We say that $\alpha$ is $\Sigma\_n$-extendable (to $\beta$), if there is $\beta>\alpha$ such that $L\_\alpha$ is a $\Sigma\_n$ elementary submodel of $L\_\beta$.


First ordinal: the least $\alpha\_0$ such that there are ordinals $\alpha\_0<\alpha\_1<...<\alpha\_{\omega^2}$, such that each $\alpha\_i$ is $\Sigma\_2$ extendable, and for any $i<j$, $\alpha\_i$ is $\Sigma\_1$ extendable to $\alpha\_j$.


Second ordinal: the least $\alpha\_0$ such that there are ordinals $\alpha\_0<\alpha\_1<...<\alpha\_{\omega}$, such that each $\alpha\_i$ is $\Sigma\_2$ extendable, and for any $i<j$, $\alpha\_i$ is $\Sigma\_1$ extendable to $\alpha\_j$, and $\alpha\_\omega$ is the limit of $\alpha\_i, i<\omega$.


Which one of these two ordinals is larger?


I'm able to prove that if we change $\omega^2$ to any smaller ordinal, then the second one is larger, but I don't know how to go further.



Which one of the following two ordinals is larger?
The second one is larger.


Because of the conflict of terminology with "extendible", I will say that an ordinal $\alpha$ is
*somewhere $\Sigma\_n$-stable*
if there is $\beta>\alpha$ such that $L\_\alpha\preccurlyeq\_n L\_\beta$.
Given a somewhere $\Sigma\_2$-stable $\alpha$,
let $\beta\_2(\alpha)$ be the least $\beta>\alpha$ such that $L\_\alpha\preccurlyeq\_2L\_\beta$.


Let $\left<\alpha\_n\right>\_{n\leq\omega}$
be as in the second situation.
So each $\alpha\_n$ (for $n\leq\omega$)
is somewhere $\Sigma\_2$-stable,
and for all $m\leq n\leq\omega$,
$L\_{\alpha\_m}\preccurlyeq\_{1}L\_{\alpha\_n}$,
and $\alpha\_m<\alpha\_{m+1}$ and $\alpha\_\omega=\sup\_{m<\omega}\alpha\_m$.


Now note that for $m<n\leq\omega$,
we have $\beta\_2(\alpha\_m)\neq\beta\_2(\alpha\_n)$,
because otherwise letting $\beta=\beta\_2(\alpha\_m)=\beta\_2(\alpha\_n)$,
we would have $L\_{\alpha\_m}\preccurlyeq\_2 L\_{\beta}$ and $L\_{\alpha\_n}\preccurlyeq\_2 L\_\beta$,
and $\alpha\_m<\alpha\_n<\beta$.
But then actually $L\_{\alpha\_m}\preccurlyeq\_2 L\_{\alpha\_n}$,
so in fact $\beta\_2(\alpha\_m)\leq\alpha\_n$, contradiction.


Now by refining the sequence $\left<\alpha\_n\right>\_{n<\omega}$
if necessary, we may assume that $\beta\_2(\alpha\_m)<\alpha\_{m+1}$ for all $m<\omega$ (though we might lose the minimality of $\alpha\_0$). For let $\gamma\_0=\inf\_{n<\omega}\beta\_2(\alpha\_n)$ and let $n\_0<\omega$ be the unique $n$
such that $\beta\_2(\alpha\_n)=\gamma\_0$.
Now let $\gamma\_1=\inf\_{n\in(n\_0,\omega)}\beta\_2(\alpha\_n)$ and let $n\_1\in(n\_0,\omega)$ be such that $\beta\_2(\alpha\_n)=\gamma\_1$. Etc,
producing $\left<\gamma\_i,n\_i\right>\_{i<\omega}$
with $n\_i<n\_{i+1}<\omega$ and $\gamma\_i<\gamma\_{i+1}$ and $\gamma\_i=\beta\_2(\alpha\_{n\_i})$.
By reindexing, we may assume that $n\_i=i$ for all $i<\omega$, so in other words,
$\beta\_2(\alpha\_n)<\beta\_2(\alpha\_{n+1})$ for all $n<\omega$. So
$$ L\_{\alpha\_n}\preccurlyeq\_2 L\_{\beta\_2(\alpha\_n)}\in L\_{\beta\_2(\alpha\_{n+1})}, $$
and
$$L\_{\alpha\_{n+1}}\preccurlyeq\_2 L\_{\beta\_2(\alpha\_{n+1})},$$
and the existence of $\beta\_2(\alpha\_{n})$
is then a $\Sigma\_1$ statement about $\alpha\_n$ which is true in $L\_{\beta\_2(\alpha\_{n+1})}$,
and therefore true in $L\_{\alpha\_{n+1}}$,
and so $\beta\_2(\alpha\_n)<\alpha\_{n+1}$.


Let $X$ be the set of all $\alpha<\alpha\_\omega$ such that $L\_\alpha\preccurlyeq\_1 L\_{\alpha\_\omega}$
and $L\_{\alpha\_\omega}\models$"$\alpha$ is somewhere $\Sigma\_2$-stable"; note the latter condition is equivalent to saying that $\alpha$ is somewhere $\Sigma\_2$-stable
(in $V$) and $\beta\_2(\alpha)<\alpha\_\omega$).
So we have (after thinning out as above)
$\alpha\_n\in X$ for all $n<\omega$,
and in particular, $X$ is cofinal in $\alpha\_\omega$.


I claim that $X$ has ordertype $\alpha\_\omega$.
For suppose not and let $\eta<\alpha\_\omega$
be the ordertype of $X$.
Let $\beta=\beta\_2(\alpha\_\omega)$.
Then $L\_\beta\models$"There is $\widetilde{\alpha}>\eta$ such that $L\_{\widetilde{\alpha}}\preccurlyeq\_1 L$ and $\widetilde{\alpha}$
is a limit of ordinals $\alpha'<\widetilde{\alpha}$
such that $L\_{\alpha'}\preccurlyeq\_1 L\_{\widetilde{\alpha}}$ and $L\_{\widetilde{\alpha}}\models$"$\alpha'$ is somewhere $\Sigma\_2$-stable", and the ordertype of the set $\widetilde{X}$ of those ordinals is $\eta$''
(in fact this is witnessed by $\widetilde{\alpha}=\alpha\_\omega$).
This statement satisfied by $L\_\beta$
is a $\Sigma\_2$ assertion about $\eta$.
(It is not a $\Sigma\_1$ assertion about $\eta$, because of the clause "$L\_{\widetilde{\alpha}}\preccurlyeq\_1 L$".)
Therefore $L\_{\alpha\_\omega}$ satisfies
the same statement about $\eta$.
Let $\widetilde{\alpha}<\alpha\_\omega$ and $\widetilde{X}$
be a witness. In particular, $L\_{\widetilde{\alpha}}\preccurlyeq\_1 L\_{\alpha\_\omega}$.
Note then that $\widetilde{X}\subseteq X$
(in particular since if $\alpha'\in\widetilde{X}$
then $L\_{\alpha'}\preccurlyeq\_1 L\_{\widetilde{\alpha}}\preccurlyeq\_1 L\_{\alpha\_\omega}$, so $L\_{\alpha'}\preccurlyeq\_1 L\_{\alpha\_\omega}$).
But $\widetilde{X}\subseteq\widetilde{\alpha}<\alpha\_\omega$,
and it follows that $X$ has ordertype $>\eta$, a contradiction.


So $X$ has ordertype $\alpha\_\omega$.
Now let $\alpha^{(1)}\_0$ be the first ordinal you defined and $\alpha^{(2)}\_0$ the second. Maybe $\alpha^{(2)}\_0<\alpha\_0$ (because of the thinning out we did), but anyway $L\_{\alpha^{(2)}\_0}\preccurlyeq\_1 L\_{\alpha\_\omega}$. From this and the fact that $X$ has ordertype $\alpha\_\omega$, it follows that there are chains of the form required for the definition of $\alpha^{(1)}\_0$ which appear in $L\_{\alpha^{(2)}\_0}$. Therefore $\alpha^{(1)}\_0<\alpha^{(2)}\_0$.


