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| = |\...

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 $...

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,...

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) \...

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 ...

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 ...

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 $\{ ...

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 ...

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 ...

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 ...

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 ...

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 $\...

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 $\kappa$-...

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 $\kappa$-...

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 $\kappa$,...

It seems unlikely. Let me use the following consequence of partial global square (which appears in my joint paper with Garti). This partial sequence is consistent with the existence of very large ...

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 ...