# $\mathcal{M} \models \Box \phi \rightarrow \Box \Box \phi$ for all $\phi$ if and only if $\mathcal{M}$ is transitive

Exercise 1.8.2 in Fitting and Mendelson's "First Order Logic" asks to show that $$\mathcal{M} \models \Box \phi \rightarrow \Box \Box \phi$$ for all $$\phi$$ if and only if the accessibility relation of $$\mathcal{M}$$ is transitive. This is reiterated in the answer to this question but I have been unable to prove the only if part and believe I have a counterexample:

Let the universe of $$\mathcal{M}$$ be $$\{\Gamma_i: I\in\mathbb{N}\}$$ with relation $$R = \{(i,i+1): i \in \mathbb{N}\}$$. Clearly, $$R$$ is not transitive. Let $$\Vdash$$ be identical for each $$\Gamma_i$$. Then, each for each $$i,j,\phi$$ we have that $$\Gamma_i \Vdash \phi$$ if and only if $$\Gamma_j \Vdash \phi$$, so $$\Box \phi \rightarrow \Box\Box \phi$$ is valid in $$\mathcal{M}$$. The proof uses standard induction on height of $$\phi$$.

What am I missing here?

Fix $$\Gamma,\Delta,\Omega$$ in the universe if $$\mathcal{M}$$ with $$\Gamma R \Delta$$ and $$\Delta R \Omega$$. Pick a valuation $$\Vdash$$ so that for $$\Gamma R \Lambda$$, $$\Lambda \Vdash P$$ and for $$\Gamma \not\text{R} \Lambda$$, $$\Lambda \not\Vdash P$$ for some Propositional Variable $$P$$. Then, $$\Gamma \Vdash \Box P$$, so $$\Gamma \Vdash \Box\Box P$$, so $$\Omega \Vdash P$$. By construction of $$\Vdash$$, $$\Gamma R\Omega$$.