sign in sign up
<<
Home , Metric (mathematics)

Metric Space

Definition: Take a X . A d : X × X → R is called a Metric or function on E if d satisfies the following properties: (a) d(p, q) ≥ 0 for all p, q ∈ X . Moreover d(p, q) = 0 iff p = q. (b) For all p, q ∈ X , d(p, q) = d(q, p) (c) () d(p, q) + d(q, r) ≥ d(p, r) for all p, q, r ∈ X . d(p, q) is called the distance between p and q. The set X along with d, is called a Metric space. Observation: Take X 0 ⊂ X . Then X 0 along with d is also a metric space. Examples: Distance Function 1. X = Set of integers; d(x, y) = 1 if x 6= y and d(x, x) = 0 2. X = R; d(x, y) = |x − y| q l Pl 2 3. X = R ; d(x, y) = k=1(xk − yk ) l 4. X = R ; d(x, y) = maxk |xk − yk | 5. X = {f | f :[a, b] → R and f is continuous}; d(f , g) = maxt∈[a,b]|f (t) − g(t)| Metric Space

On R, we used the distance function d(p, q) = |p − q|. We can now generalize many concepts, introduced earlier on R, in terms of distance function. Let (X , d) be a metric space. • A ⊆ X is bounded if there exists q ∈ A and M > 0 such that d(p, q) ≤ M for all p ∈ A. Examples: Bounded metric space 2 2 2 1. X = R and A = {x ∈ X | x1 + x2 < 1}. Here A is bounded. 2 2. X = R and A = {x ∈ X | x1 = x2}. A is not bounded. • For any a ∈ X and  > 0, the -neighbourhood of a is the set of all points in X whose distance from a is strictly less than . Formally, B(a) = {x ∈ X | d(x, a) < }. ∞ • A sequence in X , {an}n=1, converges to a: If for every  > 0, there exists N ∈ N such that ∀n ≥ N, an ∈ B(a). Examples: Convergent q 2 P2 2 1 1. X = R ; d(x, y) = k=1(xk − yk ) ; an = ( n , 1). This converges to (0, 1). t 2. X = {f | f : [0, 1] → R}; d(f , g) = maxt∈[a,b]|f (t) − g(t)|; fn(t) = n . This sequence converges to f (t) = 0 ∀t. ∞ • A sequence {an}n=1 is bounded if there exists M > 0 such that d(a1, an) ≤ M ∀n

Some of our results are easily generalized with minor modifications. Some results will require a generalized notion of addition and multiplication. Definition: Vector Space A linear space or vector space V over scalar R is a set of elements (called vectors), along with two binary operations L (vector addition) and J (scalar multiplication) that satisfies following properties. Let v, w, z ∈ V and c, d ∈ R 1. V close under L: v L w ∈ V , 2. Commutativity of L: v L w = w L v, 3. Associativity of L:(v L w) L z = v L(w L z) 4. Existence of additive identity: ∃ a vector called 0 such that v L 0 = v ∀v 5. Existence of additive inverse: For each v, ∃ w ∈ V such that v L w = 0 6. V close under J: c J v ∈ V , 7. Associativity of J:(cd) J x = c J(d J x) 8. Multiplicative identity: 1 J v = v 9. Distributive property: c J(v L w) = (c J v) L(c J w), (c + d) J v = (c J v) L(d J v)

From now on, we shall use the usual addition and multiplication for vector space as well. Additive inverse shall be denoted by usual ‘−’ sign. That is inverse of vector z ∈ V is denoted by −z. Moreover for y, z ∈ V , y + (−z) will be simply denoted by y − z. Check: Additive inverse is unique. Definition: Normed Vector Space A normed vector space is a vector space V with a function || || : V → R. The norm function has to satisfy properties similar to distance function. (a) ||x|| ≥ 0 for all x ∈ V . Moreover ||x|| = 0 iff x = 0. (b) For all x, y ∈ V , ||x − y|| = ||y − x|| (c) For all c ∈ R and x ∈ V , ||cx|| = |c| ||x|| (d) (Triangle Inequality) For all x, y ∈ V , ||x + y|| ≤ ||x|| + ||y|| A norm can be easily extended to a distance function in the following manner, d(x, 0) = ||x|| and d(x, y) = d(x − y, 0) = ||x − y||. Check: d satisfies all properties of distance function. Observation: A normed vector space is a metric space with generalized addition and multiplication. Examples: Normed Vector Spaces q l Pl 2 1. X = R ; ||x|| = k=1 xk 2. X = {f | f :[a, b] → R and f is continuous}; ||f || = maxt∈[a,b]|f (t)| Sequence in Normed Vector Space

Lets check whether our results remain the same in normed vector space. Result 5 - Sequence: Every convergent sequence is bounded. ∞ (Proof): Take any sequence {an}n=1 which converges to a. We can find N such that for all n ≥ N, d(an, a) < 1. Hence for all n ≥ N, d(an, a1) ≤ d(a1, a) + d(a, an) < d(a1, a) + 1 (Using Triangle inequality). Now choose M = max{d(a1, a2), d(a1, a3),..., d(a1, aN−1), (d(a1, 1) + 1)}. Check (Result 7-Sequence): Subsequence of a convergent sequence converges to the same .

∞ ∞ Let {an}n=1 and {bn}n=1 are sequences in a normed vector space V . lim an = a and lim bn = b. ∞ (Result 1-Sequence) Take the sequence {can}n=1, where c ∈ R. lim(can) = ca. ∞ (Result 2-Sequence) Take the sequence {an + bn}n=1. lim(an + bn) = a + b. Open and Closed Set

The following discussion is applicable for metric spaces and hence for normed vector spaces as well. Let (X , d) be a metric space and E ⊆ X . Definitions: p ∈ E is an interior point of E: There exists  > 0 such that B(p) ⊆ E. E is an : All p ∈ E are interior points of E.

p ∈ X is a limit point of E if for all  > 0 B(p) contains an element of E other than p. Note, that p may or may be an element of E. Alternative definition of limit point: p ∈ X is a limit point of E if there is a non ∞ trivial sequence in E that converges to p. That is we can find a sequence {an}n=1 such that (i) an ∈ E ∀n,(ii) an 6= p ∀n,(iii) lim an = p. E is closed if all limit points of E belong to E. Observations: 1. Finite sets are closed set. 2. A set can be both closed and open. Example: X = N , E = {1, 2, 3}, d(x, y) = |x − y|. 3. A set can be neither open nor closed. q 2 P2 2 Example: X = R , E = {x | 0 < x1 < 1, x2 = 0}, d(x, y) = k=1(xk − yk ) 4. A set can be open but not closed.Example: X = R, E = (0, 1), d(x, y) = |x − y|. 5. A set can be closed but not open. Example: X = R, E = [0, 1], d(x, y) = |x − y|. Open and Closed Set

Results 1. (X , d) is a metric space and E ⊆ X . E is open ⇔ (X \ E) is closed.

2. If E1, E2, E3,... are open sets in (X , d). Then ∞ (i) ∪k=1Ek (infinite union) is an open set. n (ii) ∪k=1Ek (finite union) is an open set. n (iii) ∩k=1Ek (finite intersection) is an open set. ∞ (iv) However ∩k=1Ek (infinite intersection) may not be an open set. 3. If E1, E2, E3,... are closed sets in (X , d). Then ∞ (i) ∩k=1Ek (infinite intersection) is a closed set. m (ii) ∩k=1Ek (finite intersection) is a closed set. m (iii) ∪k=1Ek (finite union) is a closed set. ∞ (iv) However ∪k=1Ek (infinite union) may not be a closed set. ∞ Proof of 2:(i) Take any p ∈ ∪k=1Ek , which implies p ∈ Ek for some k. Since Ek is ∞ open there exists  > 0 such that B(p) ⊆ Ek . This implies B(p) ⊆ ∪k=1Ek . Hence p ∞ is an interior point of ∪k=1Ek . (ii) Same as (i). m (iii) Take any p ∈ ∩k=1Ek , which implies p ∈ Ek for all k. Since Ek is open there m exists k > 0 such that Bk (p) ⊆ Ek . Take  = mink=1 k . Then B(p) ⊆ Ek for all k. m m Hence B(p) ⊆ ∩k=1Ek and p is an interior point of ∩k=1Ek . 1 1 ∞ (iv) Take X = R, d(x, y) = |x − y|, Ek = (− k , k ). ∩k=1Ek = {0}, which is not an open set. Open and Closed Set

∞ ∞ Proof of 3:(i) Take any convergent sequence {an}n=1 in ∩k=1Ek , which implies ∞ {an}n=1 is a sequence in Ek for all k. Since Ek is closed, lim an ∈ Ek for all k. Hence ∞ lim an ∈ ∩k=1Ek . (ii) Same as (i). ∞ m (iii) Take any convergent sequence {an}n=1 in ∪k=1Ek , which implies there is a ∞ subsequence of {an}n=1 in Ek , for some k. Since Ek is closed, the subsequence ∞ converges to an element p ∈ Ek . However {an}n=1 is a convergent sequence so it ∞ m should also converge to p ∈ Ek . Thus {an}n=1 converges to a point in ∪k=1Ek . 1 ∞ (iv) X = R, d(x, y) = |x − y|, Ek = [ k , 1]. ∪k=1Ek = (0, 1], which is not closed. Proof of 1: E is open ⇒ (X \ E) is closed: Take a limit point p of (X \ E). We want to show that p ∈ (X \ E). We shall prove this by contradiction. Suppose p ∈ E. Since E is an open set p must be an interior point of E. Then there exists  > 0, such that B(p) ⊆ E. So for that  there is no point of (X \ E), which is in B(p). This contradicts the fact that p is a limit point of (X \ E). (X \ E) is closed ⇒ E is open: Take any p ∈ E. We want to show that p is an interior point of E. Once again we prove by contradiction. Suppose the contrary is true and for every  > 0, B(p) is not a subset of E. Hence each B(p) contains some element of (X \ E) other than p. Hence p is a limit point of (X \ E). Since (X \ E) is closed, p must be in (X \ E), which contradicts with the fact that p ∈ E. Compact Set

Definition: Compact Set: Let (X , d) be a metric space and E ⊆ X . E is a compact set if every sequence has a subsequence that converges to a point in E. Heine-Borel Theorem: Take S ⊆ Rn with Euclidian norm. S is compact iff S is closed and bounded. Proof: We shall prove just one part of this result, S compact ⇒ S is closed and bounded. ∞ ∞ Take a converging sequence {an}n=1 in S. Since S is compact, {an}n=1 has a ∞ converging subsequence that converges to a point in S. But {an}n=1 converges, hence it must converge to the same point in S. Thus S is closed. Now suppose that S is not bounded then for all M > 0, there is a aM ∈ S such that ∞ ∞ ||aM || > M. Take the sequence {aM }M=1. There is no subsequence of {aM }M=1 which is bounded and hence there is no subsequence which converges. We have reached a contradiction. The converse follows a proof similar to Bolzano-Weierstrass Theorem.

∞ n Observation: A sequence {an}n=1 in R converges to a ⇔ for every k = 1, 2,... n; k ∞ k {an }n=1 converges to a .