# Wydawnictwa / Czasopisma IMPAN / Fundamenta Mathematicae / Wszystkie zeszyty

## On ordinals accessible by infinitary languages

### Tom 186 / 2005

Fundamenta Mathematicae 186 (2005), 193-214 MSC: 03C75, 03E35, 03C55. DOI: 10.4064/fm186-3-1

#### Streszczenie

Let $\lambda$ be an infinite cardinal number. The ordinal number $\delta(\lambda)$ is the least ordinal $\gamma$ such that if $\phi$ is any sentence of $L_{\lambda^+\omega}$, with a unary predicate $D$ and a binary predicate $\prec$, and $\phi$ has a model ${\cal M}$ with $\langle D^{\cal M},\prec^{\cal M}\rangle$ a well-ordering of type $\ge\gamma$, then $\phi$ has a model ${\cal M}'$ where $\langle D^{{\cal M}'},\prec^{{\cal M}'}\rangle$ is non-well-ordered. One of the interesting properties of this number is that the Hanf number of $L_{\lambda^+\omega}$ is exactly $\beth_{\delta(\lambda)}$. It was proved in \cite{BarwiseKunen1971} that if $\aleph_ 0 < \lambda < \kappa$ are regular cardinal numbers, then there is a forcing extension, preserving cofinalities, such that in the extension $2^ \lambda = \kappa$ and $\delta (\lambda) < \lambda^{{+}{+}}$. We improve this result by proving the following: Suppose $\aleph_ 0 < \lambda < \theta \leq \kappa$ are cardinal numbers such that

$\bullet$ $\lambda^{< \lambda} = \lambda$;

$\bullet$ ${\rm cf}( \theta) \geq \lambda^+$ and $\mu^\lambda < \theta$ whenever $\mu < \theta$;

$\bullet$ $\kappa^\lambda = \kappa$.

Then there is a forcing extension preserving all cofinalities, adding no new sets of cardinality $< \lambda$, and such that in the extension $2^ \lambda = \kappa$ and $\delta( \lambda) = \theta$.

#### Autorzy

• Saharon ShelahInstitute of Mathematics
The Hebrew University
Jerusalem, Israel
and
Department of Mathematics
Rutgers University
New Brunswick, NJ, U.S.A.
e-mail
• Pauli VäisänenDepartment of Mathematics
University of Helsinki
P.O. Box 4
FIN-00014, Finland
e-mail
• Jouko VäänänenDepartment of Mathematics
University of Helsinki
P.O. Box 4
FIN-00014, Finland
e-mail

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