1st December 2004

Munkres §26

Ex. 26.1 (Morten Poulsen).

(a). Let T and T′ be two topologies on the set X. Suppose T′ ⊃ T .

If (X, T′) is compact then (X, T ) is compact: Clear, since every open covering if (X, T ) is an open covering in (X, T′).

If (X, T ) is compact then (X, T ) is in general not compact: Consider [0, 1] in the standard topology and the discrete topology.

(b).

Lemma 1. If (X, T ) and (X, T′) are compact Hausdorff spaces then either T and T′ are equal or not comparable.

Proof. If (X, T ) compact and T′ ⊃ T then the identity map (X, T′) → (X, T ) is a bijective continuous map, hence a homeomorphism, by theorem 26.6. This proves the result. □

Finally note that the set of topologies on the set X is partially ordered, c.f. ex. 11.2, under inclusion. From the lemma we conclude that the compact Hausdorff topologies on X are minimal elements in the set of all Hausdorff topologies on X.

Ex. 26.2 (Morten Poulsen).

(a). The result follows from the following lemma.

Lemma 2. If the set X is equipped with the finite complement topology then every subspace of X is compact.

Proof. Suppose A ⊂ X and let A be an open covering of A. Then any set A_{0} ∈ A will covering all but a finite number of points. Now choose a finite number of sets from A covering A − A_{0}. These sets and A_{0} is a finite subcovering, hence A compact. □

(b). Lets prove a more general result: Let X be an uncountable set. Let

T_{c} = { A ⊂ X | X − A countable or equal X }.

It is straightforward to check that T_{c} is a topology on X. This topology is called the countable complement topology.

Lemma 3. The compact subspaces of X are exactly the finite subspaces.

Proof. Suppose A is infinite. Let B = {b_{1}, b_{2}, …} be a countable subset of A. Set

A_{n} = (X − B) ∪ {b_{1}, …, b_{n}}.

Note that {A_{n}} is an open covering of A with no finite subcovering. □

The lemma shows that [0, 1] ⊂ R in the countable complement topology is not compact.

Finally note that (X, T_{c}) is not Hausdorff, since no two nonempty open subsets A and B of X are disjoint: If A∩B = ∅ then X −(A∩B) = (X −A)∪(X −B), hence X countable, contradicting that X uncountable.

Ex. 26.3 (Morten Poulsen).

Theorem 4. A finite union of compact subspaces of X is compact.

Proof. Let A_{1}, …, A_{n} be compact subspaces of X. Let A be an open covering of ⋃_{i=1}^{n} A_{i}. Since A_{j} ⊂ ⋃_{i_{1}}^{n} A_{i} is compact, 1 ≤ j ≤ n, there is a finite subcovering A_{j} of A covering A_{j}. Thus ⋃_{j=1}^{n} A_{j} is a finite subcovering of A, hence ⋃_{i=1}^{n} A_{i} is compact. □

Ex. 26.5. For each a ∈ A, choose [Lemma 26.4] disjoint open sets U_{a} ∈ a and V_{a} ⊃ B. Since A is compact, A is contained in a finite union U = U_{1} ∪ ⋯ ∪ U_{n} of the U_{a}s. Let V = V_{1} ∩ ⋯ V_{n} be the intersection of the corresponding V_{a}s. Then U is an open set containing A, V is an open set containing B, and U and V are disjoint as U ∩ V = ⋃ U_{i} ∩ V ⊂ ⋃ U_{i} ∩ V_{i} = ∅.

Ex. 26.6. Since any closed subset A of the compact space X is compact [Thm 26.2], the image f(A) is a compact [Thm 26.5], hence closed [Thm 26.3], subspace of the Hausdorff space Y.

Ex. 26.7. This is just reformulation of The tube lemma [Lemma 26.8]: Let C be a closed subset of X × Y and x ∈ X a point such that the slice {x} × Y is disjoint from C. Then, since Y is compact, there is a neighborhood W of x such that the whole tube W × Y is disjoint from C.

In other words, if x ∉ π_{1}(C) then there is a neighborhood W of x which is disjoint from π_{1}(C). Thus The tube lemma says that π_{1} : X × Y → X is closed when Y is compact (so that π_{1} is an example of a perfect map [Ex 26.12]). On the other hand, projection maps are always open [Ex 16.4].

Ex. 26.8. Let G ⊂ X × Y be the graph of a function f: X → Y where Y is compact Hausdorff. Then

G is closed in X × Y ⇔ f is continuous

⇐: (For this it suffices that Y be Hausdorff.) Let (x, y) ∈ X × Y be a point that is not in the graph of f. Then y ≠ f(x) so by the Hausdorff axiom there will be disjoint neighborhoods V ∋ y and W ∋ f(x). By continuity of f, f(U) ⊂ W ⊂ Y − V. This means that (U × V) ∩ G = ∅. ⇒: Let V be a neighborhood of f(x) for some x ∈ X. Then G ∩ (X × (Y − V)) is closed in X × Y so [Ex 26.7] the projection π_{1}(G ∩ (X × (Y − V))) is closed in X and does not contain x. Let U be a neighborhood of X such that U × Y does not intersect G ∩ (X × (Y − V)). Then f(U) does not intersect Y − V, or f(U) ⊂ V. This shows that f is continuous at the arbitrary point x ∈ X.

Ex. 26.12. (Any perfect map is proper; see the January 2003 exam for more on proper maps.) Let p: X → Y be closed continuous surjective map such that p^{−1}(y) is compact for each y ∈ Y. Then p^{−1}(C) is compact for any compact subspace C ⊂ Y.

For this exercise we shall use the following lemma.

Lemma 5. 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. □

We shall not need point (2) here.

Let C ⊂ Y be compact. Consider a collection {U_{α}}_{α∈J} of open sets covering of p^{−1}(C). For each y ∈ C, the compact space p^{−1}(y) is contained in a the union of a finite subcollection {U_{α}}_{α∈J(y)}. There is neighborhood W_{y} of y such that p^{−1}(W_{y}) is contained in this finite union. By compactness of C, finitely many W_{y_{1}}, …, W_{y_{k}} cover Y. Then the finite collection ⋃_{i=1}^{k}{U_{α}}_{α∈J(y_{i})} cover p^{−1}(C). This shows that p^{−1}(C) is compact.

Ex. 26.13. Let G be a topological group and A and B subspaces of G.

(a). A closed and B compact ⇒ AB closed Assume c ∉ AB = ⋃_{b∈B} Ab. The regularity axiom for G [Suppl Ex 22.7] implies that there are disjoint open sets W_{b} ∋ c and U_{b} ⊃ Ab separating c and Ab for each point b ∈ B. Then A^{−1}U_{b} is an open neighborhood of b. Since B is compact, it can be covered by finitely many of these open sets A^{−1}U_{b}, say

B ⊂ A^{−1}U_{1} ∪ ⋯ ∪ A^{−1}U_{k} = A^{−1}U

where U = U_{1} ∪ ⋯ ∪ U_{k}. The corresponding open set W = W_{1} ∩ ⋯ ∩ W_{k} is an open neighborhood of c that is disjoint from AB since W ∩ AB ⊂ ⋃ W ∩ U_{i} ⊂ ⋃ W_{i} ∩ U_{i} = ∅.

(b). H compact subgroup of G ⇒ p: G → G/H is a closed map The saturation AH of any closed subset A ⊂ G is closed by (a).

(c). H compact subgroup of G and G/H compact ⇒ G compact The quotient map p: G → G/H is a perfect map because it is a closed map by (b) and has compact fibres p^{−1}(gH) = gH. Now apply [Ex 26.12].

References