Motivated by this question. For us "$\kappa$ is weakly compact" means any tree that has underlying set $\kappa$, height $\kappa$ and levels $<\kappa$ has a branch.

The strength of $\mathsf{ZF+}$ "$\omega_1$ is weakly compact" is exactly that of a weakly compact; the forcing direction is the same construction as Jech's making $\omega_1$ measurable. But what if we consider $L(\mathbb{R})$? Feel free to add $\mathsf{DC}$ if that matters.

More broadly, is there some statement $\varphi$ such that the strength of $L(\mathbb{R})\models\varphi$ is exactly a weakly compact? Note that

$\varphi=$ there exists a countable transitive model of $\mathsf{ZFC+}$ "there is a weakly compact cardinal"

does not work, and is cheating anyway. If there turns out to be some other way of cheating, my next question would be whether there is a natural statement $\varphi$.

Motivated by this question. For us "$\kappa$ has tree property" means any tree that has underlying set $\kappa$, height $\kappa$ and levels $<\kappa$ has a branch.

The strength of $\mathsf{ZF+}$ "$\omega_1$ has tree property" is exactly that of a weakly compact; the forcing direction is the same construction as Jech's making $\omega_1$ measurable. But what if we consider $L(\mathbb{R})$? Feel free to add $\mathsf{DC}$ if that matters.

More broadly, is there some statement $\varphi$ such that the strength of $L(\mathbb{R})\models\varphi$ is exactly a weakly compact? Note that

$\varphi=$ there exists a countable transitive model of $\mathsf{ZFC+}$ "there is a weakly compact cardinal"

does not work, and is cheating anyway. If there turns out to be some other way of cheating, my next question would be whether there is a natural statement $\varphi$.

Motivated by this question. The strength of $\mathsf{ZF+}$ "$\omega_1$ is weakly compact" is exactly that of a weakly compact; the forcing direction is the same construction as Jech's making $\omega_1$ measurable. But what if we consider $L(\mathbb{R})$? Feel free to add $\mathsf{DC}$ if that matters.

More broadly, is there some statement $\varphi$ such that the strength of $L(\mathbb{R})\models\varphi$ is exactly a weakly compact? Note that

$\varphi=$ there exists a countable transitive model of $\mathsf{ZFC+}$ "there is a weakly compact cardinal"

does not work, and is cheating anyway. If there turns out to be some other way of cheating, my next question would be whether there is a natural statement $\varphi$.

Motivated by this question. For us "$\kappa$ is weakly compact" means any tree that has underlying set $\kappa$, height $\kappa$ and levels $<\kappa$ has a branch.

The strength of $\mathsf{ZF+}$ "$\omega_1$ is weakly compact" is exactly that of a weakly compact; the forcing direction is the same construction as Jech's making $\omega_1$ measurable. But what if we consider $L(\mathbb{R})$? Feel free to add $\mathsf{DC}$ if that matters.

More broadly, is there some statement $\varphi$ such that the strength of $L(\mathbb{R})\models\varphi$ is exactly a weakly compact? Note that

$\varphi=$ there exists a countable transitive model of $\mathsf{ZFC+}$ "there is a weakly compact cardinal"

does not work, and is cheating anyway. If there turns out to be some other way of cheating, my next question would be whether there is a natural statement $\varphi$.