Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Introduction 0-dimensional space X is 0-dimensional iff for each open cover U of X there is a disjoint refinement that is still an open cover of X. Screenability (R.H.Bing) A topological space X is a screenable if: For each open cover U of X there is a sequence (Vn: nЄN ) such that 1. 2. 3. Vn is pairwise disjoint family of open sets For each n,Vn refines U UVn is a cover of X C-space (W.Haver) A metric space X is a C-space if: For each sequence (en: nЄN ) of positive real numbers there is a sequence (Vn: nЄN ) such that 1) Vn is a pairwise disjoint family of open sets 2) For each n, if VЄVn then diam(V) < en 3) U Vn is a cover of X C-space (D.F.Addis and G.J. Gresham) A topological space X is a C-space if: For each sequence of open covers (Un: nЄN ) of X there is a sequence (Vn:nЄN ) such that Vn is a pairwise disjoint family of open sets 2) For each n, Vn refines Un and 3) U Vn is a cover of X 1) PART I Selection principle Sc (A,B) Selection principle Sc (A,B) Let A and B be collections of families of subsets of X. A topological space X has the Sc(A,B) property if for each sequence (Un: nЄN ) of elements of A there is a sequence (Vn: nЄN ) such that 1) 2) Vn is a pairwise disjoint family of sets For each n, Vn refines Un and 3) U Vn is an element of B Types of open covers O - open cover Λ: An open cover C is a λ-cover if for each x in X the set {UЄ C: x Є U} is infinite Ω: An open cover C is an ω-cover if each finite subset of X is a subset of some element of C and X doesn’t belong to any element of C Г: An open infinite cover C is a γ-cover if for each x in X, the set {U Є C: x is not in U} is finite Гsubset of Ω subset of Λ subset of O Sc(A,B) - NEW selection property Hilbert cube and Baire space Sc(A,B)≠ Sfin(A,B) Sfin(A,B): For each sequence (Un:nЄN) of elements in A, exists a sequence (Vn:neN), such that for each nЄN, Vn is a finite subset of Un and UVn is an element of B. Sc - property Sc(O,O)=Sc(Λ, Λ)=Sc(Λ,O) Sc(Ω,O)=Sc(Ω,Λ) Sc(Ω, Λ)=Sc(Λ,Λ) Sc - property If the topological space satisfies Sc(Ω, Ω), then for each m, Xm satisfies Sc(Ω, Ω). Sc(O,O)->Sc(Ω,O) Sc(Г, Ω)->Sc(Ω,Owgp) Part 2 Games and covering dimension The game Gkc(A,B) The players play a predetermined number k of innings. In the n-th inning ONE chooses any On from A, TWO responds with a refinement Tn. A play ((Oj,Tj): j< k) is won by TWO if U{Tj : j < k } is in B; else ONE wins. NOTE: k is allowed to be any ordinal > 0. Finite dimension Theorem 1: For metrizable spaces X, for finite n the following are equivalent: 1. dim (X) = n. 2. TWO has a winning strategy in Gn+1c(O,O) Countable dimension A metrizable space X is countable dimensional if it is a union of countably many zero-dimensional subsets. (Hurewicz, Wallman). Alexandroff’s problem Addis and Gresham observed: countable dimensional -> Sc(O,O) -> weakly infinite dimensional. Alexandroff’s problem: Does weakly infinite dimensional imply countable dimensional? Pol (1981): No. There is a compact metrizable counterexample. Theorem 2: For metrizable space X the following are equivalent: 1) X is countable dimensional. 2) TWO has a winning strategy in Gwc(O,O). The Gc-type of a space. For any space there is an ordinal a such that TWO wins within a innings. gctp(X) = min{a :TWO wins in a innings}. Theorem: In any metrizable space X, gctp(X) ≤ w1. The Gc-type of a space. Theorem: For space X, if gctp(X) < w1, then X has Sc(O,O). Theorem: For X Pol’s counterexample for the Alexandroff problem, gctp(X)= w +1. Theorem: If X is metric space with Sfin(O,O), the following are equivalent: 1) X has Sc(O,O). 2) ONE has no winning strategy in Gwc(O,O). The Gc-type of a space. Pol’s example is Sc(O,O). 1) Since it is compact, it is Sfin(O,O), and so ONE has no winning strategy in Gwc(O,O). 2) Since it is not countable dimensional, TWO has no winning strategy in Gwc(O,O). Thus in Pol’s example the game Gwc(O,O) is undetermined.