MATH 411, HW 4 SOLUTIONS 2.18.9. Let {A Α} Be a Collection of Subsets

MATH 411, HW 4 SOLUTIONS

S 2.18.9. Let {Aα} be a collection of subsets of X, such that X = α Aα. Let f : X → Y ; suppose that f|Aα is continuous for each α.

(a) Show that if the collection {Aα} is finite and each set Aα is closed, then f is continuous. We will us use the characterization of continuity given by Theorem 18.1: a function is continuous iff preimages of closed sets are closed. −1 For each α, note (f|Aα) (B) is closed in Aα. Since Aα is closed, this implies that −1 −1 (f|Aα) (B) is closed in X as well (Theorem 17.3). At the same time, we have f (B)∩Aα = −1 −1 S −1 −1 (f|Aα) (B), so f (B) = α(f|Aα) (B). Thus, f (B) is equal to a finite union of closed sets, so it is closed. (b) Find an example where the collection {Aα} is countable and each Aα is closed, but f 1 is not continuous. Let X = R, and consider the sets A0 = (−∞, 0] and An = [ n , ∞) for S∞ n ∈ Z+. All of these are closed, and X = n=0 An. The function f : R → R given by ( 1 x ≤ 0 f(x) = 0 x > 0 is constant, hence continuous, when restricted to each An, but it is not continuous on all of R. (c) Show that if the collection {Aα} is locally finite and each set Aα is closed, then f is continuous. If U is an open set in X that intersects only finitely many of the Aα, then part (a) implies that f|U is continuous. Since X is equal to a union of such sets, the local formulation of continuity (Theorem 18.2(f)) implies that f is continuous.

2.18.10. Let f : A → B and g : C → D be continuous functions. Deﬁne a map f × g : A × C → B × D by (f × g)(a × c) = f(a) × g(c). Prove that f × g is continuous. Let U ⊂ B and V ⊂ D be open sets. Then (f × g)−1(U × V ) = f −1(U) × g−1(V ), which is open since f and g are continuous. Thus, the preimages of basic open sets in B × D are open, so f × g is continuous.

2.18.11. Let F : X × Y → Z. Show that if F is continuous, then F is continuous in each variable separately. First, note that for any y0 ∈ y, the map i: X → X × Z, i(x) = x × y0, is continuous, −1 since for any basic open set U × V , i (U × V ) equals U if y0 ∈ V and ∅ if y0 6∈ V . The map h: X → Z deﬁned by h(x) = F (x × y0) is then the composition F ◦ i0, so it is continuous. A similar argument applies for the maps k : Y → Z deﬁned by k(y) = F (x0 × y).

2.18.12. Let F : R × R → R be deﬁned by the equation ( xy if x × y 6= 0 × 0 F (X × Y ) = x2+y2 0 if x × y = 0 × 0.

1 2 MATH 411, HW 4 SOLUTIONS

(a) Show that F is continuous in each variable separately. xy0 If y0 6= 0, then the function h(x) = 2 2 is a quotient of two polynomials in which the x +y0 denominator never vanishes, hence continuous. If y0 = 0, then h(x) is identically 0. In either case, h is continuous. A similar argument applies interchanging x and y. (b) Compute the function g : R → R given by g(x) = F (x × x). ( 1 x 6= 0 g(x) = 2 0 x = 0. (c) Show that F is not continuous. Note that g(x) = F ◦ d, where d: X → X × X is the function d(x) = x × x. Since d is continuous in each variable, it is continuous. If F were continuous, then g would be the composition of two continuous functions, hence continuous; contradiction. Q 2.19.3. Prove Theorem 19.4: If each space Xα is a Hausdorff space, then Xα is a Haus- dorff space in both the box and product topologies. Q Let x = (xα) and y = (yα) be distinct elements of Xα. Then there is some α0 such that xα 6= yα . Choose disjoint open sets A, B ⊂ Xα such that xα ∈ A and yα ∈ B. Define 0 Q 0 Q 0 0 0 U = Uα and V = Vα, where ( ( A α = α0 B α = α0 Uα = Vα = Xα α 6= α0, Xα α 6= α0. Then U and V are open in both the box and product topologies, and they are disjoint since Q they are disjoint in the Xα0 factor. Thus, we see that Xα is a Hausdorf space. 2.19.5. One of the implications in Theorem 19.6 holds for the box topology. Which? Q Let f : A → α∈J Xα be given by f(a) = (fα(a))α∈J , where fα : A → Xα. Observe that each function fα is equal to the composition πα ◦ f. Therefore, if f is continuous, we deduce that fα is as well. (The converse need not hold, however.) Q 2.19.6. Let x1, x2,... be a sequence of points in the product space Xα. Show that this sequence converges to the point x iff for all α, the sequence πα(x1), πα(x2),... converges to πα(x). Is this fact true if one uses the box topology instead of the product topology? For one direction, assume that x1, x2,... converges to x in the product topology. For any −1 α, and any open set Uα ⊂ Xα containing πα(x), the preimage πα (Uα) is open and contains −1 x. Therefore, there is some N such that for all n > N, xn ∈ πα (Uα), so πα(xn) ∈ Uα. This proves that the sequence πα(x1), πα(x2),... converges to πα(x). The same argument goes through verbatim using the box topology. For the other direction, assume that for all α, the sequence πα(x1), πα(x2),... converges to πα(x), and let U be an open set (in the product topology) containing x. Choose a Q basic open set Uα containing x, where Uα ⊂ Xα, and Uα 6= Xα for at most finitely many values α1, . . . , αk. For each i = 1, . . . , k, there exists Ni such that for all n > Ni,

παi (xn) ∈ Uαi . (And for any α other than α1, . . . , αk, πα(xn) ∈ Uα automatically.) Let N = max{N1,...,Nk}; then for all n > N, πα(xn) ∈ Uα for all α, so xn ∈ U. Thus, the sequence x1, x2,... converges to x. The preceding paragraph used the product in an essential way, but we must still exhibit ω 1 1 1 a counterexample. In R , let x1, x2,... be the sequence deﬁned by xn = ( n , n , n ,... ). 1 Then for each α ∈ N, πα(xn) = n , so these sequences converge to 0. On the other hand, MATH 411, HW 4 SOLUTIONS 3

we claim that the sequence x1, x2,... does not converge to the 0 sequence. Take U = 1 1 1 1 (−1, 1) × (− 2 , 2 ) × (− 3 , 3 ) × ... , which contains the 0 sequence. For any n, xn cannot be th 1 1 1 contained in U, since the n entry of xn is n , which is not contained in (− n , n ). Thus, the sequence does not converge. 2.19.7. Let R∞ denote the set of sequences that are eventually zero. What is the closure of R∞ in Rω in the box and product topologies? In the product topology, we claim that the closure of R∞ is all of Rω. Given a sequence ω x = (x1, x2,... ) ∈ R , and a basic open set U containing x, write U = U1 × U2 × ... . There is some N such that Un = R for all n > N. Then the sequence y = (x1, . . . , xN , 0, 0, 0,... ) is in R∞ ∩ U. This shows that x is a limit point of R∞. On the other hand, in the box topology, we claim that R∞ is closed in Rω. Given a ω ∞ sequence x ∈ R − R , let U = U1 × U2 × ... , where  (0, ∞) x > 0  i Ui = R xi = 0  (−∞, 0) xi < 0. Then U contains x and is open in the box topology. However, because x 6∈ R∞, there are inﬁnitely many values of i for which xi 6= 0, and therefore for which Ui does not contain 0. Therefore, U cannot contain element of Rω. Thus, Rω − R∞ is open, so R∞ is closed.

2.19.8. Given sequences (a1, a2,... ) and (b1, b2,... ) of real numbers with ai > 0 for all i, deﬁne h: Rω → Rω by

h((x1, x2,... )) = (a1x1 + b1, a2x2 + b2,... ) Show that if Rω is given the product topology, then h is a homeomorphism of Rω with itself. What if Rω is given the box topology? First, observe that h is a bijection; its inverse is given by −1 y1 − b1 y2 − b2 h ((y1, y2,... )) = , ,... . a1 a2 Note that h−1 also has the same form as stated in the problem. Thus, when we show that h is continuous, it will follow immediately that h is a homeomorphism. The basic open sets in the product topology are of the form U = (c1, d1) × · · · × (cn, dn) × R × R × ... . It is easy to check that −1 c1 − b1 d1 − b1 cn − bn dn − bn h (U) = , × · · · × , × R × R × ... a1 a1 an an which is open in the product topology. Thus, h is continuous. Likewise, the basic open sets in the box topology are of the form U = (c1, d1)×(c2, d2)×... , and c − b d − b c − b d − b h−1(U) = 1 1 , 1 1 × 2 2 , 2 2 × ..., a1 a1 a2 a2 which is open in the box topology. Thus h is continuous. (Note: Observe that h is also open if we take the box topology on the domain and the product topology on the range, but not vice versa.)