Yair Hayut
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Claim: $|V_\alpha \cap L| = |\alpha|$ for every $\alpha \geq \omega$ implies that $0^\#$ exists. Proof: Let's assume, toward contradiction, that $0^\#$ doesn't exist that $|V_\alpha \cap L| = |\... View answer 13 votes This is a partial answer. I will show that if$\delta$is Woodin then$\diamondsuit_\delta$holds. Claim: Any Woodin cardinal is subtle. Proof: Let$\delta$be a Woodin cardinal. Let$\vec{A} = \...

Assuming $0^\#$ doesn't exist, it's consistent to get a negative answer to those questions: Assume $(2^{\aleph_0})^V > \aleph_2$ and that ${\aleph_n}^V = {\aleph_n}^L$ for at least $n=1,2,3$. I ...

The existence of such trees is independent of ZFC. On one hand, if $CH$ holds then $\omega_1^{<\omega_1}$ (and in fact - any $\sigma$-closed $\omega_1$-branching tree) cannot be partitioned into $... View answer 11 votes There is no such forcing that preserves$\lambda^+$. Since$\lambda$is measurable,$2^{<\lambda} = \lambda$and therefore$\square_{\lambda,\lambda}$holds in$V$. Since$\lambda^{+}$is preserved,... View answer Accepted answer 10 votes The consistency strength of the failure of$(\dagger)$is an inaccessible cardinal. Building on the comment of Mohammad, if$\omega_2^V$is a successor cardinal in$L$then there is a set$X \subseteq ...

This answer addresses only the consistency strength of $\omega+1$-strategically measurable. Claim: If there is a $\omega+1$-strategically measurable cardinal then there is an inner model with a ...

Theorem: Let $\kappa$ be strongly inaccessible in $V$, such that $V \models ZFC$. If $M\models ZF$, then $L(M) \cap V_\kappa = M$. Proof: Let us prove by induction on $\alpha < \kappa$ that $L(M) \... View answer Accepted answer 10 votes As Mohammad Golshani remarked, it is possible to control the cofinality of$j(\kappa)$by iterating the forcing that adds a function$f\colon \kappa \to \kappa$which is eventually larger than any ... View answer Accepted answer 10 votes In the paper "On Lowenheim-Skolem-Tarski numbers for extension of first order logic", by Magidor and Vaananen, in Theorem 21 they state that it is consistent, relative to the existence of a ... View answer 10 votes For every$\kappa$of uncountable cofinality there is a tree$T\subseteq 2^{<\kappa}$such that$[T]=\kappa$. The tree$T$is the tree of all binary sequences$f\colon \alpha \to 2$,$\alpha <\...

The answer is yes. Force a generic $\square(\omega_1)$ sequence. This poset, $S$, is $\sigma$-closed, so it doesn't collapse $\omega_1$. By Todorcevic, in the $\omega_1$-tree obtained from the ...

This should be a comment - but it is too long: Assume that $U = U_1 = U_2$. I want to show that $\mathbb{P}_U ^2 \cong \mathbb{P}_U\times \mathbb{C}$ where $\mathbb{C}$ is the Cohen forcing. Let $\{ ... View answer Accepted answer 9 votes If$\mathbb{P}$is$\kappa$-strategically closed then there is a projection from$Col(\kappa,\mathbb{P})$onto$\mathbb{P}$. The proof is very similar to the$\kappa$-closed case: Let$\sigma$be ... View answer Accepted answer 8 votes It is consistent that the answer is positive and it is consistent that the answer is negative. Claim: There is a generic extension,$V[G]$by a weakly homogeneous forcing notion in which there is a ... View answer Accepted answer 8 votes The answer to the second question is yes, without any large cardinals assumptions. Claim: if$2^{\aleph_0}$is singular then every non-trivial$\sigma$-closed forcing of size$2^{\aleph_0}$collapses ... View answer 7 votes If there is a subtle cardinal then such$\phi$doesn't exist. An subtle cardinal is a cardinal$\kappa$such that for every sequence of sets$\langle S_\alpha \mid \alpha < \kappa\rangle$such ... View answer 7 votes Claim: NGB +$\exists j\colon V\to V$is equiconsistent with NGB +$\exists j\colon V\to V + \neg RR$. In order to prove the left to right direction, we will start with a model of NGB +$\exists j\...

The consistency strength of the failure of $\square(\kappa)$ for non-weakly Mahlo inaccessible cardinal $\kappa$ (in particular, for the first strongly inaccessible) is higher than weakly compact. For ...

The answer is yes (at least for $\kappa = \omega_1$): Shelah and Harrington showed that you can force that every stationary subset of $S_{\omega}^{\omega_2}$ reflects starting with a Mahlo cardinal. ...

Here are a few observations about CORE. Claim: It is consistent that CORE is not pairwise upwards directed. Proof: Let $\mathbb{P}_0$ be the class forcing for the Easton product of the Cohen forcing $\... View answer 6 votes In my paper "Partial strong compactness and squares", I investigate those questions. The main results of this paper are: If every$\kappa$-complete filter on$\kappa$can be extended to a$\kappa$-... View answer Accepted answer 6 votes About question 1: If$\kappa$is not weakly compact in$L$, then there is an$\aleph_2$-Suslin tree in$L[G]$. In$L$there is a$\kappa$-Suslin tree,$T$, and since Mitchell's forcing is$\kappa$-... View answer Accepted answer 6 votes There is no forcing that satisfies$(*)$. Let$r$be a new real,$r \in M[G]\cap \mathcal{P}(\omega) \setminus M$. Take$t = \{n < \omega \mid n + 1\in r\}$, and let$a, b\in M$such that$t = (a\...

Similarly to Asaf's comment, let $\kappa = \kappa(\omega_1^L)$ (we assume that it exists). Then in $L_\kappa$, there is $\alpha$-Erdős cardinal for every $L$-countable $\alpha$ (since being $\alpha$-...

GCH implies that $\mathbb{Q}_\alpha$ is $\alpha$-distributive in the generic extension by $\mathbb{P}_\alpha$. Note that $\Vdash_{\mathbb{Q}_\alpha} \check{\mathbb{P}}_\alpha$ is $\check\alpha$.c....

Assuming that there is an $\omega_1$-dense $\sigma$-complete ideal, $I$, on $\omega_1$ (such that every positive set can be partitioned into $\omega_1$ positive sets), we can generalize Joel's ...

Assume that a regular cardinal $\kappa > \omega_1$ has the property that for every $\kappa$.c.c. forcing notion $\mathbb{P}$ and name of set of ordinals $x \in V^{\mathbb{P}}$ smaller than $\kappa$,...

Let me start by observing that this property is equivalent to a more standard property: Claim: Let $M\subseteq V$ be a transitive model of $\mathrm{ZFC}$ and let $\mu \in M$ be a cardinal in $V$. The ...