1st December 2004

Munkres §31

Ex. 31.1 (Morten Poulsen). Let a and b be distinct points of X. Note that X is Hausdorff, since X is regular. Thus there exists disjoint open sets A and B such that a ∈ A and b ∈ B. By lemma 31.1(a) there exists open sets U and V such that

a ∈ U ⊂ \overline{U} ⊂ A and b ∈ V ⊂ \overline{V} ⊂ B.

Clearly \overline{U} ∩ \overline{V} = ∅.

Ex. 31.2 (Morten Poulsen). Let A and B be disjoint closed subsets of X. Since X normal there exists disjoint open sets U_{0} and U_{1} such that A ⊂ U_{0} and B ⊂ U_{1}. By lemma 31.1(b) there exists open sets V_{0} and V_{1} such that

A ⊂ V_{0} ⊂ \overline{V_{0}} ⊂ U_{0} and B ⊂ V_{1} ⊂ \overline{V_{1}} ⊂ U_{1}

Clearly \overline{U} ∩ \overline{V} = ∅.

Ex. 31.3 (Morten Poulsen).

Theorem 1. Every order topology is regular.

Proof. Let X be an ordered set. Let x ∈ X and let U be a neighborhood of x, may assume U = (a, b), −∞ ≤ a < b ≤ ∞. Set A = (a, x) and B = (x, b). Using the criterion for regularity in lemma 31.1(b) there are four cases:

(1) If u ∈ A and v ∈ B then x ∈ (u, v) ⊂ \overline{(u, v)} ⊂ [u, v] ⊂ (a, b).

(2) If A = B = ∅ then (a, b) = {x} is open and closed, since X Hausdorff, c.f. Ex. 17.10.

(3) If A = ∅ and v ∈ B then x ∈ (a, v) ⊂= [x, v) ⊂ \overline{[x, v)} ⊂ [x, v] ⊂ (a, b).

(4) If u ∈ A and B = ∅ then x ∈ (u, b) ⊂ (u, x] ⊂ \overline{(u, x]} ⊂ [u, x] ⊂ (a, b).

Thus X is regular. □

Ex. 31.5. The diagonal Δ ⊂ Y × Y is closed as Y is Hausdorff [Ex 17.13]. The map (f, g) : X → Y × Y is continuous [Thm 18.4, Thm 19.6] so

{x ∈ X | f(x) = g(x)} = (f, g)^{−1}(Δ)

is closed.

Ex. 31.6. Let p: X → Y be closed continuous surjective map. Then X normal ⇒ Y normal. For this exercise and the next we shall use the following lemma from [Ex 26.12].

Lemma 2. Let p: X → Y be a closed map.

(1) If p^{−1}(y) ⊂ U where U is an open subspace of X, then p^{−1}(W) ⊂ U for some neighborhood W ⊂ Y of y.

(2) If p^{−1}(B) ⊂ U for some subspace B of Y and some open subspace U of X, then p^{−1}(W) ⊂ U for some neighborhood W ⊂ Y of B.

Proof. Note that

p^{−1}(W) ⊂ U ⇔ [p(x) ∈ W ⇒ x ∈ U] ⇔ [x ∉ U ⇒ p(x) ∉ W ] ⇔ p(X − U) ⊂ Y − W
⇔ p(X − U) ∩ W = ∅

(1) The point y does not belong to the closed set p(X − U). Therefore a whole neighborhood W ⊂ Y of y is disjoint from p(X − U), i.e. p^{−1}(W) ⊂ U.

(2) Each point y ∈ B has a neighborhood W_{y} such that p^{−1}(W_{y}) ⊂ U. The union W = ⋃ W_{y} is then a neighborhood of B with p^{−1}(W) ⊂ U. □

Since points are closed in X and p is closed, all points in p(X) are closed. All fibres p^{−1}(y) ⊂ X are therefore also closed. Let y_{1} and y_{2} be two distinct points in Y. Since X is normal we can separate the disjoint closed sets p^{−1}(y_{1}) and p^{−1}(y_{1}) by disjoint neighborhoods U_{1} and U_{2}. Using Lemma 2.(1), choose neighborhoods W_{1} of y_{1} and W_{2} of y_{2} such that p^{−1}(W_{1}) ⊂ U_{1} and p^{−1}(W_{2}) ⊂ U_{2}. Then W_{1} and W_{2} are disjoint. Thus Y is Hausdorff.

Essentially the same argument, but now using Lemma 2.(2), shows that we can separate disjoint closed sets in Y by disjoint open sets. Thus Y is normal.

Alternatively, see [Lemma 73.3].

Example: If X is normal and A ⊂ X is closed, then the quotient space X/A is normal.

Ex. 31.7. Let p: X → Y be closed continuous surjective map such that p^{−1}(y) is compact for each y ∈ Y (a perfect map).

(a). X Hausdorff ⇒ Y Hausdorff. Let y_{1} and y_{2} be two distinct points in Y. By an upgraded version [Ex 26.5] of [Lemma 26.4] we can separate the two disjoint compact subspaces p^{−1}(y_{1}) and p^{−1}(y_{2}) by disjoint open subspaces U_{1} ⊃ p^{−1}(y_{1}) and U_{2} ⊃ p^{−1}(y_{2}) of the Hausdorff space X. Choose (Lemma 2) open sets W_{1} ∋ y_{1} and W_{2} ∋ y_{2} such that p^{−1}(W_{1}) ⊂ U_{1} and p^{−1}(W_{2}) ⊂ U_{2}. Then W_{1} and W_{2} are disjoint. This shows that Y is Hausdorff as well.

(b). X regular ⇒ Y regular. Y is Hausdorff by (a). Let C ⊂ Y be a closed subspace and y ∈ Y a point outside C. It is enough to separate the compact fibre p^{−1}(y) ⊂ X and the closed set p^{−1}(C) ⊂ X by disjoint open set. (Lemma 2 will provide open sets in Y separating y and C.) Each x ∈ p^{−1}(y) can be separated by disjoint open sets from p^{−1}(C) since X is regular. Using compactness of p^{−1}(y) we obtain (as in the proof [Thm 26.3]) disjoint open sets U ⊃ p^{−1}(y) and V ⊃ p^{−1}(C) as required.

(c). X locally compact ⇒ Y locally compact [1, 3.7.21]. Using compactness of p^{−1}(y) and local compactness of X we construct an open subspace U ⊂ X and a compact subspace C ⊂ X such that p^{−1}(y) ⊂ U ⊂ C. In the process we need to know that a finite union of compact subspaces is compact [Ex 26.3]. By Lemma 2, there is an open set W ∋ y such that p^{−1}(y) ⊂ p^{−1}(W) ⊂ U ⊂ C. Then y ∈ W ⊂ p(C) where p(C) is compact [Thm 26.5]. Thus Y is locally compact.

(d). X 2nd countable ⇒ Y 2nd countable. Let {B_{j}}_{j∈Z_{+}} be countable basis for X. For each finite subset J ⊂ Z_{+}, let U_{J} ⊂ X be the union of all open sets of the form p^{−1}(W) with open W ⊂ Y and p^{−1}(W) ⊂ ⋃_{j∈J} B_{j}. There are countably many open sets U_{J}. The image p(U_{J}) is a union of open sets in Y, hence open. Let now V ⊂ Y be any open subspace. The inverse image p^{−1}(V) = ⋃_{y∈V} p^{−1}(y) is a union of fibres. Since each fibre p^{−1}(y) is compact, it can be covered by a finite union ⋃_{j∈J(y)} B_{j} of basis sets contained in p^{−1}(V). By Lemma 2, there is an open set W ⊂ Y such that p^{−1}(y) ⊂ p^{−1}(W) ⊂ ⋃_{j∈J(y)} B_{j}. Taking the union of all these open sets W, we get p^{−1}(y) ⊂ U_{J(y)} ⊂ ⋃_{j∈J(y)} B_{j} ⊂ p^{−1}(V). We now have p^{−1}(V) = ⋃_{y∈V} U_{J(y)} so that V = pp^{−1}(V) = ⋃_{y∈V} p(U_{J(y)}) is a union of sets from the countable collection {p(U_{J})} of open sets. Thus Y is 2nd countable.

Example: If Y is compact, then the projection map π_{2} : X × Y → Y is perfect. (Show that π_{2} is closed!)

Ex. 31.8. It is enough to show that p: X → G\X is a perfect map [Ex 31.6, Ex 31.7]. We show that

(1) The saturation GA of any closed subspace A ⊂ X is closed. (The map p is closed.)

(2) The orbit Gx of any point x ∈ X is compact. (The fibres p^{−1}(Gx) = Gx are compact.)

(1) Let y ∈ X be any point outside GA = ⋃_{g∈G} gA. For any g ∈ G, g^{−1}y is outside the closed set A ⊂ X. By continuity of the action G × X → X,

U_{g}^{−1}V_{g} ⊂ X − A

for open sets G ⊃ U_{g} ∋ g and X ⊂ V_{g} ∋ y. The compact space G can be covered by finitely many of the open sets U_{g}, say G = U_{1} ∪ ⋯ ∪ U_{n}. Let V = V_{1} ∩ ⋯ ∩ V_{n} be the intersection of the corresponding neighborhoods of y. Then

G^{−1}V = ⋃_{i} U_{i}^{−1}V ⊂ ⋃_{i} U_{i}^{−1}V_{i} ⊂ X − A

so y ∈ V ⊂ G(X − A) = X − GA.

(2) The orbit Gx of a point x ∈ X is compact because [Thm 26.5] it is the image of the compact space G under the continuous map G → X: g → gx.

References

[1] Ryszard Engelking, General topology, second ed., Sigma Series in Pure Mathematics, vol. 6, Heldermann Verlag, Berlin, 1989, Translated from the Polish by the author. MR 91c:54001