First category set and set-theoretical assumption

Question

Let consider the following set-theoretical assumptions

${A(\mathfrak c)}:$ The union of less than $\mathfrak c$ many first category subsets of $\mathbb R$ is of the first category again.

${B(\mathfrak c)}:$ $\Bbb R$ is not a union of less than $\mathfrak c$ many first category subsets of $\mathbb R.$

It is well known that these conditions follow from Martin's Axiom. These assumptions are not equivalent. ${A(\mathfrak c)}$ implies ${B(\mathfrak c)}$ but not the other way around. Is that right?

Answer 1

Indeed.

If $X$ is a family of sets of first category such that $\bigcup X$ is not of first category, then the least cardinality of such a set $X$ is called the additivity number of the ideal of first category sets (also called the meagre ideal). If instead we assume that $\bigcup X=\Bbb R$, then the least cardinality of such an $X$ is called the covering number of the ideal of first category sets.

I'll write the additivity number as $\mathrm{add}(\mathcal B)$ and the covering number as $\mathrm{cov}(\mathcal B)$, where $\mathcal B$ stands for the ideal of first category sets. Then these two cardinalities are cardinal characteristics of the continuum, of which it is known that they can be consistently different from both $\aleph_1$ and $\mathfrak c$. They are well-known as being part of Cichoń's diagram.

To link this to your question, $A(\mathfrak c)$ is the statement that $\mathrm{add}(\mathcal B)=\mathfrak c$ and $B(\mathfrak c)$ is the statement that $\mathrm{cov}(\mathcal B)=\mathfrak c$. Since $\Bbb R$ itself is not of first category, we have $\mathrm{add}(\mathcal B)\leq \mathrm{cov}(\mathcal B)$, but it is also consistent that these two cardinals are different.

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Probably the most elementary models for separating these cardinalities is by using the method of forcing. In particular, if we start with a model of the continuum hypothesis $\aleph_1=\mathfrak c$, and ...

• ... we add $\aleph_2$ many Cohen reals, then this has the effect of making $\mathrm{add}(\mathcal B)=\aleph_1$ and $\mathrm{cov}(\mathcal B)=\aleph_2=\mathfrak c$.
• ... we add $\aleph_2$ many random reals, then both cardinals will be equal to $\aleph_1$, while $\mathfrak c=\aleph_2$.

As you note in your question, in a model of Martin's Axiom, both of these cardinals are equal to $\mathfrak c$. This implication does not work in the other direction, since one could use forcing again to get a model where $\mathrm{add}(\mathcal B)=\mathfrak c$ but Martin's Axiom fails, for example by adding Hechler reals.