Zero sharp: Difference between revisions
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* \(0^\sharp\) exists.
* There is an uncountable set \(X\) of ordinals with no \(Y \in L\) so that \(X \subseteq Y\) and \(|X| = |Y|\) (Jensen's covering theorem fails).<ref>Any text about Jensen's covering theorem</ref>
* For all \(\alpha\), \(|V_\alpha \cap L| = |L_\alpha|\).
* Every game whose payoff set is a \(\Sigma^1_1\) subset of Baire space is [[Axiom of determinacy|determined]].
* \(\aleph_\omega^V\) is regular in \(L\).
* There is a nontrivial elementary embedding \(j: L \to L\).<ref>Many papers about 0 sharp</ref>
* There is a proper class of nontrivial elementary embedding \(j: L \to L\), all with different critical points.
* For some \(\alpha, \beta\), there is a nontrivial elementary embedding \(j: L_\alpha \to L_\beta\) with critical point below \(|\alpha|\).
* Every uncountable cardinal is inaccessible in \(L\). (Possible source? <ref>W. H. Woodin, [https://arxiv.org/abs/1605.00613 The HOD dichotomy], p.1</ref>)
* There is a singular cardinal \(\gamma\) so that \((\gamma^+)^L < \gamma\). (Possible source? <ref>W. H. Woodin, [https://arxiv.org/abs/1605.00613 The HOD dichotomy], p.1</ref>)
While "\(0^\sharp\) exists" does not at face value seem to imply the failure of the axiom of constructibility, clauses 2, 3, 5 and 6 more clearly show that this is the case. Also, "\(0^\sharp\) exists" strictly implies the following:
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