Zero sharp: Difference between revisions

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:

 

* If \(X \in L\) and \(X\) is definable in \(L\) without parameters, then \(X \in L_{\omega_1}\).

* There are only countably many constructible reals - i.e. \(\mathbb{R} \cap L\) is countable.