5. Locally compact spaces Definition. A locally compact space is a Hausdorff topological space with the property (lc) Every point has a compact neighborhood. One key feature of locally compact spaces is contained in the following; Lemma 5.1. Let X be a locally compact space, let K be a compact set in X, and let D be an open subset, with K ⊂ D. Then there exists an open set E with: (i) E compact; (ii) K ⊂ E ⊂ E ⊂ D. Proof. Let us start with the following Particular case: Assume K is a singleton K = {x}. Start off by choosing a compact neighborhood N of x. Using the results from section 4, when equipped with the induced topology, the set N is normal. In particular, if we consider the closed sets A = {x} and B = N r D (which are also closed in the induced topology), it follows that there exist sets U, V ⊂ N, such that • U ⊃ {x}, V ⊃ B, U ∩ V = ∅; • U and V are open in the induced topology on N. The second property means that there exist open sets U0,V0 ⊂ X, such that U = N ∩ U0 and V = N ∩ V0. Let E = Int(U). By construction E is open, and E 3 x. Also, since E ⊂ U ⊂ N, it follows that (1) E ⊂ N = N. In particular this gives the compactness of E. Finally, since we obviously have E ∩ V0 ⊂ U ∩ V0 = N ∩ U0 ∩ V0 = U ∩ V = ∅, we get E ⊂ X r V0, so using the fact that X r V0 is closed, we also get the inclusion E ⊂ X r V0. Finally, combining this with (1) and with the inclusion N r D ⊂ V ⊂ V0, we will get E ⊂ N ∩ [X r V0] ⊂ N ∩ [X r (N r D)] = N ∩ D ⊂ D, and we are done. Having proven the particular case, we proceed now with the general case. For every x ∈ K we use the particular case to find an open set E(x), with E(x) compact, S and such that x ∈ E(x) ⊂ E(x) ⊂ D. Since we clearly have K ⊂ x∈K E(x), by compactness, there exist x1, . , xn ∈ K, such that K ⊂ E(x1)∪· · ·∪E(xn). Notice that if we take E = E(x1) ∪ · · · ∪ E(xn), then we clearly have K ⊂ E ⊂ E ⊂ E(x1) ∪ · · · ∪ E(xn) ⊂ D, and we are done. One of the most useful result in the analysis on locally compact spaces is the following. 25 26 CHAPTER I: TOPOLOGY PRELIMINARIES Theorem 5.1 (Urysohn’s Lemma for locally compact spaces). Let X be a locally compact space, and let K, F ⊂ X be two disjoint sets, with K compact, and F closed. Then there exists a continuous function f : X → [0, 1] such that f = 1 K and f F = 0. Proof. Apply Lemma 5.1 for the pair K ⊂ X rF and find an open set E, with E compact, such that K ⊂ E ⊂ E ⊂ X r F . Apply again Lemma 5.1 for the pair K ⊂ E and find another open set G with G compact, such that K ⊂ G ⊂ G ⊂ E. Let us work for the moment in the space E (equipped with the induced topol- ogy). This is a compact Hausdorff space, hence it is normal. In particular, using Urysohn Lemma (see section 1) there exists a continuous function g; E → [0, 1] such that g = 0 and g = 0. Let us now define the function f : X → [0, 1] by K ErG g(x) if x ∈ E f(x) = 0 if x ∈ X r E Notice that f = g , so f is continuous. If we take the open set A = X r G, E E E then it is also clear that f = 0. So now we have two open sets E and A, with A A∪E = X, and f and f both continuous. Then it is clear that f is continuous. A E The other two properties f K = 1 and f F = 0 are obvious. We now discuss an important notion which makes the linkage between locally compact spaces and compact spaces Definition. Let X be a locally compact space. By a compactification of X one means a pair (θ, T ) consisting of a compact Hausdorff space T , and of a continuous map θ : X → T , with the following properties (i) θ(X) is a dense open subset of T ; (ii) when we equip θ(X) with the induced topology, the map θ : X → θ(X) is a homeomorphism. Notice that, when X is already compact, any compactification (θ, T ) of X is nec- essarily made up of a compact space T , and a homeomorphism θ : X → T . Examples 5.1. A. Take [−∞, ∞] = R ∪ {−∞, ∞}, with the “usual” topology, in which a set D ⊂ [−∞, ∞] is open if D = D0 ∪ D1 ∪ D2, where D0 is open in R and D1,D2 ∈ ∅ ∪ (a, ∞]: a ∈ R ∪ [−∞, a): a ∈ R . Then [−∞, ∞] is a compactification of R B. (Alexandrov compactification) Suppose X is a locally compact space, which is not compact. We form a disjoint union with a singleton Xα = X t {∞}, and we equip the space Xα with the topology in which a subset D ⊂ Xα is declared to be open, if either D is an open subset of X, or there exists some compact subset K ⊂ X, such that D = (XrK)t{∞}. Define the inclsuion map ι : X,→ Xα. Then (ι, Xα) is a compactification of X, which is called the Alexandrov compactification. The fact that ι(X) is open in Xα, and ι : X → ι(X) is a homeomorphism, is clear. The density of ι(X) in Xα is also clear, since every open set D ⊂ Xα, with D 3 ∞, is of the form (X r K) t {∞}, for some compact set K ⊂ X, and then we have D ∩ ι(X) = ι(X r K), which is non-empty, because X is not compact. §5. Locally compact spaces 27 Remark that, if X is already compact, we can still define the topological space Xα = X t {∞}, but this time the singleton set {∞} will be also be open (equiv- alently ∞ is an isolated point in Xα). Although ι(X) will still be open in Xα, it will not be dense in Xα. Remark 5.1. Let K be some compact Hausdorff space, and let p ∈ K be some point with the property that the set X = K r {p} is non-compact. Of course, since X is open, it follows that X is a locally compact space, when equipped with the induced topology. If we denote by Xα the Alexandrov compactification of X, then the map Ψ : Xα → K, defined by x if x ∈ X Ψ(x) = p if x = ∞ is a homeomorphism. Indeed (see Corollary 4.1), since Xα is compact, and Ψ is obviously bijective, all we need to prove is the fact that Ψ is continuous. Start with some open set D ⊂ K, and let us show that Ψ−1(D) is open in Xα. If D 63 p, then D is open in X, so Ψ−1(D) = D is open in Xα. If D 3 p, then −1 Ψ (D) = [D r {p}] ∪ {∞} = [X r (K r D)] ∪ {∞}, Of course, since K is compact and D ⊂ K is open, it follows that K r D is closed, hence compact in K. Since K r D ⊂ X, it follows that K r D is also compact in X, so the above equality shows that f −1(D) is indeed open in Xα. One should regard the Alexandrov compactification as a minimal one. More precisely, one has the following. Proposition 5.1. Suppose X is a locally compact space which is non-compact. Let (θ, T ) be a compactification of X, and let Xα = X t {∞} be the Alexandrov compactification. Then there exists a unique continuous map Ψ: T → Xα, such that (Ψ◦θ)(x) = x, ∀ x ∈ X. Moreover, the map Ψ has the property that Ψ(y) = ∞, ∀ y ∈ T r θ(X). Proof. The uniqueness part is pretty obvious, since θ(X) is dense in T . For the existence, we use the map θ : X → T to identify X with an open dense subset of T , and we define Ψ : T → Xα by x if x ∈ X Ψ(x) = ∞ if x ∈ T r X so that all we have to prove is the fact that Ψ is continuous. Start with some open set D in Xα, and let us prove that Ψ−1 is open in T . There are two cases. Case I: D ⊂ X. This case is trivial, Φ−1(D) = D, and D is open in X, hence also open in T . Case II: D 6⊂ X. In this case D 3 ∞, so there exists some compact set K in X, such that D = Xα r K = (X r K) ∪ {∞}. We then have −1 −1 −1 Ψ (D) = Ψ (X r K) ∪ Ψ ({∞}) = (X r K) ∪ (T r X) = T r K. Since K is compact in X, it will be compact in T as well. In particular, K is closed −1 in T , hence the set Ψ (D) = T r K is indeed open in T . It turns out that there exists another compactification which is described below, which can be regarded as the largest. 28 CHAPTER I: TOPOLOGY PRELIMINARIES Theorem 5.2 (Stone-Cech)˘ . Let X be a locally compact space. Consider the set F = {f : X → [0, 1] : f continuous }, and consider the product space Y T = [0, 1], f∈F equipped with the product topology, and define the map θ : X → T by θ(x) = f(x) f∈F , ∀ x ∈ X.