MA651 Topology. Lecture 6. Separation Axioms.This text is based on the following books:• ”Fundamental concepts of topology” by Peter O’Neil• ”Elements of Mathematics: General Topology” by Nicolas Bourbaki• ”Counterexamples in Topology” by Lynn A. Steen and J. Arthur Seebach, Jr.• ”Topology” by James DugundgjiI have intentionally made several mistakes in this text. The first homework assignment is to findthem.36 Separation Axioms.So far, our only requirements for a topology has been that it satisfy the axioms. From now on,we will impose increasingly more severe additional conditions on it. With each new condition, wewill determine the invariance properties of the resulting topology: by this we mean:• Whether the topology is invariant under open or closed maps rather than only homeomor-phisms.• Whether the additional properties are inherited by each subspace topology.• Whether the additional properties are transmitted to cartesian products.In this lecture we will require of a topology that it ”separate” varying types of subsets. Theseparation axioms Tistipulate the degree to which distinct points or closed sets may be separatedby open sets. These axioms are statements ab out the richness of topology. They answer questionslike, ”Are there enough open sets to tell points apart?” and ”Are there enough open sets to tellpoints from closed sets?”.Definition 36.1. (Tiaxioms). Let (X, T ) be a topological space.1T0axiom : If a, b are two distinct elements in X, there exists an open set U ∈ T such that eithera ∈ U and b 6∈ U, or b ∈ U and a 6∈ U (i.e. U containing exactly one of these points).T1axiom : If a, b ∈ X and a 6= b, there exist open sets Ua, Ub∈ T containing a, b respectively, suchthat b 6∈ Ua, and a 6∈ Ub.T2axiom : If a, b ∈ X, a 6= b, there exist disjoint open sets Ua, Ub∈ T containing a, b respectively.T3axiom : If A is a closed set and b is a point in X such that b 6∈ A, there exist disjoint open setsUA, Ub∈ T containing A and b respectively.T4axiom : If A and B are disjoint closed sets in X, there exist disjoint open sets UA, UB∈ Tcontaining A and B respectively.T5axiom : If A and B are separated sets in X, there exist disjoint open sets UA, UB∈ T containingA and B respectively.If (X, T ) satisfies a Tiaxiom, X is called a Tispace. A T0space is sometimes called a Kolmogorovspace and a T1space, a Fr´echet space. A T2is called a Hausdorff space.Each of axioms in Definition (36.1) is independent of the axioms for a topological space; in factthere exist examples of topological spaces which fail to satisfy any Ti. But they are not indepen-dent of each other, for instance, axiom T2implies axiom T1, and axiom T1implies T0.More importantly than the separation axioms themselves is the fact that they can be employedto define successively stronger properties. To this end, we note that if a space is both T3and T0it is T2, while a space that is both T4and T1must be T3. The former spaces are called regular,and the later normal.Specifically a space X is said to be regular if and only if it is both a T0and a T3space; to benormal if and only if it is both a T1and T4space; to be completely normal if and only if it is botha T1and a T5space. Then we have the following implications:Completely normal ⇒ Normal ⇒ Regular ⇒ Hausdorff ⇒ T1⇒ T0The use of terms ”regular” and ”normal” is not uniform throughout the literature. While someauthors use these terms interchangeably with ”T3space” and ”T4space” respe ctively, others(especially in Russian textbooks) refer to our T3space as a ”regular” space and vice versa, andsimilarly permute ”T4space” and ”normal”. This allows the succ essively stronger properties tocorrespond to increasing Tiaxioms.Example 36.1. An antidiscrete space X containing more than one point does not satisfy to axiomT0.2Example 36.2. Sierpinski space, X = {0, 1} and T = {Ø, {0}, {0, 1}}, satisfies the axiom T0, butdoes not satisfy T1: there does not exist an open set U containing the point 1 and not containingthe point 0.Example 36.3. This example is important for algebraic geometry. Let A be a commutative ring(i.e. A has two binary operations, such that addition makes A a commutative group (i.e. a setwith a binary associative and commutat ive operation such that the operation admits an identityelement and each element of the set has an inverse element for the operation.) and multiplicationis associative and distributes over addition and the commutative law also holds for multiplication(a × b = b × a.) with a unit and X the set of all prime ideals of A (i.e. such ideals which have thefollowing two properties: whenever a, b are two elements of A such that their product a × b lies inthe prime ideal P , then a is in P or b is in P , and at the same time P is not equal to the wholering A). For any a ∈ A, let Xadenote the set of all prime ideals in A which do not contain a. Itis clear that Xa∩ Xb= Xabfor all a, b ∈ X, X0= Ø and X1= X. Consequently, the collectionB = {Xa| a ∈ A} is a base of a topology T on A. This topology is called the spectral or Zariskitopology.The topological space (X, T ) is called the prime spectrum of the ring A and is denoted Spect(A).The closure of a one point set {x} in Spec(A) consists of all prime ideals y ∈ X =Spec(A)containing x. It follows that the space (X, T ) satisfies the separation axiom T0, but not T1, sincethe only closed points in X are the maximal ideals of the ring A.Example 36.4. Let X be an infinite set and let the topology T consist of the empty set and allsubsets of X whose complements are finite. Any two nonempty open sets in this set intersect. Atthe same time all one point sets in (X, T ) are closed. Hence the space (X, T ) satisfies axiom T1but nit axiom T2.37 Hausdorff Spaces.Theorem 37.1. Let (X, T ) be a topological space. Then the following statements are equivalent:1. (T2axiom). Any two distinct points of X have disjoint neighborhoods.2. The intersection of the closed neighborhoods of any point of X consists of that point alone.3. The diagonal of the product space X × X is a closed set.4. For every set I, the diagonal of the product space Y + XIis closed in Y .5. No filter on X has more than one limit point.6. If a filter F on X converges to x, then x is the only cluster point of F .3Proof. We will proof the implications:(1) ⇒ (2) ⇒ (6) ⇒ (5) ⇒ (1)(1) ⇒ (4) ⇒ (3) ⇒ (1)(1) ⇒ (2) : If x 6= y