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No universal group in a cardinal

Saharon Shelah

Abstract

For many classes of models, there are universal members in any cardinal $λ$ which "essentially satisfies GCH", i.e. $λ= 2^{< λ}$, in particular for the class of a complete first order $T$ (well, if at least $λ> |T|$). But if the class is "complicated enough", e.g. the class of linear orders, we know that if $λ$ is "regular and not so close to satisfying GCH" then there is no universal member. Here we find new sufficient conditions (which we call the olive property), not covered by earlier cases (i.e. fail the so-called SOP$_4$). The advantage of those conditions is witnessed by proving that the class of groups satisfies one of those conditions.

No universal group in a cardinal

Abstract

For many classes of models, there are universal members in any cardinal which "essentially satisfies GCH", i.e. , in particular for the class of a complete first order (well, if at least ). But if the class is "complicated enough", e.g. the class of linear orders, we know that if is "regular and not so close to satisfying GCH" then there is no universal member. Here we find new sufficient conditions (which we call the olive property), not covered by earlier cases (i.e. fail the so-called SOP). The advantage of those conditions is witnessed by proving that the class of groups satisfies one of those conditions.

Paper Structure

This paper contains 9 sections, 2 theorems, 11 equations.

Key Result

Theorem 1.9

1) If $T$ is complete, with the $3$-olive property and $\lambda > \kappa^+$ and $\lambda,\kappa$ are regular, $2^\kappa > \lambda \ge \kappa^{++} + |T|,$then$T$ has no universal model in $\lambda$ (for $\prec$). 2) If $T$ is complete, with the $(\eta,\bar{k},m)$-olive property and $\lambda = {\rm cf

Theorems & Definitions (42)

  • Definition 8
  • Definition 11
  • Definition 12
  • Definition 13
  • Definition 14
  • Definition 1.1
  • Remark 1.2
  • Definition 1.3
  • Remark 1.4
  • Claim 1.5
  • ...and 32 more