Definition of HYP in $L_{\omega_1^{CK}}[a]$? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T12:50:58Z http://mathoverflow.net/feeds/question/118706 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/118706/definition-of-hyp-in-l-omega-1cka Definition of HYP in $L_{\omega_1^{CK}}[a]$? Peter Gerdes 2013-01-12T04:35:52Z 2013-01-14T02:27:19Z <p>The structure $L_{\omega_1^{CK}}$ consists of only HYP sets (I believe) and HYP in this structure is the same as the actual hyperaritmetic sets. Now if I move to the structure $L_{\omega_1^{CK}}[a]$ where $a$ is some non-hyperarithmetic real I add my guess is that HYP remains unaltered simply because HYP has a bottom up definition and won't be affected by the addition of new sets as long as one doesn't move to a non-$\omega$ model or the like.</p> <p>Is this correct? Any more formal demonstration?</p> http://mathoverflow.net/questions/118706/definition-of-hyp-in-l-omega-1cka/118732#118732 Answer by Liang Yu for Definition of HYP in $L_{\omega_1^{CK}}[a]$? Liang Yu 2013-01-12T14:38:14Z 2013-01-14T02:27:19Z <p>The following is not the answer of your question. But I think it is what you really want.</p> <p>I guess you might be figuring out Leo's proof of McLaughlin's conjecture and his answer to Question 65 in Harvey's problem collection paper. The point is that by applying a nonstandard ordinal, Leo obtained a nonstandard $\Pi^0_1$-singleton so that it is not hyperarithmetic. </p> <p>There are several ways to see above. One is by applying Barwise compactness. Another is to use Gandy's basis. By either way, you may obtain a nonstandard $\omega$-model $M$ of KP with $\omega_1^M=\omega_1^{CK}$ in which there is a nonhyperarithmetic (in the real sense) $\Pi^0_1$-singleton $x$ which has the property below.</p> <p>To see it by Gandy's basis theorem. Just apply it to obtain an $\omega$-model $M\models KP$ in which $\omega_1^{CK}$ is nonstandard. In $M$, fix a nonstandard recursive ordinal $\alpha$, we can perform Leo's proof to produce a nonhyperarithmetic $\Pi^0_1$-singleton $x$so that for any $\beta&lt;\alpha$, those reals computed by $x^{\beta}$ and $\emptyset^{\alpha}$ are precisely those computed by $\emptyset^{\beta}$.</p> <p>Now take $N=L_{\omega_1^{CK}}[x]$. Since $x\leq_h M$, we have that $N\models KP$. It is not difficult to see that $N\models$ ''$x$ is a $\Pi^0_1$-singleton$"$. </p> <p>The left is to show $N\models$ $x$ is not hyperarithmetic".</p> <p>$\bf{Proof}$: Otherwise, there must be some nonstandard ordinal $\gamma_0$ in $N$ so that $N\models \emptyset^{\gamma_0}$ exists. We may assume that $\gamma_0&lt;\alpha$. Since $\omega_1^x=\omega_1^{CK}$, there must be some standard recursive ordinal $\gamma_1&lt;\gamma_0$ so that $x^{\gamma_1}\geq_T \emptyset^{\gamma_0}$. Via some absoluteness (see below), it is not difficult to see $\emptyset^{\gamma_0}&lt;_T \emptyset^{\alpha}$. Then by the property of $x$, $\emptyset^{\gamma_0}&lt;\emptyset^{\gamma_1}$, a contradiction.[]</p> <p>To see that $\emptyset^{\gamma_0}&lt;_T \emptyset^{\alpha}$. There is recursive tree $T$ so that $N\models \emptyset^{\gamma_0} \mbox{ is the unique path in } T$. Note that there is also a real $z$, which is $\emptyset^{\gamma_0}$ in $M$, so that $M\models z \mbox{ is the unique path in } T$. If $\emptyset^{\gamma_0}=z$, then we have the conclusion. Otherwise, $\emptyset^{\gamma_0}$ does not belong to $M$. But $\emptyset^{\gamma_0}$ is hyperarithemtic in $x$ and so must belong to $M$, a contradiction. </p> <p>In fact, by the proof above, every real which is hyperarithmetic in $N$ is actually hyperarithmetic.</p>