3
$\begingroup$

Let $\kappa$ be regular and $\lambda\geq\kappa$. For $f, g\in\kappa^\lambda$ say that $f\le^* g$ if the set $\{\gamma<\lambda:f(\gamma)>g(\gamma)\}$ has size less than $\kappa$. Set

$\mathfrak{b}_\kappa^\lambda:=\min\{|F|:F\subseteq \kappa^\lambda\text{ and }\neg\exists y\in \kappa^\lambda\forall x\in F(x\leq^* y)\}$,

$\mathfrak{d}_\kappa^\lambda:=\min\{|D|:D\subseteq \kappa^\lambda\text{ and }\forall x\in \kappa^\lambda\exists y\in D(x\leq^* y)\}$.

I'm studying the cardinals $\mathfrak{b}_\kappa^\lambda$ and $\mathfrak{d}_\kappa^\lambda$. My question what forcing should I use to use to increase $\mathfrak{b}_\kappa^\lambda$ and $\mathfrak{d}_\kappa^\lambda$?

For example: If $\lambda\geq\kappa$, $\mu>\lambda$ and $\mathrm{cf}(\mu)>\lambda$. What forcing can I use for $\mathfrak{d}_\kappa^\lambda\geq\mu$?

$\endgroup$
5
$\begingroup$

Short version: an appropriate version of Cohen forcing will increase $\mathfrak d^\lambda_\kappa$.

With more details: Consider the forcing $Q=Q(\mu,\kappa,\mathord<\kappa)$, the set of all partial functions from $\mu$ (equivalently, from $\mu\times \lambda$) into $\kappa$ of size $<\kappa$. The generic function $g:\mu\times\lambda\to \kappa$ induces a family $(g_i:i<\mu)$, each $g_i$ a function from $\lambda $ to $\kappa$.

Assuming $\kappa^{<\kappa} = \kappa$, the forcing notion $Q$ has the $\kappa^+$-cc, and adds no bounded sets to $\kappa$, hence preserves all cardinals.

I claim that $Q$ forces $ {\mathfrak d^\kappa_\lambda \ge \mu}$. Indeed, any function $f\colon \lambda\to \kappa$ in the extension already lives in a $Q(A, \kappa, \mathord<\kappa)$-extension, for some $A$ of size $\lambda$, as the values of $f$ are decided by a family of $\lambda$ many (labelled) antichains. Such a function $f$ cannot be forced to dominate any $g_i$ for $i\notin A$. Hence if you have fewer than $\mu$ many functions, you will find an index $i<\mu$ such that these functions cannot dominate $g_i$.

$\endgroup$
3
$\begingroup$

A partial answer concerning $\mathfrak{b}_\kappa^\lambda$:

Assume $\lambda > \kappa$, regular. I claim $\mathfrak{b}_\kappa^\lambda =\kappa$:

It is obvious, why $\mathfrak{b}_\kappa^\lambda <\kappa$ is impossible.

The family $(f_\alpha \equiv \alpha)_{\alpha < \kappa}$ is unbounded. Assume $g$ dominates all $f_\alpha$. As $\lambda$ is regular, there exists $\beta < \kappa$ such that $\{i < \lambda \colon g(i) = \beta\}$ has size $\lambda$. So $g$ does not dominate $f_{\beta +1}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.