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Show that the canonical model for $\mathbf{S5}$ is not universal.

We know that the canonical model $M$ for $\mathbf{S5}$ is based on the class of frames $\mathscr{F}$ where $R$ is an equivalence relation.

And since every universal relation is an equivalence relation, any validity on an equivalence frame, $\mathscr{F}\vDash A$, will be a validity on a universal frame $\mathscr{U}\vDash A$.

The task then is to show that $M\notin\mathscr{U}$.

My thought is that, whereas within each partition of the domain given by an equivalence relation we have a situation indistinguishable from a universal relation; when we consider the whole domain, it won't be the case that, under an equivalence relation, every world $w\in M$ sees every world (including itself). And so, we won't have a model based on a truly universal relation.

A bit more formally I am considering a frame $F=\langle W,R\rangle$ where

  • $W=\{a,b,c,d\}$; and
  • $\{\langle aa \rangle, \langle bb \rangle, \langle cc \rangle,\langle dd \rangle,\langle ab\rangle,\langle ba\rangle,\langle ac \rangle,\langle ca\rangle,\langle bc \rangle\,\langle cb\rangle\}\in R$

Now, I think $F\in\mathscr{F}$ but plainly there are worlds that cannot access other worlds and so $F\notin\mathscr{U}$.

But I cannot figure out how to make an analogous argument regarding the canonical model for $\mathbf{S5}$. How would I show that one of the maximally consistent sets of wffs that comprise the domain of the canonical model cannot access one of the other maximally consistent sets of wffs?

Any and all help is greatly appreciated!

beachtrees
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    As shown here We can drop the accessibility clause from the latter stipulation because in such total frames it is trivially true of all w and u that w R u. But note that this does not have to be the case in all S5 frames, which can still consist of multiple parts that are fully connected among themselves but still disconnected from each other. So let p is any consistent wff, so p must be true in some world in S5's canonical model, if R is universal, then $\Diamond p$ must be true in every world, then it must be a theorem thus arrive at $\bot$... – cinch Nov 01 '21 at 05:12
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    In your above constructed example R, it can be just such a valid splitted frame of S5's canonical model since world $d$ is clearly isolated. Or you can add 2 new worlds $e, f$, and only connect $d,e,f$ like what you did for $a,b,c$... – cinch Nov 01 '21 at 05:23

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