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 ∈ 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 ∈ A there exist Ux ∈ T such that x ∈ 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 x∈A 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 T∞ = {U|X − U is infinite or empty or all of X} a topology on X. Proof: PART (1) From example 4 Tc = {U|X − U is countable or is all of X}. 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 ∅ ∈ Tc. Second, let Uα ∈ Tc for α ∈ 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, α∈I α∈I S Uα ∈ 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 α∈I α∈I S subsets of countable sets are countable, Uα ∈ Tc. Third, we must show finite intersections are in Tc. Let Uα ∈ 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 ∈ 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 ∈ 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 T∞ = {U|X − U is infinite or empty or all of X} is NOT a topology on X. By counterexample let X = R and let U1 = (−∞, 0) and U2 = (0, ∞). U1 and U2 are in T∞ since X − U1 = [0, ∞) and X − U2 = (−∞, 0] are both infinite. Condition 2 of the definition of topology fails. That is, U1 ∪ U2 6∈ T∞ since X − (U1 ∪ U2) = {0}.  T 13.4a) If {Tα} 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 {Tα} 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 = {a, b, c}, let T1 = {∅,X, {a}, {a, b}} and T2 = {∅,X, {a}, {b, c}}. 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 ∅ ∈ Tα. For arbitrary unions, let Uϕ ∈ Tα for S T ϕ ∈ I. Then Uϕ ∈ Tα for all α and thus Uϕ ∈ Tα. Finally, for finite intersections, n n T T T T let Uj ∈ Tα for 1 ≤ j ≤ n. So Uj ∈ Tα for all α and hence Uj ∈ 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 ∈ Tϕ and U2 ∈ Tβ,U1 ∩ U2 are not neces- S sarily in Tα. As a counter example, let X = {a, b, c}, Tϕ = {X, ∅, {a, b}}, Tβ = S {X, ∅, {b, c}},U1 = {a, b}, and U2 = {b, c}. Then Tα = Tϕ ∪ Tβ but, U1 ∩ U2 = S {b} 6∈ Tα.  Proof: 13.4b.1 T For existence, let T = Tβ , where {Tβ} 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 {∅,X, {a}, {b}, {a, b}, {b, c}}. The largest topology contained in both T1 and T2 is {∅,X, {a}}. 

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 {TAα } be the collection of all topologies containing A. T First, TAα ⊆ TA since TA ∈ {TAα }. For the reverse inclusion let U ∈ TA. Then for all x ∈ U we have Ax ∈ A such that S S T Ax ⊆ U and U = Ax. Now since TAα contains A for all alpha, U = Ax ∈ TAα . x∈U x∈U 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 {TA } be the collection T α of all topologies containing A. Then as above, TAα ⊆ TA since TA ∈ {TAα }. α S Tn Conversely, let U ∈ TA. So U = ( Ai), where Aα1 ,...,Aαn ∈ A. Since α∈I i=α1 α T T Tn T Aα1 ,...,Aαn ∈ TAα and TAα is a topology, we have Ai ∈ TAα for all α. i=α1 α S Tn T T Now we have ( Ai) = U ∈ TAα and hence TA ⊆ TAα . α∈I 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) = {x|a ≤ x ≤ b}.

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 ∈ Z .

We must show TRl 6⊂ TRK and TRK 6⊂ TRl . That is, there exists an element Ul ∈ TRl such that Ul 6∈ TRK and an element UK ∈ TRK such that UK 6∈ TRl .

c Steven T Riggs 2014 Page 3 of 6 Compiled: 09/15/2014 7:34pm Notice that [0, 1) ∈ 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) 6∈ 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 ∈ N. 1 1 That is, 0 ∈ (− n , n ) − K but there is no interval [a, b) such that 0 ∈ [a, b) and 1 1 1 1 [a, b) ⊂ (− n , n ) − K. Hence (− n , n ) − K 6∈ 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. That is, O = U ∩A = (V ∩Y )∩A = V ∩ A and the topology A inherits as a subspace of Y is the same as the topology it inherits as a subspace of X.  16.5) Let X and X0 denote a single set in the topologies T and T0, re- spectively; let Y and Y 0 denote a single set in the topologies U and U0, respectively. Assume these sets are nonempty. (a) Show that if T0 ⊃ T and U0 ⊃ U, then the product topology on X0 ×Y 0 is finer than the product topology on X × Y . (b) Does the converse of (a) hold? Justify your answer. Proof: PART A Let W be in the product topology on X × Y . Then for all (x, y) ∈ W , there exist a basis element T × U such that T × U ⊂ W where T ∈ T and U ∈ U. Now since T0 and U0 are finer than T and U respectively, T can be written as a union of elements of T0 and U can be written as a union of elements of U0. That is, T and U are elements of T0 and U0 respectively. So T × U is a basis element for X0 × Y 0 and hence is an element of the product topology on X0 × Y 0. Thus the product topology on X0 × Y 0 is finer than the product topology on X × Y .  PART B The converse is as follows: If the product topology on X0 × Y 0 is finer than the product topology on X × Y , then T0 ⊃ T and U0 ⊃ U. Claim: The converse is true. Proof: PART B Let T ∈ T and U ∈ U. We must show T ∈ T0 and U ∈ U0. We have that T × U is a basis element for the topology on X × Y and hence is open in X × Y . Since the topology on X0 × Y 0 is finer, T × U is open in X0 × Y 0. Thus T ∈ T0 and U ∈ U0 as desired. 

c Steven T Riggs 2014 Page 4 of 6 Compiled: 09/15/2014 7:34pm 16.8) If L is a straight line in the plane, describe the topology L inherits as a subspace of Rl × R and as a subspace of Rl × Rl. In each case it is a familiar topology. PART A: Rl × R All vertical lines inherit the standard topology. All diagonal lines inherit the standard topology union the lower limit topology. All horizontal lines inherit the lower limit topology.

PART B: Rl × Rl All vertical, horizontal, and positive sloped diagonal lines inherit the lower limit topology. All negative sloped diagonal lines inherit the discrete topology.

 16.9) Show that the dictionary order topology on the set R × R is the same as the product topology Rd × R, where Rd denotes R in the discrete 2 topology. Compare this topology with the standard topology on R . Proof:PART 1 We must show for the dictionary topology T on R × R and the product topology Td on Rd × R, as stated, we have T ⊆ Td and Td ⊆ T. We will show that the bases element of each contains a bases element of the other. Let (a × b, a × d) be a bases element of T. Then for all x ∈ (a×b, a×d) we have x is contained in a bases element of the form {a} × (b, d) ⊂ (a × b, a × d) of Td. That is, T ⊆ Td. Conversely, let {a} × (b, d) be a basis element of Td. Then for all x ∈ {a} × (b, d) we have x ∈ (a × b, a × d) ⊂ {a} × (b, d) of T. That is, Td ⊆ T. Hence, T = Td.

PART 2 2 Claim: The topology T from above is finer than the standard topology Ts on R . Let (a, c) × (b, d) be a bases element in Ts. Then for all x ∈ (a, c) × (b, d) we have x ∈ ({a} × (b, d) ∪ ... ∪ {c} × (b, d)) ⊂ ((a, c) × (b, d)). That is, bases elements of Ts can be written as the union of bases elements of T. Thus, Ts ⊂ T as desired.

 ADDITIONAL PROBLEM(S)

13.2) Consider the nine topologies on the set X = {a, b, c} indicated in example 1 (pg 76). Compare them; that is, for each pair of topologies, determine whether they are comparable, and if so, which is finer. Name each topology in example 1 by its row × column address (i.e. Trc). Then we have the following: T11 is the trivial topology and NOT finer than any (coarser than all others). T12 is finer than T11, T31 T13 is finer than T11, T21, T31 T21 is finer than T11

c Steven T Riggs 2014 Page 5 of 6 Compiled: 09/15/2014 7:34pm T22 is finer than T11 T23 is finer than T11, T13, T21, T31 T31 is finer than T11 T32 is finer than T11, T12, T21, T31 T33 is the discrete topology and finer than all others. 

c Steven T Riggs 2014 Page 6 of 6 Compiled: 09/15/2014 7:34pm