The following proposition has a straightforward proof that is omitted.Proposition 8.13. Continuous images of compact spaces are compact.This then implies one of the most useful facts about compact spaces:Proposition 8.14. Let X be a compact space and f : X → R a continuousfunction. Then f is bounded and there are points x, y ∈ X realizing theextreme values of f : for all z ∈ X, we have f(x) ≤ f(z) ≤ f(y) .Proof. We know f(X) is a compact subset of R and is therefore closed andbounded. Let a be the least upper bound of f(X). Of course a is in theclosure¯f(X) (either a ∈ f(x) or else a is certainly a limit point of f (x)), sosince f(X) is closed, a ∈ f(X). Similarly f(X) contains its great e st lowerbound, and therefore points in x map to both of t he se extremal valu es . Ofcourse, this means that there are points in X mapping to these values.Another special property of maps from compact spaces:Proposition 8.15. Let X be compact and let Y be Hausdorff. Then anycontinuous map f : X → Y is closed. Further, f : X → f(X) is a quotientmap. Finally, if f : X → f(X) is injective, it is a homeomorphism.Proof. Let A ⊂ X be closed. Then A i s compact. Then the image f(A)is compact. Since Y is Haus dor ff, f (A) is closed. The map f : X → f(X)is a closed surjection, so it is a quotient map. Bijective qu ot i ent maps arehomeomorphisms, so the last statement also follows.Definition 8.16. A surjective, closed, continuous map of spaces f : X → Yis perfect if for each y ∈ Y , the set f−1({y}) is compact in X.Lemma 8.17. Suppose f : X → Y is a perfect map. Let y ∈ Y and let Ube an open subset of X containing f−1({y}). Then there is an open subsetV of Y with f−1({y}) ⊂ f−1(V ) ⊂ U.Proof. The complement X\U is a closed set that does not intersect f−1({y}).Then sin ce f is c l osed , f (X\U) is closed and does not contain y. LetV = Y \f(X\U). Then y ∈ V , so f−1({y}) ⊂ f−1(V ). Of course f−1(V ) isopen, and it is not hard to see that f−1(V ) ⊂ U.Proposition 8.18. Suppose f : X → Y is a perfect map and Y is compact.Then X is compact.40Proof. Suppose C is an open cover of Y . For each y ∈ Y , the set f−1({y})is compact, and so there is a finite subset Cyof C that c overs f−1({y}). LetUy=SCy. By the lemma, for each y, there is an op en set Vy⊂ Y withf−1({y}) ⊂ f−1(Vy) ⊂ Uy. Then the collection {Vy}y∈Yis an open cover ofY and has a finite subcover {Vy1, . . . , Vy1}. Then the col l ec ti on Cy1∪· · ·∪Cynis a finite subcover of C covering Y .Corollary 8.19. A product of two compact spaces is compact.Proof. Let X and Y be compact spaces and consider the coordinat e projec-tion p: X × Y → Y . Of course p is continuous and surjective, and for anyy ∈ Y the preimage p−1(y) is homeomorphic to X and is therefore compact.So to show that p i s a perfect map, we just need to show that it is closed.Let A ⊂ X × Y be closed. Suppose y ∈ Y is a limit point of p(A) and s up -pose for contradiction that y /∈ p(A). Then X × Y \A is open and containsp−1(y) as a subset. Since X × Y \A is open, we can write it as a union ofbasis elements of the form Ui× Vifor Uiopen in X and Viopen in Y . Thiscollection of basis elements then covers p−1(y); since p−1(y) is compact, wecan find a finite subcover U1× V1, . . . , Un× Vn. We can assume each Ui× Viintersects p−1(Y ), so that V = V1∩ · · · ∩ Vnis an open neighborhood of y inY . Then p−1(V ) contains p−1(y) but is a subset of X × Y \A. This impliesthat y ∈ V but V ∩ p(A) = ∅, contradicting ou r hypothese that y was alimit point of p(A). So p is closed and therefore perfect, and therefore X ×Yis compact.The following needs no further proof.Corollary 8.20. A subset of Rnis compact if a nd only if it is closed andbounded.There is also a characterization of compactness in terms of intersectionsof closed sets.Definition 8.21. A collection of sets C has the finite intersection propertyif for every finite subcollection C′, we haveTC′6= ∅.Proposition 8.22. A space X is compact if and only if for every collectionC of closed sets with the finite intersection property, we have that the totalintesectionTC is nonem pty .Proof. Given a collection of closed sets C, we can define a collection of opensets by taking the complements: B = {X\A|A ∈ C}. Note thatTC = ∅if and only ifSB = X. If a space X is compact and C is a set of closed41sets withTC = ∅, then its collection of complements B covers X and has afinite subcover B′. The coll ec ti on of complements C′of B′is a finite subset ofC withTC′= ∅. So if C has the finite intersection prop er ty, then certainlyTC 6= ∅.Now suppose that every collection of cl os ed sets with th e finite inter-section property has nonemp ty intersection. Let B be an open cover of X.Then the col l ec t ion of complements C has empty intersection, and thereforehas a finite subcollection C′with empty intersection. Then the collecti on ofcomplements of C′is a finite subcover of