SOLUTIONS TO EXERCISES FOR MATHEMATICS 205A | Part 3 Fall 2008 III. Spaces with special properties III.1 : Compact spaces { I Problems from Munkres, 26, pp. 170 172 x − 3. Show that a finite union of compact subspaces of X is compact. SOLUTION. Suppose that A X is compact for 1 i n, and suppose that is a family of open i ⊂ ≤ ≤ U subsets of X whose union contains i Ai. Then for each i there is a finite subfamily i whose union ∗ [ U contains Ai. If we take to be the union of all these subfamilies then it is finite and its union contains A . ThereforeU the latter is compact. [i i 7. If Y is compact, show that the projection π : X Y X is a closed map. X × ! SOLUTION. We need to show that if F X Y is closed then π [F ] is closed in X, and as usual it is ⊂ × X enough to show that the complement is open. Suppose that x πX[F ]. The latter implies that x Y is contained in the open subset X Y F , and by the62 Tube Lemma one can find an opf gen×set V X such that x V and V ×Y −X Y F . But this means that the open set x ⊂ 2 x × ⊂ × − Vx X lies in the complement of πX [F ], and since one has a conclusion of this sort for each such x it⊂follows that the complement is open as required. 8. Let f : X Y be a set-theoretic map of topological spaces, and assume Y is compact ! Hausdorff. Prove that f is continuous if and only if its graph Γf (defined previously) is a closed subset of X Y . × SOLUTION. ( = ) We shall show that the complement of the graph is open, and this does not use the compactness) condition on Y (although it does use the Hausdorff property). Suppose we are given (x; y) such that y = x. Then there are disjoint open sets V and W in Y such that y V and f(x) W . Since f6 −1[W ] is open in X and contains x, there is an open set U containing2 x such 2 that f[U] W . It follows that U V is an open subset of X Y that is disjoint from Γf . Since we have suc⊂h a subset for each poin×t in the complement of Γ ×it follows that X Y Γ is open f × − f and that Γf is closed. ( = ) As in a previous exercises let γ : X Γf be the graph map and let j : Γf X Y ( ! o ! × be inclusion. General considerations imply that the map πX j is continuous and 1{1 onto, and o γ is the associated set-theoretic inverse. If we can prove that πX j is a homeomorphism, then γ will also be a homeomorphism and then f will be continuous by a previous exercise. The map in question will be a homeomorphism if it is closed, and it suffices to check that it is a composite of 1 closed mappings. By hypothesis j is the inclusion of a closed subset and therefore j is closed, and the preceding exercise shows that πX is closed. Therefore the composite is a homeomorphism as claimed and the mapping f is continuous. Problem from Munkres, 27, pp. 177 178 x − 2. Let A X be a nonempty subset of a metric space X. ⊂ (a) Show that d(x; A) = 0 if and only if X A. 2 SOLUTION. The function of x described in the problem is continuous, so its set of zeros is a closed set. This closed set contains A so it also contains A. On the other hand if x A then there is an " > 0 such that N (x) X A, and in this case it follows that d(x; A) " >620. " ⊂ − ≥ (b) Show that if A is compact then d(x; A) = d(x; a) for some a A. 2 SOLUTION. The function f(a) = d(x; a) is continuous and d(x; A) is the greatest lower bound for its set of values. Since A is compact, this greatest lower bound is a minimum value that is realized at some point of A. (c) Define the "-neighborhood U(A; ") to be the set of all u such that d(u; A) < ". Show that this is the union of the neighborhoods N (a) for all a A. " 2 SOLUTION. The union is contained in U(A; ") because d(x; a) < " implies d(x; A) < ". To prove the reverse inclusion suppose that y is a point such that δ = d(y; A) < ". It then follows that there is some point a A such that d(y; a) < " because the greater than the greatest lower bound of all possible distances.2 The reverse inclusion is an immediate consequence of the existence of such a point a. (d) Suppose that A is compact and that U is an open set containing A. Prove that there is an " > 0 such that A U(A; "). ⊂ SOLUTION. Let F = X U and consider the function g(a) = d(a; F ) for a A. This is a continuous function and it is−always positive because A F = . Therefore it takes2a positive minimum value, say ". If y A U(A; ") then d(a; y) < " \ d(a; F; ) implies that y F , and therefore U(A; ") is contained in2 the⊂complement of F , which is≤U. 62 (e) Show that the preceding conclusion need not hold if A is not compact. SOLUTION. Take X to be all real numbers with positive first coordinate, let A be the points of X satisfying y = 0, and let U be the set of all points such that y < 1= x . Then for every " > 0 there is a point not in U whose distance from A is less than ". For example,j j consider the points (2n; 1=n). 2 Additional exercises 1. Let X be a compact Hausdorff space, and let f : X X be continuous. Define X = X ! 1 and Xn+1 = f(Xn), and set A = n Xn. Prove that A is a nonempty subset of X and f[A] = A. SOLUTION. T Each of the subsets Xn is compact by an inductive argument, and since X is Hausdorff each one is also closed. Since each set in the sequence contains the next one, the intersection of finitely many sets Xk(1); ; Xk(n) in the collection is the set Xk(m) where k(m) is the maximum of the k(i). Since X is compact,· · · the Finite Intersection property implies that the intersection A of these sets is nonempty. We need to prove that f(A) = A. By construction A is the set of all points that lie in the image of the k-fold composite kf of f with itself. To see that f maps this set into itself k ◦ k note that if a = [ f](xk) for each positive integer k then f(a) = [ f]( f(xk) ) for each k. To see that f maps this ◦set onto itself, note that a = [ kf](x ) for each positiv◦ e integer k implies that ◦ k a = f [ kf](x ) ◦ k+1 for each k. 2. A topological space X is said to be a k-space if it satisfies the following condition: A subset A X is closed if and only if for all compact subsets K X, the intersection A K is closed. ⊂ It turns out that a large number of the topological spaces⊂ one encounters in topology\ , geometry and analysis are k-spaces (including all metric spaces and compact Hausdorff spaces), and the textbooks by Kelley and Dugundji contain a great deal of information about these k-spaces (another important reference is the following paper by N. E. Steenrod: A convenient category of topological spaces, Michigan Mathematical Journal 14 (1967), pp. 133{152). (a) Prove that if (X; T) is a Hausdorff topological space then there is a unique minimal topology Tκ containing T such that (X) = (X; Tκ) is a Hausdorff k-space. K SOLUTION. In this situation it is convenient to work with topologies in terms of their closed subsets. Let be the family of closed subsets of X associated to T and let ∗ be the set of all subsets E sucFh that E C is closed in X for every compact subset C X. If EF belongs to then E C is always closed\in X because C is closed, so ∗. We claim⊂ that ∗ defines a topFology on\X and X is a Hausdorff k-space with respect to thisF ⊂topF ology. F ∗ The empty set and X belong to because they already belong to . Suppose that Eα belongs ∗ F F to for all α; we claim that for each compact subset C X the set C α Eα is -closed in X. ThisF follows because ⊂ \ \ F C E = (C E ) \ α \ α α α \ \ and all the factors on the right hand side are -closed (note that they are compact). To conclude ∗ F ∗ the verification that is a topology, suppose that E1 and E2 belong to . Once again let C X be compact, and observF e that the set-theoretic equation F ⊂ C (E E ) = (C E ) (C E ) \ 1 [ 2 \ 1 [ \ 2 implies the right hand side is -closed if E1 and E2 are. Therefore ∗ defines theFclosed subspaces of a topological space; let Tκ be the associated family of open sets.F It follows immediately that the latter contains T, and one obtains a Hausdorff space by the following elementary observation: If (X; T) is a Hausdorff topological space and T∗ is a topology for X containing T, then (X; T∗) is also Hausdorff.