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Let $\mathcal M$ be a premouse. $\mathcal M$ is said to be $\omega$-small if and only if whenever we have that $\kappa=crit(E)$ for some extender $E$ on the sequence of $\mathcal M$, then $\mathcal J^{\mathcal M}_{\kappa} \not \models$ There are $\omega$-many Woodins. $\mathcal M$ is said to be properly small if and only if $\mathcal M$ is $\omega$-small and $\mathcal M \models$ "there is a largest cardinal and $ZF^-$ (i.e without the power set axiom) and there are no Woodins".

Now if $\mathcal M$ is $\omega$-small then so is any of the $\mathcal J^{\mathcal M}{\beta}$ for any $\beta$ but if $\mathcal M$ is properly small then it could be that there is some $\beta$ we have that $\mathcal J^{\mathcal M}_{\beta} \models$ there are $\omega$ Woodins.

I am confused about how that can be, since in the definition of properly small we also have that the premouse has to be $\omega$-small. Can someone show me an example? Thx

Let $\mathcal M$ be a premouse. $\mathcal M$ is said to be $\omega$-small if and only if whenever we have that $\kappa=crit(E)$ for some extender $E$ on the sequence of $\mathcal M$, then $\mathcal J^{\mathcal M}_{\kappa} \not \models$ There are $\omega$-many Woodins. $\mathcal M$ is said to be properly small if and only if $\mathcal M$ is $\omega$-small and $\mathcal M \models$ "there is a largest cardinal and $ZF^-$ (i.e without the power set axiom) and there are no Woodins".

Now if $\mathcal M$ is $\omega$-small then so is any of the $\mathcal J^{\mathcal M}{\beta}$ for any $\beta$ but if $\mathcal M$ is properly small then it could be that there is some $\beta$ for which we have $\mathcal J^{\mathcal M}_{\beta} \models$ there are $\omega$ Woodins.

I am confused about how that can be, since in the definition of properly small we also have that the premouse has to be $\omega$-small. Can someone show me an example? Thx

Let $\mathcal M$ be a premouse. $\mathcal M$ is said to be $\omega$-small if and only if whenever we have that $\kappa=crit(E)$ for some extender $E$ on the sequence of $\mathcal M$, then $\mathcal J^{\mathcal M}_{\kappa} \not \models$ There are $\omega$-many Woodins. $\mathcal M$ is said to be properly small if and only if $\mathcal M$ is $\omega$-small and $\mathcal M \models$ there is a largest cardinal and $ZF^-$ (i.e without the power set axiom) and there are no Woodins.

If $\mathcal M$ is $\omega$-small then so is any of the $\mathcal J^{\mathcal M}{\beta}$ for any $\beta$ but if $\mathcal M$ is properly small then it could be that there is some $\beta$ we have that $\mathcal J^{\mathcal M}_{\beta} \models$ there are $\omega$ Woodins.

I am confused about how that can be, since in the definition of properly small we also have that the premouse has to be $\omega$-small. For instance one could take $\beta$ to be the ordinal height of a properly small premouse $\mathcal M$ and then $\mathcal J^{\mathcal M}_{\beta}= \mathcal M$ doesn't think there are any Woodin cardinals at all. Can someone show me an example? Thx

Let $\mathcal M$ be a premouse. $\mathcal M$ is said to be $\omega$-small if and only if whenever we have that $\kappa=crit(E)$ for some extender $E$ on the sequence of $\mathcal M$, then $\mathcal J^{\mathcal M}_{\kappa} \not \models$ There are $\omega$-many Woodins. $\mathcal M$ is said to be properly small if and only if $\mathcal M$ is $\omega$-small and $\mathcal M \models$ "there is a largest cardinal and $ZF^-$ (i.e without the power set axiom) and there are no Woodins".

Now if $\mathcal M$ is $\omega$-small then so is any of the $\mathcal J^{\mathcal M}{\beta}$ for any $\beta$ but if $\mathcal M$ is properly small then it could be that there is some $\beta$ we have that $\mathcal J^{\mathcal M}_{\beta} \models$ there are $\omega$ Woodins.

I am confused about how that can be, since in the definition of properly small we also have that the premouse has to be $\omega$-small. Can someone show me an example? Thx

Let $\mathcal M$ be a premouse. $\mathcal M$ is said to be $\omega$-small if and only if whenever we have that $\kappa=crit(E)$ for some extender $E$ on the sequence of $\mathcal M$, then $\mathcal J^{\mathcal M}_{\kappa} \not \models$ There are $\omega$-many Woodins. $\mathcal M$ is said to be properly small if and only if $\mathcal M$ is $\omega$-small and $\mathcal M \models$ there is a largest cardinal and $ZF^-$ (i.e without the power set axiom) and there are no Woodins.

If $\mathcal M$ is $\omega$-small then so is any of the $\mathcal J^{\mathcal M}{\beta}$ for any $\beta$ but if $\mathcal M$ is properly small then it could be that there is some $\beta$ we have that $\mathcal J^{\mathcal M}_{\beta} \models$ there are $\omega$ Woodins.

I am confused about how that can be, since in the definition of properly small we also we that the premouse has to be $\omega$-small. Can someone show me an example? Thx

Let $\mathcal M$ be a premouse. $\mathcal M$ is said to be $\omega$-small if and only if whenever we have that $\kappa=crit(E)$ for some extender $E$ on the sequence of $\mathcal M$, then $\mathcal J^{\mathcal M}_{\kappa} \not \models$ There are $\omega$-many Woodins. $\mathcal M$ is said to be properly small if and only if $\mathcal M$ is $\omega$-small and $\mathcal M \models$ there is a largest cardinal and $ZF^-$ (i.e without the power set axiom) and there are no Woodins.

If $\mathcal M$ is $\omega$-small then so is any of the $\mathcal J^{\mathcal M}{\beta}$ for any $\beta$ but if $\mathcal M$ is properly small then it could be that there is some $\beta$ we have that $\mathcal J^{\mathcal M}_{\beta} \models$ there are $\omega$ Woodins.

I am confused about how that can be, since in the definition of properly small we also have that the premouse has to be $\omega$-small. For instance one could take $\beta$ to be the ordinal height of a properly small premouse $\mathcal M$ and then $\mathcal J^{\mathcal M}_{\beta}= \mathcal M$ doesn't think there are any Woodin cardinals at all. Can someone show me an example? Thx