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Search options questions only not deleted user 1682
9 votes
2 answers
462 views

Relation between $\neg \square(\kappa)$ and the tree property at $\kappa$.

If $\kappa$ is an inaccessible cardinal then the tree property at $\kappa$ is equivalent to weak compactness of $\kappa$, which implies that $\square(\kappa)$ fails---that is, that every coherent sequ …
Trevor Wilson's user avatar
5 votes
2 answers
613 views

A weak (?) form of Shelah cardinals

The following definition of a large cardinal property combines parts of the definitions of "Shelah cardinal" and "Woodin cardinal": A cardinal $\kappa$ is weakly Shelah if for all $f : \kappa \to \ka …
Trevor Wilson's user avatar
10 votes
0 answers
331 views

Absoluteness of "$\kappa$-homogeneously Suslin" for sets of reals

What is known about the absoluteness, or lack thereof, of the notion of "$\kappa$-homogeneously Suslin" for sets of reals? For example, if $A$ is $\kappa$-homogeneously Suslin and $\lambda > \kappa$ …
Trevor Wilson's user avatar
13 votes
1 answer
437 views

Does this consequence of measurability in terms of games of length $\omega+1$ imply measurab...

For any two structures $\mathcal{M}$ and $\mathcal{N}$ in the same first-order language $\mathcal{L}$ and any ordinal $\theta$, let $G_\theta(\mathcal{M},\mathcal{N})$ be the two-player game of perfec …
Trevor Wilson's user avatar
9 votes
1 answer
581 views

Can $\omega_1$ be supercompact?

Is "ZF + $\omega_1$ is supercompact" consistent relative to "ZFC + there is a supercompact cardinal"? In particular, if $\delta$ is supercompact, does it remain so in $V(\mathbb{R} \cap V[G])$ where …
Trevor Wilson's user avatar
9 votes
0 answers
291 views

ZF + "every Suslin set of reals is ${\bf \Sigma}^1_2$"

What is known about the theory ($\ast$) ZF + "every Suslin set of reals is ${\bf \Sigma}^1_2$"? By "reals" I mean elements of the Baire space $\omega^\omega$. For a cardinal $\kappa$, a set of rea …
Trevor Wilson's user avatar
6 votes
2 answers
386 views

Measures that are not OD

Is anything known about the consistency strength of the statement: "There is a normal measure (on a cardinal) that is not ordinal-definable"? In particular, is it consistent relative to the existenc …
Trevor Wilson's user avatar
6 votes
2 answers
434 views

Stationary many subsets of $\kappa^+$ whose order type is a cardinal and whose intersection ...

Is anything known about the consistency strength of the following statement? $\kappa$ is a Mahlo cardinal and there is a stationary set of $a \in \mathcal{P}_\kappa(\kappa^+)$ such that $a \cap \kap …
Trevor Wilson's user avatar
17 votes
2 answers
1k views

Can measures be added by forcing?

The Lévy-Solovay theorem says that small forcings do not create measures. J.D. Hamkins has generalized this to a larger class of forcings called gap forcings. I would assume this cannot be generaliz …
Trevor Wilson's user avatar
9 votes
1 answer
624 views

Homogeneous Namba-like forcing

Let $\kappa \ge \aleph_3$ be a regular cardinal that is countably closed ($\alpha^\omega < \kappa$ for every $\alpha < \kappa$.) I'm mostly interested in the case that $\kappa$ is strongly inaccessib …
Trevor Wilson's user avatar
7 votes
0 answers
238 views

Countable choice in $L(\mathbb{R}^*_G)$

Let $\lambda$ be a singular strong limit cardinal and let $G \subset \text{Col}(\omega,\mathord{<}\lambda)$ be a $V$-generic filter. Let $\mathbb{R}^*_G = \bigcup_{\alpha < \lambda} \mathbb{R}^{V[G \ …
Trevor Wilson's user avatar
6 votes
0 answers
446 views

Inaccessible cardinals and the perfect set property for coanalytic sets

I am wondering who proved the following fact: ($\ast$) If $\omega_1$ is not inaccessible in $L$, then there is an uncountable coanalytic set of reals without a perfect subset. I have been unable to …
Trevor Wilson's user avatar
4 votes
1 answer
196 views

A version of the Martin–Solovay tree for $H_\kappa$

Consider a fixed $\Pi^1_2$ property of reals, $A(x)$. Is it true that relative to a regular cardinal $\kappa$, one can define a version of the Martin–Solovay tree $T_2$ for $A$ with the following pro …
Trevor Wilson's user avatar