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Chapter6

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Department
Economics
Course
ECON 2P30
Professor
Lester Kwong
Semester
Winter

Description
Chapter 6 Euclidean Topology ∗ Lester M.K. Kwong Department of Economics Brock University Version: 3.1.2.o Revised: October 9, 2012 ∗ Copyright ' 2004-2012 Lester M.K. Kwong. Department of Economics, Brock Uni- versity, 500 Glenridge AveCatharines, Ontario, L2S 3A1, CanEmail: lk- [email protected] Tel: +1 (905) 688-5550, Ext. 5137. This document was typeset with the LE XDocumentation System usingYthe X -picE the LT XDraw, and the MnSymbol packages. 1 Introduction Our discussion thus far has examined objects in arbitrary spaces without taking on much structure. For example, when we talked about a set X from some universe, the structure of the universe, for the most part, and hence, the structure of X was quite abstract in the sense that no structure was assumed. The universe could have been a discrete set of objects, such as N, or it could have been a continuous set of objects like R. In either case, one of the most common spaces of study in economics is the Euclidean Space. Hence, we devote some time understanding its structure as well as the contents within. The remainder of this chapter is as follows. In Section 2 the concept of an Euclidean Space is introduced. More importantly, we will introduce the concept of a metric specify the standard Euclidean metric used. In Section 3, properties of objects, namely, sets, from the Euclidean Spaces are examined in detail. Finally some review exercises follow. 2 Euclidean n-Space Given R, the set of real numbers, the Euclidean n-space refers to the Carte- sian product R . Hence, any object x from a Euclidean n-space is a n-tuple given by x = (x 1:::;xn). The de▯ning characteristic of the Euclidean n- space is that the Euclidean metric exists. In general, a metric d on a space X is a function that maps from X × X to R so that for all x;y;z ∈ X: (i) d(x;y) = d(y;x) (ii) d(x;y) = 0 ⇔ x = y (iii) d(x;y) + d(y;z) ≥ d(x;z) The concept of a metric d is sometimes commonly referred to as the notion of distance. So, for example, taking two points x and y from the space X, the requirement that d(x;y) = d(y;x) implies a symmetry in the measurement of distance from points x and y. That is, it does not matter whether you are talking about the distance from x to y or from y to x. The resulting \distance" should be the same. Consider, for example, the illustration in Figure 1. The distance between points x = (−4;−2) and y = (1;2) in Figure 1 is given by the length of the line connecting the points x and y given by d(x;y) = ((−4 − 1) + (−2 + 1 5 4 3 w y 2 dˆw,z) dˆx,y) 1 −5 −4 −3 −2 −1 1 2 3 4 5 −1 x −2 z −3 −4 −5 2 Figure 1: Euclidean metric in R ” 2) )1~2 = 41. Similarly, the distance between w = (2;3) and z = (4;−2) is 2 2 1~2 ” d(w;z) = ((2 − 4) + (3 − (−2)) ) = 29. The third property in the conditions for a metric is sometimes referred to as the triangle inequality. That is, for a given concept of a metric, the direct measurement between x and z should be smaller than or equal to some indirect measurement (from x to y and subsequently from y to z). Notice that this, in the context of R , suggests that the hypothenuse of a triangle is shorter than the length of its two sides. Naturally, the triangle inequality breaks down to the fact that the shortest distance between two points is a straight-line. This is illustrated in Figure 2. If a metric d exists over a space X then the pair (X;d) is said to be a metric space. The following immediately follows. Theorem 2.1. If d(x;y) is a metric on a space X, then ƒx;y ∈ X, d(x;y) ≥ 0. Proof. If d(x;y) is a metric on a space X, it then follows, by the trian- gle inequality, that d(x;y) + d(y;x) ≥ d(x;x) = 0. Since by symmetry of the metric function, d(x;y) = d(y;x), it now follows that d(x;y) + d(y;x) = d(x;y) + d(x;y) = 2d(x;y) ≥ 0 if and only if d(x;y) ≥ 0. ∎ 2 5 x 4 3 dˆx,z) dˆx,y) 2 1 y z dˆx,z) 0 0 1 2 3 4 5 6 Figure 2: The triangle inequality: d(x;y) + d(y;z) ≥ d(x;z) Example 2.1. Consider the metric d(x;y) de▯ned over X so that for all x;y ∈ X: d(x;y) = œ 0 if x = y (1) 1 if x ≠ y To verify that this is, indeed, a metric, take any x;y ∈ X. If x = y then by de▯nition d(x;y) = d(y;x) = 0. Conversely, if x ≠ y then d(x;y) = d(y;x) = 1. Hence, d(x;y) satis▯es the ▯rst condition above and is symmetric. The second condition that d(x;y) = 0 ⇔ x = y is trivially satis▯ed by de▯nition. Hence, we only have the triangle inequality remaining to check. If x = y = z then d(x;y) = d(y;z) = d(x;z) = 0. If x = y ≠ z then d(x;y) = 0 and d(y;z) = d(x;z) = 1. If (x ≠ y = z then d(x;y) = d(x;z) = 1 and d(y;z) = 0. Lastly, if x ≠ y, y ≠ z and x ≠ z, then d(x;y) = d(y;z) = d(x;z) = 1. In either case, the triangle inequality is satis▯ed. Therefore, the function d(⋅) as de▯ned in Eq. (1) is a metric. ◻ It is important to note that in the above example, the space X was arbi- trary chosen. Hence, it follows immediately that for any arbitrary nonempty space X, there is always a metric as given by the one in Eq. (1). Example 2.2. Consider d(x;y) = max{Sx − y S;Sx − 1 S} 1or p2ints2x;y ∈ R . To see that the function d(x;y) is a metric let us begin by considering d(y;x) = max{Sx − y2S;Sx2− y 1}. C1early, we have d(x;y) = d(y;x). Next, d(x;y) = 0 if and only if max{Sx − y S;Sx − y S} = 0. Since Sx − y S ≥ 0 for all 1 1 2 2 i i i, it follows that d(x;y) = 0 if and only if Sx − 1 S =1Sx − y2S = 2. This is the case if and only if x = y. 3 Lastly, to check the triangle inequality, take x;y;z ∈ R . We need to show that: max{Sx − 1 S;S1 − y2S} +2max{Sy − z S;S1 − z1S} 2 max{2x − z S;Sx − 1 S} 1 2 2 First note that: max{Sx − 1 S;S1 − y2S} ≥2Sx − y i; i i {1;2} and: max{Sy − 1 S;S1 − z2S} ≥2Sy − z i; ii∈ {1;2} since Sx iy S ind Sy −ziS ari nonnegative numbers. Employing the fact that for real numbers a and b, SaS+SbS ≥ Sa+bS, and setting a = Sx −y S, b 1 Sy 1z S 1 1 1 we have Sx −1 S+Sy1−z S 1 Sx 1y +y −1 S =1Sx −1 S. 1atura1ly, 1t follows that: max{Sx −y1S;Sx1−y S2+max2Sy −z S;Sy −z1S} ≥1Sx −2 S+S2 −z S ≥1Sx −1 S 1 1 1 1 By the same token: max{Sx −y S;Sx −y S}+max{Sy −z S;Sy −z S} ≥ Sx −y S+Sy −z S ≥ Sx −z S 1 1 2 2 1 1 2 2 2 2 2 2 2 2 Since max{Sx −y S;Sx −y S}+max{Sy −z S;Sy −z S} is greater than or equal 1 1 2 2 1 1 2 2 to both Sx −1 S 1nd Sx −z 2, it2must be greater than or equal to the larger of the two, given by max{Sx −z S;1x −1 S}.2Henc2, we have derived the triangle inequality for the metric. ◻ In the Euclidean n-space, however, the concept of a distance, or a metric, between two points x and y is merely equal to the length of the line segment connecting these two points. If x;y ∈ R so that x = (x ;:::;x ) and y1= n (y 1:::;y )nthen: » d(x;y) = (x1− y )1+ (x − 2 ) + 2 + (x − y ) n n 2 n 1~2 2 = Œ Q (xi− y i ‘ (2) i=1 Eq. (2) is, in essence, the Euclidean metric. 1 Note that for a and b, SaS+SbS ≥ Sa+bS is the subadditivity property of the absolute value. Also note since d(x;y) = Sx;yS for x;y ∈ R is the Euclidean metric on R, the subadditivity property that we have employed is actually the triangle inequality in R using the Euclidean metric. 4 Theorem 2.2. For all x;y ∈ R and for all ▯ ∈ R, d(▯x;▯y) = S▯Sd(x;y). n Proof. Take any x;y ∈ R and ▯ ∈ R. It follows that: n 1~2 2 d(▯x;▯y) = Œ Q (▯x i ▯y )i‘ i=1 n 1~2 2 2 = Œ Q ▯ (x i y i ‘ i=1 1~2 2 n 2 = Œ▯ Q (x i y i ‘ i=1 1~2 n = S▯SŒ Q (x i y i ‘ i=1 ∎ n 3 Properties of Sets in R Given the Euclidean n-space, and the associated Euclidean metric, an open ball with radius ▯ centred around x, denoted as B (x)▯is de▯ned as: n B ▯x) = {y ∈ R ∶ d(x;y) < ▯} (3) The reason why B (x)▯is referred to as a ball, in this case open, is because, by de▯nition, B (x) contains all the points y for which the distance between ▯ such y’s from x is strictly less than ▯. 2 For example, all the objects within the dashed circle in Figure 3 is a ball with radius ▯ around the point x. In R space, for any given point x, the collection of all other points within ▯ 3 distance from x essentially maps out a circle. In R space, such points will map out a sphere with radius ▯. Conversely, a closed ball with radius ▯ centred around x, denoted as C (▯) is: C ▯x) = {y ∈ R ∶ d(x;y) ≤ ▯} (4) It should be pointed out that while the concept of a ball, open or closed, is de▯ned, here, in the Euclidean n-space. However, as can be seen in its de▯ni- n tion, one could easily swap out the Euclidean n-space, R , with an arbitrary 2 The open ball B▯(x) is also sometimes referred to as an ▯-neighborhood at x. 5 e e e e b x e e B ǫx) 2 Figure 3: An open ball B (x) ▯ R . space X and replace the corresponding metric d(x;y) for one appropriate in X to de▯ne similar balls. With the aid of the concept of an open ball, we can now de▯ne an open set in R . A set X ⊆ R is said to be open if for all x ∈ X, there exists ▯ > 0, so that B (x) ⊆ X. For example, in Figure 4, the ball B (x) is contained in ▯ ▯ the set X. Example 3.1. The interval (0;1) ‘ R is an open set in R. To see that this set is, indeed, open, we must show that for every x ∈ (0;1), we can ▯nd some ▯ > 0 so that B (▯) ⊆ (0;1). Heuristically, this can be easily accomplished since for every object x ∈ (0;1), the distance between point x and the closest boundary of the set is strictly positive. It follows that choosing a radius, for the open ball, to be any distance smaller than this distance will always keep the open ball within the set. That is, ƒx ∈ (0;1), x−0 = x > 00and 1−x = x > 0. Le1 x = min{x ;0 }.1 Clearly, x provides us with the shortest distance x is to one of the \bound- aries" of the set (0;1). It now follows that since x > 0, for every 0 < ▯ < x so that B (▯) ‘ (0;1). To demonstrate this explicitly, we could, for example, partition (0;1) into two subsets (0;1~2) and [1~2;1). Clearly x = x if x ∈ (0;1~2) and x = 1 − x 3Note that the inclusion of 1~2 to the subset [1~2;1) was arbitrarily done. We could have included it into (0;1~2) to form (0;1~2] instead without changing the results. Al- 6 X B ǫx) x ǫ Figure 4: An open ball B (x▯ around x in set X. x 1 0 x − ǫ x + ǫ Figure 5: Construction of B (▯) for x < 1~2. if x ∈ [1~2;1). We begin by ▯rst considering some x ∈ (0;1). Let ▯ = x~2. We need to show that B (x) ▯ (0;1). Note that since x ∈ (0;1) implies x > 0 therefore ▯ = x~2 > 0. By construction, B x~2x) = (x − x~2;x + x~2) = (x~2;3x~2). It su▯ces to show that the lower bound of B x~2(x) is larger than the lower bound of (0;1) and that the upper bound of B (x) is smaller than x~2 the upper bound of (0;1) to conclude that B x~2(x) ⊆ (0;1). See, for example Figure 5. Clearly, the lower bound of B (x), given by x~2, is strictly greater x~2 than 0 if and only if x > 0. Since x ∈ (0;1~2) this is true. The upper bound of B x~2(x), given by 3x~2 is strictly less than 1 if and only if x < 2~3. Since x is assumed to come from (0;1~2) this is also the case. Therefore B x:~2x) ⊆ (0;1) for every x ∈ (0;1~2). We are left to show that for x ∈ [1~2;1) that this also holds. Since for ternatively, we could have partitioned (0;1) into three subsets instead, namely, (0;1~2), {1~2}, and (1~2;1). 7 x ∈ [1~2;1), x = 1 − x let ▯ = (1 − x)~2. Note that ▯ > 0 if and only if x < 1 which is true since x ∈ [1~2;1) implies x < 1 which we have here. It follows 1−x 1−x 3x−1 x+1 that B (1−x)~2x) = (x− 2 ;x+ 2 ) = ( 2 ; 2 ). We are left to show that the lower bound of B (x) is strictly larger than 0 and that the upper bound (1−x)~2 of B (1−x)~2x) is strictly smaller than 1. For the former, (3x−1)~2 > 0 if and only if x > 1~3 which is true since x ∈ [1~2;1), and for the latter (x+1)~2 < 1 if and only if x < 1 which again is true. Therefore, B (1−x)~2(x) ‘ (0;1). Therefore, we have shown using the combination of x ∈ (0;1~2) in the ▯rst part and x ∈ [1~2;1) in the second part that for every x ∈ (0;1), there exists some ▯ > 0 so that B (▯) ‘ (0;1). Therefore, (0;1) is an open set. ◻ Theorem 3.1. The sets R and ∅ are open sets in R . n n n Proof. For the case of R , this is rather intuitive. Since for every x ∈ R , every B (▯) ⊆ R , it follows that R is open in R . For the case of ∅, note that since x ∈ ∅ is a false statement, it trivially satis▯es the requirement that ƒx ∈ ∅, ∃▯ > 0 so that B (▯) ⊆ ∅. That is, since there are no objects in ∅, it holds that every object in ∅, such an open ball exists. ∎ Consider some set X in R . If a contained open ball B (x) ▯xists for some point x ∈ X, then x is said to be an interior point of X. That is, x ∈ X is an interior point of X if there exists some ▯ > 0 so that B (x) ⊆ X. The interior ▯ of a set, on the other hand, is the set of all interior points of X denoted int(X). Clearly, int(X) ⊆ X. Theorem 3.2. For any set X in R , the set int(X) is an open set. Proof. Consider some set X in R and suppose int(X) is not open. This implies that there exists some x ∈ int(X) so that ¤▯ > 0 for which B (x) ▯ X. But then x ∉ int(X), a contradiction. ∎ n Theorem 3.3. Let X and Y be two open sets in R . Then X ∪Y is an open set in R . n Proof. Let X and Y be two open sets in R . It follows, by de▯nition, that ƒx ∈ X, ∃▯ > 0 so that B (x) ⊆ X. It also follows that ƒx ∈ Y , ∃▯ > 0 so ▯ that B ▯x) ⊆ Y . We need to show that for every x ∈ X ∪ Y , ∃▯ > 0 so that B (x) ⊆ (X ∪ Y ). But this result readily follows since for every x ∈ X ∪ Y , ▯ x ∈ X or x ∈ Y . If x ∈ X, we have established that an ▯ > 0 exists so that B ▯x) ⊆ X. Since X ⊆ (X ∪ Y ), by the transitivity of ⊆, it follows that B ▯x) ⊆ (X ∪ Y ). A similar argument for if x ∈ Y completes the proof. ∎ 8 The proof of Theorem 3.3 relied on the idea that for any sets X and Y , X ⊆ (X ∪Y ). Naturally, if an open ball for every point exists in X, then that same open ball will exist in X ∪ Y . Naturally, this logic, when extended to any family of sets X still holds. We state this formally without proof in the next theorem. Theorem 3.4. Let X be any family of sets in R . Then ⋃ X∈X X is an open set. n Theorem 3.5. Let X and Y be two open sets in R . Then X ∩Y is an open set in R . Proof. Let X and Y be two open sets in R . It follows then that ƒx ∈ X, ∃▯X > 0 so that B (▯X ⊆ X. It, again, follows that ƒx ∈ Y , ∃▯ Y > 0 so that B (x) ⊆ Y . Take any x ∈ X ∩ Y . It follows that x ∈ X and x ∈ Y . ▯Y Hence, ▯ X and ▯ Y exists. Without loss of generality, suppose ▯ X ▯ . Yhen, B ▯Y) ⊆ B (x▯X⊆ X. Furthermore, B (x) ⊆ YY. Hence, B (x) ⊆ (X ▯YY ). Therefore, X ∩ Y is open. ∎ The intuition behind Theorem 3.5 is straightforward. Given an element x ∈ X∩Y , any two open balls around x, say B (x▯Xand B (x) w▯Yl necessar- ily have B (x) ⊆ B (x) or B (x) ⊆ B (x). By the transitivity of ⊆, the ▯X ▯Y Y X smaller of the two open balls must be fully contained in both sets. However, unlike the case of the union between open sets, the intersection of sets, when extended over an in▯nite number may not hold. To see this, consider the following example. “ Example 3.2. Let X = {X } i i=1be a family of sets where X i (−1~i;1~i) ‘ R. “ Consider ⋂ i=1X i Note that X ∩1 = 2−1;1)∩(−1~2;1~2) = (−1~2;1~2) = X . 2 Similarly, X 1X ∩X2= (−3;1)∩(−1~2;1~2)∩(−1~3;1~3) = (−1~3;1~3) = X . 3 n Extending this argument, we can see that ⋂ i=1X i X =n(−1~n;1~n). Since both −1~n and 1~n converge to 0 as n approaches “ it follows, by the above argument that ⋂i=1X i {0}. The set, {0}, since it only contains a single element, is not an open set in R. That is, no open balls exist that can be fully contained in {0}. Hence, “ X is not an open set, despite each X ⋂ i=1 i i being open. ◻ While the property of openness fails when generalized over an in▯nite family of sets, its generalization, restricted to a ▯nite family of sets still holds. We state this in the following. 9 m Theorem 3.6. Consider a family of sets X = {X ;X ;:1:;X2} in R n where X is an open set for all i ∈ {1;:::;n}. Then ⋂n X is an open set. i i=1 i The proof of this result is analogous to that of Theorem 3.5 and we only provide a brief sketch here. When we consider an arbitrary point x ∈ ⋂ n X i i=1 and note the existence of n open balls, we must restrict our attention to the smallest of the n open balls. That is, let ▯ = min{▯ ;:::;▯ } where ▯ is X 1 Xn X i the corresponding open ball for the set X . ihis, along with the transitivity n of ⊆ ensures that B (▯) will be contained in ⋂i=1X i Note that this proof fails, when n is extended to “, because the ▯, as de▯ned in this sketch, is not well de▯ned. While Example 3.2 demonstrated that an in▯nite intersection of open sets may not be open, one could readily construct examples of in▯nite intersec- tions of open sets to be open as well. For example, a simple, yet trivial, tweak to Example 3.2 by de▯ning X = (−i~i;1~i) ∪ (−1~z;1~z) for any z ∈ Z makes X open for all i ∈ Z yet, ⋂ “ X = (−1~z;1~z) for the chosen z, hence i i=1 i open. This follows since for a given z, the set X fir all i > z are identically given by the set (−1~z;1~z). Hence, the convergence towards 0, as was the case in Example 3.2 does not occur. Example 3.3. To better understand the concept of open sets, it will be an insightful exercise if we momentarily stepped away from the Euclidean n-space and considered an arbitrary space X with a metric as de▯ned by Example 2.1. That is, suppose we have: d(x;y) = œ 0 if x = y 1 if x ≠ y Using the de▯nition of openness as de▯ned above, obviously substituting the set X for R in the de▯nition of B (x), Eq. (3), it follows that every subset ▯ of X is an open set. Since ▯ can be arbitrary chosen. To demonstrate this, suppose X = {x;y;z} and consider the set {x;y}. To show that {x;y} is an open set in X, we need to show that for both x and y, there exists some ▯, not necessarily the same for both x and y, so that an open ball is a subset of {x;y}. For x ∈ {x;y}, if we take ▯ = 1~2, then B 1~2x) = {x} ‘ {x;y}. Similarly, B 1~2y) = {y} ‘ {x;y}. Hence, {x;y} is an open set in X = {x;y;z}. ◻ n With the concept of an open set de▯ned in R , we can now de▯ne a closed set. A set X in R is said to be closed if X is open. We have the following. 10 n n Theorem 3.7. The sets R and ∅ are closed sets in R . n n c c n Proof. Since in R , {R } = ∅ and ∅ = R , the result follows immediately from Theorem 3.1. ∎ The combination of Theorems 3.1 and 3.7 demonstrate a common miscon- ception; openness and closeness are not mutually exclusive properties. That is, as the two theorems have shown, a set can be both open and closed. It follows that statements such as \the set X is not open because it is closed" or its counterpart \the set X is not closed because it is open" are not logically valid arguments. They fail precisely on the assumption that such properties are mutually exclusive. Consider, for example, the set [0;“) = R = {+ ∈ R ∶ x
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