Can any poset of cardinality $\leq 2^{\aleph_0}$ be embedded in ${\cal P}(\omega)/(\text{fin})$?

Question

We endow ${\cal P}(\omega)$ with an equivalence relation by saying that $A\simeq_{\text{fin}} B$ iff the symmetric difference $A\Delta B$ is finite. The resulting set of equivalence classes is denoted ${\cal P}(\omega)/(\text{fin})$, and by $$[A]_{\simeq_{\text{fin}}} \leq [B]_{\simeq_{\text{fin}}} \text{ if and only if } A\setminus B \text{ is finite } $$ for $A,B\in {\cal P}(\omega)$, we get a partial order on ${\cal P}(\omega)/(\text{fin})$, resulting in an atomless Boolean algebra.

Can any poset $(P,\leq)$ with $|P|\leq 2^{\aleph_0}$ be order-embedded into ${\cal P}(\omega)/(\text{fin})$?

Answer 1

Here is an attempt at a 'definitive summary'.

To begin with positive results: $\mathsf{CH}$ implies a “yes” answer to this question. The fastest way to see this is to first embed a given partial order $(P,\le)$ (of cardinality at most $\mathfrak{c}$) into its power set via $p\mapsto\{x:x\le p\}$ and then take the Boolean subalgebra generated by the image under this embedding. An application of Parovichenko's theorem that every Boolean algebra of cardinality at most $\aleph_1$ embeds into $\mathcal{P}(\omega)/\mathit{fin}$ finishes the argument.

To elaborate a bit on Will Brian's answer to this question: Laver's proof of the consistency with $\neg\mathsf{CH}$ with the statement that was published in Linear orders in $(\omega)^\omega$ under eventual dominance, in Logic colloquium ’78.

The proof and the proof of the generalization to Boolean algebras that Will mentioned both proceed by embedding a $\mathfrak{c}$-saturated linear order or Boolean algebra of cardinality $\mathfrak{c}$ respectively into $\mathcal{P}(\omega)/\mathit{fin}$. The saturated order/algebra contains all orders/algebras of cardinality at most $\mathfrak{c}$. This establishes the consistency of a ``yes'' answer with $\neg\mathsf{CH}$.

It is easy to see that every separable compact space is a continuous image of $\beta\omega\setminus\omega$ so that, dually, every $\sigma$-centered Boolean algebra embeds into $\mathcal{P}(\omega)/\mathit{fin}$.

There are many consistent counterexamples: Kunen's result that establishes that in the Cohen model (with any allowable value of $2^{\aleph_0}$) the ordinal $\omega_2$ is not embeddable into $\mathcal{P}(\omega)/\mathit{fin}$ shows that Parovichenko's result is sharp. It also shows that the result on $\sigma$-centered Boolean algebras does not generalize to partial orders: a linear order is ($\sigma$-)centered.

Other negative results:

1. The Open Colouring Axiom implies that the Measure Algebra does not embed into $\mathcal{P}(\omega)/\mathit{fin}$ and, in fact, that many 'natural candidates' are not continuous images of $\beta\omega\setminus\omega$ and so, dually, many 'natural candidates' cannot be embedded into $\mathcal{P}(\omega)/\mathit{fin}$.
2. It is even consistent that there is no universal Boolean algebra of cardinality $\mathfrak{c}$. The proof is topological and yields, given a compact space of weight $\mathfrak{c}$, a linearly ordered compact space that is not a continuous image of the given space; it dualizes to yield given a Boolean algebra of cardinality $\mathfrak{c}$ a linearly ordered set that cannot be embedded into the given Boolean algebra.