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Lecture 3 Outline Lecture 3 Econ 2001 2015 August 12 Lecture 3 Outline 1 Metric and Metric Spaces 2 Norm and Normed Spaces 3 Sequences and Subsequences 4 Convergence 5 Monotone and Bounded Sequences Announcements: - Friday’sexam will be at 3pm, in WWPH 4716; recitation will be at 1pm. The exam will last an hour. Vector Space Over the Reals Definition A vector space V over R is a 4-tuple (V , R, +, ) where V is a set, + : V V V is vector addition, : R V V is scalar multiplication,· satisfying: ! · ! 1 Closure under addition: x, y V , (x + y) V 8 2 2 2 Associativity of +: x, y, z V , (x + y) + z = x + (y + z) 8 2 3 Commutativity of +: x, y V , x + y = y + x 8 2 4 Existence of vector additive identity: !0 V s.t. x V , x + 0 = 0 + x = x 9 2 8 2 5 Existence of vector additive inverse: x V !y V s.t. x + y = y + x = 0. Denote this unique y as ( x), and define8 2 x 9y as2 x + ( y). 6 Distributivity of scalar multiplication over vector addition: R x, y V , (x + y) = x + y 8 2 8 2 · · · 7 Distributivity of scalar multiplication over scalar addition: , R x V , ( + ) x = x + x 8 2 8 2 · · · 8 Associativity of : , R, x V , ( ) x = ( x) · 8 2 8 2 · · · · 9 Multiplicative identity: x V , 1 x = x 8 2 · Fun With Vector Spaces Supose we have a vector space V over R and x V . 2 What is 0 x =? · distributivity of scalar muliplication existence of vector additive identity 0 x = (0+0) x = 0 x+0 x = 0 · · · · where the first zero is a scalar while the last is a vector. z}|{ z}|{ Using this result: 0 = 0 x = (1 1) x = 1 x + ( 1) x = x + ( 1) x · · · · · So we conclude ( 1) x = ( x). · which is not something implied by the nine properties of a vector space. Distance We work with abstract sets (typically vector spaces) and our main objective is to be able to talk about their properties as well as properties of functions defined over them. For example, we would like to define notions like limits and convergence, continuity of a function from one of these sets to another, and so on. Continuity of a function f : R R is defined as ! " > 0 > 0 such that x y < f (x) f (y) < " 8 9 j j ) j j How can we construct a similar definition when the domain is some abstract set X ? ", , and f ( ) are scalars, but we do not know how to measure the distance between two· points. So, next we define function that measures “distance” between any two points in a set. using this function, one can express the idea that two points are “close”(say within of each other). Notice that the even more general case in which f is a function from X to Y does not complicate things. First, we define distance for abstract settings; next, specialize to vector spaces. Metric Spaces and Metrics Definition A metric d on a set X is a function d : X X R+ satisfying ! 1 d(x, y) 0, with d(x, y) = 0 x = y x, y X . ≥ , 8 2 2 d(x, y) = d(y, x) x, y X (symmetry). 8 2 3 triangle inequality: d(x, z) d(x, y) + d(y, z) x, y, z X ≤ 8 2 A metric space (X , d) is a set X with a metric d defined on it. The value d(x, y) is interpreted as the distance between x and y. This definition generalizes the concept of distance to abstract spaces. Why “triangle inequality”? y x %& z ! Notice that a metric space need not be a vector space. Metrics: Examples Example The discrete metric on any given set X is defined as 0 if x = y d(x, y) = 1 if x = y 6 It trivially satisfies 1. and 2., and as for triangle inequality: 1 2. ≤ Example 1 2 2 2 The l metric (taxicab) on R is defined as d : R R R+: ! d(x, y) = x1 y1 + x2 y2 j j j j Example 2 2 2 The l 1 metric on R is defined as d : R R R+: ! d(x, y) = max x1 y1 , x2 y2 fj j j jg Metrics and Applications Definition In a metric space (X , d), a subset S X is bounded if x X and R such that s S, d(s, x) 9 2 2 8 2 ≤ Given a metric space, boundedness is a propety of subsets (it is relative!). Definition In a metric space (X , d), we define the diameter of a subset S X by diam (S) = sup d(s, s0): s, s0 S f 2 g A metric can be used to define a “ball of a given radius around a point ”. Definition In a metric space (X , d), we define an open ball of center x and radius " as B"(x) = y X : d(y, x) < " f 2 g and a closed ball of and center x and radius " as B"[x] = y X : d(y, x) " f 2 ≤ g Distance Between Sets We can also define the distance from a point to a set: d(A, x) = inf d(a, x) a A 2 and the distance between sets: d(A, B) = inf d(B, a) a A 2 = inf d(a, b): a A, b B f 2 2 g Note that d(A, B) is not a metric (why?). Norms and Normed Spaces The following defines a vector’slength (its size). Definition Let V be a vector space. A norm on V is a function : V R+ satisfying k · k ! 1 x 0 x V . k k ≥ 8 2 2 x = 0 x = 0 x V . k k , 8 2 3 triangle inequality: x + y x + y x, y V k k ≤ k k k k 8 2 4 x = x R, x V k k j jk k 8 2 8 2 x x y Why is this also called “triangle” inequality? 0 %&x + y y ! x &%y Definition A normed vector space is a vector space over R equipped with a norm. There are many different norms: if is a norm on V , so are 2 and 3 and k for any k > 0. k · k k · k k · k k · k Normed Spaces Can one relate norm and distance on a given normed vector space? In Rn, the standard norm is defined as 1 n 2 z = z2 k k i i=1 ! X while the Euclidean distance between two vectors x and y n 2 2 2 2 d(x, y) = (x1 y1) + (x2 y2) + ... + (xn yn) = (xi yi ) v u i=1 p uX t Norm and Euclidean distance Here, the distance between x and y is the length of the vector given by their difference: d(x, y) = x y k k This idea generalizes. Norm and Distance In any normed vector space, the norm can be used to define a distance. Theorem Let (V , ) be a normed vector space. Let d : V V R+ be defined by k · k ) d(v, w) = v w k k Then (V , d) is a metric space. In a normed vector space, we define a metric using the norm. Remember that x : V R+ is the lenght of a vector x V . k k ! 2 Here x = v w. Hence, the distance between two vectors is the length of the difference vector. Proof that (V , d) is a Metric SpaceI We must verify that the d defined satisfies the three properties of a metric. First property: d(x, y) 0, with d(x, y) = 0 x = y x, y X ≥ , 8 2 Proof. Let v, w V . By definition, d(v, w) = v w 0. Moreover 2 k k ≥ d(v, w) = 0 v w = 0 , k k v w = 0 by def of , k · k (v + ( w)) + w = w by vector additive inverse , v + (( w) + w) = w by associative property , v + 0 = w by vector additive inverse , v = w , Proof that (V , d) is a Metric Space II For symmetry, we need to show that d(x, y) = d(y, x) x, y X 8 2 Proof. Let v, w V . 2 d(v, w) = v w k k = 1 v w j jk k = ( 1)(v + ( w)) by def of k k k · k = ( 1)v + ( 1)( w) k k = v + w by the first slide k k = w + ( v) k k = w v k k = d(w, v) Proof that (V , d) is a Metric Space III The triangle inequality is d(x, z) d(x, y) + d(y, z) x, y, z X . ≤ 8 2 Proof. Let u, w, v V . 2 d(u, w) = u w k k = u + ( v + v) w k k = u v + v w k k u v + v w by triangle for ≤ k k k k k · k = d(u, v) + d(v, w) Summary Since we have shown that d( ) = satisfies the three properites of a metric on V , we conclude that (V , d) is· a metrick · k space. Normed Vector Spaces: Examples Rn admits many norms The n-dimensional Euclidean space Rn supports the following norms the standard (Euclidean) norm: n 2 x 2 = (xi ) = px x k k v · u i=1 uX where (the “dot product” of vectors)t is defined as: · n x y = x1y1 + x2y2 + ..
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