Canonical bases in superstable theories The following claim appears on
https://modeltheory.fandom.com/wiki/Stationary_type_(stability_theory)

If $T$ is superstable, then the canonical base of any stationary type is in the definable closure of a finite set. Moreover, this property characterizes superstability.

There is no proof.
The same claim appears in this question again without a proof.
Can anyone help with a hint, or a sketch of the proof?
 A: Theorem: Let $T$ be a complete theory. The following are equivalent:

*

*$T$ is stable, and for all sets $B$ and all types $p\in S(B)$, there is a finite set $A\subseteq B$ such that $p$ does not fork over $A$.

*$T$ is superstable, i.e., $T$ is stable in $\kappa$ for all $\kappa\geq 2^{|T|}$.

*$T$ is stable in $\kappa$ for some $\kappa$ such that $\kappa^{\aleph_0} > \kappa$.

I'll sketch a proof.
$(1)\Rightarrow (2)$: Let $\kappa$ be such that $\kappa\geq 2^{|T|}$. To show that $T$ is stable in $\kappa$, it suffices to show that for every model $M$ with $|M| = \kappa$, we have $|S^1(M)| \leq \kappa$. By $(1)$, for every $p\in S^1(M)$, there is some finite set $A_p\subseteq M$ such that $p$ does not fork over $A_p$. Note that the number of finite subsets of $M$ is $\kappa$. By stationarity (since $T$ is stable), $p$ is the unique non-forking extension of its restriction to $\mathrm{acl}^{\mathrm{eq}}(A_p)$. Since $A_p$ is finite, $|\mathrm{acl}^{\mathrm{eq}}(A_p)| \leq |T|$, so $|S^1(\mathrm{acl}^{\mathrm{eq}}(A_p))|\leq 2^{|T|}$. Now each type $p\in S^1(M)$ is uniquely determined by (a) the finite subset $A_p$ and (b) the restrction $p|_{\mathrm{acl}^{\mathrm{eq}}(A_p)}$, so $|S^1(M)| \leq \kappa\cdot 2^{|T|} = \kappa$.
$(2)\Rightarrow (3)$: Recall that for any infinite cardinal $\kappa$, $\kappa^{\mathrm{cf}(\kappa)} > \kappa$. So for this direction, it suffices to note that there are arbitrarily large cardinals (larger  than $2^{|T|}$) with cofinality $\omega$. For example, take $\kappa = \beth_{\omega}(|T|)$.
$(3)\Rightarrow (1)$: Let $\kappa$ be such that $\kappa^{\aleph_0}>\kappa$ and $T$ is stable in $\kappa$. Then $T$ is stable.
Assume for contradiction that there is some set $B$ and some type $p\in S(B)$ such that $p$ forks over every finite subset of $B$. Let $A_0 = \varnothing$. Given $A_n$, since $p$ forks over $A_n$, there is some finite set $A_{n+1}$ containing $A_n$ such that the restriction $p|_{A_{n+1}}$ forks over $n$. In this way, we build a "forking chain" $(A_n)_{n\in \omega}$ of finite subsets of $B$ by induction.
This is the point where I have to get a bit sketchy. Turn the forking chain $(A_n)_{n\in \omega}$ of finite sets into a $\kappa$-branching tree of height $\omega$, $(A_\eta)_{\eta\in \kappa^{<\omega}}$, such that a path through the tree $(A_{\nu|_n})_{n\in \omega}$  for $\nu\in \kappa^\omega$ has the same type over the empty set as the forking chain $(A_n)_{n\in \omega}$, and such that sets in the tree are independent whenever possible. We do this by induction as follows: let $A_{\langle\rangle} = A_0$. Given $A_\eta$, define $A_{\eta\alpha}$ for $\alpha\in \kappa$ by induction, picking $A_{\eta\alpha}$ so that $\mathrm{tp}(A_{\eta\alpha}/(A_{\eta|_m})_{m\leq|\eta|})$ matches
$\mathrm{tp}(A_{n+1}/(A_m)_{m\leq n})$ and so that $A_{\eta\alpha}$ is free over $A_\eta$ from everything picked so far.
Now let $A^*$ be the union of all the sets in the tree. Since all these sets are finite, $|A^*| = |\kappa^{<\omega}| = \kappa$. For each path through the tree, $\nu\in \kappa^\omega$, let $A_\nu = (A_{\nu|_n})_{n\in \omega}$. Since $A_\nu$ has the same type over the empty set as the forking chain $(A_n)_{n\in \omega}$, there is a type $p_\nu\in S^1(A_\nu)$ such that $p_\nu|_{A_{\nu|_{n+1}}}$ forks over $A_{\nu|_n}$ for all $n\in \omega$. Let $q_\nu\in S^1(A^*)$ be a non-forking extension of $p_\nu$.
Now the number of paths through the tree is $\kappa^{\aleph_0}>\kappa = |A^*|$, so to contradict the stability of $T$ in $\kappa$, it remains to show that $q_\nu \neq q_{\nu'}$ when $\nu\neq \nu'$. The idea is that $q_\nu$ can be distinguished from $q_{\nu'}$ by the fact that $q_\nu$ forks at every step along the path $\nu$.
If you'd like to see more details, I wrote them up long ago in this document (on p. 8), which could be helpful to you. [Note that I wrote that this implication requires countable language, but that's because for some reason I wrote the statement for $\kappa$ such that $\kappa^{|T|}>\kappa$. The argument with $\kappa^{\aleph_0}>\kappa$ goes through as written in an arbitrary language.]
