A Comment on ``$\mathfrak{p} < \mathfrak{t}$''

http://dx.doi.org/10.4153/CMB-2009-033-4
Canad. Math. Bull. 52(2009), 303-314

  Published:2009-06-01
 Printed: Jun 2009

Dealing with the cardinal invariants ${\mathfrak p}$ and ${\mathfrak t}$ of the continuum, we prove that ${\mathfrak m}={\mathfrak p} = \aleph_2\ \Rightarrow\ {\mathfrak t} =\aleph_2$. In other words, if ${\bf MA}_{\aleph_1}$ (or a weak version of this) holds, then (of course $\aleph_2\le {\mathfrak p}\le {\mathfrak t}$ and) ${\mathfrak p}=\aleph_2\ \Rightarrow\ {\mathfrak p}={\mathfrak t}$. The proof is based on a criterion for ${\mathfrak p}<{\mathfrak t}$.