In Devlin – Constructibility on page 158f. he defines $\square_\kappa$ and $\square_\kappa'$ and then proceeds to show that these two are equivalent.
$\square_\kappa$ holds if there is a sequence $(C_\alpha\mid\alpha<\kappa^+\wedge\alpha\in\text{Lim})$ s.t.
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$C_\alpha$ is a club of $\alpha$,
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$\operatorname{cf}(\alpha)<\kappa\implies|C_\alpha|<\kappa$ and
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if $\beta$ is a limit point of $C_\alpha$, then $C_\beta=\beta\cap C_\alpha$.
$\square_\kappa'$ holds if there is a sequence $(B_\alpha\mid\alpha<\kappa^+\wedge\alpha\in\text{Lim})$ s.t.
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$B_\alpha$ is a closed subset of $\text{Lim}\cap\alpha$,
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$\operatorname{cf}(\alpha)>\omega\implies B_\alpha$ is unbounded in $\alpha$,
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$\operatorname{otp}(B_\alpha)\leq\kappa$ and
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$\beta\in B_\alpha\implies B_\beta=\beta\cap B_\alpha$
Assuming that $(B_\alpha\mid\alpha<\kappa^+\wedge\alpha\in\text{Lim})$ is a $\square_\kappa'$-sequence, he defines inductively
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$C_\alpha=\bigcup\lbrace C_\gamma\mid \gamma\in B_\alpha\rbrace$, if $\sup B_\alpha=\alpha$
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$C_\alpha=\bigcup\lbrace C_\gamma\mid \gamma\in B_\alpha\rbrace\cup\lbrace\theta_n^\alpha\mid n<\omega\rbrace$ else, where the $\theta_n^\alpha$ are strictly increasing, bigger than $\sup B_\alpha$, and cofinal in $\alpha$.
Then he says that if $\operatorname{otp}(C_\alpha)>\kappa$ then $\operatorname{otp}(B_\alpha)>\kappa$ just because $B_\alpha$ is the set of all limit points of $C_\alpha$ below $\alpha$ and because $\kappa$ is uncountable. Why is $\operatorname{otp}(C_\alpha)=\kappa+\omega$ and $\operatorname{otp}(B_\alpha)=\kappa$ not possible? Maybe i'm just being dumb here.
Maybe its useful to state the rest of the proof since it's also not that long anymore.
If $\kappa$ is regular we need that $\operatorname{otp}(B_\alpha)\leq\kappa$, because we then can assume that it's equal and have a contradiction, since then, if $\operatorname{cf}(\alpha)<\kappa$ then $\operatorname{cf}(\kappa)\leq\operatorname{cf}(\alpha)<\kappa$.
(So maybe in this case we use
3'. $\operatorname{otp}(B_\alpha)\leq\kappa$ and if $\operatorname{cf}
(\alpha)=\omega$ then $\operatorname{otp}(B_\alpha)<\kappa$
instead of 3. in the first definition.)
In the singular case, we define a sequence $(C_\alpha'\mid\alpha<\kappa^+)$ as follows.
Let $(\theta_\nu\mid\nu<\operatorname{cf}(\kappa))$ be a strictly increasing, continuous sequence sequence of limit ordinals, which is cofinal in $\kappa$.
Set $\theta_{\operatorname{cf}(\kappa)}=\kappa$. (And i think one could set $\theta_{\operatorname{cf}(\kappa)+n}=\kappa+n$ for all $n<\omega$ and $\theta_{\operatorname{cf}(\kappa)+\omega}=\kappa+\omega$)
Now let
$C_\alpha'=\lbrace\gamma\in C_\alpha\mid\operatorname{otp}(C_\alpha\cap\gamma)\geq\theta_\nu\rbrace$,
if there is a $\nu<\operatorname{cf}(\kappa)$ s.t. $\theta_\nu<\operatorname{otp}(C_\alpha)\leq\theta_{\nu+1}$. Otherwise $\operatorname{otp(C_\alpha)}=\theta_\nu$ for a limit $\nu$, then we set
$C_\alpha'=\lbrace\gamma\in C_\alpha\mid(\exists\tau<\nu)\operatorname{otp}(C_\alpha\cap\gamma)=\theta_\tau\rbrace$
Then $(C_\alpha'\mid\alpha<\kappa^+)$ is a $\square_\kappa'$-sequence (and i think this is still true even with $\theta_{\operatorname{cf}(\kappa)+n}$ and $\theta_{\operatorname{cf}(\kappa)+\omega}$ defined).
Since Devlin only uses $\square_\kappa'$ to show that $\square_\kappa$ holds in $L$, i think that it would be okay to change 3. from the definition of $\square_\kappa'$ to 3' in the regular case, because in that proof we get clubs $B_\alpha$ which have order type $\leq\kappa$ and if $\kappa$ is regular, then we should get 3'.
My question is, is everything stated here correct?
Best Answer
If $\operatorname{otp}(C_\alpha)=\kappa+\omega$, then $\operatorname{otp}(\operatorname{acc}(C_\alpha))=\kappa+1$ and $\operatorname{acc}(C_\alpha)=B_\alpha$. So i think what Devlin stated is true.