Metric Space Definition: Metric Space Take a set X . A function d : X × X !R is called a Metric or distance function on E if d satisfies the following properties: (a) d(p; q) ≥ 0 for all p; q 2 X . Moreover d(p; q) = 0 iff p = q. (b) For all p; q 2 X , d(p; q) = d(q; p) (c) (Triangle Inequality) d(p; q) + d(q; r) ≥ d(p; r) for all p; q; r 2 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) = jx − yj q l Pl 2 3: X = R ; d(x; y) = k=1(xk − yk ) l 4: X = R ; d(x; y) = maxk jxk − yk j 5: X = ff j f :[a; b] !R and f is continuousg; d(f ; g) = maxt2[a;b]jf (t) − g(t)j Metric Space On R, we used the distance function d(p; q) = jp − qj. 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 2 A and M > 0 such that d(p; q) ≤ M for all p 2 A. Examples: Bounded metric space 2 2 2 1: X = R and A = fx 2 X j x1 + x2 < 1g. Here A is bounded. 2 2: X = R and A = fx 2 X j x1 = x2g. A is not bounded. • For any a 2 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) = fx 2 X j d(x; a) < g. 1 • A sequence in X , fangn=1, converges to a: If for every > 0, there exists N 2 N such that 8n ≥ N; an 2 B(a). Examples: Convergent Sequences q 2 P2 2 1 1: X = R ; d(x; y) = k=1(xk − yk ) ; an = ( n ; 1). This sequence converges to (0; 1). t 2: X = ff j f : [0; 1] ! Rg; d(f ; g) = maxt2[a;b]jf (t) − g(t)j; fn(t) = n . This sequence converges to f (t) = 0 8t. 1 • A sequence fangn=1 is bounded if there exists M > 0 such that d(a1; an) ≤ M 8n Vector Space 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 2 V and c; d 2 R 1. V close under L: v L w 2 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: 9 a vector called 0 such that v L 0 = v 8v 5. Existence of additive inverse: For each v, 9 w 2 V such that v L w = 0 6. V close under J: c J v 2 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) Normed Vector Space From now on, we shall use the usual addition and multiplication sign for vector space as well. Additive inverse shall be denoted by usual `−' sign. That is inverse of vector z 2 V is denoted by −z. Moreover for y; z 2 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 norm function jj jj : V !R. The norm function has to satisfy properties similar to distance function. (a) jjxjj ≥ 0 for all x 2 V . Moreover jjxjj = 0 iff x = 0. (b) For all x; y 2 V , jjx − yjj = jjy − xjj (c) For all c 2 R and x 2 V , jjcxjj = jcj jjxjj (d) (Triangle Inequality) For all x; y 2 V , jjx + yjj ≤ jjxjj + jjyjj A norm can be easily extended to a distance function in the following manner, d(x; 0) = jjxjj and d(x; y) = d(x − y; 0) = jjx − yjj. 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 ; jjxjj = k=1 xk 2: X = ff j f :[a; b] !R and f is continuousg; jjf jj = maxt2[a;b]jf (t)j 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. 1 (Proof): Take any sequence fangn=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 = maxfd(a1; a2); d(a1; a3);:::; d(a1; aN−1); (d(a1; 1) + 1)g. Check (Result 7-Sequence): Subsequence of a convergent sequence converges to the same limit. 1 1 Let fangn=1 and fbngn=1 are sequences in a normed vector space V . lim an = a and lim bn = b. 1 (Result 1-Sequence) Take the sequence fcangn=1, where c 2 R. lim(can) = ca. 1 (Result 2-Sequence) Take the sequence fan + bngn=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 2 E is an interior point of E: There exists > 0 such that B(p) ⊆ E. E is an open set: All p 2 E are interior points of E. p 2 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 2 X is a limit point of E if there is a non 1 trivial sequence in E that converges to p. That is we can find a sequence fangn=1 such that (i) an 2 E 8n,(ii) an 6= p 8n,(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 = f1; 2; 3g, d(x; y) = jx − yj. 3. A set can be neither open nor closed. q 2 P2 2 Example: X = R , E = fx j 0 < x1 < 1; x2 = 0g, 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) = jx − yj. 5. A set can be closed but not open. Example: X = R, E = [0; 1], d(x; y) = jx − yj. Open and Closed Set Results 1: (X ; d) is a metric space and E ⊆ X . E is open , (X n E) is closed. 2: If E1; E2; E3;::: are open sets in (X ; d). Then 1 (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. 1 (iv) However \k=1Ek (infinite intersection) may not be an open set. 3: If E1; E2; E3;::: are closed sets in (X ; d). Then 1 (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. 1 (iv) However [k=1Ek (infinite union) may not be a closed set. 1 Proof of 2:(i) Take any p 2 [k=1Ek , which implies p 2 Ek for some k. Since Ek is 1 open there exists > 0 such that B(p) ⊆ Ek . This implies B(p) ⊆ [k=1Ek . Hence p 1 is an interior point of [k=1Ek . (ii) Same as (i). m (iii) Take any p 2 \k=1Ek , which implies p 2 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 1 (iv) Take X = R, d(x; y) = jx − yj, Ek = (− k ; k ). \k=1Ek = f0g, which is not an open set. Open and Closed Set 1 1 Proof of 3:(i) Take any convergent sequence fangn=1 in \k=1Ek , which implies 1 fangn=1 is a sequence in Ek for all k. Since Ek is closed, lim an 2 Ek for all k. Hence 1 lim an 2 \k=1Ek . (ii) Same as (i). 1 m (iii) Take any convergent sequence fangn=1 in [k=1Ek , which implies there is a 1 subsequence of fangn=1 in Ek , for some k.