Incomparable Cardinal Invariants of the Continuum

Question

For most cardinal invariants of the continuum we know one to always be greater or equal to the other. Consider for example $\mathfrak p \leq \mathfrak t \leq \mathfrak b$ which is somewhat elementary. Are there interesting invariants - say $\mathfrak f, \mathfrak g$ - where both $\mathfrak f < \mathfrak g$ and $\mathfrak g > \mathfrak f$ are consistent with ZFC?

Answer 1

There are; see Cichoń’s diagram and the remarks following it. For instance, it is consistent with $\mathsf{ZFC}$ that $\operatorname{non}(\mathscr{K})=\omega_1$ and $\operatorname{non}(\mathscr{L})=\omega_2$, but it is also consistent that $\operatorname{non}(\mathscr{K})=\omega_2$ and $\operatorname{non}(\mathscr{L})=\omega_1$. Here $\mathscr{K}$ is the $\sigma$-ideal of meagre subsets of the real line, $\mathscr{L}$ is the $\sigma$-ideal of subsets of $\Bbb R$ of Lebesgue measure $0$, and if $\mathscr{I}$ is an ideal of subsets of an infinite set $X$ that contains all of the finite subsets of $X$, $\operatorname{non}(\mathscr{I})$ is the minimum cardinality of a subset of $X$ that is not in $\mathscr{I}$. In particular, it is consistent that the smallest non-meagre subset of $\Bbb R$ be larger in cardinality than the smallest subset not of measure $0$, and the opposite inequality is also consistent.

As you can see from the diagram, it is also consistent both that $\mathfrak{b}<\operatorname{cov}(\mathscr{K})$ and that $\operatorname{cov}(\mathscr{K})<\mathfrak{b}$, where $\mathfrak{b}$ is the bounding number, and $\operatorname{cov}(\mathscr{K})$ is the smallest number of meagre subsets of $\Bbb R$ whose union is $\Bbb R$.