# How large are measurable cardinals of higher orders?

For each ordinal $\alpha$ define the notions of $\alpha$ - measurable cardinals and $\alpha$ - normal measures as follows:

A measure $\mu$ on a measurable cardinal $\kappa$ is a $0$-normal measure iff $\mu$ is a normal measure on $\kappa$.

A cardinal $\kappa$ is $0$-measurable iff there is a $0$-normal measure on $\kappa$.

For any ordinal $\alpha$ if the notions are defined for $\alpha$ then define them for $\alpha+1$ as follows,

A measure $\mu$ on a measurable cardinal $\kappa$ is a $(\alpha +1)$-normal measure iff $\mu$ is a normal measure on $\kappa$ such that $\mu(\{\lambda <\kappa~|~\lambda~\text{is} ~\alpha - \text{measurable}\})=1$

A cardinal $\kappa$ is $(\alpha+1)$-measurable iff there is a $(\alpha+1)$-normal measure on $\kappa$.

For any limit ordinal $\alpha$ if the notions are defined for each $\beta <\alpha$ then define them for $\alpha$ as follows,

A measure $\mu$ on a measurable cardinal $\kappa$ is a $\alpha$-normal measure iff $\mu$ is a normal measure on $\kappa$ such that for each $\beta <\alpha$ we have $\mu(\{\lambda <\kappa~|~\lambda~\text{is} ~\beta - \text{measurable}\})=1$

A cardinal $\kappa$ is $\alpha$-measurable iff there is a $\alpha$-normal measure on $\kappa$.

Question 1. How large are the $\alpha$-measurable cardinals in comparison with usual large cardinals like superstrongs, supercompacts, etc?

Question 2. Is there any ordinal $\alpha$ such that within ZFC one can prove non-existence of a $\alpha$-measurable cardinal?

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## migrated from mathoverflow.netFeb 7 '14 at 1:35

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This looks awfully like a characterization of the Mitchell order. It should be pointed that the height of the Mitchell order is at most $(2^\kappa)^+$; and of course (as in this case) in order for $\kappa$ to be even $1$-measurable it needs to have stationary many measurable below it. –  Asaf Karagila Feb 7 '14 at 0:17
1. Much smaller. 2. No. This is not research level, asking at the other site may be better. –  Andres Caicedo Feb 7 '14 at 1:10

It's pretty straightforward to show that if $\kappa$ is a 2-strong cardinal$^*$, then $V_\kappa$ thinks that for all $\alpha$, there is an $\alpha$-measurable cardinal. In particular, we can show by induction that for $\alpha<\kappa$, $\kappa$ is $\alpha$-measurable.
Let $\mathcal U$ be the normal ultrafilter on $\kappa$ defined by $X\in \mathcal U$ iff $\kappa\in j(X)$. First, note that since every ultrafilter on $\kappa$ is in $V_{\kappa+2}$, $\alpha$-measurability for $\kappa$ is absolute for $M$ - that is, $M\vDash \kappa$ is $\alpha$-measurable$"$ iff $\kappa$ is $\alpha$-measurable. So, if $\kappa$ is $\alpha$-measurable for $\alpha<\kappa$, then by the definition of $\mathcal U$, $\{\beta<\kappa: \beta$ is $\alpha$-measurable$\}\in \mathcal U$. In other words, if $\kappa$ is $\alpha$-measurable (for $\alpha<\kappa$), then it's $\alpha+1$-measurable. The limit case it similar.
This also answers your second question, since $V_\kappa\vDash ZFC$.
$^*$ a cardinal $\kappa$ is 2-strong if there is $j:V\prec M$ such that $crit(j) =\kappa$ and $V_{\kappa+2}\subseteq M$.
One can prove something significantly stronger: If $\kappa$ is $2$-strong, then it has order $(2^{\kappa})^+$ -- the largest possible. This was first observed by Solovay. –  Andres Caicedo Feb 7 '14 at 8:10