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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
$\mathfrak{p}\,<\,\mathfrak{t}$
.
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