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For $A,B\in{\cal P}(\omega)$ we say $A\subseteq^* B$ if $A\setminus B$ is finite (that is, $A$ is "almost contained" in $B$). We write $A\simeq_{\text{fin}} B$ if $A\subseteq^* B$ and $B\subseteq^* A$ (that is, the sets $A, B$ are "almost the same set" except for finitely many elements). It is easy to see that $\simeq_{\text{fin}}$ is an equivalence relation on ${\cal P}(\omega)$.

We denote the collection of equivalence classes on ${\cal P}(\omega)$ with respect to $\simeq_{\text{fin}}$ by ${\cal P}(\omega)/(\text{fin})$. Using $\subseteq^*$ on representatives of equivalence classes, it is easy to see that we can make ${\cal P}(\omega)/(\text{fin})$ into a partially ordered set.

In the Noah Schweber's beautiful post inspiring this question, we learn that there is no order-embedding from $\omega_1$ into ${\cal P}(\omega)$, but $\omega_1$ can be order-embedded in ${\cal P}(\omega)/(\text{fin})$. So within the Continuum Hypothesis ${\sf (CH)}$ we get that $2^{\aleph_0} = \omega_1$ can be order-embedded in ${\cal P}(\omega)/(\text{fin})$.

Question. Can the cardinal $2^{\aleph_0}$ (well-ordered by $\in$) be order-embedded in ${\cal P}(\omega)/(\text{fin})$ even if $\neg{\sf(CH)}$? If not: can every member of $2^{\aleph_0}$ be order-embedded in ${\cal P}(\omega)/(\text{fin})$?

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  • $\begingroup$ Yes, this is quite a trivial fact, that $2^\omega$ can be embedded into $2^\omega/\mathrm{fin}$ as a Boolean subalgebra. Use that $2^\omega$ is isomorphic to $2^{\omega^2}$ and the latter contains a copy $A$ of $2^\omega$, namely the set of sets of the form $A\times\omega$ for $A\subseteq\omega$. Then the projection $2^{\omega^2}\to 2^{\omega^2}/\mathrm{fin}$ is injective in restriction to $A$. $\endgroup$
    – YCor
    Feb 20, 2023 at 14:33
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    $\begingroup$ @YCor: I read the problem differently from you. I think Dominic is asking whether the ordinal $\mathfrak{c}$ (defined as the least ordinal with cardinality $2^{\aleph_0}$) can be order-embedded . Dominic, perhaps you can clarify? $\endgroup$
    – Will Brian
    Feb 20, 2023 at 14:35
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    $\begingroup$ @WillBrian Indeed, these are two very different orderings on $2^\omega$, and your interpretation definitely makes a more interesting question. Hopefully OP will clarify. $\endgroup$
    – YCor
    Feb 20, 2023 at 14:39
  • $\begingroup$ Apologies for the ambiguity, and thanks for your request for clarification! @WillBrian interpreted the question exactly as I had it in mind. $\endgroup$ Feb 20, 2023 at 19:52

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Yes! The fact that this is consistent is originally due to Laver. Later, Baumgartner, Frankiewicz, and Zbierski strengthened Laver's result to the following theorem:

Theorem: Is it consistent that $\mathsf{MA}_{\sigma\text{-linked}}$ holds and that every Boolean algebra of size $\mathfrak{c}$ can be order-embedded in $\mathcal P(\omega)/\mathrm{fin}$.

The theorem is proved in

Baumgartner, J.; Frankiewicz, R.; Zbierski, P., Embedding of Boolean algebras in P((\omega) )/fin, Fundam. Math. 136, No. 3, 187-192 (1990). ZBL0718.03039.

On the other hand, Kunen proved in his thesis that in the Cohen model, there is no order-preserving embedding of $\omega_2$ into $\mathcal P(\omega)/\mathrm{fin}$. So the statement "every linear order of size $\mathfrak{c}$ can be order-embedded into $\mathcal P(\omega)/\mathrm{fin}$" is independent of ZFC.

I believe it is still an open problem whether the "$\mathsf{MA}_{\sigma\text{-linked}}$" in the theorem above can be strengthened to simply "MA". I seem to remember hearing at one point that Woodin had proved something about this, but I'm not sure it was ever published and I don't remember exactly what he proved.

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