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Proposition 4.17. For any subset S of any space X, the exterior, interior,and boundary of S are disjoint, and their union is X.Proof. If p ∈ S◦, then some open neighborhood of p is a subset of S. Soclearly p is not in ∂S or in the exterior of S. Similarly, if p is in the exteri orof S, then some open neighborhood of p is disjoint from S, and p cannot bein ∂S.Now suppose p ∈ X and p is not in ∂S. Then some open neighborhoodof p is either disjoint from S or a subset of S. In the first case, p is in theexterior of S, and in the second case, it is in S◦.Proposition 4.18. For any subset S of any space X, we have¯S = ∂S ∪ S◦and ∂S =¯S\S◦.Proof. A point p is in the exterior of S if and only if there is an open set Uwith p ∈ U and U ∩ S = ∅. This is true if and only if there is a closed setA = X\U with p /∈ A and S ⊂ A. This is true if and only if p is not in¯S.So therefore¯S = ∂S ∪ S◦, the complement of the exterior of S.Then¯S\S◦is ∂S.Proposition 4.19. The following are true:• A subset S of X is open if and only i f S◦= S.• A subset S of X is clos ed if and only i f ∂S ⊂ S.• A subset S of X is open if and only i f ∂S ∩ S = ∅.The proofs are straightforward.There is a rich interplay between these set operations. In fact, just u si n gthe operations of set complementation and set closure , one can derive asmany as 14 different sets from a subset of a space. We will explore this inthe second homework assignment.4.4 Comparing topologies on the same point setDefinition 4.20. Fix a set X and let S, T be two different topologies onX. Then (X, S) is coarser than (X, T ) if S ⊂ T . In the same sit u at ion ,(X, T ) is finer than (X, S).To restate: for (X, T ) to be finer than (X, S), we need that every openset in S is also open in T , but possibly some other sets are open as well.The r el at i on ‘c oars er ’ is refl ex i ve, transitive relation on the set of alltopologies on a point set. The discrete topology is the finest topology on14a set, and the indiscrete topology is the coarsest topology on a set. Thediscrete and indiscrete topologies c an b ot h be compared to any t opology onthe set. As the following example shows, not all topologies are comparable.Example 4.21. Let X be a set and let p ∈ X. The particular point topologywith respect to p is the top ol ogy on X where a set is open if and only ifit contains p. Let p 6= q ∈ X and let T (p) and T (q) be the respectiveparticular point topologies. Then T (p) 6⊂ T (q) and T (q) 6⊂ T (p); neithertopology is coarser than t h e other.Many topologies on a space can be characterized as the coarsest topologywith a particular p roperty.Example 4.22. • Let X be a space and let S ⊂ X. Then the subspacetopology is the coarsest topology for which t he i n cl u si on S ֒→ X iscontinu ous .• The topology of finite complements is the coarsest topology in whichpoints are closed.• The topology generated by a given subbasis B is th e coarsest topologyin which every set in B is open.5 Topologies from other structure5.1 Order topologiesThe standard topology on R is a special case of the following general con-struction:Definition 5.1. A strict total ordering on a set X is a relation < satisf yi n g:• for all x, y ∈ X with x 6= y, we have x < y or y < x,• for all x, y, z ∈ X with x < y an d y < z, we have x < z, and• for all x ∈ X, we have x 6< x.A set with a strict total ordering is an ordered set.Just like in the real line, we have intervals in any ordered set. Define(x, y) = {z ∈ X|x < z < y}, and define closed and half-closed intervals to bethe union of the open interval with the indicated endpoints. We also havethe intervals (−∞, x) and (x, ∞) (these are sometimes called open rays).15If X has a strict total ord er i n g <, then the order topology on X is thetopology with a basis given by open intervals (x, y) for x, y ∈ X, along withhalf-infinite open intervals (−∞, x) and (x, ∞) for x ∈ X.Since any fin i te intersection of intervals is either empty or an interval,this set is actuall y a basis for a topology and not just a subbasis.5.2 Countable and uncountable setsRecall that a set is countable if it is finite or can be put in bijection with N.A set is uncountable if it is infinite and cannot be put in bijection with N.Int u i ti vely, countability of sets relates to the amount of informationneeded to specify a member of a set. If there is a universal bound on thenu mb er of bits needed to specify a member of the set S, then S is finite.If there is no universal bound on the number of bits needed to specify amember of the set S, but every member of S can be specified with a finitenu mb er of bits, then S is countable . If most members of the set S requirean infinite number of bits t o specify, then S is uncountable. (The prec ed in gremarks are not so much rigorous results as they are intuitive guidelines.)We put together some resul t s from set theory and analysis:Proposition 5.2. • Subsets and quotients of countable sets are count-able.• Unions of countable sets, indexed over a countable set, are countable.• The Cartesian product of a family of finitely many countable sets iscountable.• Power sets of infinite sets are uncountable.• The rational numbers Q are countable, but the real numbers R areuncountable.5.3 A pathological exampleDefinition 5.3. A well ordering on a set X is a s tr i ct total order suchthat every nonempty subset of X cont ai n s a minimal element. An initialsegment of a well ordered set is a subset that contain s all the predecessorsof any element it contains.The following theorem in set theory depends on the axiom of choice. Wequote it without proof. Halmos’s Naive Set Theory i s an excellent book andcontains this theorem.16Theorem 5.4 (Well-ordering theorem ). Every set is the domain of a wellordering.From this can deduce the followin gProposition 5.5. There is an uncountable, well ordered set (X, <), suchthat every proper initial segment of X i s count able.Proof. Suppose (Y, <) is uncou ntable and well or de re d and let 0 denotethe minimal element of Y …


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CALTECH MA 109A - Lecture notes

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