Definition : a cardinal is called ineffable if for every sequence of sets $\left<A_\alpha\ \big|\ \alpha<\kappa\right>$ s.t $A_\alpha \subset\alpha$ there is a set $B\subset\kappa$ s.t
$$S=\left\{\alpha<\kappa\ \bigg|\ B\cap\alpha=A_\alpha\right\}$$
is stationary.
I want to show that I can assume that $S$ contains only strong inaccessible cardinals. I have trouble showing that there is even a strong inaccessible below $\kappa$.
Any help would be appreciated.
The previous proof has a critical error.
First, I claim that every ineffable cardinal $\kappa$ is inaccessible. By definition, we assume that $\kappa$ is regular. Thus it suffices to show that $\kappa$ is a strong limit.
Assume the contrary that there is $\lambda<\kappa$ such that $2^\lambda\ge\kappa$. Let $\langle A_\alpha\mid \alpha<\kappa\rangle$ be $\kappa$ distinct subsets of $\lambda$. (We may assume that $A_\alpha\subseteq\alpha$ for all $\alpha$.) Take $B$ such that $S=\{\alpha<\kappa\mid B\cap \alpha = A_\alpha\}$ is stationary. Then the sequence $$\langle A_\alpha\mid \alpha\in S\rangle$$ is a strictly increasing $\kappa$-sequence of subsets of $\lambda$, but such a sequence cannot exist: consider the sequence $\beta_\xi=\min(A_{\xi'}\setminus A_\xi)$ for $\xi \in S$, where $\xi'$ is the next element of $\xi$ in $S$.
Now we will prove the main problem: For each $\langle A_\alpha \mid\alpha<\kappa\rangle$, take $\overline{A}_\alpha$ such that $\overline{A}_\alpha=A_\alpha$ if $\alpha$ is regular, and choose any cofinal subset of $\alpha$ of ordertype $\operatorname{cf}\alpha$ if $\alpha$ is not regular.
Let $B\subseteq \kappa$ be such that $S=\{\alpha<\kappa\mid B\cap \alpha = \overline{A}_\alpha\}$ is stationary. We claim that the set $\overline{S}$ of all regular cardinals in $S$ is stationary. If it holds, then by the inaccessibility of $\kappa$, the set of all strong limit cardinals $C$ is a club. Hence $\overline{S}\cap C$ is the desired set.
Assume the contrary that $\overline{S}$ is non-stationary. Then $S\setminus \overline{S}$ is stationary, and its elements are non-regular cardinals. Now consider the function $f:S\setminus \overline{S}\to\kappa$ defined by $f(\alpha)=\operatorname{cf}\alpha$. Then $f$ is regressive, so there is a stationary $T\subseteq S\setminus \overline{S}$ such that $f\upharpoonright T$ is constant with value, say, $\gamma$.
Note that for every $\alpha\in T$, $\alpha\cap B$ is a cofinal subset of ordertype $\gamma$. Hence $\langle \alpha\cap B\mid \alpha\in T\rangle$ forms a strictly increasing sequence of subsets, all of them has ordertype $\gamma$, which is impossible since $\alpha\cap B$ is an initial segment of $\beta\cap B$ if $\alpha<\beta$.
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