MATH 411 HOMEWORK 2 SOLUTIONS

ADAM LEVINE

2.13.3. Let X be a set, and let

Tc = {U ⊂ X | X − U is countable or all of X}

T∞ = {U ⊂ X | X − U is infinite or empty or all of X}.

Show that Tc is a topology. Is T∞ a topology? For Tc, it is clear that ∅ and X are in Tc. If Uα is any collection of sets in Tc, then [ \ X − Uα = (X − Uα), α α which is an intersection of countable sets, hence countable. Likewise, if U1,...,Uk are sets in Tc, then k k \ [ X − Uα = (X − Uα), i=1 i=1 which is a finite union of countable sets, hence countable. (Cf. Example 3 on page 77.) We claim that T∞ is not a topology provided that X is an infinite set. (If X is finite, then T∞ is just the indiscrete topology.) Choose some x0 ∈ X, and consider all of the 1-point sets {x} for x 6= x0. These sets all have infinite complement. On the other hand, the union S {x} equals X − {x }, which has complement {x }, x6=x0 0 0 so it is not open.

2.13.6. Show that the topologies of R` and RK are not comparable. The set [2, 3) is open in R`, but it is not open in RK ; any open set in RK containing 2 must contain an interval (2 − , 2 + ) for some  > 0. The set (−1, 1) − K is open in RK , but it is not open in R`, since any open set in R` containing 0 must contain an interval [0, ) for some  > 0, and therefore contain elements of K.

2.13.7. We have already established that T2 and T4 are strictly finer than T1 (anal- ogous to the argument for the lower limit topology). A basic open set (a, b)−K in T2 is either a (possibly infinite) union of open intervals (if a ≥ 0 or b ≤ 0), or the union of (a, 0] with some infinite collection of open intervals (if a < 0 < b). That is, (a, b) − K is a union of sets that are open in the upper limit topology T4. Thus, T2 ⊂ T4. On the other hand, the set (1, 2] is open in T4 but not in T2. Thus, T4 is strictly finer than T2. Any open set in the finite complement topology is a union of finitely many open intervals: if U = R r {a1, . . . , an}, where a1 < ··· < an, we can write

U = (−∞, a1) ∪ (a1, a2) ∪ ... (an−1, an) ∪ (an, ∞). 1 2 ADAM LEVINE

On the other hand, an open interval (a, b) is not open in the finite complement topology. Thus, T1 is strictly finer than T3. (Therefore, T2 and T4 are also strictly finer than T3.) Any set of the form (−∞, a) is open in the standard topology on R. On the other hand, an open interval (a, b) does not contain any infinite ray. Therefore, T1 is strictly finer than T5. (Therefore, T2 and T4 are also strictly finer than T5.) Finally, we show that T3 and T5 are incomparable. If U = R r {0, 1}, then there is 1 no basic open set of T5 containing 2 and contained in U, since such a set would have to contain 0. Thus, U is not open in T5. Likewise, the set V = (−∞, 0) is open in T5 but not in T3 because its complement is not finite. 2.13.8. (a) Show that the collection B = {(a, b) | a, b ∈ Q, a < b} is a basis for the standard topology on R. The key fact is that for any real numbers a, b, we may find a rational number c such that a < c < b. We will not prove this here. We apply Lemma 13.2. For any open set U ∈ R, and any x ∈ U, choose  > 0 such that (x − , x + ) ⊂ U. By the above principle, we may find rational numbers a and b with x −  < a < x < b < x + . Therefore, x ∈ (a, b) ⊂ U, as required. (b) Show that the collection C = {[a, b) | a, b ∈ Q, a < b} is a basis for a topology other than the lower limit topology on R. First, note that if C1 = [a, b) and C2 = [c, d) are elements of C, then C1 ∩ C2 is either empty or an element of C, depending on the ordering of a, b, c, and d. (You can check all the cases.) Since every element of C is open in the lower limit topology, the topology generated by C √is contained in the lower limit topology. On the other hand, consider the set U = [ 2,√2), which is open in the lower limit√ topology.√ For any rational numbers a, b with a ≤ 2 < b, we must in fact have a < 2 since 2 is irrational, so [a, b) cannot be contained in U. Thus, U is not open in the topology generated by C. 2.16.1. 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 as a subspace of X. Let TX be the topology A inherits as a subspace of X, and TY the topology A inherits as a subspace of Y . If U ⊂ A belongs to TX , then there is an open set V ⊂ X such that U = V ∩ A = (V ∩ Y ) ∩ A. (The latter equality holds because A ⊂ Y .) Hence U is in TY . If U ⊂ A belongs to TY , then there is a set V ⊂ Y which is open in Y such that U = V ∩ A. V being open in Y means that there is an open set W ⊂ X such that V = W ∩ Y . Then U = (W ∩ Y ) ∩ A = W ∩ (Y ∩ A) = W ∩ A. Hence U belongs to TX .

2.16.4. Show that π1 : X × Y → X and π2 : X × Y → Y are open maps. We will just prove the statement about π1; the statement about π2 follows iden- tically. Let U ⊂ X × Y be an open set; we must show that π1(U) is open. Let x ∈ π1(U), meaning that there is a y ∈ Y for which x × y ∈ U. Since U is open, there are open sets V ⊂ X and W ⊂ Y such that x ∈ V , y ∈ W , and V × W ⊂ U. For MATH 411 HOMEWORK 2 SOLUTIONS 3

0 0 0 any x ∈ V , we have π1(x , y) = x , so V ⊂ π1(U). Thus, π1(U) contains an open set around each point, so it is open. 2.16.8. If L is a straight line in the plane, describe the topology L inherits as a subspace of R` × R and as a subspace of R` × R`. First of all, we can parametrize the line L by a map f : R → R2 given by f(t) = (t, at + b) (if L is not vertical) and by f(t) = (c, t) (if L is vertical). The preimages of the basic open sets in R2 (with either of the specified topologies) give a basis for the induced topology on R, so we simply need to understand what these are. For R` × R, the basic open sets are boxes of the form B = [a, b) × (c, d) for all real numbers a, b, c, d with a ≤ b and c < d. If L is non-vertical, then the intersection of such a box with L is either a half-open interval (if L passes through the left side of B and one of the other three sides), an open interval (if L passes through two of the sides other than the left side), or the empty set. Therefore, the induced topology on L is the lower-limit topology. If L is vertical, then it can only intersect B in an open interval, so the induced topology is the standard topology on L. For R` × R`, the basic open sets are boxes of the form B = [a, b) × [c, d). If L is vertical or has non-negative slope, then the intersection of L with any such box must be a half-open interval (since it contains a point on the left or bottom side of B). Hence the induced topology is the lower-limit topology. However, if L has negative slope, then for any point x × y ∈ L, we can find a box B that intersects L in only that point, namely B = [x, x + ) × [y, y + ) for  > 0. Thus, one-point sets in L are open, so L has the discrete topology. 2.16.9. Show that the dictionary order topology on R × R is the same as the product topology on Rd×R, where Rd denotes R in the discrete topology. Compare this topology with the standard topology on R2. A basis for the product topology on Rd × R is B = {{x} × (y1, y2) | x, y1, y2 ∈ R, y1 < y2}. (These are just products of the basic open sets for Rd and R.) Every such set is open in the dictionary order topology on R × R, since it is just the open interval (x × y1, x × y2). Conversely, notice that any set of the form {x}×R, {x}×(y, ∞), or {x}×(−∞, y) (i.e. vertical lines and rays) is an infinite union of sets of the form {x} × (y1, y2), so it is open in Rd × R. Any open interval in the dictionary order of the form U = (x1 × y1, ∪x2 × y2) can be written as a union [ U = ({x1} × (y1, ∞)) ∪ {x} × R ∪ ({x2} × (−∞, y2)) x1

2.16.10. Let I = [0, 1]. Compare the product topology on I × I, the dictionary order topology on I × I, and the topology I × I inherits as a subspace of R × R in the dictionary order topology. Let us denote the three topologies by T1, T2, and T3. A basis for T1 consists of “boxes” having any of the following nine forms: (a, b) × (c, d), [0, b)×(c, d), (a, 1]×(c, d), (a, b)×[0, d), [0, b)×[0, d), (a, 1]×[0, d), (a, b)×(c, 1], [0, b) × (c, 1], or (a, 1] × (c, 1]. (Here 0 ≤ a < b ≤ 1 and 0 ≤ c < d ≤ 1.) Using Problem 9, a basis for T3 consists of products {a} × (c, d), {a} × [0, d), and {a} × [c, 1). (That is, we used the alternate description of the dictionary order topology on R × R as the product topology on Rd × R to find a smaller basis.) It follows just as above that T3 is strictly finer than T1. A basis for T2 consists of all intervals (x1 × y1, x2 × y2), where x1, y1, x2, y2 ∈ I and (x1, y1) < (x2, y2) in the dictionary order topology. Any such set is open in T2, since it is the intersection of an open interval in the dictionary order topology on R × R with I × I. However, the key property of T2 is that both endpoints need to be in I × I. In particular, if U = (x1 × y1, x2 × y2) contains a point of the form a × 0 (where 0 < a ≤ 1), we must have x1 < a, and therefore U contains (a, x1) × [0, 1]. This implies neither T1 nor T3 is contained in T2. Thus, T2 is strictly finer than T3. And a set {x1} × (y1, y2) is open in T3 but not T1, so T1 and T3 are incomparable.