A non-metrizable space whose countable power is -metrizable

We answer a question of A.V. Arhangel’skii by finding a nonmetrizable space X such that Xω is the countable union of metrizable spaces. A space is σ-metrizable if it is the countable union of metrizable subspaces. A.V. Arhangel’skii (see [2] and [3]) asked if a space X must be metrizable if X is σmetrizable. In [2], M.G. Tkacenko showed that the answer is “yes” if X is separable or countably compact. In [3], V.V. Tkachuk proved that X must be metrizable if X is the union of finitely many metrizable spaces. But here we shall show that in general the answer to Arhangel’skii’s question is “no”. Let ω1 denote the set of countable ordinals, and L the limit ordinals in ω1. For each α ∈ L, choose a sequence α0, α1, ... of non-limits converging to α. Let X be the space whose set is ω1, with points of ω1\L isolated, and the k basic neighborhood of α ∈ L having the form {αn : k < n < ω} ∪ {α}. Note that L is a closed discrete subset of X (i.e., every subset of L is closed in X). Such an X is an example of what is often called a ladder space on ω1. It is known that X is not metrizable or even paracompact. For the benefit of the reader, we give an elementary proof of this fact here. Note that X has an open cover each member of which contains at most one point of L. If X were paracompact, then there would be neighborhoods U(α) of α for α ∈ L such that the collection {U(α) : α ∈ L} is locally finite. For each δ ∈ ω1, [0, δ] would meet at most countably many U(α)’s, so we could find δ0 < δ1 < ... such that [0, δn]∩U(α) 6= ∅ implies α < δn+1. But we arrive at a contradiction by noting that if δω = supn<ωδn, then U(δω) ∩ [0, δn] 6= ∅ for some n. In [1], we answered a question of Tkachuk by showing that every finite power of X is the union of two metrizable spaces. Here we will show that if X has the additional property that L is a Gδ-set in X (which can be ensured by, e.g., always choosing αn to have the form β+n for some limit β), then X ω will be the countable union of metrizable spaces. Theorem. Let X be a ladder space on ω1 in which the set L of limit ordinals is a Gδ-set in X. Then X ω is σ-metrizable. Proof. In this proof, letters i, j, k, l,m and n will always denote natural numbers. Of course, X itself is the union of two metrizable (discrete) subspaces, L and ω1 \ L. So for each n, X × L is the union of finitely many metrizable spaces. It Received by the editors March 6, 1996. 1991 Mathematics Subject Classification. Primary 54E35, 54B10, 54B05. c ©1997 American Mathematical Society