Uso de energía renovable

18. Let X be a locally compact space and K a compact subset of X. For each x ∈ K, there is an open set Oxcontaining x with Oxcompact. Since K is compact, there is a finite subcollection {Ox1, . . . , Oxn} that covers K. Let O = !ⁿ_i=1Oxi. Then O ⊃ K and ¯O =!n

i=1Oxi is compact.

*19a. Let X be a locally compact Hausdorff space and K a compact subset. Then K is closed in

X^∗. By Q8.23a, there exists a closed set D containing ω with D ∩ K = ∅. By Urysohn’s Lemma, there is a continuous function g on X^∗ with 0 ≤ g ≤ 1 that is 1 on K and 0 on D. Define f(x) = min(2(g(x) −1/2), 0). Then {x : f(x) > 0} = {x : g(x) > 1/2}, which is compact because g is continuous on X^∗.

(*) Alternatively, use Q16 to regard K as a closed subset of the compact Hausdorff space Q.

*19b. Let K be a compact subset of a locally compact Hausdorff space X. Then by Q18, there is an open set O ⊃ K with ¯O compact. Now X^∗\ O and K are disjoint closed subsets of X^∗ so by Urysohn’s Lemma, there is a continuous function g on X^∗ with 0 ≤ g ≤ 1 that is 1 on K and 0 on X^∗\ O. Then f = g|^X is the required function.

20a. Let X^∗ be the Alexandroff one-point compactification of a locally compact Hausdorff space X.

Consider the collection of open sets of X and complements of compact subsets of X. Then ∅ and X^∗are in the collection. If U1 and U2 are open in X, then so is U1∩ U². If K1 and K2 are compact subsets of X, then (X^∗\ K¹) ∩ (X^∗\ K²) = X^∗\ (K¹∪ K²) where K1∪ K² is compact. Also, U1∩ (X^∗\ K¹) = U1∩ (X \ K¹), which is open in X. Thus the collection of sets is closed under finite intersection. If {Uα} is a collection of open sets in X, then ! Uα is open in X. If {K^β} is a collection of compact subsets of X, then !(X^∗\ K^β) = X^∗\"

Kβ. Since each Kβ is compact in the Hausdorff space X, each Kβ

is closed in X and " Kβ is closed in X. Thus " Kβ is closed in each Kβ so " Kβ is compact. Finally,

!Uα∪!(X^∗\ K^β) = U ∪ (X^∗\ K) = X^∗\ (K \ U), where U = !

Uα and K = " Kβ. Since K \ U is closed in the compact set K, K \ U is compact. Thus the collection of sets is closed under arbitrary union. Hence the collection of sets forms a topology for X^∗.

20b. Let id be the identity mapping from X to X^∗\ {ω}. Clearly id is a bijection. If U is open in X^∗\ {ω}, then U = (X^∗\ {ω}) ∩ U^! for some U^! open in X^∗. If U^! is open in X, then U = U^! so U is open in X. If U^! = X^∗\ K for some compact K ⊂ X, then U = X \ K, which is open in X.

Thus id is continuous. Also, any open set in X is open in X^∗ so id is an open mapping. Hence id is a homeomorphism.

20c. Let U be an open cover of X^∗. Then U contains a set of the form X^∗\ K for some compact K ⊂ X.

Take the other elements of U and intersect each of them with X to get an open cover of K. Then there is a finite subcollection that covers K. The corresponding finite subcollection of U together with X^∗\ K then covers X^∗. Hence X^∗ is compact.

Let x, y be distinct points in X^∗. If x, y ∈ X, then there are disjoint open sets in X, and thus in X^∗, that separate x and y. If x ∈ X and y = ω, then there is an open set O containing x with ¯O compact.

The sets O and X^∗\ ¯O are disjoint open sets in X^∗separating x and y. Hence X^∗ is Hausdorff.

*21a. Let Sⁿ denote the unit sphere in Rⁿ⁺¹. Let p = %0, . . . , 0, 1& ∈ Rⁿ⁺¹. Define f : Sⁿ− p → Rⁿ by f(x) = _1−x¹

n+1%x¹, . . . , xn&. The map g : Rⁿ → Sⁿ− p defined by g(y) = %t(y)y¹, . . . , t(y)yn, 1− t(y)&

where t(y) = 2/(1+||y||²) is the inverse of f. Thus Rⁿis homeomorphic to Sⁿ−p and the Alexandroff one-point compactification of Rⁿ is homeomorphic to the Alexandroff one-point compactification of Sⁿ− p, which is Sⁿ.

21b. Let X be the space in Q11 and Y be the space in Q12. Define f : X^∗→ Y by f(x) = x for x ∈ X and f(ω) = ω1. Then f is clearly a bijection. Consider the basic sets in Y . Now f⁻¹[{x : x < a}] = {x : x < a}, which is open in X and thus open in X^∗. Similarly for sets of the form {x : a < x < b}.

Also, f⁻¹[{x : a < x}] = {x ∈ X : a < x} ∪ {ω}, whose complement {x : x ≤ a} is compact. Thus f⁻¹[{x : a < x}] is open in X^∗. Since f is a continuous bijection from the compact space X^∗ to the Hausdorff space Y , f is a homeomorphism. Hence the one-point compactification of X is Y .

22a. Let O be an open subset of a compact Hausdorff space X. By Q8.23a, for any x ∈ O, there is an open set U such that x ∈ U and ¯U ⊂ O. Then the closure of U in X is the same as the closure of U in O and ¯U is compact, being closed in a compact space. Hence O is locally compact.

22b. Let O be an open subset of a compact Hausdorff space X. Consider the mapping f of X to the one-point compactification of O which is identity on O and takes each point in X \ O to ω. If U is open in O, then f⁻¹[U] = U is open in X. If K ⊂ O is compact, then f⁻¹[O^∗\ K] = X \ K, which is open in X. Hence f is continuous.

23. Let X and Y be locally compact Hausdorff spaces, and f a continuous mapping of X into Y . Let X^∗ and Y^∗ be the one-point compactifications of X and Y , and f^∗ the mapping of X^∗ into Y^∗ whose restriction to X is f and which takes the point at infinity in X^∗ to the point at infinity in Y^∗. Suppose

f is proper. If U is open in Y , then (f^∗)⁻¹[U] = f⁻¹[U], which is open in X and thus open in X^∗. If K⊂ Y is compact, then (f^∗)⁻¹[Y^∗\ K] = f⁻¹[Y \ K] ∪ {ωX} = X^∗\ f⁻¹[K], which is open in X^∗since f⁻¹[K] ⊂ X is compact. Hence f^∗ is continuous. Conversely, suppose f^∗ is continuous. Then f = f^∗|^X is continuous. Also, for any compact set K ⊂ Y , (f^∗)⁻¹[Y^∗\ K] = X^∗\ f⁻¹[K] is open in X^∗. Hence f⁻¹[K] ⊂ X is compact.

*24a. Let X be a locally compact Hausdorff space. Suppose F is a closed subset of X. For each closed compact set K, F ∩ K is closed. Conversely, suppose F is not closed. Take x ∈ ¯F \ F . There is an open set O containing x with ¯O compact. For any open set U containing x, (O∩ U) ∩ F != ∅. Thus U∩ (F ∩ ¯O)!= ∅. Then x ∈ F ∩ ¯O\ (F ∩ ¯O) so F∩ ¯O is not closed.

*24b. Let X be a Hausdorff space satisfying the first axiom of countability. Suppose F is a closed subset of X. For each closed compact set K, F ∩K is closed. Conversely, suppose F is not closed. Take x ∈ ¯F\F . Since X is first countable, there is a sequence %xn& in F converging to x. Let K = {xⁿ : n ∈ N} ∪{ x}.

Then K is compact in the Hausdorff space X and thus closed. Now F ∩ K = {xⁿ: n ∈ N} is not closed.

25. Let F be a family of real-valued continuous functions on a locally compact Hausdorff space X, and suppose that F has the following properties: (i) If f, g ∈ F, then f + g ∈ F. (ii) If f, g ∈ F, then f/g ∈ F provided that supportf ⊂ {x ∈ X : g(x) != 0}. (iii) Given an open set O ⊂ X and x0 ∈ O, there is an f ∈ F with f(x⁰) = 1, 0 ≤ f ≤ 1 and supportf ⊂ O.

Let {Oλ} be an open covering of a compact subset K of a locally compact Hausdorff space X. Let O be an open set with K ⊂ O and ¯O compact. For each x0∈ K, there is an f^x0 ∈ F with f^x0(x0) = 1, 0 ≤ fx0 ≤ 1 and supportfx₀ ⊂ O∩O^λfor some λ. For each x0∈ ¯O\K, there is a g^x0 ∈ F with g^x0(x0) = 1, 0 ≤ g^x0 ≤ 1 and supportgx₀ ⊂ K^c. By compactness of ¯O, we may choose a finite number f1, . . . , fn, g1, . . . , gm of these functions such that the sets where they are positive cover ¯O. Set f =#n

i=1fi and g = #^m_i=1gi. Then f, g ∈ F, f > 0 on K, supportf ⊂ O, f + g > 0 on ¯O and g ≡ 0 on K. Thus f/(f + g) ∈ F is continuous and ≡ 1 on K. The functions ϕi = fi/(f + g) ∈ F, i = 1, . . . , n form a finite collection of functions subordinate to the collection {Oλ} and such that ϕ¹+ · · · ϕn ≡ 1 on K.

*26. Lemma: Let X be a locally compact Hausdorff space and U be an open set containing x ∈ X.

Then there is an open set V containing x such that ¯V is compact and ¯V ⊂ U.

Proof: There is an open set O ⊂ X containing x such that ¯O is compact. Then U ∩ O is open in ¯O and contains x. Thus there is an open set O^! ⊂ ¯O such that x ∈ O^! and ¯O! ¯O ⊂ U ∩ O. Note that O¯! ¯O= ¯O∩ ¯O^! = ¯O^!. Thus ¯O^!⊂ U ∩ O. Let V = O ∩ O^!. Then x ∈ V . Furthermore, since O^! is open in O, it is open in O and thus open in X so V is open in X. Also, ¯¯ V ⊂ ¯O^! ⊂ U ∩ O ⊂ U. Since ¯V is closed in ¯O, ¯V is compact.

Let X be a locally compact Hausdorff space and {On} a countable collection of dense open sets. Given an open set U, let x1be a point in O1∩ U. Let V¹be an open set containing x1such that V1is compact and V1⊂ O¹∩ U. Suppose x¹, . . . , xn and V1, . . . , Vn have been chosen. Let xn+1∈ Oⁿ⁺¹∩ Vⁿ and let Vn+1

be an open set containing xn+1such that Vn+1is compact and Vn+1⊂ Oⁿ⁺¹∩ Vⁿ. Then V1⊃ V²⊃ · · · is a decreasing sequence of closed sets in the compact set V1. This collection of closed sets has the finite intersection property so " Vn != ∅. Let y ∈"

Vn. Then y ∈"

On∩ U. Hence"

On is dense in X.

(*) Assume that X is Hausdorff.

*27. Let X be a locally compact Hausdorff space and let O be an open subset contained in a countable union ! Fn of closed sets. Note that O is a locally compact Hausdorff space (see Q29b). Also, O =

!(O ∩ Fⁿ), which is a union of sets closed in O. If !(O ∩ Fn)^◦ = ∅, then (O ∩ Fn)^◦ = ∅ for all n (with the interior taken in O) so O \ Fn is dense and open in O for all n. By Q26, "(O \ Fn) is dense in O.

But "(O \ Fn) = O \ (!

Fn)^c = ∅. Contradiction. Hence!(O ∩Fn)^◦!= ∅. Now O ∩!

F_n^◦⊃!(O ∩Fn)^◦ so O ∩!

F_n^◦!= ∅ and !

F_n^◦ is an open set dense in O.

(*) Assume that X is Hausdorff.

*28. Let Y be a dense subset of a Hausdorff space X, and suppose that Y with its subspace topology is locally compact. Given y ∈ Y , there is an open set U ⊂ Y containing y with ¯UY = Y ∩ ¯U compact. Since X is Hausdorff, Y∩ ¯U is closed. Then since U ⊂ Y ∩ ¯U , we have ¯U ⊂ Y ∩ ¯U ⊂ Y . Now U = V ∩Y for some open set V ⊂ X. Note that since Y is dense, ¯V = V ∩ Y . Then x ∈ V and V ⊂ ¯V = V ∩ Y = ¯U ⊂ Y . Hence Y is open in X.

29a. Suppose F is closed in a locally compact space X. Given x ∈ F , there is an open set O ⊂ X

containing x with ¯O compact. Then O∩ F is an open set in F containing y and O ∩ F^F = F ∩ O ∩ F is closed in ¯O and thus compact. Hence F is locally compact.

*29b. Suppose O is open in a locally compact Hausdorff space X. Take x ∈ O. By the lemma in Q26, there is an open set U containing x such that ¯U is compact and ¯U ⊂ O. Now U ∩ O is an open set in O containing x with U ∩ OO = O ∩ U ∩ O = U ∩ O being compact since it is closed in the compact set ¯U.

Hence O is locally compact.

29c. Suppose a subset Y of a locally compact Hausdorff space X is locally compact in its subspace topology. Then Y is dense in the Hausdorff space ¯Y . By Q28, Y is open in ¯Y . Conversely, suppose Y is open in ¯Y . By part (a), ¯Y is locally compact since ¯Y is closed in X. By part (b), Y is locally compact since Y is open in ¯Y .