POINT SET TOPOLOGY De¯nition 1 A topological structure on a set X is a family (X) called open sets and satisfying O ½ P (O ) is closed for arbitrary unions 1 O (O ) is closed for ¯nite intersections. 2 O De¯nition 2 A set with a topological structure is a topological space (X; ) O ; = 2;Ei = x : x Eifor some i = [ [i f 2 2 ;g ; so is always open by (O ) ; 1 ; = 2;Ei = x : x Eifor all i = X \ \i f 2 2 ;g so X is always open by (O2). Examples (i) = (X) the discrete topology. O P (ii) ; X the indiscrete of trivial topology. Of; g These coincide when X has one point. (iii) =the rational line. Q =set of unions of open rational intervals O De¯nition 3 Topological spaces X and X 0 are homomorphic if there is an isomorphism of their topological structures i.e. if there is a bijection (1-1 onto map) of X and X 0 which generates a bijection of and . O O e.g. If X and X are discrete spaces a bijection is a homomorphism. (see also Kelley p102 H). De¯nition 4 A base for a topological structure is a family such that B ½ O every o can be expressed as a union of sets of 2 O B Examples (i) for the discrete topological structure x x2X is a base. f g (ii) for the indiscrete topological structure ; X is a base. f; g (iii) For , topologised as before, the set of bounded open intervals is a base.Q 1 (iv) Let X = 0; 1; 2 f g Let = (0; 1); (1; 2); (0; 12) . Is this a base for some topology B f g on X? i.e. Do unions of members of X satisfy (O2)? (0; 1) (1; 2) = (1)- which is not a union of members of , so is not a\base for any topology on X. B B Theorem 1 A necessary and su±cient condition for to be a base for a 0 B 00 topology on X = o2Bo is that for each O and O and each x O0 O00 O [such that x O O0 O00. 2 B 2 \ 9 2 B 2 ½ \ Proof Necessary: If is a base for ; O0 O00 and if x O0 O00, since O0 O00 is a unionB of sets of O O \ suc2hOthat x O2 O\0 O00. \ B 9 2 B 2 ½ \ Su±cient: let be the family of unions of sets of . O B (O1) is clearly satis¯ed. (O2) ( Ai) ( Bj) = (Ai Bj) so that it is su±cient to prove that the intersection[ \ [of two [sets of\ is a union of sets of . B B Let x O0 O00. Then Ox such that x O x O0 O00 so that 0 200 \ 9 2 B 2 ½ \ O O = x2O0\O00 Ox. \ [ Theorem 2 If is a non-empty family of sets the family of their ¯nite S B intersections is a base for a topology on S [ Proof Immediate veri¯cation of O1 and O2. The topology generated in this way is the smallest topology including all the sets of . S De¯nition 4 A family is a sub base for a topology if the set of ¯nite intersections is a baseS for the topology. e.g. a a2Q and ( a) a2Q are sub bases for f 1g f ¡1 g Q De¯nition 5 If a topology has a countable base it satis¯es the second axiom of countability. De¯nition 6 In a topological space a neighbourhood of a set A is a set which contains an open set containing A. A neighbourhood of a point X is a neighbourhood of x . f g Theorem 3 A necessary and su±cient condition that a set be open is that it contains (is) a neighbourhood of each of its points. Proof Necessary: De¯nition of a neighbourhood Su±cient: Let OA = open subsets of A. OA is open (O ) and OA A. [ 1 ½ 2 If x A A a neighbourhood of x open set x therefore x OA 2 ⊃ ⊃ 3 2 therefore A OA. ½ Let V (x) denote the family of neighbourhoods of x. Then V (x) has the following properties. (V1) Every subset of X which contains a member of V (x) is a member of V (x) (V2) V (x) is closed for ¯nite intersections (V3) x belongs to every member of V (x). (V ) If v V (x) W V (x) such that v V (y) for all y W . 4 2 9 2 2 2 (Take W to be an open set x and V .) 3 ½ Theorem 4 If for each point x of X there is given a family V (x) of subsets of X, satisfying V ¡ then a unique topology on X for which the 1 4 9 sets of neighbourhoods of each point x are precisely the given V (x). (Hausdor®). Proof If such a topology exists theorem 3 shows that the open sets must be the O such that for each x O O V (x) and so there is at most one 2 2 such topology. Consider the set O so de¯ned. (O ) Suppose x O. Then x some O0 O0 V (x) O0 cupO so 1 2 [ 2 2 ½ j O V (x). [ 2 (O ) Suppose x F o x each O. each O V (x) therefore 2 2 \ ) 2 2 F O V (x) by V . \ 2 2 Now consider the system U(x) of neighbourhoods of x de¯ned by this topology. (i) U(x) V (x). Let U U(x). Then U O x. But O V (x) so by V½U V (x). 2 ⊃ 3 2 1 2 (ii) V (x) U(x). Let V V (x). It is su±cient to prove that O such that½ x O V2. 9 2 O 2 ½ Let O = y : V V (y) x O since V V (x). f 2 g 2 2 O V since y V for all y O by V . ½ 2 2 3 To prove that O it is su±cient to prove that V V (y) for all y O. 2 O 2 2 If t O V V (y) by de¯nition of O therefore W V (y) such 2 2 9 2 that V V (z) for all z W by V 4. 2 2 ¡ Therefore W O (de¯nition of O) ½ 3 Therefore O V (y) by V 2 1 e.g. R1 : V (x)= sets which contain interval (a; b) a < x < b. De¯nition 7 A metric for a set X is a function ½ form X X to R+ (non- negative reals) such that £ M1 ½(x; y) = ½(y; x) for all x; y M ½(x; z) ½(x; y) + ½(y; z) for all x; y; z 2 · M3 ½(x; x) = 0 for all x. This is sometimes called a pseudo metric and for a metric we have (M 0 ) ½(x; y) 0; = 0 x = y. 3 ¸ , De¯nition 8 The open r- Ball about x = y : ½(x; y) < r and is denoted by B(r; x) f g The closed r-ball around x = y : ½(x; y) R and is denoted by B(r; x). f · g 1 V (x) = sets which contain one of B n ; x n = 1; 2; : : : satis¯es V1¡4 and so de¯nesf a topology on X. This³ topology´ is the metricg topology de¯ned on X by ½. The development of topology from the neighbourhood point of ve=iew is due to H. Weyl and Hausdor®. That from the open sets aspect is due to Alexandro® and Hopf. De¯nition 9 The closed sets G of a topological space are the complements of the open sets. (G1) G is closed for arbitrary intersections (G ) G is closed for ¯nite unions. and X are both closed and open. 2 ; Clearly given a family G satisfying G1 and G2 the family of comple- ments is a topology for which the closed sets are the sets of G. De¯nition 10 a point x is an interior point of a set A if A is a neighbourhood of x. The set of interior points of A is the interior A0 of A. An exterior point of A is an interior point of cA (i.e. a neighbourhood of x which does not meet A, or x is isolated, separated9 from A.) Theorem 5 (i) The interior of a set is open and 4 (ii) is the largest open subset. (iii) A necessary and su±cient condition for a set to be open is that it coincides with its interior. Proof (i) x A0 open O such that x O A. 2 ) 9 2 ½ y O y A0 therefore x O A0 therefore A0 is a neigh- bourho2 o)d of2each of its points2and ½so it is open. o o 0 (ii) Let O A O A therefore O½A0 O A , but A is such an ½ ) ½ 0 [ ½ O therefore O½AO = A . [ (iii) Su±cient condition from (i). Necessary condition from (ii). 0 0 A B = A0 B0 but A B = A0 B0 z }|\ { \ z }|[ { 6 [ e.g. X = R1 with metric topology, A=rationals, B=irrationals. 0 0 0 0 A = B = therefore A B = . But A B = R1 and so 0 ; ; [ ; [ A B = R1. z }|[ { 0 However A0 B0 A B always. [ ½ z }|[ { De¯nition 11 A point x is adherent to a set A if every neighbourhood of x meets A.