Embedding the bicyclic semigroup into countably compact topological semigroups

Abstract

We study algebraic and topological properties of topological semigroups containing a copy of the bicyclic semigroup C(p, q). We prove that a topological semigroup S with pseudocompact square contains no dense copy of C(p, q). On the other hand, we construct a (consistent) example of a pseudocompact (countably compact) Tychonoff semigroup containing a copy of C(p, q).2000 Mathematics Subject Classification. 22A15, 54C25, 54D35, 54H15.Although all non-trivial applications concern infinite D, we do not restrict ourselves by infinite spaces and formulate our results for any (not necessarily infinite) discrete space D.Let D be a discrete topological space. If D is infinite, then let αD = D ∪ {∞} be the Aleksandrov compactification of D. If D is finite, then let αD = D ∪ {∞} be the topological sum of D and the singleton {∞} for some point ∞ / ∈ D. Given a map π : D → M to a T 1 -topological space M , consider the closed subspaceof the product αD×M . We shall identify the space D with the open discrete subspace {(x, π(x)) : x ∈ D} and M with the closed subspace {∞} × M of D ∪ π M . Letπ = π ∪ id M : D ∪ π M → M denote the projection to the second factor. Observe that the topology of the space D ∪ π M is the weakest T 1topology that induces the original topologies on the subspaces D and M of D ∪ π M and makes the map π continuous.The following (almost trivial) propositions describe some elementary properties of the space D ∪ π M .Proposition 1.1. If for some i ≤ 3 1 2 the space M satisfies the separation axiom T i , then so does the space D ∪ π M . Proposition 1.2. If M is (separable) metrizable and D is countable, then the space D∪ π M is (separable) metrizable too.We recall that a topological space X is countably compact if each countable open cover of X has a finite subcover. This is equivalent to saying that the space X contains no infinite closed discrete subspace.Proposition 1.4. If some power M κ of the space M is countably compact, then the power (D ∪ π M ) κ is countably compact too.Proof. Since D ∪ π M is a closed subspace of αD × M , the power (D ∪ π M ) κ is a closed subspace of (αD × M ) κ . So, it suffices to check that the latter space is countably compact. Since the product of a countably compact space and a compact space is countably compact [10, 3.10.14], the product M κ ×(αD) κ is countably compact and so is its topological copy (αD × M ) κ .

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