Topology Class 1: Metric Spaces

TOPOLOGY CLASS 1: METRIC SPACES PROF. SEBASTIAN VATTAMATTAM 1. Definition and Examples Denition 1.1. Metric Space Let X be a non-empty set. A metric on X is a function d: X X ! R such that, for x; y; z 2 X; (1) d(x; y) 0 (2) d(x; y) = 0 , x = y (3) d(x; y) = d(y; x)(symmetry) (4) d(x; y) d(x; z) + d(z; y)(triangle inequality) (X; d) is called a metric space. Example 1.2. For x 2 R dene jxj = x if x 0; x if x < 0: Then d : R R ! R dened by d(x; y) = jx yj is a metric(called usual metric) on R and (R; d) is a metric space. It is denoted by (R; j j) also. Example 1.3. For x = x1 + ix2 2 C dene jxj = qx21 + x22 Then d : C C ! R dened by d(x; y) = jx yj is a metric(called usual metric) on C and (C; d) is a metric space. It is denoted by (C; j j) also. 12 PROF. SEBASTIAN VATTAMATTAM Example 1.4. X 6= ; d: X X ! R dened by d(x; y) = 0 if x = y; 1 if x 6= y: Then, (X; d) is a metric space. d is called the discrete metric. Proof (1) d(x; y) = 0 or 1 ) d(x; y) 0 (2) It follows from denition that d(x; y) = 0 , x = y (3) It follows from denition that d(x; y) = d(y; x) (4) Let x; y; z 2 X If x = y then d(x; y) = 0 d(x; z) + d(z; y) If x 6= y then we have three cases: (a) z = x; z 6= y, then d(x; y) = 1; d(x; z) = 0; d(z; y) = 1 ) d(x; y) d(x; z)+d(z; y) (b) z 6= x; z = y, then d(x; y) = 1; d(x; z) = 1; d(z; y) = 0 ) d(x; y) d(x; z)+d(z; y) (c) z 6= x; z 6= y, then d(x; y) = 1; d(x; z) = 1; d(z; y) = 1 ) d(x; y) d(x; z)+d(z; y) Hence the conclusion. Theorem 1.5. If (X; d) is a metric space and x; y; z 2 X then show that jd(x; z) d(z; y)j d(x; y)TOPOLOGY 3 Proof By the triangle inequality, d(z; y) d(z; x)+d(x; y) ) d(x; y) d(z; x)d(z; y)::::(1) Again by the triangle inequality, d(x; z) d(x; y)+d(y; z) ) d(x; y) d(x; z)d(z; y)::::(2) From (1) and(2), d(x; y) d(z; x)d(z; y) d(x; y) ) jd(x; z)d(z; y)j d(x; y) Denition 1.6. Subspace of a Metric Space. Let (X; d) be a metric space and Y X: For x; y 2 Y dene dY (x; y) = d(x; y) Then (Y; dY ) is a metric space, called a subspace of (X; d): 2. Open Sets Denition 2.1. Let (X; d) be a metric space. If a 2 X; > 0 then dene U(a) := fx 2 X : d(x; a) < g U(a) is called the neighborhood of a in X: It is also called an open sphere or open ball. Example 2.2. Consider (R; j j); a 2 R; 0 < r 2 R; Ur(a) = fx 2 R : jx aj < rg It is the open interval ]a r; a + r[; denoted also by (a r; a + r): Example 2.3. Consider (C; j j); a 2 C; 0 < r 2 C; Ur(a) = fx 2 C : jx aj < rg If x = x1 + ix2; a = a1 + ia2;then jx aj = p(x1 a1)2 + (x2 a2)2 It is the interior of the sphere (x1 a1)2 + (x2 a2)2 = r24 PROF. SEBASTIAN VATTAMATTAM Example 2.4. If (X; d) is a discrete metric space, then given a 2 X; 0 < 1, the neighborhood U(a) = fag Denition 2.5. Let (X; d) be a metric space. A set U X is a neighborhood of a 2 X if 9> 0 such that U(a) U Theorem 2.6. 1:3:2 in [1] If X is a metric space, a 2 X and U is a neighborhood of a then (1) a 2 U (2) Any superset of U is also a neighborhood of a: (3) The union of a nite collection of neighborhoods of a is also a neighborhood of a: Proof (1) Obvious. (2) Let M U Since U is a neighborhood of a, 9> o : U(a) U ) U(a) M (3) Let U1; U2 be two neighborhoods of a 91; 2 > 0 : U1(a) U;U2(a) U Let := minf1; 2g U(a) U1; U(a) U2 ) U(a) U1[U2 Denition 2.7. Open Set If (X; d) is a metric space, a set M X is open if M is a neighborhood of every element in it. That is M is open if for x 2 M; 90 such that U(a) MTOPOLOGY 5 Theorem 2.8. TheoremA; p:60 in [2] In a metric space (X; d); the empty set and the whole space X are open. Proof (1) Obvious (2) For a 2 X and any > 0; U(a) X: Theorem 2.9. ThmB; p:61 in [2] In metric space (X; d), each open sphere is open. Proof (1) This part is evident. (2) Let for r > 0; Ur(a) = fx 2 X : d(x; a) < rg. Let x 2 Ur(a) ) d(x; a) < r Let r1 = r d(x; a) Let y 2 Ur1(x) ) d(y; x) < r1 Then d(y; a) d(y; x) + d(x; a) < r1 + d(x; a) = r ) y 2 Ur(a) Thus, Ur1(x) Ur(a) Hence the conclusion. Theorem 2.10. ThmC; p:61 in [2] Let (X; d) be a metric space. A set M X is open iit is a union of open balls. Proof (1) Suppose M is open. Let a 2 M9> 0 such that U(a) M6 PROF. SEBASTIAN VATTAMATTAM Then M = [a2M U(a) (2) Suppose M = [a2M U(a) x 2 M ) x 2 U(a) for some > 0 1 := d(x; a) > 0; U1(x) U(a) M Hence the conclusion. Theorem 2.11. ThmD; p:61 in [2] Let (X; d) be a metric space. (1) Any union of open sets in X is open. (2) Any nite intersection of open sets in X is open Proof (1) Let fGg be an arbitrary family of open sets in X; and G :=[Gx 2 G ) x 2 G; for some Then Gis a nbd of x. Since G G;G is a nbd of x. So, G is open. (2) Let fGi : i = 1; 2; ::; ng be a nite family of open sets in X; and G := n \i=1Gi x 2 G ) x 2 Gi; for all i Then Gi is a nbd of x for all i. 9i > 0; i = 1; 2; :::; n such that Ui(x) Gi; 8iTOPOLOGY 7 Let := minfi; i = 1; 2; :::; ng U(x) Ui(x); 8i ) U(x) G Hence the conclusion. Problem 2.12. 1:3:3 in [1] If (X; d) is a metric space, b 2 X; 0 letM := fx 2 X : d(x; b) > g Then show that M is open. Let a 2 M ) d(a; b) > Set ~:= d(a; b) For x 2 U~(a); d(x; a) < ~) d(x; b) jd(x; a) d(a; b)j d(a; b) d(x; a) > d(a; b) ~= ) x 2 M So M is open. Denition 2.13. HausdorSpace A metric space (X; d) is a Hausdorspace if given a 6= b in X there exist a neighborhood U of a and a neighborhood V of b such that U TV = . That is, distinct points are separated by disjoint neighborhooods Theorem 2.14. Every metric space is a Hausdorspace. Proof (X; d) a metric space. Let a; b 2 X; a 6= b ) d(a; b) = r > 0 U := Ur2 (a); V := Ur2 (b)8 PROF. SEBASTIAN VATTAMATTAM nbds of a and b:x 2 U\V ) x 2 U; x 2 V ) d(a; x) < r2; d(x; b) < r2 d(a; b) d(a; x) + d(x; b) < r This is a contradiction. Therefore U\V = 3. Problems Problem 3.1. For x; y 2 R dene d(x; y) = jx yj 1 + jx yj Show that d is a metric on R. Solution Lemma: Deneh : [0;1) ! R by t ! t 1 + t Then the function h is monotone increasing. It follows from the fact that h0(t) > 0 (1) d(x; y) 0 (2) d(x; y) = 0 , jx yj = 0 , x = y (3) Since jx yj = jy xj; d(x; y) = d(y; x) (4) jx yj jx zj + jz yj Since jx yj; jx zj + jz yj 2 [0;1) h(jxyj) h(jxzj+jzyj) ) jx yj 1 + jx yj jx zj + jz yj 1 + jx zj + jz yj = jx zj 1 + jx zj + jz yj + jz yj 1 + jx zj + jz yj jx zj 1 + jx zj + jz yj 1 + jz yjTOPOLOGY 9 ) d(x; y) d(x; z) + d(z; y) Hence the conclusion. Problem 3.2. Let (X; d) be a metric space. For x; y 2 X dene d1(x; y) = d(x; y) 1 + d(x; y) Show that (X; d1) is a metric space. Solution Lemma: Deneh : [0;1) ! R by t ! t 1 + t Then the function h is monotone increasing. It follows from the fact that h0(t) > 0 (1) d1(x; y) 0 (2) d1(x; y) = 0 , d(x; y) = 0 , x = y (3) Since d(x; y) = d(y; x); d1(x; y) = d1(y; x) (4) d(x; y) d(x; z) + d(z; y) Since d(x; y); d(x; z) + d(z; y) 2 [0;1) h(d(x; y)) h(d(x; z)+d(z; y)) ) d(x; y) 1 + d(x; y) d(x; z) + d(z; y) 1 + d(x; z) + d(z; y) = d(x; z) 1 + d(x; z) + d(z; y) + d(z; y) 1 + d(x; z) + d(z; y) d(x; z) 1 + d(x; z) + d(z; y) 1 + d(z; y) ) d1(x; y) d1(x; z) + d1(z; y) Hence the conclusion. To be continued For online classes in Mathematics vattamattam@gmail.com10 PROF. SEBASTIAN VATTAMATTAM References [1] W. Beekmann,Analysis 2,Fern University in Hagen [2] George F. Simmons, Topology and Modern Analysis,International Student Edition, McGraw- Hill, 1963.