Math 553 - Topology Todd Riggs Assignment 1 Sept 10, 2014 Problems: Section 13 - 1, 3, 4, 5, 6; Section 16 - 1, 5, 8, 9; 13.1) Let X be a topological space and let A be a subset of X. Suppose that for each x 2 A there is an open set U containing x such that U ⊂ A. Show that A is open in X. Proof: Since for each x 2 A there exist Ux 2 T such that x 2 Ux and Ux ⊂ A, we have S A = Ux. That is, A is the union of open sets in X and thus by definition of the x2A topology on X; A is open. 13.3) Show that the collection Tc given in example 4 (pg 77) is a topology on the set X. Is the collection T1 = fUjX − U is infinite or empty or all of Xg a topology on X. Proof: PART (1) From example 4 Tc = fUjX − U is countable or is all of Xg. To show that Tc is a topology on X, we must show we meet all 3 conditions of the definition of topology. First, since X −X = ; is countable and X −; = X is all of X we have X and ; 2 Tc. Second, let Uα 2 Tc for α 2 I. We must look at two cases: (1) X − Uα = X for all α and (2) There exist α such that X − Uα is countable. Case (1): X − Uα = X S T This implies Uα = ; for all α. So X − ( Uα) = (X − Uα) = X and thus, α2I α2I S Uα 2 Tc. Case (2): There exist α such that X − Uα is countable. S T We have X − ( Uα) = (X − Uα) ⊆ X − Uα which is countable. Thus, since α2I α2I S subsets of countable sets are countable, Uα 2 Tc. Third, we must show finite intersections are in Tc. Let Uα 2 Tc for 1 ≤ α ≤ n. Again, we must check two cases. Case (1): There exists an α such that X − Uα = X. n n n T S T So X − ( Uα) = (X − Uα) = X and thus, Uα 2 Tc. α=1 α=1 α=1 Case (2): X − Uα is countable for all α. n n T S We have X − ( Uα) = (X − Uα). α=1 α=1 n T Since a finite collection of countable sets is countable, we again have Uα 2 Tc. α=1 Thus Tc is a topology on X. c Steven T Riggs 2014 Page 1 of 6 Compiled: 09/15/2014 7:34pm Proof: PART (2) The collection T1 = fUjX − U is infinite or empty or all of Xg is NOT a topology on X. By counterexample let X = R and let U1 = (−∞; 0) and U2 = (0; 1). U1 and U2 are in T1 since X − U1 = [0; 1) and X − U2 = (−∞; 0] are both infinite. Condition 2 of the definition of topology fails. That is, U1 [ U2 62 T1 since X − (U1 [ U2) = f0g. T 13.4a) If fTαg is a family of topologies on X, show that Tα is a topology S on X. Is Tα a topology on X? 13.4b) Let fTαg be a family of topologies on X. Show that there is a unique smallest topology on X containing all the collections Tα, and a unique largest topology contained in all Tα. 13.4c) If X = fa; b; cg, let T1 = f;; X; fag; fa; bgg and T2 = f;; X; fag; fb; cgg. Find the smallest topology containing T1 and T2, and the largest topology contained in T1 and T2. Proof: 13.4a.1 T T Since each Tα is a topology, X and ; 2 Tα. For arbitrary unions, let U' 2 Tα for S T ' 2 I. Then U' 2 Tα for all α and thus U' 2 Tα. Finally, for finite intersections, n n T T T T let Uj 2 Tα for 1 ≤ j ≤ n. So Uj 2 Tα for all α and hence Uj 2 Tα. Thus j=1 j=1 T Tα is a topology on X. Proof: 13.4a.2 S Tα is NOT a topology on X. For U1 2 T' and U2 2 Tβ;U1 \ U2 are not neces- S sarily in Tα. As a counter example, let X = fa; b; cg; T' = fX; ;; fa; bgg; Tβ = S fX; ;; fb; cgg;U1 = fa; bg, and U2 = fb; cg. Then Tα = T' [ Tβ but, U1 \ U2 = S fbg 62 Tα. Proof: 13.4b.1 T For existence, let T = Tβ , where fTβg is the collection of all topologies containing S T Tα. We claim that T = Tβ is the uniquely smallest topology on X containing all the collections Tα. First, by part a, we know that T is a topology and Tβ contains S S Tα for all β. Since T is the intersection of all topologies containing Tα, it must be the smallest. 0 S For uniqueness, assume there exists a topology T that contains Tα and is such that T0 ⊆ T. Since T is the smallest, we must have T ⊆ T0. Thus T0 = T as desired T and T = Tβ is the uniquely smallest topology on X containing all the collections Tα. c Steven T Riggs 2014 Page 2 of 6 Compiled: 09/15/2014 7:34pm Proof: 13.4b.2 T As for the largest topology contained in all Tα, consider T = Tα. It is contained in Tα for all alpha and, by part a, is a topology. Since it is the intersection of all 0 Tα, it must be the largest. Now for uniqueness, assume there is another topology T 0 contained in each Tα and such that T ⊇ T. Since T is the largest, we must have 0 0 T T ⊇ T . Thus, again, T = T as desired and T = Tα is the uniquely largest topology on X contained in all Tα. Prof: 13.4c The smallest topology containing T1 and T2 is f;; X; fag; fbg; fa; bg; fb; cgg. The largest topology contained in both T1 and T2 is f;; X; fagg. 13.5) Show that if A is a basis for a topology on X, then the topol- ogy generated by A equals the intersection of all topologies on X that contain A. Prove the same if A is a subbasis. Proof: PART (1) Let TA be the topology generated by the basis A and let fTAα g be the collection of all topologies containing A. T First, TAα ⊆ TA since TA 2 fTAα g. For the reverse inclusion let U 2 TA. Then for all x 2 U we have Ax 2 A such that S S T Ax ⊆ U and U = Ax. Now since TAα contains A for all alpha, U = Ax 2 TAα . x2U x2U T T That is, TA ⊆ TAα and TA = TAα as desired. Proof: PART (2) Let TA be the topology generated by the subbasis A and let fTA g be the collection T α of all topologies containing A. Then as above, TAα ⊆ TA since TA 2 fTAα g. α S Tn Conversely, let U 2 TA. So U = ( Ai), where Aα1 ;:::;Aαn 2 A. Since α2I i=α1 α T T Tn T Aα1 ;:::;Aαn 2 TAα and TAα is a topology, we have Ai 2 TAα for all α. i=α1 α S Tn T T Now we have ( Ai) = U 2 TAα and hence TA ⊆ TAα . α2I i=α T 1 Thus TA = TAα as desired. 13.6) Show that the topologies of Rl and RK are not comparable. Proof: The lower limit topology, TRl , is the topology generated on R by the collection of all half-open intervals of the form [a; b) = fxja ≤ x ≤ bg. The K-topology, TRK , on R is the topology generated by the collection of all open intervals (a; b), along with the sets of the form (a; b) − K. Here K is the is the set of 1 + all positive rationals of the form n , for n 2 Z . We must show TRl 6⊂ TRK and TRK 6⊂ TRl . That is, there exists an element Ul 2 TRl such that Ul 62 TRK and an element UK 2 TRK such that UK 62 TRl . c Steven T Riggs 2014 Page 3 of 6 Compiled: 09/15/2014 7:34pm Notice that [0; 1) 2 TRl . No bases element from RK exists that contains 0 and is a subset of [0; 1). That is, there is no open interval (a; b) or interval (a; b) − K that is a subset of [0; 1) and contains 0. Thus [0; 1) 62 TRK and hence TRl 6⊂ TRK . 1 1 Conversely, note that (− n ; n ) − K is an element of RK. Additionally, notice for 1 any neighborhood [a; b) containing 0 that it will also contain n for some n 2 N. 1 1 That is, 0 2 (− n ; n ) − K but there is no interval [a; b) such that 0 2 [a; b) and 1 1 1 1 [a; b) ⊂ (− n ; n ) − K. Hence (− n ; n ) − K 62 TRl and TRK 6⊂ TRl . Thus the topologies of Rl and RK are not comparable. 16.1) Show that if Y is a subspace of X, and A is a subset of Y , then the topology A inherits as a subspace of Y is the same as the topology it inherits as a subspace of X. Proof: By the subspace topology on Y; U ⊆ Y is open in Y if U = V \Y for some V open in X. Similarly, O ⊆ A is open in A if O = U \A.