How to show that MA implies $\mathfrak{d}=\mathfrak{c}$?

Question

I can't find a proof for this online. Can somebody help me out?

We have the dominating number $\mathfrak{d} = \min\{|\mathcal{F}|: \mathcal{F} \text{ dominating family} \}$.

Now I am supposed to use Martins Axiom to show that $\mathfrak{d}=\mathfrak{c}$.

We know that $\mathfrak{d}\leq \mathfrak{c}$, so my idea was to assume $\mathfrak{d}<\mathfrak{c}$ and find a contradiction via MA. Maybe taking partial functions of functions in $\mathcal{F}$ as poset $P$ with $p\leq q$ when $p\subseteq q$. Then we know that $P$ satisfies ccc, but from here my attempts failed.

Or is there maybe a way to show that every free filter can be extended to a P-point with MA?

Answer 1

There's no need to phrase this as a proof by contradiction. "$\mathfrak{d}=\mathfrak{c}$" is the same thing as saying "if $A\subseteq\omega^\omega$ has cardinality $<\mathfrak{c}$, then there is some $f\in\omega^\omega$ which escapes every $g\in A$." (That is, no "small" set is "sufficient.")

To prove this, you need to show how you can find such an $f$ for a given $A$. The most direct way to do this is to find some poset $\mathbb{P}$ such that:

• $\mathbb{P}$ is c.c.c.

• A (sufficiently) $\mathbb{P}$-generic filter $G$ can be viewed as an element of $\omega^\omega$ (or if you prefer, has an associated $\gamma\in\omega^\omega$).

• For all $A\subseteq\omega^\omega$ of cardinality $<\mathfrak{c}$, there is an associated set $\mathcal{D}$ of dense subsets of $\mathbb{P}$ such that

  • $\mathcal{D}$ has cardinality $<\mathfrak{c}$, and

  • any $\mathcal{D}$-generic filter through $\mathbb{P}$ is (or yields) a function escaping every element of $A$.

Thinking of dense sets as requirements that need to be met, $\mathcal{D}$ should have an element (or a family of elements) for each $f\in A$, corresponding to the "escapes $f$" requirement, as well has a family of dense sets corresponding to the "builds a function" requirement. So:

Can you find a $\mathbb{P}$ which builds a function $\omega\rightarrow\omega$, which it is "easy" to make escape a given function $\omega\rightarrow\omega$?

With luck, the $\mathbb{P}$ you come up with will be c.c.c. and will lead you quickly to a solution to the problem. If not, try a different one!

HINT: In this case, our $\mathbb{P}$ is quite simple, so don't try to make things too complicated for yourself.