Are $\mathfrak a$ and $\mathfrak d$ your cup of tea? Revisited

Saharon Shelah

Published 2021-08-08Version 1

This is a revised version (of late 2020) of [Sh:700], which is arXiv:math/0012170 . First point is noting that the proof of Theorem 4.3 in [Sh:700], which says that the proof giving the consistency $ \mathfrak{b} = \mathfrak{d} = \mathfrak{u} < \mathfrak{a} $ also gives $ \mathfrak{s} = \mathfrak{d} $. The proof uses a measurable cardinal and a c.c.c. forcing so it gives large $ \mathfrak{d} $ and assumes a large cardinal. Second point is adding to the results of \S2,\S3 which say that (in \S3 with no large cardinals) we can force $ {\aleph_1} < \mathfrak{b} = \mathfrak{d} < \mathfrak{a}$. We like to have $ {\aleph_1} < \mathfrak{s} \le \mathfrak{b} = \mathfrak{d} < \mathfrak{a} $. For this we allow in \S2,\S3 the sets $ K_t $ to be uncountable; this requires non-essential changes. In particular, we replace usually $ {\aleph_0}, {\aleph_1} $ by $ \sigma , \partial $. Naturally we can deal with $ \mathfrak{i} $ and similar invariants. Third we proofread the work again. To get $ \mathfrak{s} $ we could have retained the countability of the member of the $ I_t$-s but the parameters would change with $ A \in I_t$, well for a cofinal set of them; but the present seems simpler. We intend to continue in [Sh:F2009].