem 8.10Let C be a compact subset of n, and let be an increasing sequence of open subsets of n such that . Then there exists a positive integer k such that .
PROOFTo the contrary suppose that, for each , there exists a point xk of C that is not in Vk. Then no one of the sets contains infinitely many of the points (why ?).
Since C is compact, we may assume (taking a subsequence if necessary) that the sequence converges to a point . But then for some k, and since the set Vk is open, it must contain infinitely many elements of the sequence . This contradiction proves the theorem.
Exercises
8.1Verify that the collection of all open subsets of n satisfies conditions (i)—(iii).
8.2Verify that the collection of all closed subsets of n satisfies conditions (i′)—(iii′). Hint: If {Aα} is a collection of subsets of n, then and .
8.3Show, directly from the definitions of open and closed sets, that open and closed balls are respectively open and closed sets.
8.4Complete the proof of Theorem 8.2.
8.5The point a is called a boundary point of the set A if and only if every open ball centered at a intersects both A and n − A. The boundary of the set A is the set of all of its boundary points. Show that the boundary of A is a closed set. Noting that the sphere Sr(p) is the boundary of the ball Br(p), this gives another proof that spheres are closed sets.
8.6Show that n is the only nonempty subset of itself that is both open and closed. Hint: Use the fact that this is true in the case n = 1 (see the Appendix), and the fact that n is a union of straight lines through the origin.
8.7Show that A is compact if and only if every sequence of points of A has a subsequence that converges to a point of A.
8.8If bn > n for each n, show that the sequence has no limit.
8.9Prove that the union or intersection of a finite number of compact sets is compact.
8.10Let be a decreasing sequence of compact sets (that is, for all n). Prove that the intersection is compact and nonempty. Give an example of a decreasing sequence of closed sets whose intersection is empty.
8.11Given two sets C and D in n, define the distance d(C, D) between them to be the greatest lower bound of the numbers a − b for and . If a is a point of n and D is a closed set, show that there exists such that d(a, D) = a − d. Hint: Let B be an appropriate closed ball centered at a, and consider the continuous function defined by f(x) = x − a.
8.12If C is compact and D is closed, prove that there exist points and such that d(C, D) = c − d. Hint: Consider the continuous function f: C → n defined by f(x) = d(x, D).