On some cardinal invariants of hyperspace with ( C 0 , C 1 ) Ochan type topology

. In the article are examined some cardinal invariants of hyperspace with Ochan type topology.


Introduction
By P * (X) we denote the space of all closed subsets of topological space X (or simply of some set X). By C 0 (X) or C 0 we note family of all finite subsets of X and C 1 (X) or C 1 family of all closed subsets of space X. For family M ⊂ P * (X) we provide (C 0 , C 1 ) topology by giving base {[A, B] M : A ∈ C 0 , B ∈ C 1 } there [A, B] M = {(P ) ∈ M: P ⊂ A and P ∩B = ∅}. If M is cleare from context we simply note [A, B]. exp X denote hyperspace of space X. All topological notions and terms wich we use can be found in [2]. The first time Ochan type topology was mentioned in [4] and later examined in generale by R. Kasuba [3]. All definitions of cardinal invariants and their basic properties are in [1]. Also all spaces we assumed be Hausdorff. Now let discus some cardinal invariants starting by.

Lemma 1.
Character of the point (F ) in exp X not exceed τ (there τ some infinite cardinal number) if and only if character χ(F, X) of subset F in X don't exceed τ and |F | τ .
and consequently F ⊂ U α * ⊂ OF and it means that U is base of F at X. Because | U| |U| than in the case of fulfilment assumption |U| τ we conclude χ(F, X) τ . Let D ⊂ F and |D| < ℵ 0 . Than open set [D, ∅] is neighborhood of (F ) at exp X. We can find β ∈ A, that [A β , B β ] ⊂ [D, ∅] and consequently D ⊂ A β , but than {A α : α ∈ A} = F and |F | τ if |U| τ , because A α are finite for each α ∈ A. Let prove sufficiency of conditions. Let U = {U α : α ∈ A} some base of set F at X. Then U = {[P, X \ U α ]: P ∈ C 0 (F ), α ∈ A} base of (F ) at exp X and |U| |F | · | U |, consequently |U | τ · τ = τ if |F | τ and | U| τ . The proof is over.
Proof. Let space exp X is I countable. Then conditions (i) and (iii) follow from Lemma 1. Prove, that K = X \ I(X) is compact. Because X is countable is sufficient to prove that K is countable-compact. Let K is not countably compact, than in K exist discrete closed subset Let prove sufficiency of conditions (i), (ii), (iii). For that, using Lemma 1, is sufficient to show that character of each closed set F in X is countable. Let form Than αβ is countable base of neighborhoods of F in X. Proof is over.
Proof. Tightness not exceed character of space for each space. So is sufficient to prove that t(exp X) χ(exp X). Using Lemma 1 at first we prove t((F ), exp X) |F | for Now let prove sufficiency of conditions (i) and (ii). Let conditions (i), (ii) are satisfied for some (F ) ∈ exp X. Than we prove, that Ψ ((F ), exp X) τ . For that pick A ⊂ F , thatĀ = F and |A| τ , and family of neighborhoods {O α : α ∈ I} of set F power wich do not exceed τ , that {O α : α ∈ I} = F . Than family of basec subsets {[c, X \ O α ]: α ∈ I, c ∈ C 0 (A)} is pseudobasis of (F ) in exp X and it's power do not exceed τ · τ = τ . Proof is over. ⊓ ⊔ Corollary 3. Pseudocharacter Ψ (exp X) do not exceed network weight nω(X) of regular space X. Let P is closed and P = F . If P ∩ (X \ F ) = ∅, than because X is regular we can find W ∈ B and D =W ∈ B, that P ∩ D = ∅, and D ∩ F = ∅. Let V 0 ∈ B and V 0 ∩ F = ∅ and V 0 ∩ D = ∅ than (P ) / ∈ U 0 ∈ PB, but (F ) ∈ U 0 . If P F , we can find V 1 ∈ B, that P ∩ V 1 = ∅, and V 1 ∩ F = ∅. Assume U 1 = {[y, ∅]: y ∈ V 1 }. Than U 1 ∈ PB, (F ) ∈ U 1 , but (P ) / ∈ U 1 . Consiquently PB is pseudobasis of exp X. Let |B| = ω(X) than we get |PB| ω(X). Proof is over. ⊓ ⊔ Corollary 5. Let X is regular space and K ⊂ exp X is compact. Than ω(K) ω(X). Remark 1. Evaluation of Proposition 3 to improve is not possible. Let Y is one point compactification of discrete space power τ . Then exp Y include Cantor discontinuum D τ weight τ consequently pω(exp Y ) = ω(Y ).
Proof. Let N is network of the space X such, that |N | = nω(X). Assume that N consists of closed X subsets. We prove, that N = {( N * ): N * ⊂ N , |N * | < ℵ 0 } is close in the space exp X. Let [A, B] is set of standart base. Thus A is finite we can find N * * ⊂ N , that A ⊂ N * * ⊂ X \ B. Consequently ( N * * ) ∈ [A, B]. Thus [A, B] is chose freelly we conclude N is network of exp X. Conclude nω(X) d(exp X). Now let prove inverse d(exp X) nω(X). Let M is dense subset of the space exp X and |M| = d(exp X). For each point x ∈ X and each neighborhood V we can find (F ) ∈ M, such that (F ) ∈ [{x}, X \ V ], consequently x ∈ F ⊂ V . Because x is freelly chose we conclude that M = F : (F ) ∈ M is netwok of the space X and |M| = |M|. Consequently nω(X) d(exp X). Proof is over.
Corollary 6. Pseudocharacter of hyperspace of regular space X don't exceed it's density.
Theorem 2. Let X is Hausdorft topological space and it's hyperspace exp X with (C 0 , C 1 ) topology. Then following are equivalent: Proof. At first we note, that from conditions (i), (ii), (iii), (v) follow countability of tightness of the space exp X, but than X countable, because (X) ∈ {(A): A ⊂ X, |A| < ℵ 0 }. From condition (iv) follows countability of X, because {({x}): x ∈ X} is discrete subset of hyperspace exp X. So to prove that (vi) follows from each of condition (i), (ii), (iii), (iv), (v) is sufficient to prove that in the case of satisfaction at lest of one of condition (i), (ii), (iii), (iv), (v) space X is countable compact. Assume that from condition (i), (ii), (iii), (iv), (v) don't follow countable compactness of X. Than X contens countable, closed discrete subset D and exp D is embeddible into exp X an exp D topology concide with Vietoris topology. But exp D is not simmetrizable space and contents discrete subset power 2 ℵ0 . We get contradiction.
Because conditions (i), (iii), (iv), (v) follow from condition (ii) to conclude the proof is sufficient to prove that from (iv) follows (ii). Let B is countable base of countable and compact space X. Familly B = { B * , |B * | < ℵ 0 } is also countable, consequently countable is and family {[A, B]: A ∈ C 0 (X), B ∈ B} from base of exp X, because X is compact. Proof is over.
The large role in evaluation of cardinal invariants of exp X plays power of the space X: Proposition 5. Let cardinal number τ 1 is equal to minimal number of ω(exp X), nω(exp X), S(exp X), l(exp X), χ(exp X) and cardinal number τ 2 is equal to largest of numbers ψ(exp X), pω(exp X), d(exp X), c(exp X). Than τ 1 |X| τ 2 .