By E. Ramirez De Arellano

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For example, if S has three points and :! consists of Sand the three doubletons, then :! = k:F. We now consider family constructions associated with a monoid action on S. First recall that we regard a subset G of sl X s2 as a relation from SJ to s2 with c-I = {(y,x): (x,y) E G} the inverse relation from S2 to S1. For A c S1, G(A) = {y : (x,y) E G for some x E A} is the image of A under G. 23) since all of these say there exists (x,y) E G with x E A andy E B. 26) For a map g: S1 ....... 27) Then for a map g, BE g:F iff g- 1 (B)::::> A for some A E :F, so iff g- 1(B) E Thus for a map g: :r.

F2). 86). 7). (d) It is an easy exercise to prove that the operator u commutes with union. Fa). Fa). 88) follow from duality. 89) is clear from the definition of the product uniformity. (f) For every symmetric V E 'Us, uniform continuity implies there exists a symmetric V E 'Us such that go V c V o g, so V o g-I c g-I o V. F and V E 'Us. Js, P contains g(V(F)) = g(V(F)) ng(S). 91). f. [. 82). 92). The remaining identities follow from duality [cf. 31)]. 91). 92). 87) we see that the collection of open families is closed under finite intersections and arbitrary unions.

For A c S 1 observe that: PROOF. 34) 32 Chapter 2 So (I) implies (4) in the injective case. f2 implies Bng(S1) E :F2 . f2. 29). The other two are simple exercises. JI). Then 1i. FI). 29). 28) the family operator g is (g- 1) *. 29). The third equation follows from g- 1(B nB) = g- 1(B) ng- 1(B). 32). JJ. f1. f1. f2 · [g(SI)] c gg- 1:F2. 30). 33). f2. 35), • g(A) = g(g- 1(B)). fz. REMARK. f2. f2 · [g(SI)]. BI) c k'lh, then g( SI) E k'lh. Fa a family for Sa (a= 1' 2). f2}. JI). 36) Now we assume that T is a monoid that acts on S.