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

TL;DR

This paper revisits and refines results on cardinal invariants of the continuum, particularly focusing on the relationships between alnd al, using forcing and large cardinals, and extends previous work with new forcing techniques.

Contribution

It provides new forcing constructions to manipulate the relationships between alnd al without large cardinals, and revises earlier proofs for clarity and generality.

Findings

Established consistency results for alnd al with al.

Extended forcing techniques to include uncountable sets in the construction.

Clarified the proof of al=al=al and its reliance on large cardinals.

Abstract

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].