A few years ago, in this paper, I introduced the following reflection principle:
Definition. $R_2(\theta,\kappa)$ asserts that for every function $f:E^\theta_{<\kappa}\rightarrow\kappa$, there exists some $j<\kappa$ for which the following set is nonstationary: $$A_j:=\{\delta\in E^\theta_\kappa\mid f^{-1}[j]\cap\delta\text{ is nonstationary}\}.$$
I wrote there that by a theorem of Magidor, $R_2(\aleph_2,\aleph_1)$ is consistent modulo the existence of a weakly compact cardinal, and at the end of that paper, I asked (Question 3) what is the consistency strength of $R_2(\aleph_2,\aleph_1)$.
People I asked about this mentioned Magidor’s other result that if there exist two stationary subset of $E^{\aleph_2}_{\aleph_0}$ that do not reflect simultaneously, then $\aleph_2$ is weakly compact in $L$, however, to address $R_2(\aleph_2,\aleph_1)$, a more complicated counterexample is needed.
In this post, I will answer my own question, proving that the consistency strength of $R_2(\aleph_2,\aleph_1)$ is exactly that of a weakly compact cardinal.
Proposition 1. If $R_2(\aleph_2,\aleph_1)$ holds, then $\aleph_2$ is weakly compact in $L$.
Proof. As mentioned in a previous blog post, if $\aleph_2$ is not weakly compact in $L$, then $\square(\aleph_2)$ holds. Now, appeal to the next proposition. $\blacksquare$
Proposition 2. If $\square(\theta)$ holds, then $R_2(\theta,\kappa)$ fails for every regular uncountable cardinal $\kappa<\theta$.
Proof. Let $\kappa<\theta$ be arbitrary regular uncountable cardinals. By Lemma 3.2 of this paper, we may fix a sequence $\langle C_\delta\mid\delta<\theta\rangle$ such that:
- $C_\delta$ is a club in $\delta$ for all limit $\delta<\theta$;
- if $\alpha\in\text{acc}(C_\delta)$, then $C_\alpha=C_\delta\cap\alpha$;
- $S_\gamma:=\{\delta\in E^\theta_\kappa\mid \min(C_\delta)=\gamma\}$ is stationary for all $\gamma<\theta$.
(here, $\text{acc}(C)=\{\alpha\in C\mid \sup(C\cap\alpha)=\alpha>0\}$.)
Now, define $f:E^\theta_{<\kappa}\rightarrow\kappa$ by stipulating:$$f(\alpha):=\begin{cases}\min(C_\alpha),&\text{if }\min(C_\alpha)<\kappa\\0,&\text{otherwise.}\end{cases}$$
Finally, let $j<\kappa$ be arbitrary. To prove that $A_j$ is not nonstationary, let us show that it contains the stationary set $S_{j}$.
Towards a contradiction, suppose that $\delta\in S_{j}\setminus A_j$. Then $f^{-1}[j]\cap\delta$ is stationary, and we may pick some $\alpha\in f^{-1}[j]\cap\text{acc}(C_\delta)$.
- By $\alpha\in f^{-1}[j]$, we have either $\min(C_\alpha)=f(\alpha)< j$ or $\min(C_\alpha)\ge\kappa$.
- By $\alpha\in\text{acc}(C_\delta)$ and $\delta\in S_j$, we have $\min(C_\alpha)=\min(C_\delta\cap\alpha)=\min(C_\delta)=j$.
This is a contradiction. $\blacksquare$
