Generación a 25 años: Balance económico y ambiental

18a. Given two distinct points x, y in a metric space (X, ρ), let r = ρ(x, y)/2. Then the open balls Bx,r

and By,r are disjoint open sets with x ∈ Bx,r and y ∈ By,r. Hence every metric space is Hausdorff.

18b. Given two disjoint closed sets F1, F2 in a metric space (X, ρ), let O1 = {x : ρ(x, F1) < ρ(x, F2)}

and let O2= {x : ρ(x, F2) < ρ(x, F1)}. The sets O1 and O2are open since the functions f(x) = ρ(x, F1) and g(x) = ρ(x, F2) are continuous. Thus O1 and O2 are disjoint open sets with F1⊂ O¹ and F2⊂ O². Hence every metric space is normal.

19. Let X consist of [0, 1] and an element 0^!. Consider the sets (α, β), [0, β), (α, 1] and {0^!}∪(0, β) where α, β∈ [0, 1]. Clearly, any x ∈ [0, 1] belongs to one of the sets. Note that (α, β) ⊂ [0, β) ∩ (α, 1]. If x ∈ (α, β)∩[0, β^!), then x ∈ (α, min(β, β^!)) ⊂ (α, β)∩[0, β^!). If x ∈ (α, β)∩(α^!, 1], then x∈ (max(α, α^!), β) ⊂ (α, β) ∩ (α^!, 1]. If x ∈ (α, β) ∩ ({0^!} ∪ (0, β^!)), then x ∈ (α, min(β, β^!)) ⊂ (α, β) ∩ ({0^!} ∪ (0, β^!)). If x∈ [0, β) ∩ ({0^!} ∪ (0, β^!)), then x ∈ (0, min(β, β^!)) ⊂ [0, β) ∩ ({0^!} ∪ (0, β^!)). If x ∈ (α, 1] ∩ ({0^!} ∪ (0, β)), then x ∈ (α, β) ⊂ (α, 1] ∩ ({0^!} ∪ (0, β)). Hence the sets form a base for a topology on X.

Given two distinct points x, y ∈ X, if neither of them is 0’, then [0, (x + y)/2) is an open set containing one but not the other. If y = 0^!, then {0^!} ∪ (0, x/2) is an open set containing y but not x. Hence X is T1. Consider the distinct points 0 and 0’. If O is an open set containing 0 and O^! is an open set containing 0’, then 0 ∈ [0, β) ⊂ O and 0^! ∈ {0^!} ∪ (0, β^!) ⊂ O^! for some β, β^!. Since [0, β) ∩ (0, β^!) != ∅, O∩ O^! != ∅. Hence X is not Hausdorff.

20. Let f be a real-valued function on a topological space X. Suppose f is continuous. Then for any real number a, {x : f(x) < a} = f⁻¹[(−∞, a)] and {x : f(x) > a} = f⁻¹[(a, ∞)]. Since (−∞, a) and (a, ∞) are open, the sets {x : f(x) < a} and {x : f(x) > a} are open. Conversely, suppose the sets {x : f(x) < a} and {x : f(x) > a} are open for any real number a. Since any open set O ⊂ R is a union of open intervals and any open interval is an intersection of at most two of the sets, f⁻¹[O] is open so f is continuous.

Since {x : f(x) ≥ a}^c= {x : f(x) < a}, we also have f is continuous if and only if for any real number a, the set{x : f(x) > a} is open and the set {x : f(x) ≥ a} is closed.

21. Suppose f and g are continuous real-valued functions on a topological space X. For any real number a, {x : f(x) + g(x) < a} = !

q∈Q({x : g(x) < q} ∩{ x : f(x) < a − q}), which is open, and {x : f(x) + g(x) > a} =!

q∈Q({x : g(x) > q} ∩{ x : f(x) > a − q}), which is open, so f + g is continuous.

{x : f(x)g(x) < a} =!

q∈Q({x : g(x) < q}) ∩ {x : f(x) < a/q}, which is open, and {x : f(x)g(x) > a} =

!

q∈Q({x : g(x) > q}) ∩ {x : f(x) > a/q}, which is open, so fg is continuous. {x : (f ∨ g)(x) < a} = {x : f (x) < a} ∩{ x : g(x) < a}, which is open, and {x : (f ∨ g)(x) > a} = {x : f(x) > a} ∪{ x : g(x) > a}, which is open, so f ∨ g is continuous. {x : (f ∧ g)(x) < a} = {x : f(x) < a} ∪ {x : g(x) < a}, which is open, and {x : (f ∧ g)(x) > a} = {x : f(x) > a} ∩{ x : g(x) > a}, which is open, so f ∧ g is continuous.

22. Let %fn& be a sequence of continuous functions from a topological space X to a metric space Y . Suppose %fn& converges uniformly to a function f. Let ε > 0 and let x ∈ X. There exists N such that ρ(fn(x), f(x)) < ε/3 for all n ≥ N and all x ∈ X. Since fN is continuous, there is an open set O containing x such that fN[O] ⊂ BfN(x),ε/3. If y ∈ O, then ρ(f(y), f(x)) ≤ ρ(f(y), fN(y)) + ρ(fN(y), fN(x)) + ρ(fN(x), f(x)) < ε. Thus f[O] ⊂ Bf (x),ε. Since any open set in Y is a union of open balls, f is continuous.

23a. Let X be a Hausdorff space. Suppose X is normal. Given a closed set F and an open set O containing F , there exist disjoint open sets U and V such that F ⊂ U and O^c ⊂ V . Now ¯U is disjoint from O^c since if y ∈ O^c, then V is an open set containing y that is disjoint from U. Thus ¯U ⊂ O.

Conversely, suppose that given a closed set F and an open set O containing F , there is an open set U such that F ⊂ U and ¯U ⊂ O. Let F and G be disjoint closed sets. Then F ⊂ G^c and there is an open set U such that F ⊂ U and ¯U ⊂ G^c. Equivalently, F ⊂ U and G ⊂ ¯U^c. Since U and ¯U^c are disjoint open sets, X is normal.

*23b. Let F be a closed subset of a normal space contained in an open set O. Arrange the rationals in (0, 1) of the form r = p2⁻ⁿ in a sequence %rn&. Let U¹ = O. By part (a), there exists an open set U0 such that F ⊂ U⁰ and U0⊂ O = U¹. Let Pn be the set containing the first n terms of the sequence.

Since U0 ⊂ U¹, there exists an open set Ur₁ such that U0 ⊂ U^r1 and Ur₁ ⊂ U¹. Suppose open sets Ur

have been defined for rationals r in Pn such that Up⊂ U^q whenever p < q. Now rn+1 has an immediate predecessor ri and an immediate successor rj (under the usual order relation) in Pn+1∪ {0, 1}. Note that Uri⊂ U^rj. Thus there is an open set Urn+1 such that Uri⊂ U^rn+1 and Urn+1 ⊂ U^rj. Now if r ∈ Pn, then either r ≤ ri, in which case Ur ⊂ U^ri ⊂ U^rn+1, or r ≥ rj, in which case Urn+1 ⊂ U^rj ⊂ U^r. By induction, we have a sequence {Ur} of open sets, one corresponding to each rational in (0, 1) of the form p = r2⁻ⁿ, such that F ⊂ Ur⊂ O and U^r⊂ U^sfor r < s.

23c. Let {Ur} be the family constructed in part (b) with U¹= X. Let f be the real-valued function on X defined by f (x) = inf{r : x ∈ U^r}. Clearly, 0 ≤ f ≤ 1. If x ∈ F , then x ∈ U^r for all r so f(x) = 0.

If x ∈ O^c, then x /∈ U^r for any r < 1 so f(x) = 1. Given x ∈ X and an open interval (c, d) containing f (x), choose rationals r1 and r2 of the form p2⁻ⁿsuch that c < r1< f (x) < r2< d. Consider the open set U = Ur₂\ U^r1. Since f(x) < r2, x ∈ Ur⊂ U^r2 for some r < r2. If x ∈ Ur₁, then x ∈ Ur for all r > r1

so f(x) ≤ r1. Since f(x) > r1, x /∈ U^r1. Thus x ∈ U. If y ∈ U, then y ∈ Ur2 so f(y) ≤ r2 < d. Also, y /∈ U^r1 so f(y) ≥ r1> c. Thus f [U ]⊂ (c, d). Hence f is continuous.

23d. Let X be a Hausdorff space. Suppose X is normal. For any pair of disjoint closed sets A and B on X, B^c is an open set containing A. By the constructions in parts (b) and (c), there is a continuous real-valued function f on X such that 0 ≤ f ≤ 1, f ≡ 0 on A and f ≡ 1 on (B^c)^c= B.

(*) Proof of Urysohn’s Lemma.

24a. Let A be a closed subset of a normal topological space X and let f be a continuous real-valued function on A. Let h = f/(1 + |f|). Then |h| = |f|/(1 + |f|) < 1.

24b. Let B = {x : h(x) ≤ −1/3} and let C = {x : h(x) ≥ 1/3}. By Urysohn’s Lemma, there is a continuous function h1 which is −1/3 on B and 1/3 on C while |h1(x)| ≤ 1/3 for all x ∈ X. Then

|h(x) − h¹(x)| ≤ 2/3 for all x ∈ A.

24c. Suppose we have continuous functions hn on X such that |hn(x)| < 2ⁿ⁻¹/3ⁿ for all x ∈ X and

|h(x) −#n

i=1hi(x)| < 2ⁿ/3ⁿ for all x ∈ A. Let B^! = {x : h(x) −#n

i=1hi(x) ≤ −2ⁿ/3ⁿ⁺¹} and let C^! = {x : h(x) −#n

i=1hi(x) ≥ 2ⁿ/3ⁿ⁺¹}. By Urysohn’s Lemma, there is a continuous function hn+1 which is −2ⁿ/3ⁿ⁺¹ on B^! and 2ⁿ/3ⁿ⁺¹ on C^! while |hn+1(x)| ≤ 2ⁿ/3ⁿ⁺¹ for all x ∈ X. Then

|h(x) −#n+1

i=1 hi(x)| < 2ⁿ⁺¹/3ⁿ⁺¹for all x ∈ A.

24d. Let kn = #ⁿ_i=1hi. Then each kn is continuous and |h(x) − kn(x)| < 2ⁿ/3ⁿ for all x ∈ A. Also,

|kⁿ(x) −kn−1(x)| < 2ⁿ⁻¹/3ⁿfor all n and all x ∈ X. If m > n, then |km(x) −kn(x)| = |#m

i=n+1hi(x)| ≤

#m

i=n+12ⁱ⁻¹/3ⁱ < (2/3)ⁿ. Let k(x) = #^∞_i=1hi(x). Then |k(x) − kⁿ(x)| ≤ (2/3)ⁿ for all x ∈ X. Thus

%kⁿ& converges uniformly to k. i.e. %hⁿ& is uniformly summable to k and since each kⁿis continuous, k is continuous. Also, since |kn| ≤ 1 for each n, |k| ≤ 1. Now |h(x) − kⁿ(x)| < 2ⁿ/3ⁿ for all n and all x ∈ A so letting n → ∞, |h(x) − k(x)| = 0 for all x ∈ A. i.e. k = h on A.

24e. Since |k| = |h| < 1 on A, A and {x : k(x) = 1} are disjoint closed sets. By Urysohn’s Lemma, there is a continuous function ϕ on X which is 1 on A and 0 on {x : k(x) = 1}.

24f. Set g = ϕk/(1 − |ϕk|). Then g is continuous and g = k/(1 − |k|) = h/(1 − |h|) = f on A.

(*) Proof of Tietze’s Extension Theorem.

25. Let F be a family of real-valued functions on a set X. Consider the sets of the form {x : |fi(x) − fi(y)| < ε for some ε > 0, some y ∈ X, and some finite set f1, . . . , fn of functions in F}. The weak topology on X generated by F has {f⁻¹[O] : f ∈ F, O open in R} as a base. Now if x^! ∈ f⁻¹[O] for some f ∈ F and O open in R, we may assume O is an open interval (c, d). If x ∈ {x : |f(x) − f(x^!)| <

min(f(x^!) − c, d − f(x^!))}, then f(x) ∈ (c, d). i.e. x ∈ f⁻¹[O]. Thus x^! ∈ {x : |f(x) − f(x^!)| <

min(f(x^!) − c, d − f(x^!))} ⊂ f⁻¹[O]. Hence the sets of the form {x : |fⁱ(x) − fⁱ(y)| < ε for some ε >

0, some y ∈ X, and some finite set f¹, . . . , fn of functions in F} is a base for the weak topology on X generated by F.

Suppose this topology is Hausdorff. For any pair {x, y} of distinct points in X, there are disjoint open sets Oxand Oysuch that x ∈ Oxand y ∈ Oy. Then there are sets Bxand Byof the form {x : |fi(x)−fi(z)| <

ε for some ε > 0, some z∈ X, and some finite set f¹, . . . , fn of functions in F} such that x ∈ Bx ⊂ O^x and y ∈ By ⊂ O^y. Suppose f(x) = f(y) for all f ∈ F. Then x, y ∈ Bx∩ B^y. Contradiction. Hence there is a function f ∈ F such that f(x) != f(y). Conversely, suppose that for each pair {x, y} of distinct points in X there exists f ∈ F such that f(x) != f(y). Then x ∈ {x^! : |f(x^!) − f(x)| < |f(y) − f(x)|/2}

and y ∈ {x^! : |f(x^!) − f(y)| < |f(y) − f(x)|/2} with the two sets being disjoint open sets. Hence the topology is Hausdorff.

26. Let F be a family of real-valued continuous functions on a topological space (X, T ). Since f is continuous for each f ∈ F, the weak topology generated by F is contained in T . Suppose that for each closed set F and each x /∈ F there is an f ∈ F such that f(x) = 1 and f ≡ 0 on F . Then for each O ∈ T and each x ∈ O, there is an f ∈ F such that f(x) = 1 and f ≡ 0 on O^c. i.e. x ∈ f⁻¹[(1/2, 3/2)] ⊂ O.

Thus O is in the weak topology generated by F.

27. Let X be a completely regular space. Given a closed set F and x /∈ F , there is a function f ∈ C(X) such that f(x) = 1 and f ≡ 0 on F . Then F ⊂ f⁻¹[(−∞, 1/2)], x ∈ f⁻¹[(1/2, ∞)], and the sets f⁻¹[(−∞, 1/2)] and f⁻¹[(1/2, ∞)] are disjoint open sets. Hence X is regular.

28. Let X be a Hausdorff space and let Y be a subset of X. Given two distinct points x and y in Y , there are disjoint open sets O1 and O2 in X such that x ∈ O1 and y ∈ O2. Then x ∈ O1∩ Y and y∈ O²∩ Y , with O¹∩ Y and O²∩ Y being disjoint open sets in Y . Hence Y is Hausdorff.

*29. In Rⁿ let B be the family of sets {x : p(x) != 0} where p is a polynomial in n variables. Let T be the family of all finite intersections O = B1∩ · · · ∩ B^k from B. By considering the constant polynomials, we see that ∅ and X are in T . It is also clear that if O1, O2∈ T , then O¹∩ O²∈ T . Now if O^α∈ T , Given two distinct points x, y ∈ Rⁿ, say x = %x1, . . . , xn& and y = %y¹, . . . , yn&, let p be the polynomial

#n

i=1(Xi− xⁱ)². Then p(x) = 0 and p(y) != 0. Thus there is an open set containing y but not x. Hence T is T¹.

Any two open sets are not disjoint since for any finite collection of polynomials there is an x that is not a root of any of the polynomials. Hence T is not T2.

30. Let A ⊂ B ⊂ ¯A be subsets of a topological space, and let f and g be two continuous maps of B into a Hausdorff space X with f(u) = g(u) for all u ∈ A. For any v ∈ B, we have v ∈ ¯A. Suppose f (v)!= g(v).

Then there are disjoint open sets O1 and O2 such that f(v) ∈ O1 and g(v) ∈ O2. Since f and g are continuous, there are open sets U1 and U2 such that v ∈ U1, f[U1] ⊂ O1, v ∈ U2, g[U2] ⊂ O2. Now there exists u ∈ A with u ∈ U1∩ U². Then f(u) ∈ O1 and g(u) ∈ O2 but f(u) = g(u) so O1∩ O²!= ∅.

Contradiction. Hence f(v) = g(v). i.e. f ≡ g on B.

31. Omitted.