Let $\mathbb S\_\kappa$ be the standard forcing for $\square\_\kappa$ by initial segments. This is $(\kappa+1)$-strategically closed.



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> **Observation:** Let $T \subseteq \kappa^+$ be stationary. If $T$ concentrates on $\mathrm{cof}({<}\kappa)$,then $\mathbb S\_\kappa$ forces $\diamondsuit(T)$. If $T$ concentrates on $\mathrm{cof}(\kappa)$ and $T$ is approachable, then $\mathbb S\_\kappa$ forces $\diamondsuit(T)$.
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The proof is like the argument that adding a Cohen subset of a regular cardinal $\kappa$ forces $\diamondsuit\_\kappa$. The coding goes by fixing bijections $f\_\alpha : \alpha \to \kappa$ for $\alpha <\kappa^+$ and taking the diamond sequence to be $a\_\alpha = \{ \beta<\alpha: \alpha+f\_\alpha(\beta)+1 \in C\_{\alpha+\kappa} \}$. It works because of the freedom we have in choosing $C\_{\alpha+\kappa}$. In the $\mathrm{cof}(\kappa)$-concentration case, we use an elementary submodel argument plus approachability so that we can work with a chain of conditions of length only $\kappa+1$.


**Question 1:** This seems like it might have been observed before. Does it appear in a paper?


**Question 2:** Can we eliminate the approachability assumption?



Forcing square introduces diamond
Regarding Question 2, the assumption of approachability is necessary (or, more precisely, the assumption that $T$ has a stationary subset that is approachable), at least if we have $2^\kappa = \kappa^+$ in the ground model. The reason is that, if $T$ does not have a stationary subset that is approachable, then it becomes non-stationary after forcing with $\mathbb{S}\_\kappa$, so $\diamondsuit(T)$ necessarily fails.


To see this, note that if $2^\kappa = \kappa^+$, then we can fix in $V$ an enumeration $\vec{a} = \langle a\_\alpha \mid \alpha < \kappa^+ \rangle$ of all bounded subsets of $\kappa^+$. Then a set $S \subseteq \kappa^+$ is in $I[\kappa^+]$ if and only if there is a club $C \subseteq \kappa^+$ such that $\gamma$ is approachable with respect to $\vec{a}$ for every $\gamma \in S \cap C$. This is because, if $\vec{b} = \langle b\_\alpha \mid \alpha < \kappa^+ \rangle$ is any other sequence of bounded subsets of $\kappa^+$, then there is a club of $\delta < \kappa^+$ such that all entries in $\langle b\_\alpha \mid \alpha < \delta \rangle$ appear in $\langle a\_\alpha \mid \alpha < \delta \rangle$. But now the assumption that $T$ does not have a stationary subset in $I[\kappa^+]$ means that there is in fact a club $C \subseteq \kappa^+$ such that, for every $\gamma \in T \cap C$, $\gamma$ is *not* approachable with respect to $\vec{a}$. Forcing with $\mathbb{S}\_\kappa$ does not add any new bounded subsets to $\kappa^+$, so if $G$ is generic, then $\vec{a}$ remains an enumeration of all bounded subsets of $\kappa^+$ in $V[G]$ and every ordinal $\gamma < \kappa^+$ is approachable with respect to $\vec{a}$ in $V[G]$ if and only if it is approachable with respect to $\vec{a}$ in $V$. However, $\square\_\kappa$ holds in $V[G]$, so also $\mathrm{AP}\_\kappa$ holds. Then, in $V[G]$, $T \cap C \in I[\kappa^+]$, but every ordinal in this set is not approachable with respect to $\vec{a}$ so, there must be a club $D \subseteq \kappa^+$ disjoint from $T \cap C$. Then $C \cap D$ is disjoint from $T$, so $T$ is non-stationary.


Regarding Question 1, I made a similar but slightly different observation in my paper "[Aronszajn trees, square principles, and stationary reflection](http://people.vcu.edu/~cblambiehanso/squares_and_reflection.pdf)" (see Lemma 3.11). I was forcing $\square(\lambda)$ rather than $\square\_\kappa$, which slightly simplifies things. I just state the lemma for $\diamondsuit(\lambda)$, but the same argument works for $\diamondsuit(T)$ for any stationary $T \subseteq \lambda$. Approachability is not needed there, since the forcing notion for $\square(\lambda)$ is fully $\lambda$-strategically closed. I'm sure this had been recognized before, but I haven't seen it elsewhere in print.


