POINT TOPOLOGY

Definition 1 A topological structure on a set X is a family (X) called open sets and satisfying O ⊂ P

(O ) is closed for arbitrary unions 1 O (O ) is closed for finite intersections. 2 O Definition 2 A set with a topological structure is a (X, ) O

∅ = ∈∅Ei = x : x Eifor some i = ∪ ∪i { ∈ ∈ ∅} ∅ so is always open by (O ) ∅ 1

∅ = ∈∅Ei = x : x Eifor all i = X ∩ ∩i { ∈ ∈ ∅}

so X is always open by (O2). Examples (i) = (X) the discrete topology. O P (ii) , X the indiscrete of . O{∅ } These coincide when X has one point. (iii) =the rational line. Q =set of unions of open rational intervals O Definition 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).

Definition 4 A base for a topological structure is a family such that B ⊂ O every o can be expressed as a union of sets of ∈ O B

Examples (i) for the discrete topological structure x x∈X is a base. { } (ii) for the indiscrete topological structure , X is a base. {∅ } (iii) For , topologised as before, the set of bounded open intervals is a base.Q

1 (iv) Let X = 0, 1, 2 { } Let = (0, 1), (1, 2), (0, 12) . Is this a base for some topology B { } 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 sufficient condition for to be a base for a 0 B 00 topology on X = o∈Bo is that for each O and O and each x O0 O00 O ∪such that x O O0 O00. ∈ B ∈ ∩ ∃ ∈ B ∈ ⊂ ∩ 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 ∩ suc∈hOthat x O∈ O∩0 O00. ∩ B ∃ ∈ B ∈ ⊂ ∩ Sufficient: let be the family of unions of sets of . O B (O1) is clearly satisfied.

(O2) ( Ai) ( Bj) = (Ai Bj) so that it is sufficient 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 ∈00 ∩ ∃ ∈ B ∈ ⊂ ∩ O O = x∈O0∩O00 Ox. ∩ ∪ Theorem 2 If is a non-empty family of sets the family of their finite S B intersections is a base for a topology on S ∪

Proof Immediate verification of O1 and O2. The topology generated in this way is the smallest topology including all the sets of . S Definition 4 A family is a sub base for a topology if the set of finite intersections is a baseS for the topology.

e.g. a a∈Q and ( a) a∈Q are sub bases for { ∞} { −∞ } Q Definition 5 If a topology has a countable base it satisfies the second axiom of countability.

Definition 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 . { } Theorem 3 A necessary and sufficient condition that a set be open is that it contains (is) a neighbourhood of each of its points.

Proof Necessary: Definition of a neighbourhood

Sufficient: Let OA = open subsets of A. OA is open (O ) and OA A. ∪ 1 ⊂

2 If x A A a neighbourhood of x open set x therefore x OA ∈ ⊃ ⊃ 3 ∈ therefore A OA. ⊂ 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 finite 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 ∈ ∃ ∈ ∈ ∈ (Take W to be an open set x and V .) 3 ⊂ 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 ∃ sets of neighbourhoods of each point x are precisely the given V (x). (Hausdorff).

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 ∈ ∈ such topology. Consider the set O so defined. (O ) Suppose x O. Then x some O0 O0 V (x) O0 cupO so 1 ∈ ∪ ∈ ∈ ⊂ | O V (x). ∪ ∈ (O ) Suppose x F o x each O. each O V (x) therefore 2 ∈ ∩ ⇒ ∈ ∈ F O V (x) by V . ∩ ∈ 2 Now consider the system U(x) of neighbourhoods of x defined by this topology.

(i) U(x) V (x). Let U U(x). Then U O x. But O V (x) so by V⊂U V (x). ∈ ⊃ 3 ∈ 1 ∈ (ii) V (x) U(x). Let V V (x). It is sufficient to prove that O such that⊂ x O V∈. ∃ ∈ O ∈ ⊂ Let O = y : V V (y) x O since V V (x). { ∈ } ∈ ∈ O V since y V for all y O by V . ⊂ ∈ ∈ 3 To prove that O it is sufficient to prove that V V (y) for all y O. ∈ O ∈ ∈ If t O V V (y) by definition of O therefore W V (y) such ∈ ∈ ∃ ∈ that V V (z) for all z W by V 4. ∈ ∈ − Therefore W O (definition of O) ⊂ 3 Therefore O V (y) by V ∈ 1 e.g. R1 : V (x)= sets which contain interval (a, b) a < x < b.

Definition 7 A metric for a set X is a function ρ form X X to R+ (non- negative reals) such that ×

M1 ρ(x, y) = ρ(y, x) for all x, y M ρ(x, z) ρ(x, y) + ρ(y, z) for all x, y, z 2 ≤ M3 ρ(x, x) = 0 for all x.

This is sometimes called a pseudo metric and for a metric we have (M 0 ) ρ(x, y) 0, = 0 x = y. 3 ≥ ⇔ Definition 8 The open r- Ball about x = y : ρ(x, y) < r and is denoted by B(r, x) { } The closed r-ball around x = y : ρ(x, y) R and is denoted by B(r, x). { ≤ } 1 V (x) = sets which contain one of B n , x n = 1, 2, . . . satisfies V1−4 and so defines{ a topology on X. This³ topology´ is the metric} topology defined on X by ρ. The development of topology from the neighbourhood point of ve=iew is due to H. Weyl and Hausdorff. That from the open sets aspect is due to Alexandroff and Hopf.

Definition 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 finite 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.

Definition 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, separated∃ from A.)

Theorem 5 (i) The interior of a set is open and

4 (ii) is the largest open subset. (iii) A necessary and sufficient 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. ∈ ⇒ ∃ ∈ ⊂ y O y A0 therefore x O A0 therefore A0 is a neigh- bourho∈ o⇒d of∈each of its points∈and ⊂so it is open. o o 0 (ii) Let O A O A therefore O⊂A0 O A , but A is such an ⊂ ⇒ ⊂ 0 ∪ ⊂ O therefore O⊂AO = A . ∪ (iii) Sufficient 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. ∪ ⊂ z }|∪ { Definition 11 A point x is adherent to a set A if every neighbourhood of x meets A. The set of points adherent to a set A is called the adherence () A of A. An of A is an of A if there is a neighbour- hood of A which contains no point of A other than x; otherwise it is a point of accumulation ( point) of A. Examples (i) A = A = R Q ⊂ R (ii) In a no set has accumulation points, every point is isolated. (iii) In an indiscrete space every non-empty set has X as its adherence. Theorem 5 (i) The adherence of a set is closed and (ii) is the smallest containing the given set (iii) A is closed A = A A its accumulation points. ⇔ ⇔ ⊃

0 cA = cA z}|{ cA0 = cA

5 Corollary A = A A B = A B ∪ ∪ Definition 12 The set of accumulation points of A is its derived set A0. A perfect set is a closed set without isolated points. Suppose we map (X) (X) where A A then P → P → (C ) = 1 ∅ ∅ (C ) A A 2 ⊂ (C ) A A (with C A = A) 3 ⊂ 2 ⇒ (C ) A B = A B. 4 ∪ ∪ Theorem 6 If we are given an operation mapping (X) into (X) which P P has the properties C1−4 then the set of complements of the sets G such that G = G is a topology on X for which AS is the closure of A for all A X (Kunatowski). ⊂ Definition 13 The frontier or of a set A is A cA and is the set of points adherent to A and to cA, or is the set in∩terior to neither A nor cA, or is the set neither interior or exterior to A. It is a closed set. A set is closed it contains its boundary. ⇔ A set is open it is disjoint from its boundary. ⇔ Definition 14 A set is dense in X if A = X. A is dense in itself if all its points are accumulation point, i.e. A A0. 0 ⊂ A set is nowhere dense if cA is dense i.e. A = . ∅ Induced topologies Given (X ) and Y X , can we use to get a topology for Y . O ⊂ O

Theorem 7 Y O O∈O is a topology OY on Y called the topology induced on Y by{(x∩ ) } O

Proof (O ) Y O = Yn O 1 ∪ ∩ ∪ (O ) F YnO = Yn F O 2 ∩ ∩ If Z Y X then Y and induce the same topology on Z. ⊂ ⊂ O O A (sub) base for induces a (sub) base Y for Y . B O B O “O open in (relative to )Y ” means“O open in (Y OY ).

6 A necessary and sufficient condition that every set open in Y be open in X is that Y be open in X. Theorem 8 (i) G Y is closed in Y G = Y G0 where G0 is closed in X. ⊂ ⇔ ∩

(ii) VY (x) = Y V ∈ { ∩ }v V (x) (iii) If z Y X then ZmY = YnZmx ⊂ ⊂ Proof (i) G is closed in Y Y cG is open in Y ⇔ ∩ Y cG = Y O; O open in X ↔ ∩ ∩ Y G = Y cO (take comp. in Y ) ⇔ ∩ ∩ G = Y cO ⇔ ∩ take G0 = cO.

(ii) Y U VY (x) ⊃ ∈ U O x, O open in Y ⇔ ⊃ 3 U O0 Y x O0 open in X ⇔ ⊃ ∩ 3 U = V Y where V O0 x ⇔ ∩ ⊃ 3 i.e. U = Y V where V V (x). ∩ ∈ (iii) ZmY = closed sets in Y which Z T ⊃ = (Y∩ closed sets Z) ⊃ = YT closed sets Z ∩ ⊃ = Y ZT ∩ Continuous functions Definition 15 A map F of a topological space X into a topological space Y is continuous at x X if given a neighbourhood V of f(x ) in Y 0 ∈ 0 ∃ a neighbourhood U of x 0 in X such that f(U) V . + ⊂ −1 f is continuous at x0 if for every neighbourhood V of f(x0) f (V ) is a neighbourhood of x0. Theorem 9 If f : X Y is continuous at X and x A then f(x) f(A) → ∈ ∈ Proof Let V V (f(x0) then f −1(V ) V (x) therefore f −1(V ) A = . Therefore∈f(f −1(V )) f(A) = f( ∈) = . ∩ 6 ∅ ∩ 6 ∅ ∅ f f −1 is the identity therefore V f(A) = ◦ ∩ 6 ∅

Theorem 10 if f : X Y is continuous at x0 and g : Y Z is continuous at f(x ) Then g →f is continuous at x . → 0 ◦ 0 7 Proof Let W V (g f(x )) then g−1(W ) V (f(x )) ∈ ◦ 0 ∈ 0 f −1 g−1(W ) V (x ) (g f)−1(W ) V (x ) ◦ ∈ 0 ◦ ∈ 0 Definition 16 A map f : X is continuous (on X) if it is continuous at each point of X. →

Theorem 11 Let f : X Y . Then the following properties are equivalent. → (i) f continuous on X (ii) f(A) f(A) for all A (X) ⊂ ∈ P (iii) the inverse image of a closed set is closed (iv) the inverse image of an open set is open.

Proof (i) (ii) by theorem 9. ⇒ (ii) (iii) Let G0 be closed in Y and f −1(G0) = G. ⇒ f(G) f(G) by (ii) G0 = G0. ⊂ ⊂ Therefore G f −1(G0) = G G therefore G = G therefore f −1(G0) is closed. ⊂ ⊂ (iii) (iv) ⇒ cf −1(A) = f −1(cA) A Y ⊂ (iv) (i) Let x Xv V (f(x)). ⇒ ∈ ∈ x O V O open in Y . f −1(O) is open in Y and contains x so is a ∃neigh∈ bourho⊂ od of x in Xf(f −1(O)) V ⊂ Note The image of an open set under a continuous map is not necessarily open. 1 e.g. f : R R x 2 → 7→ 1+x f(R) = (0 1] not open.

Comparison of Topologies

Definition 17 If 1 and 2 are topologies on a set X, 1 is finer than ( is coarserO thenO ) if (strictly finer if notO equal). O2 O2 O O1 ⊃ O2 [Topologies on X are not necessarily complete]

e.e. R1 : 1:usual topology 2: open sets are open sets of 1 which contains OOand o . O O { } These topologies are not comparable.

8 Theorem 12 If 1 and 2 are topologies on X, the following properties are equivalent: O O

(i) O1 ⊃ O2 (ii) - closed sets are closed O2 O1 (iii) V (x) V (x) for all x X 1 ⊃ 2 ∈ (iv) The identity map (X ) (X ) is continuous. O → O2 (1) (2) (v) A A for all A X ⊂ ⊂ We also have the following qualitative results: the discrete topology on a set is the finest topology on the set, and the indiscrete is the coarsest. The finer the topology, the more open sets, closed sets, neighbourhoods of a point, the smaller the adherence, the larger the interior of a set, the fewer the dense sets. If we refine the topology of X we get more continuous functions. If we refine the topology of Y we get fewer continuous functions.

Final Topologies

Theorem 13 Let X be a set. Let (Yi, i = Oij j∈Ji )i∈I be a family of topological spaces. O { } −1 Let fi : Yi X. Let = O X : f (o) open in Yi for all i I . → O { ⊂ i ∈ } Then is a topology on X and is the finest for which the fi are continuous.O If g : X Z is a map into a topological space Z, g is continuous from → (X ) Z g fi are all continuous. O → ⇔ ◦ −1 Proof is non-empty: f (X) = Yi O −1 −1 −1 (O1) Let fi (Ok) 1, Ok . Then fi ( kOk) = K fi (Ok) 1 for all i. ∈ O ∈ O ∪ ∪ ∈ O −1 −1 −1 (O ) Let f (O k) i, Ok . Then f ( F Ok) = F f (Ok) 2 i − ∈ O ∈ O i ∩ ∩ i ∈ i for all i. O A necessary and sufficient condition for fi to be continuous is that −1 fi (O) be open in Yi for all i. A finer topology than will not satisfy this. O

Now let fi : Yi X g : X Z g fi : Yi Z. → → ◦ → 9 The necessary condition is obvious. Sufficient condition: g : X Y continuous g−1(O) open in X for all O open in Z. → ⇔ −1 g fi continuous g fi (O) open in Yi for all i ◦ ⇒ ◦ −1 −1 f g (o) open in Yi for all i ⇒ i ◦ g−1(O) is open in X ⇒ We define to be the final topology for X, the maps fi and spaces Yi. O Examples (i) X a topological space. R is an equivalence relation on X. φ : X X = Y x x˙ ( class) → r 7→ The finest topology on Y such that φ is continuous is the quotient topology of that of X by the relation R. f : X Z is continuous f φ : X Z is continuous. R → ⇔ ◦ → e.g. X + R2

R : (x1 y1) (x2 y2) x1 = x2. X ∼ ⇔ Then R = R (isomorphically).

(ii) X a set. (X i) a family of topological spaces. O φi : (X, i X x x O → 7→ The final topology on X is the finest topology coarser than all the i. O is called the lower bound of the i. = i i O O O ∩ O Initial Topologies

Theorem 14 Let X be a set. Let (Yi i)i∈I be a family of topological spaces. O −1 Let fi : X Yi. Let fi : X Yi. Let S = f (Oij) i∈I,j∈Ji . → → { i } Then S is a sub-base for a topology on X, the coarsest topology for J which all the fi are continuous.

If Z is a topological space g : Z X is continuous fi g continuous for all i I → ⇔ ◦ ∈ Proof S is non-empty, for S. ∅ ∈ −1 s∈SS = X( f (Yi)) then use theorem 2. ∪ − i A necessary and sufficient condition for fi to be continuous is the −1 f (Oij) be open in for all ij. i J the rest of the proof is similar to theorem 13.

We define as the initial topology for X, the maps fi and spaces Yi. J 10 Examples (i) X a set (Y ) a topological space f : X Y . O → the initial topology here is called the inverse image of . O (ii) If X Y f : X Y x x is the canonical injection. ⊂ → 7→ f −1(A) = A X and the open sets of the initial topology are the intersections∩with X of the open sets of Y - we have the induced topology.

(iii) (X i) φ : X X x x. O 1 → 7→ The initial topology is the coarsest topology finer than all the i O (iv) (Xi i)i∈I X = i∈I Xi O Q φi = proj : X Xi xi i∈I xi i → { } 7→ the initial topology is the coarsest for which all the projections are continuous and is called the product topology of the i. (X ) O J is the topological product of the (Xi, 1). The (Xi, i) are the Factor spaces. The open sets of the Oproduct topologyO have as −1 base the finite intersections of sets proji (Oij) where Oij is open in (Xi i). −1 O proj (Oij = Ai where Au = Xu i = j, Aj = Oij and the i i∈I 6 base consists ofQsets i∈I Ai where Ai = Xi except for a finite set of i, where Ai is openQin Xi. These are called elementary sets.

g : Z Xi is continuous proji g is continuous for all i i.e. → ⇔ ◦ all the co-ordinatesQ are continuous. Limit Processes Consider the following limit processes:

(i) lim an n→∞ (ii) lim f(x) x→a b (iii) R f(x) dx = lim (xi+1 xi)f(ξi) Za X − What have these in common.

A. A set with some order properties

(i) Z R n an → 7→ (ii) V (a) R v f(v) → 7→ (iii) Nets on (a b) N I(N, f) → B. A map of the ordered set into a topological space. C. We consider the set of images as we proceed along the order. We now unify all these.

11 The Theory of Filters (H.CARTAN 1937)

Definition 18 A filter on a set X is a family (X) such that F ⊂ P (F ) A and B A B 1 ∈ F ⊃ ⇒ ∈ F (F ) is closed for finite intersections 2 F (F ) . 3 ∅ 6∈ F F every finite An in F is non-empty. 3 ⇒ F X i.e. is non-empty. 2 ⇒ ∈ F F Examples (i) X = x is a filter on X. 6 ∅ { } (ii) = A X : Y : Y A is a filter on X. ∅ 6 ⊂ { ⊃ } (iii) X an infinite set. The complements of the finite subsets form a filter on X. (iv) If X = Z (the positive integers) the filter of (iii) is called the Fr´echet filter. (v) X a topological space. The set of neighbourhoods of = A X ∅ 6 ⊂ is a filter. A = x gives the neighbourhoods of x. { } Comparison of Filters

Definition 19 If and 0 are filters on a set X and 0 we say is coarser than F 0, 0 Fis finer than . F ⊂ F F F F F Filters are not neccessarily comparable e.g. the filters of neighbour- hoods of distinct points in a .

If n i∈I is a family of filters on X then = i is a filter. {F } F i\∈I F

Definition 20 The intersection of the i is the finest filter coarser than all F the i and is the lower bound of the i. F F Theorem 15 Let be a system of sets in X . A necessary and sufficient condition thatS a filter on X containing is that the finite intersections of members of ∃ be non-empty. S S

Proof Necessary: Immediate from F2 Sufficient: Consider the family of sets which contain a member of 0, the set of finite intersectionsFon . S S satisfies F , F , F . F 1 2 3 12 Any filter is finer than . is the coarsest filter . ⊃ S F F ⊃ S is called a system of generators of . S F Corollary 1 is a filter on X, A X. A necessary and sufficient condition that 0F such that A ⊂ 0 is that F A = F . ∃F ⊃ F ∈ F ∩ 6 ∅∀ ∈ F Corollary 2 A set Φ of filter on (non-empty) X has an upper bound in the set of all filters on X for every finite . ⇔ i i , ,...,n Φ and every Ai i i = 1, 2, . . . n Ai = . {F } =1 2 ⊂ ∈ F ∩ 6 ∅ Bases If is a system of generators for , is not, in general, the set of subsetsS of X which contain an elemenFt ofF S Theorem 16 Given (X), a necessary condition that the family of B ⊂ P subsets of X which contain an element of be a filter is that have the properties B B

(B ) The intersection of 2 sets of contains a set of . 1 B B (B ) is not empty; . 2 B ∅ 6∈ B

Definition 21 A system (X) satisfying B1, B2 is called a base for the filter it generates byBTheorem⊂ P 16. 2 filter bases are equivalent if they generate the same filter. Theorem 17 is a base for each set of contains a set of . B ⊂ F F ⇔ F B Theorem 18 A necessary and sufficient condition that 0 with base 0 be finer than with base id 0 . F B F B B ⊂ B Examples (i) Let X be a non-empty partially ordered set ( ) in which each pair of elements has an upper bound. The sections≤ x : x a of X form a filter base. The filter it defines is called{the filter≥ of} sections of X.

(ii) X=set of nets on [a, b] N = (a = x0 < x1 < . . . < xn = b) N1 N2 if N N . ≤ 1 ⊂ 2 An upper bound for N and N is N N . 1 2 1 ∪ 2 1 1 (iii) The filter of neighbourhoods of O in R has as bases , , 1 ½µ−n n¶¾ 1 1 , ( a, b) ab > 0 etc. ½·−n n¶¾ { − } (iv) In R2 the squares, discs, ellipses et. centre 0 form bases for the filter of neighbourhoods of 0.

13 (v) In R x, x > n is a filter base. 1 { | | } Definition 22 A fundamental system of neighbourhoods of a point in a topological space is a base for the filter of neighbourhoods of the point.

Definition 23 A space satisfies the first axiom of countability if every point has a countable fundamental system of neighbourhoods. 2nd axiom 1st axiom ⇒ 1st axiom 2nd axiom. 6⇒ e.g. X uncountable, with discrete topology. x is a base for the neighbourhoods of x and is countable. { }

Theorem 19 Let be a filter on X A X. A necessary and sufficient F ⊂ condition that A = F A F ∈F be a filter on A is that F A = F . F { ∩ } ∩ 6 ∅∀ ∈ F If A A is a filter on A. ∈ F F

Definition 24 A is a filter induced on A by F F e.g. X a topological space A X VA(x) is a filter on A x A ⊂ ⇔ ⊂ Let f : X Y and let be a filter on X. Then in general f(F ) F ∈F is not a filter,→ for F breaksF down. But if is a filter base{for a }filter 2 B on X then f(B) B∈B is a base for a filter i=on Y . { } Definition 25 X a topological space. a filter on X. x is a of if V (x) we say convergesF to X. F F ⊃ F x is a limit of a filter base if the filter with base converges to x B B e.g. 1 1 (i) In R , the filter with base , converges to 0, but that 1 ½µ−n n¶¾ with base x : x > n does not converge. {{ | | }} (ii)

X = x, y, z { } = , x, y , z X O {∅ { } { } } = x, y X B {{ } } V (x) = x, y X {{ } } Therefore converges to x, V (y) = (x, y) X therefore con- verges to yB. { } B

14 Definition 26 X a topological space. is a filter base on X. x is adherent to if it is adherent to every set ofB . B B x is adherent to V B = V V (x), B . B ⇔ ∩ 6 ∅∀ ∈ ∈ B Every limit point of a filter is adherent to the filter.

The set of point adherent to a filter is B∈BB and is closed. ∩ Definition 27 Let X be a set, Y a topological space. Let : X Y . Let P → be a filter on X. F y Y is a limit of along if y is adherent to ( ) ∈ F F P F

Examples (i) Fr´echet (of sections of Z) α : Z R n an. F → 7→ A set of the filter is a set m : m n = F α(F3 am : m n . ⊃ { ≥ } { ≥ } a is a limit of α along a is adherent to every set of i.e. F ⇔ F every (a ε, a + ε) meets every set of i.e. (a ε, a + ε) am : m n =− for all n. F − ∩ { ≥ } 6 ∅ (ii) X a topological space, Y a topological space. : X Y . V (a) √ → is the filter of neighbourhoods of a X. ∈ In this case we write y = lim (x) instead of lim . x→a P F P Theorem 20 X, Y topological spaces. : X Y continuous at a Y lim (x) = (a). P → ∈ ⇔ x→a P P Proof lim (x) = (a) given u U( (a)) V V (a) such that (V ) x→a P P ⇔ ∈ P ∃ ∈ P ⊂ U. let X, Y be topological spaces. Let A X and let a A. Let f : A ⊂ ∈ → Y . Let = VA(a) = A V ∈ . F { ∩ }v V (a) We write lim f(x) instead of lim f. x→a,x∈A F Definition 28 lim f(x) is a limit of f at a relative to A. x→a, x∈A

Theorem 21 Let X be a set and let Yi i∈I be a family of topological spaces. { } Let fi : X Yi. → A filter on X converges to a X in the initial topology on X F ∈ O ⇔ the filter base fi( ) converges to fi(a) i I. F ∀ ∈

15 Proof Necessary condition: The fi are continuous by Theorem 20.

Sufficient condition: Let w V (a). By definition of J Ai W −1 ∈ O∃ ∩ ⊂ where a Ai = f (Oi), Oi open in Yi and J is a finite set. ∈ −1 Since fi( ) converges to fi(a) in Yi Oi fi( ) and f (Oi) so F−1 ∈ F ∈ F that J f (Oi) i.e. V (a) F . ∩ ∈ F ⊂

Corollary A filter on a product space X = Xi converges to x the F YI ⇔ filter base pri( ) converges to xi i I. F ∀ ∈

16