Homeomorphisms of Compact Sets in Certain Hausdorff Spaces

We construct a class of Hausdorff spaces (compact and noncompact) with the property that nonempty compact subsets of these spaces that have the same cardinality are homeomorphic. Also, it is shown that these spaces contain compact subsets that are infinite. 1. Introduction In this paper, we construct a class of Hausdorff spaces with the property that nonempty compact subsets of these spaces that have the same cardinality are homeomorphic (Theorem 3.7). Conditions are given for these spaces to be compact (Corollary 2.10). Also, it is shown that these spaces contain compact subsets that are infinite (Corollary 2.10). This paper uses the Zermelo-Fraenkel axioms of set theory with the axiom of choice (see [1–3]). We let denote the finite ordinals (i.e., the natural numbers) and denotes the counting numbers (i.e., ). Also, for a given set , we denote the collection of all subsets of by , and we denote the cardinality of by . In other words, is the smallest ordinal number for which a bijection of onto exists. In this paper, we will only consider compact topologies that are Hausdorff. A topology on a set is compact if and only if and imply for some and . Therefore, compact topologies need not be Hausdorff. 2. A Class of Hausdorff Spaces Let , , and be sets such that is infinite and the collection is pairwise disjoint. For example, let , , and . Unless otherwise stated, we let Recall that for set and , we have Definition 2.1. Let be an infinite set. Define We call ？？the Fréchet filter on . Note that being infinite implies that is a filter (see [4, Definition？？3.1, page 48]). Definition 2.2. Consider the collection defined as follows: Proposition 2.3. The collection generates a Hausdorff topology on . Proof. Clearly, is a basis for a topology (see [5, Section？？13]). Let such that . If , then , , , and If , then either or . Assume that , and let . Since , and , we have Assume that . Note that . Also, , which implies and Observe that, We infer that is Hausdorff. Proposition 2.4. If , then is compact in if and only if is a finite set. Proof. Note that finite sets are compact in any topological space. So, assume that is an infinite, and let which implies Let be a nonempty, finite subcollection of . Therefore, there exists , for some , such that which implies If , then we would have , contradicting being an infinite set. Consequently, infinite subsets of are not compact in the topological space . Corollary 2.5. The set is not compact in . Corollary 2.6. The set is compact in if and only if is finite. Proposition 2.7. Let . The set is compact in if and only if is a