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We work in $\mathsf{ZFC+V=L}$.

Define a plausible theory to be a theory $T\subseteq\mathcal{L}_{\omega_1,\omega}$ in an $\omega_1$-finite language which is $\omega_1$-c.e. and $\omega_1$-finitely satisfiable; that is, $T$ is a theory in a countable language which is $\Sigma_1$-definable over $L_{\omega_1}$ (with parameters) and every countable subtheory of $T$ is satisfiable.

Since $L_{\omega_1}$ is uncountable, Barwise compactness does not apply and plausible theories need not be satisfiable. That said, every plausible theory is "generically" satisfiable: they all become satisfiable after forcing with $\mathit{Col}(\omega,\omega_1)$, since in the generic extension we can apply Barwise compactness.

I'm interested in measuring the difficulty of "satisfiabilizing" a plausible theory via forcing. Specifically, set $T_0\trianglelefteq T_1$ iff $T_0$ is satisfiable in every generic extension in which $T_1$ is satisfiable. (Note that by absoluteness of $\mathcal{L}_{\omega_1,\omega}$-semantics we don't have to worry about theories becoming unsatisfiable.) As usual, this preorder induces an equivalence relation $\approx$ and a corresponding poset $\mathcal{Plaus}$ of plausibility degrees.

I'm generally interested in the "shape" of $\mathcal{Plaus}$, but in particular the following question seems particularly natural:

Question. Does $\mathcal{Plaus}$ have coatoms?

(See below for a proof that $\mathcal{Plaus}$ does have a top element.) I strongly suspect that the answer is negative, but I don't see how to prove that.

Very briefly, here's what I already know. $\mathcal{Plaus}$ is a bounded lattice: ${\bf 0}$ = "already satisfiable," ${\bf 1}$ = "requires $\omega_1^L<\omega_1$," $S\sqcup T$ = "disjoint union of models," and $S\sqcap T=\{\sigma\vee\tau:\sigma\in S,\tau\in T\}$. In fact $\mathcal{Plaus}$ is countably complete, by the obvious extensions of those operations. Moreover, ${\bf 0}$ is not the meet of countably many nonzero elements, so a fortiori $\mathcal{Plaus}$ has at most one atom (contra an earlier claim of mine - thanks to Farmer S). Finally, $\mathcal{Plaus}$ is not linear and has $\omega_1$-many degrees; these facts follow from some easy-but-tedious forcing arguments, whose details I'm happy to add if anyone's interested. I believe these same arguments can be used to show that ${\bf 1}$ is not a nontrivial finite join (and so $\mathcal{Plaus}$ has at most one coatom), but that's a bit messier.

Unfortunately, none of this seems particularly relevant to the coatom question (except the existence of $\mathbf 1$ which is needed to pose it in the first place).

We work in $\mathsf{ZFC+V=L}$.

Define a plausible theory to be a theory $T\subseteq\mathcal{L}_{\omega_1,\omega}$ in an $\omega_1$-finite language which is $\omega_1$-c.e. and $\omega_1$-finitely satisfiable; that is, $T$ is a theory in a countable language which is $\Sigma_1$-definable over $L_{\omega_1}$ (with parameters) and every countable subtheory of $T$ is satisfiable.

Since $L_{\omega_1}$ is uncountable, Barwise compactness does not apply and plausible theories need not be satisfiable. That said, every plausible theory is "generically" satisfiable: they all become satisfiable after forcing with $\mathit{Col}(\omega,\omega_1)$, since in the generic extension we can apply Barwise compactness.

I'm interested in measuring the difficulty of "satisfiabilizing" a plausible theory via forcing. Specifically, set $T_0\trianglelefteq T_1$ iff $T_0$ is satisfiable in every generic extension in which $T_1$ is satisfiable. (Note that by absoluteness of $\mathcal{L}_{\omega_1,\omega}$-semantics we don't have to worry about theories becoming unsatisfiable.) As usual, this preorder induces an equivalence relation $\approx$ and a corresponding poset $\mathcal{Plaus}$ of plausibility degrees.

I'm generally interested in the "shape" of $\mathcal{Plaus}$, but in particular the following question seems particularly natural:

Question. Does $\mathcal{Plaus}$ have coatoms?

(See below for a proof that $\mathcal{Plaus}$ does have a top element.) I strongly suspect that the answer is negative, but I don't see how to prove that.

Very briefly, here's what I already know. $\mathcal{Plaus}$ is a bounded lattice: ${\bf 0}$ = "already satisfiable," ${\bf 1}$ = "requires $\omega_1^L<\omega_1$," $S\sqcup T$ = "disjoint union of models," and $S\sqcap T=\{\sigma\vee\tau:\sigma\in S,\tau\in T\}$. In fact $\mathcal{Plaus}$ is countably complete, by the obvious extensions of those operations. Moreover, ${\bf 0}$ is not the meet of countably many nonzero elements (observed by Farmer S.), and has a unique atom (there is a plausible theory which becomes satisfiable exactly when we add a non-constructible subset of $\omega_1^L$, which has to happen when any unsatisfiable plausible theory is made satisfiable). Finally, $\mathcal{Plaus}$ is not linear and has $\omega_1$-many degrees; these facts follow from some easy-but-tedious forcing arguments, whose details I'm happy to add if anyone's interested. I believe these same arguments can be used to show that ${\bf 1}$ is not a nontrivial finite join (and so $\mathcal{Plaus}$ has at most one coatom), but that's a bit messier.

Unfortunately, none of this seems particularly relevant to the coatom question (except the existence of $\mathbf 1$ which is needed to pose it in the first place).

We work in $\mathsf{ZFC+V=L}$.

Define a plausible theory to be a theory $T\subseteq\mathcal{L}_{\omega_1,\omega}$ in an $\omega_1$-finite language which is $\omega_1$-c.e. and $\omega_1$-finitely satisfiable; that is, $T$ is a theory in a countable language which is $\Sigma_1$-definable over $L_{\omega_1}$ (with parameters) and every countable subtheory of $T$ is satisfiable.

Since $L_{\omega_1}$ is uncountable, Barwise compactness does not apply and plausible theories need not be satisfiable. That said, every plausible theory is "generically" satisfiable: they all become satisfiable after forcing with $\mathit{Col}(\omega,\omega_1)$, since in the generic extension we can apply Barwise compactness.

I'm interested in measuring the difficulty of "satisfiabilizing" a plausible theory via forcing. Specifically, set $T_0\trianglelefteq T_1$ iff $T_0$ is satisfiable in every generic extension in which $T_1$ is satisfiable. (Note that by absoluteness of $\mathcal{L}_{\omega_1,\omega}$-semantics we don't have to worry about theories becoming unsatisfiable.) As usual, this preorder induces an equivalence relation $\approx$ and a corresponding poset $\mathcal{Plaus}$ of plausibility degrees.

I'm generally interested in the "shape" of $\mathcal{Plaus}$, but in particular the following question seems particularly natural:

Question. Does $\mathcal{Plaus}$ have coatoms?

(See below for a proof that $\mathcal{Plaus}$ does have a top element.) I strongly suspect that the answer is negative, but I don't see how to prove that.

Very briefly, here's what I already know. $\mathcal{Plaus}$ is a bounded lattice: ${\bf 0}$ = "already satisfiable," ${\bf 1}$ = "requires $\omega_1^L<\omega_1$," $S\sqcup T$ = "disjoint union of models," and $S\sqcap T=\{\sigma\vee\tau:\sigma\in S,\tau\in T\}$. In fact $\mathcal{Plaus}$ is countably complete, by the obvious extensions of those operations. Moreover, ${\bf 0}$ is not the meet of countably many nonzero elements, so a fortiori $\mathcal{Plaus}$ has at most one atom (contra an earlier claim of mine - thanks to Farmer S). Finally, $\mathcal{Plaus}$ is not linear and has $\omega_1$-many degrees; these facts follow from some easy-but-tedious forcing arguments, whose details I'm happy to add if anyone's interested.

Unfortunately, none of this seems particularly relevant to the coatom question (except the existence of $\mathbf 1$ which is needed to pose it in the first place).

We work in $\mathsf{ZFC+V=L}$.

Define a plausible theory to be a theory $T\subseteq\mathcal{L}_{\omega_1,\omega}$ in an $\omega_1$-finite language which is $\omega_1$-c.e. and $\omega_1$-finitely satisfiable; that is, $T$ is a theory in a countable language which is $\Sigma_1$-definable over $L_{\omega_1}$ (with parameters) and every countable subtheory of $T$ is satisfiable.

Since $L_{\omega_1}$ is uncountable, Barwise compactness does not apply and plausible theories need not be satisfiable. That said, every plausible theory is "generically" satisfiable: they all become satisfiable after forcing with $\mathit{Col}(\omega,\omega_1)$, since in the generic extension we can apply Barwise compactness.

I'm interested in measuring the difficulty of "satisfiabilizing" a plausible theory via forcing. Specifically, set $T_0\trianglelefteq T_1$ iff $T_0$ is satisfiable in every generic extension in which $T_1$ is satisfiable. (Note that by absoluteness of $\mathcal{L}_{\omega_1,\omega}$-semantics we don't have to worry about theories becoming unsatisfiable.) As usual, this preorder induces an equivalence relation $\approx$ and a corresponding poset $\mathcal{Plaus}$ of plausibility degrees.

I'm generally interested in the "shape" of $\mathcal{Plaus}$, but in particular the following question seems particularly natural:

Question. Does $\mathcal{Plaus}$ have coatoms?

(See below for a proof that $\mathcal{Plaus}$ does have a top element.) I strongly suspect that the answer is negative, but I don't see how to prove that.

Very briefly, here's what I already know. $\mathcal{Plaus}$ is a bounded lattice: ${\bf 0}$ = "already satisfiable," ${\bf 1}$ = "requires $\omega_1^L<\omega_1$," $S\sqcup T$ = "disjoint union of models," and $S\sqcap T=\{\sigma\vee\tau:\sigma\in S,\tau\in T\}$. In fact $\mathcal{Plaus}$ is countably complete, by the obvious extensions of those operations. Moreover, ${\bf 0}$ is not the meet of countably many nonzero elements, so a fortiori $\mathcal{Plaus}$ has at most one atom (contra an earlier claim of mine - thanks to Farmer S). Finally, $\mathcal{Plaus}$ is not linear and has $\omega_1$-many degrees; these facts follow from some easy-but-tedious forcing arguments, whose details I'm happy to add if anyone's interested. I believe these same arguments can be used to show that ${\bf 1}$ is not a nontrivial finite join (and so $\mathcal{Plaus}$ has at most one coatom), but that's a bit messier.

Unfortunately, none of this seems particularly relevant to the coatom question (except the existence of $\mathbf 1$ which is needed to pose it in the first place).

We work in $\mathsf{ZFC+V=L}$.

Define a plausible theory to be a theory $T\subseteq\mathcal{L}_{\omega_1,\omega}$ in an $\omega_1$-finite language which is $\omega_1$-c.e. and $\omega_1$-finitely satisfiable; that is, $T$ is a theory in a countable language which is $\Sigma_1$-definable over $L_{\omega_1}$ (with parameters) and every countable subtheory of $T$ is satisfiable.

Since $L_{\omega_1}$ is uncountable, Barwise compactness does not apply and plausible theories need not be satisfiable. That said, every plausible theory is "generically" satisfiable: they all become satisfiable after forcing with $\mathit{Col}(\omega,\omega_1)$, since in the generic extension we can apply Barwise compactness.

I'm interested in measuring the difficulty of "satisfiabilizing" a plausible theory via forcing. Specifically, set $T_0\trianglelefteq T_1$ iff $T_0$ is satisfiable in every generic extension in which $T_1$ is satisfiable. (Note that by absoluteness of $\mathcal{L}_{\omega_1,\omega}$-semantics we don't have to worry about theories becoming unsatisfiable.) As usual, this preorder induces an equivalence relation $\approx$ and a corresponding poset $\mathcal{Plaus}$ of plausibility degrees.

I'm generally interested in the "shape" of $\mathcal{Plaus}$, but in particular the following question seems particularly natural:

Question. Does $\mathcal{Plaus}$ have coatoms?

(See below for a proof that $\mathcal{Plaus}$ does have a top element.) I strongly suspect that the answer is negative, but I don't see how to prove that.

Here are some easy observations:

• The already-satisfiable plausible theories constitute the least degree $\mathbf 0$, and there is a greatest degree $\mathbf 1$ as well: we can whip up a plausible theory $T_{\mathit{max}}$ describing a structure which (i) is a countable linear order and (ii) has each countable ordinal as an initial segment, and in order to make $T_{\mathit{max}}$ satisfiable we have to make $\omega_1$ countable.

• There are also intermediate degrees. For example, we can whip up a plausible theory describing $(\omega;<)$ equipped with a unary predicate which does not correspond to any constructible real, which becomes satisfiable exactly when we add a non-constructible real.

• The "$\omega$-with-a-predicate" trick can be extended to reasonably-simple forcing notions to get a lot more examples — e.g. there are plausible theories corresponding to the existence of a sufficiently Cohen generic real and to the existence of a sufficiently Sacks generic real, the pair of which show that $\trianglelefteq$ is not total. It's also not hard to show that there are exactly $\omega_1$-many plausibility degrees.

Unfortunately, none of this seems particularly relevant to the coatom question (except the existence of $\mathbf 1$ which is needed to pose it in the first place).

We work in $\mathsf{ZFC+V=L}$.

Define a plausible theory to be a theory $T\subseteq\mathcal{L}_{\omega_1,\omega}$ in an $\omega_1$-finite language which is $\omega_1$-c.e. and $\omega_1$-finitely satisfiable; that is, $T$ is a theory in a countable language which is $\Sigma_1$-definable over $L_{\omega_1}$ (with parameters) and every countable subtheory of $T$ is satisfiable.

Since $L_{\omega_1}$ is uncountable, Barwise compactness does not apply and plausible theories need not be satisfiable. That said, every plausible theory is "generically" satisfiable: they all become satisfiable after forcing with $\mathit{Col}(\omega,\omega_1)$, since in the generic extension we can apply Barwise compactness.

I'm interested in measuring the difficulty of "satisfiabilizing" a plausible theory via forcing. Specifically, set $T_0\trianglelefteq T_1$ iff $T_0$ is satisfiable in every generic extension in which $T_1$ is satisfiable. (Note that by absoluteness of $\mathcal{L}_{\omega_1,\omega}$-semantics we don't have to worry about theories becoming unsatisfiable.) As usual, this preorder induces an equivalence relation $\approx$ and a corresponding poset $\mathcal{Plaus}$ of plausibility degrees.

I'm generally interested in the "shape" of $\mathcal{Plaus}$, but in particular the following question seems particularly natural:

Question. Does $\mathcal{Plaus}$ have coatoms?

(See below for a proof that $\mathcal{Plaus}$ does have a top element.) I strongly suspect that the answer is negative, but I don't see how to prove that.

Very briefly, here's what I already know. $\mathcal{Plaus}$ is a bounded lattice: ${\bf 0}$ = "already satisfiable," ${\bf 1}$ = "requires $\omega_1^L<\omega_1$," $S\sqcup T$ = "disjoint union of models," and $S\sqcap T=\{\sigma\vee\tau:\sigma\in S,\tau\in T\}$. In fact $\mathcal{Plaus}$ is countably complete, by the obvious extensions of those operations. Moreover, ${\bf 0}$ is not the meet of countably many nonzero elements, so a fortiori $\mathcal{Plaus}$ has at most one atom (contra an earlier claim of mine - thanks to Farmer S). Finally, $\mathcal{Plaus}$ is not linear and has $\omega_1$-many degrees; these facts follow from some easy-but-tedious forcing arguments, whose details I'm happy to add if anyone's interested.

Unfortunately, none of this seems particularly relevant to the coatom question (except the existence of $\mathbf 1$ which is needed to pose it in the first place).