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. . , x m ).The geometric analogies can be extended by introducing a distance on Rm between the points x1 = (x11 , . . . , x1m ) and x2 = (x21 , . . . , x2m ) according to the formula45 m52d(x1 , x2 ) = 6x1i − x2i .(7.1)i=1The functiond : Rm × Rm → Rdefined by the formula (7.1) obviously has the following properties:a)b)c)d)d(x1 , x2 ) ≥ 0;(d(x1 , x2 ) = 0) ⇔ (x1 = x2 );d(x1 , x2 ) = d(x2 , x1 );d(x1 , x3 ) ≤ d(x1 , x2 ) + d(x2 , x3 ).This last inequality (called, again because of geometric analogies, the triangleinequality) is a special case of Minkowski’s inequality (see Sect.
5.4.2).A function defined on pairs of points (x1 , x2 ) of a set X and possessing the properties a), b), c), and d) is called a metric or distance on X.A set X together with a fixed metric on it is called a metric space.Thus we have turned Rm into a metric space by endowing it with the metric givenby relation (7.1).The reader can get information on arbitrary metric spaces in Chap.
9 (Part 2).Here we do not wish to become distracted from the particular metric space Rm thatwe need at the moment.Since the space Rm with metric (7.1) will be our only metric space in this chapter,forming our object of study, we have no need for the general definition of a metricspace at the moment. It is given only to explain the term “space” used in relation toRm and the term “metric” in relation to the function (7.1).It follows from (7.1) that for i ∈ {1, . . . , m} i√x − x i ≤ d(x1 , x2 ) ≤ m max x i − x i ,12121≤i≤m(7.2)that is, the distance between the points x1 , x2 ∈ Rm is small if and only if the corresponding coordinates of these points are close together.It is clear from (7.2) and also from (7.1) that for m = 1, the set R1 is the sameas the set of real numbers, between whose points the distance is measured in thestandard way by the absolute value of the difference of the numbers.7.1 The Space Rm and Its Subsets4117.1.2 Open and Closed Sets in RmDefinition 1 For δ > 0 the setB(a; δ) = x ∈ Rm | d(a, x) < δis called the ball with center a ∈ Rm of radius δ or the δ-neighborhood of the pointa ∈ Rm .Definition 2 A set G ⊂ Rm is open in Rm if for every point x ∈ G there is a ballB(x; δ) such that B(x; δ) ⊂ G.Example 1 Rm is an open set in Rm .Example 2 The empty set ∅ contains no points at all and hence may be regarded assatisfying Definition 2, that is, ∅ is an open set in Rm .Example 3 A ball B(a; r) is an open set in Rm .
Indeed, if x ∈ B(a; r), that is,d(a, x) < r, then for 0 < δ < r − d(a, x), we have B(x; δ) ⊂ B(a; r), sinceξ ∈ B(x; δ) ⇒ d(x, ξ ) < δ ⇒⇒ d(a, ξ ) ≤ d(a, x) + d(x, ξ ) < d(a, x) + r − d(a, x) = r .Example 4 A set G = {x ∈ Rm | d(a, x) > r}, that is, the set of points whose distance from a fixed point a ∈ Rm is larger than r, is open.
This fact is easy to verify,as in Example 3, using the triangle inequality for the metric.Definition 3 The set F ⊂ Rm is closed in Rm if its complement G = Rm \F is openin Rm .Example 5 The set B(a; r) = {x ∈ Rm | d(a, x) ≤ r}, r ≥ 0, that is, the set of pointswhose distance from a fixed point a ∈ Rm is at most r, is closed, as follows fromDefinition 3 and Example 4. The set B(a; r) is called the closed ball with center aof radius r.Proposition 1 a) The union α∈A Gα of the sets of any system {Gα , α ∈ A} of openmsets in Rm is an open setnin R .b) The intersection i=1 Gi of a finite number of open sets in Rm is an open setin Rm .a ) The intersection α∈A Fα of the sets of any system {Fα , α ∈ A} of closed setsmFα in Rm is a closednset in R .b ) The union i=1 Fi of a finite number of closed sets in Rm is a closed setin Rm .4127 Functions of Several Variables: Their Limits and ContinuityProof a) If x ∈ α∈A Gα , then there exists α0 ∈ A such that x ∈ Gα0 , and consequently there is a δ-neighborhood B(x; δ) of x such that B(x; δ) ⊂ Gα0 .
But thenB(x; δ) ⊂ α∈A Gα .b) Let x ∈ ni=1 Gi . Then x ∈ Gi (i = 1, . . . , n). Let δ1 , . . . , δn be positive numbers such that B(x; δi ) ⊂ Gi (i = 1,. . . , n). Setting δ = min{δ1 , . . . , δn }, we obviously find that δ > 0 and B(x; δ) ⊂ ni=1 Gi .a ) Let us show that the set C( α∈A Fα ) complementary to α∈A Fα in Rm isan open set in Rm .Indeed, CFα =(CFα ) =Gα ,α∈Aα∈ARm .α∈Aa )Partnow follows from a).where the sets Gα = CFα are open inb ) Similarly, from b) we obtain n nnCFi = (CFi ) =Gi .i=1i=1i=1Example 6 The set S(a; r) = {x ∈ Rm | d(a, x) = r}, r ≥ 0, is called the sphere ofradius r with center a ∈ Rm .
The complement of S(a; r) in Rm , by Examples 3and 4, is the union of open sets. Hence by the proposition just proved it is open, andthe sphere S(a; r) is closed in Rm .Definition 4 An open set in Rm containing a given point is called a neighborhoodof that point in Rm .In particular, as follows from Example 3, the δ-neighborhood of a point is aneighborhood of it.Definition 5 In relation to a set E ⊂ Rm a point x ∈ Rm isan interior point if some neighborhood of it is contained in E;an exterior point if it is an interior point of the complement of E in Rm ;a boundary point if it is neither an interior point nor an exterior point.It follows from this definition that the characteristic property of a boundary pointof a set is that every neighborhood of it contains both points of the set and pointsnot in the set.Example 7 The sphere S(a; r), r > 0 is the set of boundary points of both the openball B(a; r) and the closed ball B(a; r).Example 8 A point a ∈ Rm is a boundary point of the set Rm \a, which has noexterior points.Example 9 All points of the sphere S(a; r) are boundary points of it; regarded as asubset of Rm , the sphere S(a; r) has no interior points.7.1 The Space Rm and Its Subsets413Definition 6 A point a ∈ Rm is a limit point of the set E ⊂ Rm if for any neighborhood O(a) of a the intersection E ∩ O(a) is an infinite set.Definition 7 The union of a set E and all its limit points in Rm is the closure of Ein Rm .The closure of the set E is usually denoted E.Example 10 The set B(a; r) = B(a; r) ∪ S(a; r) is the set of limit points of the openball B(a; r); that is why B(a; r), in contrast to B(a; r), is called a closed ball.Example 11 S(a; r) = S(a; r).Rather than proving this last equality, we shall prove the following useful proposition.Proposition 2 (F is closed in Rm ) ⇔ (F = F in Rm ).In other words, F is closed in Rm if and only if it contains all its limit points./ F .
Then the open set G = Rm \FProof Let F be closed in Rm , x ∈ Rm , and x ∈is a neighborhood of x that contains no points of F . Thus we have shown that ifx∈/ F , then x is not a limit point of F .Let F = F . We shall verify that the set G = Rm \F is open in Rm . If x ∈ G, thenx∈/ F , and therefore x is not a limit point of F . Hence there is a neighborhood ofx containing only a finite number of points x1 , . . . , xn of F . Since x ∈/ F , one canconstruct,forexample,ballsaboutx,O(x),...,O(x)suchthatx∈/Oi (x). Then1niO(x) = ni=1 Oi (x) is an open neighborhood of x containing no points of F at all,that is, O(x) ⊂ Rm \F and hence the set Rm \F = Rm \F is open. Therefore F isclosed in Rm .7.1.3 Compact Sets in RmDefinition 8 A set K ⊂ Rm is compact if from every covering of K by sets that areopen in Rm one can extract a finite covering.Example 12 A closed interval [a, b] ⊂ R1 is compact by the finite covering lemma(Heine–Borel theorem).Example 13 A generalization to Rm of the concept of a closed interval is the setI = x ∈ Rm | a i ≤ x i ≤ bi , i = 1, .
. . , m ,4147 Functions of Several Variables: Their Limits and Continuitywhich is called an m-dimensional interval, or an m-dimensional block or an mdimensional parallelepiped.We shall show that I is compact in Rm .Proof Assume that from some open covering of I one cannot extract a finite covering. Bisecting each of the coordinate closed intervals I i = {x i ∈ R : a i ≤ x i ≤ bi }(i = 1, . . . , m), we break the interval I into 2m intervals, at least one of whichdoes not admit a covering by a finite number of sets from the open system westarted with.
We proceed with this interval exactly as with the original interval. Continuing this division process, we obtain a sequence of nested intervalsI = I1 ⊃ I2 ⊃ · · · ⊃ In ⊃ · · · , none of which admits a finite covering. If In = {x ∈Rm | ani ≤ x i ≤ bni , i, . . . , m}, then for each i ∈ {1, . . . , m} the coordinate closed intervals ani ≤ x i ≤ bni (n = 1, 2, . . .) form, by construction, a system of nested closedintervals whose lengths tend to zero. By finding the point ξ i ∈ [ani , bni ] common toall of these intervals for each i ∈ {1, . . . , m}, we obtain a point ξ = (ξ 1 , .
. . , ξ m )belonging to all the intervals I = I1 , I2 , . . . , In , . . . . Since ξ ∈ I , there is an openset G in the system of covering sets such that ξ ∈ G. Then for some δ > 0 we alsohave B(ξ ; δ) ⊂ G. But by construction and the relation (7.2) there exists N such thatIn ⊂ B(ξ ; δ) ⊂ G for n > N .
We have now reached a contradiction with the fact thatthe intervals In do not admit a finite covering by sets of the given system.Proposition 3 If K is a compact set in Rm , thena) K is closed in Rm ;b) any closed subset of Rm contained in K is itself compact.Proof a) We shall show that any point a ∈ Rm that is a limit point of K must belongto K. Suppose a ∈/ K. For each point x ∈ K we construct a neighborhood G(x)such that a has a neighborhood disjoint from G(x).
The set {G(x)}, x ∈ K, consisting of all such neighborhoods forms an open covering of the compact set K, fromwhich we can select a finite covering G(x1 ), . . . , G(xn ). If nowOi (a) is a neighborhood of a such that G(xi ) ∩ Oi (a) = ∅, then the set O(a) = ni=1 Oi (a) is alsoa neighborhood of a, and obviously K ∩ O(a) = ∅. Thus a cannot be a limit pointof K.b) Suppose F is a closed subset of Rm and F ⊂ K. Let {Gα }, α ∈ A, be a covering of F by sets that are open in Rm . Adjoining to this collection the open setG = Rm \F , we obtain an open covering of Rm , and in particular, an open coveringof K, from which we select a finite covering of K. This finite covering of K willalso cover the set F . Observing that G ∩ F = ∅, one can say that if G belongs tothis finite covering, we will still have a finite covering of F by sets of the originalsystem {Gα }, α ∈ A, if we remove G.Definition 9 The diameter of a set E ⊂ Rm is the quantityd(E) := sup d(x1 , x2 ).x1 ,x2 ∈E7.1 The Space Rm and Its Subsets415Definition 10 A set E ⊂ Rm is bounded if its diameter is finite.Proposition 4 If K is a compact set in Rm , then K is a bounded subset of Rm .Proof Take an arbitrary point a ∈ Rm and consider the sequence of open balls{B(a; n)} (n = 1, 2, .
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