2 SEPARATION AXIOMS Definition 2.1 A space X is a T0 space iff it satisfies the T0 axiom, i.e. for each x, y ∈ X such that x 6= y there is an open set U ⊂ X so that U contains one of x and y but not the other. Obviously the property T0 is a topological property. An arbitrary product of T0 spaces is T0. Discrete spaces are T0 but indiscrete spaces of more than one point are not T0. Definition 2.2 A space X is a T1 space or Frechet space iff it satisfies the T1 axiom, i.e. for each x, y ∈ X such that x 6= y there is an open set U ⊂ X so that x ∈ U but y∈ / U. T1 is obviously a topological property and is product preserving. Every T1 space is T0. Example 2.3 The set {0, 1} furnished with the topology {∅, {0}, {0, 1}} is called Sierpinski space. It is T0 but not T1. Proposition 2.4 X is a T1 space iff for each x ∈ X, the singleton set {x} is closed. Proof. Easy. Definition 2.5 A space X is a T2 space or Hausdorff space iff it satisfies the T2 axiom, i.e. for each x, y ∈ X such that x 6= y there are open sets U, V ⊂ X so that x ∈ U, y ∈ V and U ∩ V = ∅. T2 is a product preserving topological property. Every T2 space is T1. Example 2.6 Recall the cofinite topology on a set X defined in Section 1, Exercise 3. If X is finite it is merely the discrete topology. In any case X is T1, but if X is infinite then the cofinite topology is not T2. Proposition 2.7 Let f, g : X → Y be maps with Y Hausdorff. Then {x ∈ X / f(x) = g(x)} is closed. Proof. Let A = {x ∈ X / f(x) 6= g(x)}, and suppose x ∈ A. Since f(x) 6= g(x), there are −1 −1 open sets U, V ⊂ Y so that f(x) ∈ U, g(x) ∈ V and U ∩ V = ∅. Let W = f (U) ∩ g (V ). Then W is open and x ∈ W . Moreover, W ⊂ A. Thus A is open, so {x ∈ X / f(x) = g(x)} is closed. In particular if X is T2 and f : X → X is a map then the fixed-point set of f [i.e. the set of points x for which f(x) = x] is closed. Definition 2.8 A space X is regular iff for each x ∈ X and each closed C ⊂ X such that x∈ / C there are open sets U, V ⊂ X so that x ∈ U, C ⊂ V and U ∩ V = ∅. A regular T1 space is called a T3 space. The properties T3 and regular are both topological and product preserving. Every T3 space is T2. 2 Example 2.9 The slit disc topology on R is T2 but not regular, hence not T3. 2 2 2 Take X = R and let P = {(x, y) ∈ R / x 6= 0} and L = {(0, y) ∈ R }. Topologise X as follows (c.f. Theorem 1.12): 2 • if z ∈ P , let the open discs in R centred at z form a basis of neighbourhoods of z; 11 • if z ∈ L let the sets of the form {z} ∪ (P ∩ D) form a basis of neighbourhoods of z where D is an open disc in the plane centred at z. Clearly X is T2. However, note that L is a closed subset of X and that as a subspace, L is discrete. Thus any subset of L is closed in X; in particular, L − {(0, 0)} is closed and does not contain (0, 0), although every open set containing (0, 0) meets every open set containing L − {(0, 0)}. Thus X is not regular and hence not T3. Example 2.10 Every indiscrete space is vacuously regular but no such space (of more than 1 point!) is T0 and hence also no such space is T2. Theorem 2.11 A space X is regular iff for each x ∈ X, the closed neighbourhoods of x form a basis of neighbourhoods of x. Proof. ⇒: given x ∈ X, and a neighbourhood N of x, there is an open set O ⊂ X such that x ∈ O ⊂ N. Consider the point x and the closed set X − O, which does not contain x. By regularity, there are open sets U and V such that x ∈ U, X − O ⊂ V and U ∩ V = ∅. Thus x ∈ U ⊂ X − V ⊂ O ⊂ N, so X − V is a closed neighbourhood of X contained in the given neighbourhood N of x. ⇐: given x ∈ X and the closed set C ⊂ X − {x}, since X − C is open and contains x, there is a closed neighbourhood N of x so that N ⊂ X − C. Let V = X − N. Then V is open and C ⊂ V . Since N is a neighbourhood of x, there is an open set U such that x ∈ U ⊂ N. Then U ∩ V ⊂ N ∩ (X − N) = ∅, so X is regular. Definition 2.12 A space X is normal iff for each pair A, B of disjoint closed subsets of X, there is a pair U, V of disjoint open subsets of X so that A ⊂ U, B ⊂ V and U ∩ V = ∅.A normal T1 space is called a T4 space. The properties T4 and normal are both topological properties but, perhaps surprisingly, are not product preserving. Every T4 space is clearly a T3 space, but it should not be surprising that normal spaces need not be regular. Example 2.13 Sierpinski’s space is vacuously normal but is not regular since 0 and {1} cannot be separated. 2 Example 2.14 The tangent-discs topology on R is T3 but not normal, hence not T4. Let X, P and L be as in Example 2.9 but topologise X as follows: 2 • if z ∈ P , let the open discs in R centred at z form a basis of neighbourhoods of z; • if z ∈ L let the sets of the form {z} ∪ D form a basis of neighbourhoods of z where D is the union of two open discs in P tangent to L at z, one to the left of L and the other to the right of L. Clearly X is T1. X is regular, hence T3. Indeed, suppose z ∈ X and N ∈ N (z). If z ∈ P , then there is an open disc D centred at z so that z ∈ D ⊂ N. The closed disc of half the radius also centred at z is a closed neighbourhood of z and lies in N. If z ∈ L, then there is D as in the definition of the neighbourhood basis of z so that {z} ∪ D ⊂ N. The union of the closed discs of half the radii of those of D also tangent to L at z is a closed neighbourhood of z contained in N. Thus in either case, closed neighbourhoods form a neighbourhood basis so by Theorem 2.11, X is regular. 12 Suppose that X is normal. Let M = {(x, y) / x, y ∈ Q}. As in Example 2.9, every subset of L is a closed subset of X. Thus to each A ⊂ L, we can assign open sets UA and VA of X so that A ⊂ UA, L − A ⊂ VA and UA ∩ VA = ∅. Let MA = M ∩ UA. We show that the function sending A to MA is injective. This will lead to a contradiction because L is uncountable whereas M is countable, so it is impossible to find an injective function from the power set of L to that of M. Loosely speaking, if X were normal then there would need to be more open subsets of X than there are. To show that the function constructed in the last paragraph really is injective, let A, B ⊂ L be such that A 6= B: to show that MA 6= MB. Either A − B 6= ∅ or B − A 6= ∅: assume the former. We have ∅ 6= A − B = A ∩ (L − B) ⊂ UA ∩ VB. Since UB and VB are disjoint and open, we must have UB ∩ VB = ∅, so VB ⊂ X − UB, and hence ∅ 6= UA ∩ VB ⊂ UA − UB. Every non-empty open subset of X meets M so, since UA −UB is open, we have: ∅ 6= M ∩(UA −UB) ⊂ M ∩ (UA − UB) = MA − MB. Thus if A − B 6= ∅ then MA − MB 6= ∅. Similarly, if B − A 6= ∅ then MB − MA 6= ∅, so the function is injective as promised. Example 2.15 Let S be the real line with the right half-open interval topology of Example 1.4. Then S is normal but S × S is not normal. The space S × S is sometimes called Sorgenfrey’s square. Suppose that A and B are disjoint subsets of S. Since B is closed, for each a ∈ A, there 0 0 0 is a > a so that [a, a ) ∩ B = ∅. Let U = ∪a∈A[a, a ). Then U is an open set containing A. 0 0 0 Similarly define V = ∪b∈B[b, b ), where for each b ∈ B, b > b is chosen so that [b, b ) ∩ A = ∅. V too is open and contains B. It is claimed that U ∩ V = ∅, for suppose not. Then there 0 0 0 0 are a ∈ A, b ∈ B such that [a, a ) ∩ [b, b ) 6= ∅. Thus either b ∈ [a, a ) or a ∈ [b, b ), so either 0 0 0 0 [a, a ) ∩ B 6= ∅ or [b, b ) ∩ A 6= ∅, contrary to the choice of a and b .