It is proved in $\mathsf {ZF}$ (without the axiom of choice) that, for all infinite sets M, there are no surjections from $\omega \times M$ onto $\operatorname {\mathrm {\mathscr {P}}}(M)$.

Strong Turing Determinacy, or ${\mathrm {sTD}}$, is the statement that for every set A of reals, if $\forall x\exists y\geq _T x (y\in A)$, then there is a pointed set $P\subseteq A$. We prove the following consequences of Turing Determinacy ( ${\mathrm {TD}}$ ) and ${\mathrm {sTD}}$ over ${\mathrm {ZF}}$ —the Zermelo–Fraenkel axiomatic set theory without the Axiom of Choice: (1) ${\mathrm {ZF}}+{\mathrm {TD}}$ implies $\mathrm {wDC}_{\mathbb {R}}$ —a weaker version of $\mathrm {DC}_{\mathbb {R}}$.(2) ${\mathrm {ZF}}+{\mathrm {sTD}}$ implies that every (...) set of reals is measurable and has Baire property.(3) ${\mathrm {ZF}}+{\mathrm {sTD}}$ implies that every uncountable set of reals has a perfect subset.(4) ${\mathrm {ZF}}+{\mathrm {sTD}}$ implies that for every set of reals A and every $\epsilon>0$ :(a)There is a closed set $F\subseteq A$ such that $\mathrm {Dim_H}(F)\geq \mathrm {Dim_H}(A)-\epsilon $, where $\mathrm {Dim_H}$ is the Hausdorff dimension.(b)There is a closed set $F\subseteq A$ such that $\mathrm {Dim_P}(F)\geq \mathrm {Dim_P}(A)-\epsilon $, where $\mathrm {Dim_P}$ is the packing dimension. (shrink)

We construct a model in which MA[Formula: see text] holds and [Formula: see text] fails. This shows that MA[Formula: see text] does not imply [Formula: see text] and answers an old question of Larson and Todorcevic in [Katetov’s problem, Trans. Amer. Math. Soc. 354(5) (2002) 1783–1791]. We also investigate different strong colorings in models of MA[Formula: see text].

First, we show that every coherent tree that contains a Countryman suborder is R\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${\mathbb {R}}$$\end{document}-embeddable when restricted to a club. Then for a linear order O that can not be embedded into ω\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\omega $$\end{document}, there exists an R\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${{\mathbb {R}}}$$\end{document}-embeddable O-ranging coherent tree which is not Countryman. And for a linear order O′\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} (...) \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$O'$$\end{document} that can not be embedded into Z\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${\mathbb {Z}}$$\end{document}, there exists an R\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${\mathbb {R}}$$\end{document}-embeddable O′\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$O'$$\end{document}-ranging coherent tree which contains no Countryman suborder. Finally, we will see that this is the best we can do. (shrink)