Published online by Cambridge University Press: 20 November 2018
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 $\text{M}{{\text{A}}_{{{\aleph }_{1}}}}$ (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 $\mathfrak{p}\,<\,\mathfrak{t}$.