Problems in Metric Spaces 1

TOPOLOGY PROBLEMS 1: METRIC SPACES PROF. SEBASTIAN VATTAMATTAM Problem 0.1. Let (X; d) be a metric space and K X. Show that @K is closed. Figure 1. Problem 0.1, Case 1 Proof Let a 2 X n @K) 9> 0 such that U(a) contains either no point in K or no point in X n K Case 1: See gure 1 U(a) contains no point in K ) U(a) X n K 12 PROF. SEBASTIAN VATTAMATTAM Suppose U(a)T@K 6= Let x 2 U(a)T@K 1 := d(x; a) > 0 x 2 @K ) 9y 2 U1(x)\K d(y; a) d(y; x) + d(x; a) < 1 + d(x; a) = ) y 2 U(a);! Therefore U(a)T@K = U(a) X n @K So, X n @K is open, and hence @K is closed. Case 2: See gure 2 U(a) contains no point in X n K ) U(a) K Suppose U(a)T@K 6= Let x 2 U(a)T@K 1 := d(x; a) > 0 x 2 @K ) 9y 2 U1(x)\X n K d(y; a) d(y; x) + d(x; a) < 1 + d(x; a) = ) y 2 U(a);! Therefore U(a)T@K = X n U(a) X n @K So, X n @K is open, and hence @K is closed. Problem 0.2. Let (X; d) be a metric space and K X. Show that K is closed iK contains its boundary. ProofTOPOLOGY 3 Figure 2. Problem 0.1, Case 2 (1) Suppose K is closed. To prove @K K Let a 2 @K If a =2 K then a 2 X n K Since X n K is open, 9> 0 such that U(a) X n K ) U(a)\K = This is a contradiction since a is a boundary point. Therefore, a 2 K ) @K K (2) Suppose @K K To prove K is closed. We shall show that X n K is open. If X n K is not open there is an element a 2 XnK such that 8> 0; U(a) contains a point in K. Thus U(a) intersects K and Kc ) a 2 @K ) a 2 K This is a contradiction since a 2 X n K. Therefore X n K is open and hence K is closed. Problem 0.3. Let (X; d) be a metric space and K X. Show that K is closed , K = K.4 PROF. SEBASTIAN VATTAMATTAM Proof Suppose K is closed. Let a 2 K0 8> 0; U(a)\K 6= ) a =2 X n K; since X n K is open ) a 2 K ) K0 K ) K = K Suppose K = K ) K0 K Thus X n K contains no limit point of K Therefore X n K is open and hence K is closed. Denition 0.4. If (X; d) is a metric space and K X then the distance of a 2 X from K is dened as d(a;K) := inffd(x; a)jx 2 Kg See gure 3 Figure 3. Denition 0.4 Problem 0.5. Let (X; d) be a metric space and K X. Show that (1) X n K = Int(X n K) (2) K = fx 2 X : d(x;K) = 0gTOPOLOGY 5 Proof (1) (a) Let a 2 X n K a 2 X n K ) a =2 K and a =2 K0 Since a =2 K; a 2 X n K If a =2 Int(X n K) 8> 0; U(a) intersects K ) a 2 K0 This is a contradiction since a =2 K0 This contradiction shows that a 2 Int(X n K) Therefore X n K Int(X n K):::::(1) (b) Let a 2 Int(X n K) ) 9> 0 such that U(a) X n K ) a =2 K ) a 2 X n K ) Int(X n K) X n K::::(2) Conclusion from (1) and (2) (2) (a) Let a 2 K ) x 2 K or x 2 M0 If a 2 K then d(a;K) = 0 Suppose a =2 K ) a 2 K0 ) 8> 0; U(a) contains a point x 2 K If d(a;K) := r > 0; then x 2 K and x 2 U(a) ) d(x; a) < rg ) d(a;K) = inffd(a; y)jy 2 Kg < r;! Therefore d(a;K) = 0 K fx 2 X : d(x;K) = 0g::::(1) (b) Let a 2 X such that d(a;K) = 0 ) inffd(a; x)jx 2 Kg = 06 PROF. SEBASTIAN VATTAMATTAM Given > 0; there exists y 2 K such that d(a; y) < ) U(a) intersects K a 2 K fx 2 X : d(x;K) = 0g K::::(2) The conclusion follows from (1) and (2). Problem 0.6. prob:3; p:64 in [2] Let (X; d) be a metric space, x 2 X; r > 0; and A X is such that d(A) < r and A intersects Ur(x) Prove that A U2r(x) Proof ATUr(x) 6= Let a 2 A\Ur(x) y 2 A ) d(y; a) < r d(y; x) d(y; a) + d(a; x) < 2r ) y 2 U2r Hence the conclusion. Problem 0.7. prob:4; p:64 in [2] Let (X; d) be a metric space.Prove that every subset of X is open ievery singleton in X is open. Give an example. Suppose every set M X is open. Then for every x 2 X; fxg is open. Suppose every singleton in X is open. Then 8M X;M = [x2MfxgTOPOLOGY 7 Therefore M is open. Example The discrete metric space. Theorem 0.8. Let (X; d) be a metric space and M X M= M[M0 Theorem 0.9. Let (X; d) be a metric space and M X M= M[@M Theorem 0.10. Let (X; d) be a metric space and M X M= M[@M Theorem 0.11. 1:3:4 in [1] Let (X; d) be a metric space and M X C := fF M : F is closed inXg Then the closureM =\fF X : F 2 Cg Problem 0.12. In (R; j j) describe the interior of (1) Z (2) Q (3) ]0; 1[ (4) [0; 1] (5) [0; 1[Sf1; 2g Problem 0.13. In (C; j j) describe the interior of (1) fxjd(x; 0) < 1g (2) fxjd(x; 0) 1g (3) fx = (x1; x2)jx2 = 1g References [1] W. Beekmann,Analysis 2,Fern University in Hagen [2] George F. Simmons, Topology and Modern Analysis,International Student Edition, McGraw- Hill, 1963.