(a) Show that if A⊆X, then ¯A∩¯X−A=ø where their closures are taken in β(X).
(b) Show that if U⊆β(X) is open, then ¯U is open.
(c) Show that X is totally disconnected.
8. Show the cardinality of β(ℕ) is at least as great as II where I=[0,1].
Proof: 7. (a) Define a function f:X→ℝ by f(A)=1 and f(X−A)=0. Letting F:β(X)→ℝ extend f, we see F−1(1) and F−1(0) are disjoint closed sets in β(X) containing A and X−A respectively, so these latters' closures are disjoint.
(b) Note ¯U∩X∪¯X−U∩X is the whole space β(X) since its complement is an open set not intersecting X, so by part (a) ¯U∩X is open. Now evidently ¯U∩X⊆¯U, but also ¯U⊆¯U∩X since if there exists x∈¯U with a neighborhood V disjoint from U∩X, let y∈V∩U and now V∩U is a neighborhood of y not intersecting X hence y∉¯X, a contradiction.
(c) Let x,y∈β(X) be distinct. Since β(X) is Hausdorff let U be open such that x∈U and v∉¯U. Then by part (b) ¯U∪β(X)−¯U is a separation of β(X) disconnecting x from y.
8. As we've seen (cf. 5.31.16a), ℝI≅(0,1)I has a countable dense subset, so ¯(0,1)I=[0,1]I does as well, call it S. Letting f:ℕ→S⊆[0,1]I be a surjection that is automatically continuous, we obtain a map of β(ℕ) into II containing S, and since images of compact sets are compact hence closed in a Hausdorff space, the map is a surjection and the claim is proven. ◻
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