The Hanf number for sentences of a language L is defined to be the least cardinal κ with the property that for any sentence φ of L, if φ has a model of power ≥ κ then φ has models of arbitrarily large cardinality. We shall be interested in the language Lω1,ω (see [3]), which is obtained by adding to the formation rules for first-order logic the rule that the conjunction of countably many formulas is also a formula.Lopez-Escobar proved [4] that the Hanf number for sentences of Lω1,ω is ⊐ω1, where the cardinals ⊐α are defined recursively by ⊐0 = ℵ0 and ⊐α = Σ{2⊐β: β < α} for all cardinals α > 0. Here ω1 denotes the least uncountable ordinal.A sentence of Lω1,ω is complete if all its models satisfy the same Lω1,ω-sentences. In [5], Malitz proved that the Hanf number for complete sentences of Lω1,ω is also ⊐ω1, but his proof required the generalized continuum hypothesis (GCH). The purpose of this paper is to give a proof that does not require GCH.More precisely, we will prove the following:Theorem 1. For any countable ordinal α, there is a complete Lω1,ω-sentence σαwhich has models of power ⊐α but no models of higher cardinality.Our basic approach is identical with Malitz's. We simply use a different combinatorial fact at the crucial point.
Finite undirected graphs which are not reconstructible from their large cardinality subgraphs
