No universal group in a cardinal
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arXiv
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| Format: | Preprint |
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2013
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| _version_ | 1866915833223053312 |
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| author | Shelah, Saharon |
| author_facet | Shelah, Saharon |
| contents | 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. |
| format | Preprint |
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arxiv_https___arxiv_org_abs_1311_4997 |
| institution | arXiv |
| publishDate | 2013 |
| record_format | arxiv |
| spellingShingle | No universal group in a cardinal Shelah, Saharon Logic 03C55, 20A15, 03C45, 03E04 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. |
| title | No universal group in a cardinal |
| topic | Logic 03C55, 20A15, 03C45, 03E04 |
| url | https://arxiv.org/abs/1311.4997 |