If a poset is $\sigma$-centered, so what can we say about $P\times P$

Question

We say that a poset $P$ is $\sigma$-centered if it can be partitioned into countably-many pieces so that each piece is finite-wise compatible.

i.e. it is $\sigma$-centered if there exists a partition $\pi : P\rightarrow\omega$ such that for every $n$ and every $p_0,\ldots,p_n \in P$, if $\pi(p_i) = k$ for every $i \leq n$ (i.e. these elements of the posets belong to the same piece of the partition), then there exists $q \in P$ such that $q \leq p_i$ for all $i\leq n$.

Note that it is trivial to sho that $\sigma$-centered implies ccc. In general, it is not true that the product of ccc posets is ccc. For instance, if $T$ is a normal Suslin tree on $\omega_1$, then $T \times T$ is not ccc.

But I was wondering if we could say something about the $\sigma$-centered condition. Do we know if $P$ $\sigma$-centered implies that $P \times P$ is also $\sigma$-centered. or do we have some counter-example?

Answer 1

In fact a finitely supported product of $2^{\aleph_0}$ many $\sigma$-centered posets is still $\sigma$-centered:

Let $P_{f} = \bigcup_{n \in \omega}P_{f,n}$ be $\sigma$-centered as witnessed by $(P_{f,n})_{n \in \omega}$ for every $f \in 2^{\omega}$. The finite support product $\prod^{< \omega}_{f \in 2^\omega} P_{f}$ consits of finite partial functions $p$ on $2^{\omega}$ so that $p(f) \in P_f$ for $f$ in the domain of $p$.

For every finite antichain $A \subseteq 2^{<\omega}$ and a function $r \colon A \to \omega$ consider the set $X_{A,r}$ consiting of $p \in \prod^{< \omega}_{f \in 2^\omega} P_{f}$ so that:

• $\vert \operatorname{dom}(p) \vert = \vert A \vert$,
• $\forall f \in \operatorname{dom}(p) \exists ! a \in A (a \subseteq f)$,
• and $\forall f \in \operatorname{dom}(p) \forall a \in A ( a \subseteq f \rightarrow p(f) \in P_{f,r(a)})$

The sets $X_{A,r}$ are centered and $\prod^{< \omega}_{f \in 2^\omega} P_{f} = \bigcup X_{A,r}$.