# Sh:700a

- Shelah, S.
*Are \mathfrak a and \mathfrak d your cup of tea? Revisited.*Preprint. arXiv: 2108.03666

Revised version of [Sh:700] -
Abstract:

This was non-essentially revised in late 2020. 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 §2,§3 which say that (in §3 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 §2,§3 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].

- Version 2021-08-16 (45p)

@article{Sh:700a, author = {Shelah, Saharon}, title = {{Are $\mathfrak a$ and $\mathfrak d$ your cup of tea? Revisited.}}, note = {\href{https://arxiv.org/abs/2108.03666}{arXiv: 2108.03666} Revised version of [Sh:700]}, arxiv_number = {2108.03666}, refers_to_entry = {Revised version of [Sh:700]} }