3. Hausdorff Spaces and Compact Spaces 3.1 Hausdorff Spaces Definition A topological space X is Hausdorff if for any x, y X with x = y there ∈ 6 exist open sets U containing x and V containing y such that U V = . ∅ T n (3.1a) Proposition Every metric space is Hausdorff, in particular R is Hausdorff (for n 1). ≥ Proof Let (X, d) be a metric space and let x, y X with x = y. Let r = d(x, y). Let ∈ 6 U = B(x; r/2) and V = B(y; r/2). Then x U, y V . We claim U V = . If not there ∈ ∈ ∅ exists z U V . But then d(x, z) < r/2 and d(z, y) < r/2 so we get ∈ T T r = d(x, y) d(x, z) + d(z, y) < r/2 + r/2 ≤ i.e. r < r, a contradiction. Hence U V = and X is Hausdorff. ∅ T Reamrk In a Hausdorff space X the subset x is closed, for every x X. To see this { } ∈ let W = C ( x ). For y W there exist open set U ,V such that x U , y V and X { } ∈ y y ∈ y ∈ y Uy Vy = . Thus Vy W and W = y W Vy is open. So CX (W ) = x is closed. ∅ ⊂ ∈ { } T S Exercise 1 Suppose (X, ) is Hausdorff and X is finite. Then is the discrete topology. T T Proof Let x X. Then x is closed. If Z = x , x , . , x is any subset of X then ∈ { } { 1 2 m} Z = x x x is closed. So all subsets are closed and hence all subsets are open { 1} { 2} { n} and X hasS the discreteS topology. 34 Exercise 2 Let X be an infinite set and let be the cofinite topology on X. Then T (X, ) is not Hausdorff. T Lets suppose it is and derive a contradiction. Pick x, y X with x = y. Then there ∈ 6 exists open sets U, V such that x U, y V and U V = . Since X has the cofinite ∈ ∈ ∅ topology and U, V are nonempty, C (U) and C (V ) are finite. But then X = C ( ) = X X T X ∅ CX (U V ) = CX (U) CX (V ) is finite - contradiction. T S (3.1b) Let X be a Hausdorff space and let Z X. Then Z (regarded as a topological ⊂ space via the subspace topology) is Hausdorff. Proof Let x, y Z. Since X is Hausdorff there exist open sets U, V in X such that ∈ x U, y V and U V = . But then U ∗ = U Z and V ∗ = V Z are open in (the ∈ ∈ ∅ subspace topology on) Z, moreover x U , y V also T ∈ ∗ ∈ ∗T T U ∗ V ∗ = U Z V Z = (U V ) Z = . ∅ \ \ \ \ \ \ Hence Z is Hausdorff. (3.1c) Proposition Suppose that X, Y are topological spaces that X is homeomorphic to Y and Y is Hausdorff. Then X is Hausdorff. Proof Let f : X Y be a homeomorphism. Let x , x X with x = x . Then → 1 2 ∈ 1 6 2 f(x ), f(x ) Y and f(x ) = f(x ) (as f is a homeomorphism, in particular it is a 1 1- 1 2 ∈ 1 6 2 − map). By the Hausdorff condition there exist open sets V ,V of Y such that f(x ) V , 1 2 1 ∈ 1 f(x ) V and V V = . But now x f 1V , x f 1V and f 1V f 1V = 2 ∈ 2 1 2 ∅ 1 ∈ − 1 2 ∈ − 2 − 1 − 2 f 1(V V ) = f 1( ) = . Hence X is Hausdorff. − 1 2 − T∅ ∅ T T Definition Suppose is a property which a topological space may or may not have P (e.g. the property of being Hausdorff). We say that is a topological property if whenever P X, Y are homeomorphic topological spaces and Y has the property then X also has the P property . P So we may re-cast (3.1c) as: (3.1c)’ Hausdorffness is a topological property. 35 3.2 Compact Spaces How can we tell whether [0, 1] is homeomorphic to R? Find a topological property which [0, 1] has but R does not have. Definition Let X be a topological space. An open cover of X is a collection of open sets Ui i I such that X = i I Ui. { | ∈ } ∈ A subcover of an open cover U i I is an open cover of the form U j J , where {S i | ∈ } { j | ∈ } J is a subset of I. Examples 1. Let X = and let U = ( n, +n), for n = 1, 2,.... Then U n R n − { n | ∈ N} is an open cover of , i.e. = ∞ U . This is so because, for r we can choose a R R n=1 n ∈ R positive integer m greater than r and then r ( m, +m) = U , so r U ∞ U . |S| ∈ − m ∈ m ⊂ n=1 n Hence n∞=1 Un contains every real number, i.e. n∞=1 Un = R. S A subcover of this open cover is U n J where J is the set of even positive integers. S { n | ∈ } S Example 2 Let X = . Let U = ( , 0), U = (0, ), U = ( 1, 1), U = ( 4, 4), R 1 −∞ 2 ∞ 3 − 4 − U = ( 5, 5) and U = ( n, n), for n 4. Then U n is an open cover of and 5 − n − ≥ { n | ∈ N } R U ,U ,U is a subcover. { 1 2 3} Notice that in both Examples above X is given an open cover consisting of infinitely many sets. In Example 2 there is a finite subcover (a subcover consisting of finitely many sets) and in Example 1 there is not. Example 3 Let X = [0, 1] (with the subspace topology induced from R ). Let U1 = [0, 1/4) and U = (1/n, 1], for n = 2, 3, 4,.... Then U = [0, 1] ( 1/4, 1/4) is open in the n 1 − subspace topology and so is U = [0, 1] (1/n, 2), for n 2. Note that U n = 1, 2,... n ≥ T { n | } is an open cover and U ,U is a subcover. { 1 5} T Definition A topological space X is compact if every open cover of X has a finite subcover, i.e. if whenever X = i I Ui, for a collection of open sets Ui i I then we ∈ { | ∈ } also have X = i F Ui, for someS finite subset F of I. ∈ S (3.2a) Proposition Let X be a finite topological space. Then X is compact. 36 Proof Let X = x , x , . , x . Let U i I be an open cover of X. Then x { 1 2 n} { i | ∈ } 1 ∈ X = i I Ui so that x1 Ui1 for some i1 I. Similarly, x2 Ui2 for some i2 I, ..., ∈ ∈ ∈ ∈ ∈ x U , for some i I. n ∈ Sin n ∈ Let F = i1, i2, . , in . Then xr Uir i F Ui, for each r. Hence every x in X { } ∈ ⊂ ∈ belongs to i I Ui and so X = i F Ui, i.e. USi i F is a finite subcover of Ui i I . ∈ ∈ { | ∈ } { | ∈ } S S When is a subspace of a topological space compact? (3.2b) Lemma Let X be a topological space and let Z be a subspace. Then Z is compact if and only if for every collection Ui i I of open sets of X such that Z i I Ui there { | ∈ } ⊂ ∈ is a finite subset F of I such that Z i F Ui. S ⊂ ∈ S Proof ( ) Suppose Z is compact (regarding Z as a topological space with the subspace ⇒ topology). Let Ui i I be a collection of open sets of X with Z i I Ui. Then we { | ∈ } ⊂ ∈ have S Z = Z Z Z ( U ) = Z U . ⊂ i i i U i I \ \ [∈ [∈ \ On the other hand i I Z Ui Z so we have Z = i I Z Ui. Writing Vi = Z Ui ∈ ⊂ ∈ we thus have that all Vi are open in Z (in the subspace topology) and Z = i I Vi. By S T S T ∈ T compactness we therefore have Z = i F Vi for some finite subset F of I. NowSVi Ui so ∈ ⊂ we get S Z = V U i ⊂ i i F i I [∈ [∈ and Z i F Ui, as required. ⊂ ∈ ( ) NowS suppose that Z has the property that whenever Z i I Ui, for open sets Ui ⇐ ⊂ ∈ in X, there exists a finite subset F of I such that Z i F Ui.S We will show that Z is ⊂ ∈ compact. Let V i I be an open cover of Z. Thus each V is open in the subspace { i | ∈ } S i topology, so have the form Vi = Z Ui for some open set Ui in X. Now we have Z = i I Vi i I Ui. By the assumed propertyT we therefore have Z i F Ui for some finite ∈ ⊂ ∈ ⊂ ∈ Ssubset F ofSI. Hence we have Z = Z ( i F Ui) = i F Ui) = Si F Z Ui = i F Vi. ∈ ∈ ∈ ∈ Thus V i F is a finite subcover of U i I and we have shown that every open { i | ∈ } T {S i | ∈ } S S T S cover of Z has a finite subcover. Hence Z is compact. Is a subspace of a compact space compact ? The answer is generally no! We shall see that [0, 1] is compact, but on the other hand (0, 1) is not compact (e.g. (0, 1) = n∞=2 Un where U = (1/n, 1) but U n = 2, 3,... has no finite subcover).