most recent 30 from http://mathoverflow.net 2013-05-21T10:50:05Z http://mathoverflow.net/feeds/user/27201 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/109351/generic-extensions-and-lv-lambda1/109507#109507 X.Shi 2012-10-13T02:43:33Z 2012-10-13T05:22:23Z

<p>They are immediate consequences of Theorem 175 of W. Hugh Woodin, "Suitable extender models II: beyond $\omega$-huge", J. Math. Log., vol. 11 (2011), no. 2, pp. 115–436, which says </p> <blockquote> <p>If there is a (proper) elementary embedding $j: L(V_{\lambda+1}) \to L(V_{\lambda+1})$ with $\text{crit}(j)&lt;\lambda$ and if $G\subset \mathbb{P}$ is $V$-generic for some poset $\mathbb{P}\in V_\lambda$, then $V_{\lambda+1} \in L_\lambda(V_{\lambda+1})^{V[G]}$ if and only if $(\lambda^\omega)^V = (\lambda^\omega)^{V[G]}$.</p> </blockquote> <p>This theorem can be proved using the Large Perfect Set Theorem for subsets of $V_{\lambda+1}$ in $L_\lambda(V_{\lambda+1})$, which says assuming there is a (proper) elementary embedding $j: L(V_{\lambda+1}) \to L(V_{\lambda+1})$ with $\text{crit}(j)&lt;\lambda$, those subsets of $V_{\lambda+1}$ either has size $\leq \lambda$ or contains a large perfect subset, i.e. a homeomorphic copy of $\lambda^\omega$. Cramer recently improves the Large Perfect Set Theorem to all subsets of $V_{\lambda+1}$ in $L(V_{\lambda+1})$, so $L_\lambda(V_{\lambda+1})$ can be replaced by $L(V_{\lambda+1})$.</p>