Math 190: Fall 2014 Homework 2 Due 5:00Pm on Friday 10/17/2014 Problem 1: Let X Be a Set and Let B Be a Basis for a Topology On

Math 190: Fall 2014 Homework 2 Due 5:00pm on Friday 10/17/2014

Problem 1: Let X be a set and let B be a basis for a topology on X. Let T be the topology on X generated by B. Prove that T is the intersection of all the topologies on X which contain B. Solution: Let U denote the intersection of all the topologies that contain B. Since T consists of all unions of sets in B, we have that T contains B, so that U ⊂ T. On 0 the other hand, if T is a topology which contains B and {Bα}α∈I is a collection of 0 0 sets in B, then for all α ∈ I we have that Bα ∈ B ⊂ T . Since T is a topology, we S 0 have that α∈I Bα ∈ T . Since every set in T is a union of sets in B, this implies that T ⊂ T0. Since T0 was an arbitrary topology containing B, we must have that T ⊂ U. We conclude that T = U, as desired. Problem 2: Let K denote the set K = {1/n : n = 1, 2, 3,... }. Let B denote the set of all open intervals (a, b) in R, together with all sets of the form (a, b)−K. Prove that B is the basis for a topology on R. (This is called the K-topology and denoted RK .) Solution: Given any x ∈ R, we have that x ∈ (x − 1, x + 1), so that the sets in B cover R. On the other hand, if x ∈ (a, b) ∩ (c, d), we necessarily have that x ∈ (max(a, c), min(b, d)). If x ∈ ((a, b) − K) ∩ (c, d), then x ∈ (max(a, c), min(b, d)) − K. Finally, if x ∈ ((a, b) − K) ∩ ((c, d) − K), then x ∈ (max(a, c), min(b, d)) − K. We conclude that B is the basis for a topology on R.

Problem 3: (Exercise 13.6 in Munkres) Recall that the lower limit topology R` is the topology on R generated by the basis {[a, b): a, b ∈ R}. Prove that the topologies RK and R` on R are incomparable (that is, neither is finer than the other). Solution: Let K = {1, 1/2, 1/3,... } and consider the set U = (−1, 1)−K ⊂ R. Then U is open in RK (since it is a basis element of RK ). We claim that U is not open in R`. If U were open in R`, since 0 ∈ U there would exist some basis element [a, b) of R` such that 0 ∈ [a, b) ⊂ U. But 0 ∈ [a, b) implies that a ≤ 0 and b > 0, so that [a, b) ∩ K is nonempty, and [a, b) 6⊂ U. We conclude that U is not open in R`, so that R` is not finer than RK . Let V = [0, 1) ⊂ R. Then V is open in R` (since is it a basis element of R`). We claim that V is not open in RK . To see this, suppose that V were open in RK . Then there would exist some basis element B of the basis B of RK given in Problem 2 such that 0 ∈ B ⊂ V . This basis element B would have to have one of the forms (a, b) for a a < 0 < b or (a, b) − K for a < 0 < b. In either case, we have that 2 ∈ B − V , so that B 6⊂ V , which is a contradiction. We conclude that RK is not finer than R`. 1 2

Problem 4: (Exercise 13.7 in Munkres) Consider the following ﬁve topologies on R:

T1 = the standard topology, T2 = the topology RK , T3 = the ﬁnite complement topology,

T4 = the upper limit topology, having all sets of the form (a, b] as a basis,

T5 = the topology having all sets of the form (−∞, a) = {x ∈ R : x < a} as a basis. Determine, for each of the topologies, which of the others it contains.

Solution: We claim that the topologies T3 and T5 are incomparable, the topology T1 contains both T3 and T5, the topologies T2 contains T1, and the topology T4 contains T2. To see that T3 and T5 are incomparable, observe that T5 consists precisely of the sets T5 = {(−∞, a): a ∈ R} ∪ {∅, R}, as these are the unions of sets in the given basis. In particular, we have that R − {0} is in T3 but not T5. Moreover, we have that (−∞, 0) is in T5 but not T3. We conclude that T3 and T5 are incomparable. To see that T1 contains T3, let U ∈ T3 and let x ∈ U. We have that R − U is finite, so we can find some interval (a, b) in R such that x ∈ (a, b) ⊂ R − U. Since intervals of the form (a, b) form a basis for T1, we get that T1 contains T3. To see that T1 contains T5, let (−∞, a) be a basis element for T5 and let x ∈ (−∞, a). Then x ∈ (x − 1, a) ⊂ (−∞, a). Since intervals of the form (c, d) form a basis for T1, we get that T1 contains T5. To see that T2 contains T1, we need only observe that the basis for T2 given in Problem 2 contains the standard open interval basis for T1. Since a topology consists exactly of the unions of sets in a given basis, we get that T2 contains T1. On the other hand, we have that U = (−1, 1) − {1, 1/2, 1/3,... } is open in T2 but not in T1 (since it contains no interval (a, b) containing 0), so that the containment of T1 inside T2 is strict. To see that T4 contains T2, let B be a basis element of T2 and let x ∈ B. We have that B is of one of the forms (a, b) or (a, b) − K, where K is as in Problem 2. If B is of the form (a, b), we have that x ∈ (a, x] ⊂ (a, b). If B is of the form (a, b) − K and x ≤ 0, we have that x ∈ (a, x] ⊆ (a, b)−K. If B is of the form (a, b)−K and x > 0, we can find n minimal such that 1/n < x. Then x ∈ (1/n, x] ⊂ (a, b) − K. We conclude that T4 contains T2. To see that this containment is strict, observe that (−1, 0] is an open set in T4, but contains no basis element of T2 which contains 0. Problem 5: Endow R2 with dictionary order <. That is, we define (x, y) < (x0, y0) if and only if x < x0 or x = x0 and y < y0. Let T be the associated order topology on R2. Prove that T is strictly finer than the Euclidean topology on R2. (That is, prove that T is finer than the Euclidean topology, and also not equal to the Euclidean topology.) Solution: Let U ⊂ R2 be open in the Euclidean topology and let (x, y) ∈ U. Then there exists > 0 such that B((x, y), ) ⊂ U. In particular, we have that I = {(x, y0): y − < y0 < y + } is contained in U. But I is an open interval in the dictionary order on R2, so that I ∈ T. We conclude that T is finer than the Euclidean topology. To see 3 that T is strictly finer than the Euclidean topology, we need only observe that the line segment {0} × (−1, 1) is open in T but not open in the Euclidean topology. Problem 6: Let X be a finite ordered set. Prove that the order topology on X is the discrete topology.

Solution: We can write X = {x1 < x2 < ··· < xn}. For 2 ≤ i ≤ n − 1, we have that {xi} = (xi−1, xi+1). We also have that {x1} = [x1, x2) and {xn} = (xn−1, xn]. We conclude that every singleton subset of X is open in the order topology, and hence the order topology on X equals the discrete topology. Problem 7: (Exercise 16.4 in Munkres) A map f : X → Y of topological spaces is called an open map if whenever U ⊂ X is open, we have that f(U) = {f(x): x ∈ U} ⊂ Y is open. Let X and Y be topological spaces and endow X × Y with the product topology. Prove that the projections π1 : X × Y → X and π2 : X × Y → Y deﬁned by π1(x, y) = x and π2(x, y) = y are open maps. Solution: Let U ⊂ X × Y be an open set. For any (x, y) ∈ U, there exist open sets Vx,y ⊂ X and Wx,y ⊂ Y such that (x, y) ∈ Vx,y × Wx,y ⊂ U. It follows that S S π1(U) = π1( (x,y)∈U Vx,y × Wx,y) = (x,y)∈U Vx,y is open in X. Similarly, we have that S S π2(U) = π2( (x,y)∈U Vx,y × Wx,y) = (x,y)∈U Wx,y is open in Y . We conclude that π1 and π2 are open maps. Problem 8: Give an example of an inﬁnite ordered set X such that the order topology on X is the discrete topology. Solution: Let X = Z be the set of integers with the usual order <. For any n ∈ Z, the interval (n − 1, n + 1) = {n} is open in the order topology, so that all singleton subsets of Z are open in the order topology. We conclude that the order topology on Z is the discrete topology. Problem 9: Let X and Y be topological spaces and assume that X has the discrete topology. Prove that the open sets in the product topology on X × Y are precisely those subsets of X × Y of the form [ {x} × Ux, x∈X where for all x ∈ X, Ux is an open subset of Y .

Solution: Suppose that Ux is an open subset of Y for all x ∈ X. We show that S S {x} × Ux is open in X × Y . Indeed, let (x, y) ∈ {x} × Ux. Then {x} is open x∈X x∈X S in X, Ux is open in Y , and y ∈ Ux, so that (x, y) ∈ {x} × Ux ⊂ {x} × Ux. We S x∈X conclude that x∈X {x} × Ux is open in the product topology on X × Y . On the other hand, let V be an open set in X × Y . Let x ∈ X and deﬁne Ux := {y ∈ Y :(x, y) ∈ V }. It is enough to show that Ux is open in Y . Indeed, suppose (x, y) ∈ V . We have that there exist open sets Wy in X and Uy in Y such that (x, y) ∈ Wy × Uy ⊂ V . In particular, we have that y ∈ Uy ⊂ Ux. It follows that U = S U is open in Y . x y∈Ux y