ELEMENTARY TOPOLOGY OF R<sup>n</sup> - Euclidean Space and Linear Mappings - Advanced Calculus of Several Variables

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 Imagen possesses a topological structure (defined below). Among other things, this topological structure enables us to define and study a certain class of subsets of Imagen, 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 Imagen, and f : DImagem is continuous, then its image f(D) is a compact set in Imagem (Theorem 8.7).

(B)If C is a compact set on the real line Image, then C contains a maximal element b, that is, a number Image such that Image for all Image.

It follows immediately from (A) and (B) that, if f : DImage is a continuous real-valued function on the compact set Image, then f(x) attains an absolute maximum value at some point Image. For if b is the maximal element of the compact set Image, 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 Image of subsets, called open subsets of S, such that Image satisfies the following three conditions:

(i)The empty set Image 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 Imagen is called open if and only if, given any point Image, 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 Image 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 Imagen satisfies conditions (i)–(iii) above (Exercise 8.1).

Examples (a) An open interval is an open subset of Image, but a closed interval is not. (b) More generally, an open ball in Imagen is an open subset of Imagen (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 Imagen, then ImagenF is an open set. (d) Although Image is an open subset of itself, it is not an open subset of the plane Image2.

The subset B of Imagen is called closed if and only if its complement ImagenB is open. It is easily verified (Exercise 8.2) that conditions (i)–(iii) above imply that the collection of all closed subsets of Imagen satisfies the following three analogous conditions:

(i′)Image and Imagen 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 Image. (b) More generally, a closed ball in Imagen is a closed subset of Imagen (Exercise 8.3). (c) A finite set F of points is a closed set. (d) The real line Image is a closed subset of Image2. (e) If Ais the set of points of the sequence Image, 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 Imagen 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 ImagenA is open, a cannot be a point of ImagenA. Thus Image.

Conversely, suppose that A contains all of its limit points. If Image, then b is not a limit point of A, so there exists an open ball Br(b) which contains no points of A. Thus ImagenA is open, so A is closed.

Image

If, given Image, we denote by Image 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 = Image.

The empty set Image and Imagen itself are the only subsets of Imagen that are both open and closed (this is not supposed to be obvious—see (Exercise 8.6). However there are many subsets of Imagen that are neither open nor closed. For example, the set Q of all rational numbers is such a subset of Image.

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: ImagenImagem is continuous if and only if, given any open set Image, the inverse image f−1(U) is open in Imagen. Also, f is continuous if and only if, given any closed set Image is closed in Imagen.

PROOFThe inverse image f−1(U) is the set of points in Imagen that map under f into U, that is,

Image

We prove the “only if” part of the Theorem, and leave the converse as Exercise 8.4.

Suppose f is continuous. If Image is open, and Image, then there exists an open ball Image Since f is continuous, there exists an open ball Bδ(a) such that Image. Hence Imagethis shows that f−1(U) is open.

If Image is closed, then ImagemC is open, so f−1 (ImagemC) is open by what has just been proved. But f−1(ImagemC) = Imagenf−1(C), so it follows that f−1(C) is closed.

Image

As an application of Theorem 8.2, let f : ImagenImage be the continuous mapping defined by f(x) = Imagex − aImage, where Image 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]) = Imager(a), so the closed ball is indeed closed. Finally,

Image

so the (n − 1)-sphere of radius r, centered at a, is a closed set.

The subset A of Imagen 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 Image which converges to a point Image. (This means the same thing in Imagen as on the real line: Given Image > 0, there exists N such that Image) The equivalence of this statement and the definition is just a matter of language (Exercise 8.7).

Examples: (a) Image is not compact, because the set of all integers is an infinite subset of Image that has no limit point at all. Similarly, Imagen is not compact. (b) The open interval (0, 1) is not compact, because the sequence Image 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 Image is compact. If a is a limit point of A then, for each integer n, there is a point an such that ImageanaImage < 1/n. Then the point a is the only limit point of the sequence Image. But, since A is compact, the infinite set Image must have a limit point in A. Therefore Image, 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 Image with ImagebnImage > n. But then Image would be an infinite subset of A having no limit point (Exercise 8.8), thereby contradicting the compactness of A.

Image

Lemma 8.4A closed subset of a compact set is compact.

PROOFSuppose that A is closed, B is compact, and Image. If S is an infinite subset of A, then S has a limit point Image, because B is compact. But Image 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.

Image

In the next theorem and its proof we use the following notation. Given Image and Image, write Image. If Image and Image, then the Cartesian product

Image

is a subset of Imagem + n.

Theorem 8.5If A is a compact subset of Imagem and B is a compact subset of Imagen, then A × B is a compact subset of Imagem+n.

PROOFGiven a sequence Image 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 Image has a subsequence Image which converges to a point Image. Since B is compact, the sequence Image has a subsequence Image which converges to a point Image. Then Image is a subsequence of the original sequence Image which converges to the point Image.

Image

We are now ready for the criterion that will serve as our recognition test for compact sets.

Theorem 8.6A subset of Imagen 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 Imagen.

Image

Figure 1.8

Choose r > 0 so large that Image. (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.

Image

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 Imagen, and f : AImagem is continuous, then f(A) is a compact subset of Imagem.

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 Image of points that converges to a point Image. Since f is continuous, Image is then a sequence of points of T that converges to the point Image (see Exercise 7.8). Therefore f(A) is compact.

Image

Statement (B) will be absorbed into the proof of the maximum–minimum value theorem.

Theorem 8.8If D is a compact set in Imagen, and f : DImage 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 Image for all Image.

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 Image, and is therefore closed and bounded by (Theorem 8.6. Then f(D) has a least upper bound b, the least number such that Image for all Image (see the Appendix). We want to show that Image. Since b is the least upper bound for f(D), either Image or, for each positive integer n, there exists a point Image with b − 1/n < tn < b. But then the sequence Image 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 Image as desired. If now Image is a point such that f(b) = b, it is clear that Image for all Image.

Image

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 : DImage 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 Image for all Image, 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 : DImage is continuous if, given Image and Image > 0, there exists δ > 0 such that

Image

In general, δ will depend upon the point Image. If this is not the case, then f is called uniformly continuous on D. That is, f : DImage is uniformly continuous if, given Image > 0, there exists δ > 0 such that

Image

Not every continuous mapping is uniformly continuous. For example, the function f : (0, 1) → Image defined by f(x) = 1/x is continuous but not uniformly continuous on (0, 1) (why?).

Theorem 8.9If f : CImage is continuous, and Image is compact, then f is uniformly continuous on C.

PROOFSuppose, to the contrary, that there exists a number Image > 0 such that, for every positive integer n, there exist two points xn and yn of C such that

Image

Since C is compact we may assume, taking subsequences if necessary, that the sequences Image and Image both converge to the point Image. But then we have an easy contradiction to the continuity of f at a.

Image

The Heine–Borel theorem states that, if C is a compact set in Imagen and Image is a collection of open sets whose union contains C, then there is a finite sub-collection Image of these open sets whose union contains C. We will assume that the collection Image is countable (although it turns out that this assumption involves no loss of generality). So let Image be a sequence of open sets such that Image. If Image for each k > 1, then Image is an increasing sequence of sets—that is, Image for each Image—and it suffices to prove that Image for some integer k.

Theorem 8.10Let C be a compact subset of Imagen, and let Image be an increasing sequence of open subsets of Imagen such that Image. Then there exists a positive integer k such that Image.

PROOFTo the contrary suppose that, for each Image, there exists a point xk of C that is not in Vk. Then no one of the sets Image contains infinitely many of the points Image (why ?).

Since C is compact, we may assume (taking a subsequence if necessary) that the sequence Image converges to a point Image. But then Image for some k, and since the set Vk is open, it must contain infinitely many elements of the sequence Image. This contradiction proves the theorem.

Image

Exercises

8.1Verify that the collection of all open subsets of Imagen satisfies conditions (i)—(iii).

8.2Verify that the collection of all closed subsets of Imagen satisfies conditions (i′)—(iii′). Hint: If {Aα} is a collection of subsets of Imagen, then Image and Image.

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 ImagenA. 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 Imagen 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 Imagen 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 ImagebnImage > n for each n, show that the sequence Image has no limit.

8.9Prove that the union or intersection of a finite number of compact sets is compact.

8.10Let Image be a decreasing sequence of compact sets (that is, Image for all n). Prove that the intersection Image 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 Imagen, define the distance d(C, D) between them to be the greatest lower bound of the numbers Imagea − bImage for Image and Image. If a is a point of Imagen and D is a closed set, show that there exists Image such that d(a, D) = Imagea − dImage. Hint: Let B be an appropriate closed ball centered at a, and consider the continuous function Image defined by f(x) = Imagex − aImage.

8.12If C is compact and D is closed, prove that there exist points Image and Image such that d(C, D) = Imagec − dImage. Hint: Consider the continuous function f: CImagen defined by f(x) = d(x, D).