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5. Lecture. Compact Spaces. In the last lecture we have seen that the Cantor set is perfect, totally disconnected, homogeneous, not homeomorphic to [0, 1] and homeomorphic to the product with itself. Any infinite discrete space has the same properties. Nevertheless, the Cantor set is not homeomorphic to any discrete space. The distinguishing feature is compactness. The Cantor set is compact and an infinite discrete space is not. In this lecture we introduce the notion of compactness and discuss some of its properties. Compact spaces are the spaces of choice. If one has a non-compact space one often tries to compactify it. Here we describe two such compactifications - the one-point compactifiction and the compactifications of trees. Basics. We say that a collection of open subspaces of a topological space X is an (open) covering of X if its union equals X. A subcovering of a covering is a subset of the covering. Klaus Johannson, Topology I 72 . Topology I 5.1. Definition. A topological space X is compact if (1) X is Hausdorff, and if (2) every open cover of X has a finite subcovering. A topological space that has only property (2) is also called quasi-compact. A subset A ⊂ X is relative compact if its closure Ā is compact. Remark. In the literature a space with (1) and (2) is called compact. Some other authors say a space is compact if it has only property (2). In order to avoid ambiguity we also use the term ”compact Hausdorff” when we want to remind ourselves that the space is quasicompact and Hausdorff. 5.2. Example. Many topological spaces are not compact. For instance, the Euclidean line R is not compact. Indeed, the open covering given by {(n, n + 2) | n ∈ Z} has no finite sub-covering. An infinite tree such as the Cayley graph of a free group is not compact. Klaus Johannson, Topology I §5 Compact Spaces 73 Here is the non-trivial standard example of a compact space. 5.3. Theorem. Let a, b ∈ R be real numbers with a < b. Then the closed interval [a, b] is compact. Proof. Assume the converse. Then there is an open cover O of [a, b] which has no finite subcover. We say a closed interval J in [a, b] is bad if no finite subcollection of O covers J. In this terminology J0 = [a0 , b0 ] := [a, b] is a bad interval. If Jn := [an , bn ] ⊂ [a0 , b0 ] is a bad sub-interval, let xn = 12 (an + bn ) be its mid-point. and observe that either [an , xn ] or [xn , bn ] must be a bad sub-interval. This is immediate from the fact that the union of two compact spaces is compact again. Thus, starting with J0 and selecting bad sub-intervals through splitting along mid-points, one constructs a nested strictly decreasing sequence Jn = [an , bn ] of bad sub-intervals. The end-points an and bn define Cauchy sequence. Therefore they converge an → x and bn → y, say. Moreover, x = y since an −bn → 0. Let U ∈ O be a covering element that contains x. Then there is an open interval (c, d) with x ∈ (c, d) ⊂ U . In particular, there is an index m with Jm = [am , bm ] ⊂ (c, d) since x lies in all [an , bn ]. It follows Klaus Johannson, Topology I 74 . Topology I that Jm is not bad. But this contradicts our choice of Jm . This contradiction proves the theorem. ♦ Remark. It follows that the open interval (0, 1) is relative compact but not (0, ∞). The compactness of the closed interval is the basis for compactness properties of many other spaces. For instance the next result implies the compactness of the unit cube [0, 1]n . 5.4. Theorem. The product of two compact spaces is compact. Proof. Let X, Y be two compact spaces. We are asked to show that X × Y is compact in the product topology. First we are going to show that X × Y is Hausdorff and after that we show that every covering of X × Y has a finite subcovering. Let p1 = (x1 , y1 ) and p2 = (x2 , y2 ) be two different points of X × Y . Then x1 6= x2 or y1 6= y2 . Thus w.l.o.g. there are two disjoint neighborhoods U (x1 ), U (x2 ) for x1 , x2 in X since X is Hausdorff. Hence U (x1 ) × Y and U (x2 ) × Y are two disjoint Klaus Johannson, Topology I §5 Compact Spaces 75 neighborhoods of p1 , p2 in X × Y . This is what we needed in order to show that X × Y is Hausdorff. What about the second property? So let us be given an open covering C of X ×Y . Without loss of generality we may suppose that all the covering elements of C are actually boxes U × V since, by definition, the boxes form a basis of the product topology (i.e., all open sets are unions of boxes). Now, we need a procedure for selecting a finite subcovering. Let us proceed as follows. First observe that for every x in X the subspace x × Y is compact in X × Y (why?). Thus already finitely many of the boxes from C cover x × Y . Let us denote those finitely many boxes by U1 (x) × V1 , . . . , Un (x) × Vn . The intersection U (x) := U1 (x) ∩ . . . ∩ Un (x) Klaus Johannson, Topology I 76 . Topology I is an open set (a topology is closed under taking finite intersections). But notice that the strip U (x) × Y is still covered by the previous covering elements. Thus it remains to show that X × Y is covered by only finitely many of all the strips U (x) × Y . The collection {U (x) | x ∈ X} is a covering of the compact space X and so has a finite subcovering. In other words there are finitely many points x1 , . . . , xm such that X = U (x1 ) ∪ . . . ∪ U (xm ). Thus X × Y is covered by the finitely many strips U (x1 ) × Y, . . . , U (xm ) × Y . This is what we needed. ♦ Properties of Compact Spaces. A closed subset need not be compact as e.g. the subspace [0, ∞) ⊂ R shows. However, we have: 5.5. Theorem. Every closed subset of a compact space is compact. Every compact subspace of a Hausdorff space is closed. Proof. Let A be a closed subspace of a compact space X. LetS O be a collection of open subsets of X with A ⊂ O. Then O ∪ {X − A} is an open cover of X. Then compactness of X implies that also O has a finite sub-cover that covers A. Klaus Johannson, Topology I §5 Compact Spaces 77 Let Y ⊂ X be a compact subspace of a Hausdorff space X. To show that Y is closed, it suffices to prove that X−Y is open. So let x ∈ X−Y . For every point y ∈ Y , there are open neighborhoods U (y) of y and U (x, y) of x with U (y) ∩ U (x, y) = ∅. The collection {U (y) | y ∈ Y covers Y . Hence it has a finite subcover. In other words there are finitely many points y1 , . . . , yn ∈ Y with Y ⊂ U (y1 ) ∪ . . . ∪ U (yn ). But then the finite intersection V (x) = U (x, y1 )∩. . .∩ U (x, yn ) is an open neighborhood of x that does not meet U (y1 ) ∪ . . . ∪ U (yn ) and so it does not meet Y . Since x was arbitrary, we have shown that X − Y is open. ♦ 5.6. Theorem. Every compact space is regular. Proof. Same as the previous proof. ♦ 5.7. Theorem. The Cantor set is compact. Every compact discrete space is finite. Proof. The Cantor set is a closed subset of the unit interval. Hence it is compact. Every point in a discrete space is open. Thus the set of all points of a discrete space is an open covering of the space. It can only have a finite subcovering when the space itself is finite. ♦ Klaus Johannson, Topology I 78 . Topology I Expanding on the idea of the proof of the previous theorem somewhat, we show that all compact spaces are normal. Recall that a space is normal if disjoint closed subsets can be separated. 5.8. Theorem. Every compact space is normal. Proof. In order to prove the lemma let us be given two disjoint and closed subsets A, B ⊂ X. We already know that X is regular. Thus, for every x ∈ B, there are disjoint, open neighborhoods U (x), Vx (A) of x and A, respectively. The collection {U (x)}x∈B is an open covering of B. But B is compact as a closed subset of a compact space. Hence there are finitely many points x1 , . . . , xn ∈ B such that B ⊂ U (x1 ) ∪ . . . ∪ U (xn ). In other words, the union U (B) := U (x1 ) ∪ . . . ∪ U (xn ) is an open neighborhood of B. Moreover, the finite intersection U (A) := Vx1 (A)∩. . .∩Vxn (A) is an open neighbrhood of A. Finally, U (A) ∩ U (B) = ∅, by construction. Thus disjoint closed sets can be separated by open neighborhoods. ♦ Recall from the previous lecture that normal spaces have a sandwich property. The next theorem gives Klaus Johannson, Topology I §5 Compact Spaces 79 a stronger version for components of compact Hausdorff spaces. At this point recall that a component of a topological space is always closed but not necessarily open. 5.9. Theorem. Let X be a compact Hausdorff space. Then for every component C of X and every open neighborhood U (C), there is an open subset D with C ⊂ D = D̄ ⊂ U (C). Proof. Exercise. ♦ Remark. Observe that D must be a union of components. Compact spaces have other nice properties. One of the most basic one is 5.10. Theorem. Every infinite sequence in a compact metric space has a convergent subsequence. Klaus Johannson, Topology I 80 . Topology I Proof. Assume for a moment the converse, i.e., assume there is an infinite sequence xn with no convergent subsequence. We get a contradiction in two steps. Step 1. In this step we form the set A := {xn | n ∈ N } of all sequence elements. W.l.o.g. this set A is infinite (otherwise we find a constant subsequence which is convergent) and has no limit point x (otherwise we find a convergent subsequence through choices xm ∈ 1 A ∩ U (x, m )). The first property will be important in a moment. The second property means that for every point x not in A there is a neighborhood which does not meet A. In other words X − A is open and so A is closed and so it is compact. Step 2. No element a ∈ A is a limit point either. Thus for every such A there is a neighborhood U (a) which intersects A only in the point a. In other words A has a covering by sets each containing exactly one point of A. Now, the covering above has a finite subcovering since A is compact. It follows that A is finite. Contradiction. ♦ Klaus Johannson, Topology I §5 Compact Spaces 81 For us the above property is one of the characterizing properties for compact spaces. In fact, it is hard to think of any space that does have this property and is not compact. Amazingly, such examples exist but they are pretty weird (see e.g. [Munkres]). Another characterizing property is the finite intersection property. We say a collection C of subsets of a space X satisfies the finite intersection property if \ Ci 6= ∅, 1≤i≤n for every finite subcollection {C1 , . . . , Cn } ⊂ C. Example. A nested sequence C1 ⊃ C2 ⊃ C3 ⊃ . . . of non-empty sets is a typical example of a collection with the finite T intersection property. Note that the intersection i Ci may be empty. Here is an example. Let X be the open unit-interval X = (0, 1) (with the subspace topology). This is a non-compact space. The sets Ci := T (0, 1/i) are closed in (0, 1) and their intersection i (0, 1/i) is empty. 5.11. Theorem. Let X be a topological space. Then X is compact if and only if \ C 6= ∅, C∈C Klaus Johannson, Topology I 82 . Topology I for all collections C of closed subsets of X with the finite intersection property. Proof. X is compact ⇔ no collection O of open sets can cover X, if no finite subcollection of O covers X. ⇔ S if C is a collection of closed sets, then, if 1≤i≤n (X − Ci ) 6= S X, for all finite subcollections C1 , . . . , Cn ∈ C, then C∈C (X − C) 6= X. ⇔ T if C is a collection of closed sets, then, if 1≤i≤n Ci 6= ∅, for all finite subcollections C1 , . . . , Cn ∈ C, then T C∈C C 6= ∅. ♦ Compact Spaces and Continuous Maps. Compact spaces are well behaved with respect to continuous maps. Klaus Johannson, Topology I §5 Compact Spaces 83 5.12. Theorem. Let f : X → Y be a continuous map between topological spaces. Suppose X is compact and suppose f is surjective. Then Y is compact. Proof. Let O be an open cover of Y . Then f −1 (O) := { f −1 (U ) | U ∈ O } is an open cover of X since f is continuous. Since X is compact, there is a finite subcover. particular, S In there are U1 , . . . , Un ∈ O with X ⊂S i f −1 Ui . Since f is surjective, it follows that Y = i Ui . Hence O has a finite sub-cover. ♦ 5.13. Theorem. Let X be a topological space and let f : X → R be a continuous map. Suppose X is compact. Then there are real numbers a, b ∈ R with a, b ∈ f (X) and a ≤ f (x) ≤ b, for all x ∈ X. Proof. The image f (X) is compact since X is compact. It follows that f (X) ⊂ R must be bounded from above and from below. Moreover, f (X) is closed as a compact subset of the Hausdorff space R. Hence Klaus Johannson, Topology I 84 . Topology I it contains its least upper bound, say b, as well as its greatest lower bound, say a. In particular, f (X) ⊂ [a, b]. ♦ 5.14. Theorem. Any continuous bijection f : X → Y between compact spaces X, Y is a homeomorphism. Proof. We have to show that the inverse map f −1 is continuous. Since f is bijective, this is the same as saying that f is an open map in the sense that it maps open subsets to open subsets. Of course, a continuous bijection is an open map if and only if it is a closed map in the sense that it maps closed sets to closed sets. To continue the proof, let A ⊂ X be any closed subset of X. Then A is compact as a closed subset of a compact space. Hence the image f (A) is compact and so closed as a compact subset of a Hausdorff space. ♦ Remark. The proof allows the somewhat stronger statement that a continuous bijection f : X → Y is a homeomorphism if X is quasi-compact and Y is Hausdorff. Klaus Johannson, Topology I §5 Compact Spaces 85 We next establish an important property of compact spaces, namely the Urysohn lemma. Actually, the property is valid for normal spaces which are more general than compact spaces. 5.15. Theorem. Let X be a normal space (e.g., a compact space) and let A, B ⊂ X be two disjoint and closed subspaces of X. Then there is a continuous map h : X → [0, 1] with h(A) = 0 and h(B) = 1. Proof. We view h as a height function. The idea is to approximate this height function by a sequence of step functions. We visualize the process in X × R, i.e., we approximate the graph of the height function. The steps of the step function are given by Up × p ⊂ X × R, where Up ⊂ X are open subsets in X, indexed by rational numbers p, with Ūq ⊂ Up , iff p < q. Klaus Johannson, Topology I 86 . Topology I To any such collection of steps we associate a step function hn (x) := max { p | x ∈ Up }, where the maximum is taken over all the finitely many steps constructed so far. Here is an illustration of such a step function: h(B) 1 q q r p h(A) p 0 Up Uq Up Ur Uq Given a step function as above, we refine it as follows. First, we look for two adjacent steps Up × p, Uq × q and Uq ⊂ Up whose distance p − q is maximal among the distances of all pairs of adjacent steps. Since Ūq ⊂ Up there is an open subset U with Ūq ⊂ U ⊂ Ū ⊂ Up . For this subset U we choose the index r := 12 (p + q) and set Ur := U . Then the step Ur × r lies between Uq × q and Up × p. Klaus Johannson, Topology I §5 Compact Spaces 87 This finishes the construction. The step functions converge to the map h(x) = sup { p | x ∈ Up }, where this time the supremum is taken over an infinite collection of indices. Clearly h(A) = 0 and h(B) = 1. It therefore remains to show that h is a continuous map. So let x ∈ X be any point and let (c, d) ⊂ R be any open interval containing h(x). We are asked to find an open neighborhood U of x with h(U (x)) ⊂ (c, d). For this choose indices c < p < h(x) < q < d. They clearly exist. We claim U := Up − Ūq is one of the desired neighborhoods. First, U is open since Up is open and Ūq is closed. Moreover, x ∈ Up , for otherwise h(x) ≤ p. Similarly, x 6∈ Ūq , for otherwise h(x) ≥ q. Hence x ∈ U . Finally, h(U ) ⊂ (c, d). To see this let h(z) ∈ f (U ). Then z ∈ U = Up − Ūq . Since z ∈ Up , we have h(x) ≥ p. Since z 6∈ Ūq , implies z 6∈ Uq and so h(z) ≤ q. Thus h(z) ∈ [p, q] ⊂ (c, d). Klaus Johannson, Topology I 88 . Topology I This finishes the proof. ♦ Appendix: Compactifications. A non-compact space can often be compactified. Sometimes the compactification is unique but often it is not. In the last case it is a challenge to find a compactification (if there is one at all) that really reflects the topological properties of the space at infinity. We will give examples to illustrate the problem. First, the definition. 5.16. Definition. Let Y be a compact Hausdorff space and let X ⊂ Y be a subspace. We say Y is a compactification of X if Y = X̄. As examples we consider the one-point compactification and two compactifications for infinite trees. One-Point Compactification. Every locally-compact space can be compactified through adding a single point. Here a space X is called locally-compact if every point x ∈ X has an open neighborhood which is contained in a compact subset. Equivalently, every point x ∈ X has an open Klaus Johannson, Topology I §5 Compact Spaces 89 neighborhood U whose closure Ū is compact. Of course, every compact space is locally compact. We need local-compactness in the following theorem to make sure that the compactification is Hausdorff. 5.17. Theorem. Let X be a locally-compact Hausdorff space. Then X has a one-point compactification in the sense that there is a topological space X + with X ⊂ X + and X + − X a single point. Moreover, any two compact spaces Y, Z are homeomorphic if X ⊂ Y, Z and if Y − X, Z − X is a single point. Proof. (Idea) Let ∞ be any symbol (not contained in X). Then define X + := X ∪ {∞}. Define a topology on X + by declaring a subset U ⊂ X + to be open if (1) U ⊂ X and U is open in X, or (2) X − U is a closed compact subset of X. It is easy to verify that the set of open sets forms a topology of X + which turns X + into a compact space. Furthermore, use local-compactness to show Klaus Johannson, Topology I 90 . Topology I that X + is a Hausdorff space. To show uniqueness verify that the identity id : X → X extends to a homeomorphism Y → Z. ♦ Remark. We did not spend much time proving the above theorem because the one-point compactification is not really that useful. It is too insensitive. It destroys all the delicate features that a non-compact space often has at infinity. Compactifications of Trees. Let T be a tree which is infinite but locally finite, i.e., every vertex is end-point of only finitely many edges. Then T is locally compact and so has a onepoint compactification. But we will see that for applications this particular compactification is not very useful. Better compactifications can be obtained by looking at embeddings into the unit disk. There are two essentially different compactifications of T that one can obtain in this way - the connected compactification and the totally disconnected compactification. Here is the first construction. Klaus Johannson, Topology I §5 Compact Spaces 91 The embedding of T is described through a recursive process. Start with 6 equally spaced points on the boundary ∂D2 of the unit disk. Let T1 be the cone over three of those points with the other points in between its edges. Next, divide the intervals in the boundary ∂D2 by their midpoints and replace the tree T1 by the tree T2 on the right, Continue the process. The closure T̄ of this subspace in R2 is the union of T with the boundary ∂D of the unit disk D2 . Thus the boundary ∂T = T̄ − T = ∂D2 is connected. The boundary ∂T as well as the closure T̄ is a compact Hausdorff space. Thus T̄ is a compactification. To distinguish this compactification Klaus Johannson, Topology I 92 . Topology I from the next we say that T is compactificed by a connected boundary. Here is the second construction. The embedding of the tree T is again described through a recursive process. This time start by dividing the boundary circle into six equal intervals, three black intervals and three white intervals. Let T1 be the cone over the mid-points of the black intervals. Next, take out the middle third of every black interval and replace T1 by the tree T2 on the right with endpoints in the middle of the black intervals. Continue the process ad infinitum. In the end the tree can be viewed as a subspace of the unit disk. The closure T̄ of this subspace is the union of T with the Cantor set. Thus the boundary Klaus Johannson, Topology I §5 Compact Spaces 93 ∂T = T̄ − T = C is totally disconnected. The boundary ∂T as well as the closure T̄ is a compact Hausdorff space. Thus T̄ is a compactification. We say that T is compactificed by a totally disconnected boundary. Remark. We have carried out the construction for a homogeneous infinite tree of valence three. An easy modification yields compactifications for other trees as well. Here are two examples that indicate how both constructions come up in concrete situations. 5.18. Example. Recall the group SL2 Z of all 2 × 2matrices with entries in Z and determinant 1. We know that it acts on the simplicial complex ∆(Q), where Q = {(x, y) ∈ Z2 | gcd(x, y) = ±1 }. Next we take the quotient under the antipodal map (a, b) 7→ (−a, −b) and we get the quotient complex on the right: Klaus Johannson, Topology I . Topology I 2 1 3 _13 _10 7 3 _ 1 2 _1 4 _ 3 _8 _ -_3 3 1 7 5 -_ -8 -_ 3 2 -_2 4 1 -_8 -_5 5 -_7 3 1 1_2 1 4 1 1 -_ -2 -_ 5 4 _6 5 5 _ _9 4 7 _4 3 _11 _7 _10 8 5 7 _3 2 13 11 _ 8_ _ 7 5 8 5_ 3 _ _ _ -1 7 -3 7 -2 _-5 -3 -4 _ _ -3 _ _ _ -2 _ -1 8 5 7 _ 7 5 8 3 3 2 _-4 5 _3 -5 4 9_ 1_2 7_ 5 7 4 2_ - 9_ _5 4 _ -1 1 _-4 7_ 3 _-7 3 2 -_3 5 5 5_ 5 _1 =-1 _ 0 0 _1 4 3 _ 1 1_ 3_ 4 _4 3 11 _ _5 8 1 3 _2 5 _5 _3 _4 12 7 9 _1 2 3_ 5 5_ 4_ 7_ 9 7 12 2_ 7_ 8_ 5_ 11 13 8 7_ 8_ 5_ 11 7 0 1 5 6 _5 9 3 _1 9 2_ 5 _7 1_ 6 _4 1_ 1 _1 0_ _3 11 _2 7 _3 10 1 94 4 3 The complex on the right is obtained from the identification (±a, ±b) ↔ ab of the lattice points from Q/{±1} with the rational numbers. The rational numbers Q ⊂ R are then mapped under the stereographic projection into S 1 and two rational numbers a c b , d are joined by some circular arc iff (a, b), (c, d) are joined by some 1-simplex in ∆(Q). Now, let lim(Q) = {x ∈ S 1 | there is a sequence xn ∈ Q with xn → x be the so called limit set of the action of PSL2 Z := SL2 Z/{±id}. Then Q ⊂ S 1 as indicated in the picture on the right. The edges of the dual complex of ∆(Q) form an infinite tree, say T . The closure of this tree is a compact space. It is the compactification of T with a connected boundary. The group PSL2 Z acts on T . Moreover, the action extends continuously Klaus Johannson, Topology I §5 Compact Spaces 95 to an action on the boundary ∂T . The fixpoints on ∂T of a matrix from PSL2 (Z) correspond to the eigenlines of the matrix. Thus the induced action of PSL2 Z on the boundary contains important information about matrices. 5.19. Example. Consider the Cayley tree of the free group F2 on two letters a, b, say. This is an infinite, homogeneous tree of valence four. A modification of the above construction yields the compactifications with connected and disconnected boundaries, respectively. The action of the free group F2 on its Cayley tree extends continuously on both boundaries. However, this is different for the action of the entire automorphism group Aut(F2 ). To see this consider the compactification with connected boundary: B a b B b A a a A b B x Klaus Johannson, Topology I 96 . Topology I For simplicity, we write A = a−1 , B = b−1 for the inverses. Notice that the point x is approached by two edge paths. One starting with A and the other with b. It follows that the automorphism ϕ : F2 → F2 defined by na → 7 a b 7→ B maps those two edge paths to edge paths with different end-points. This means that ϕ does not extend to a continuous map on the connected boundary since it maps pairs of points that are close together to pairs of points that are far apart (in fact, ϕ fixes one of the two halfs given by the broken line and rotates the other half). However, one can verify that ϕ extends to the totally disconnected boundary given by the second construction. It therefore would appear that adding the totally disconnected boundary to the tree is a better idea when it comes to studying the automorphisms of the free group. We will return to this problem later. Klaus Johannson, Topology I