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 ${\cal P}(\omega)/(\text{fin})$?



Can any poset of cardinality $\leq 2^{\aleph_0}$ be embedded in ${\cal P}(\omega)/(\text{fin})$?
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](https://mathoverflow.net/questions/441234): 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*](https://zbmath.org/0464.03046), 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 $\mathcal{P}(\omega)/\mathit{fin}$*](http://matwbn.icm.edu.pl/ksiazki/fm/fm163/fm16325.pdf)
and, in fact, that many 'natural candidates' are
[*not continuous images of $\beta\omega\setminus\omega$*](http://dx.doi.org/10.1007/BF02775024) 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 $\mathfrak{c}$*](http://dx.doi.org/10.1090/S0002-9947-00-02601-5).
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.


