On a modal correspondence Is there an intuitive characterization of the correspondence for the modal logical formula $\square (\alpha \rightarrow \square \alpha) \rightarrow (\square \alpha \vee \square \lnot \alpha)$? 
In the presence of $D$ the schema is slightly weaker than the schema $\square (\square \alpha \rightarrow \alpha) \rightarrow (\square \alpha \vee \square \lnot \alpha)$ considered by Raymond Turner in his book Truth and Modality for Knowledge Representation (1991) and named Turner's schema by Andrea Cantini in the latter's book Logical Frameworks for Truth and Abstraction (1996). 
 A: The scheme you write $(*)$ can be interpreted as saying, "No world sees two distinct worlds, each of which sees no world other than itself." 
Turner's scheme $(T)$, by contrast, can be interpreted as "Any world which sees only self-seeing worlds, sees at most one world."
In the presence of no additional axioms, these two schemes are incomparable: the Kripke model $\mathcal{K}_0=(\{A, B, C\}; \{(A, B), (A, C)\})$ satisfies $(T)$ but not $(*)$, and the Kripke model $\mathcal{K}_1=(\{A, B, C\}, \{(A, A), (A, B), (A, C), (B, A), (B, B), (B, C), (C, A), (C, B), (C, C)\})$ satisfies $(*)$ but not $(T)$. [Here, a Kripke model is presented as $(\{worlds\}, \{"sees"\})$.]
Note that both schemes are valid in any Kripke frame of size $<3$, so the examples given above are minimal.
A: Let $\mathrm{Z}$ be the formula in question, which can be rewritten as:
$(\Diamond\alpha\land\Box(\alpha\rightarrow\Box\alpha))\rightarrow\Box\alpha$.
If I understand correctly "No world sees two distinct worlds, each of which sees no world other than itself"
from the previous answer, then this is a necessary,
but not sufficient, condition for a frame to validate $\mathrm{Z}$.
For example, the frame $F=(\{A,B,C,D\},\{(A,B),(B,C),(A,D)\})$ satisfies the condition,
yet $\mathrm{Z}$ is invalid on $F$ for the valuation $V(\alpha)=\{A,B,C\}$.
A necessary and sufficient condition for a frame to validate $\mathrm{Z}$
is that the following holds for every world $w$:

$C(\mathrm{Z})$: If $w$ sees $w'$ then every $w''\not=w'$ seen by $w$ can see $w'$
  in a finite number of steps through worlds seen by $w$.

Proof: Let $S(w)$ be the set of worlds seen by $w$ and $w_0\in S(w)$.
For any $w_i\in S(w)$ for which $w_i\not=w_0$,
let $L_{w_0}(w_i)>0$ be the length of the shortest path from $w_i$ to $w_0$ through worlds from $S(w)$;
we write $L_{w_0}(w_i)=\infty$ if no such path exists.
Sufficiency: Assume $C(\mathrm{Z})$ holds.
We attempt to invalidate $\mathrm{Z}$ at $w$ and show that this is impossible.
If $S(w)=\varnothing$ then $\mathrm{Z}$ is obviously valid at $w$.
If $w$ sees some world, then to invalidate $\mathrm{Z}$ at $w$ we need $V(\Diamond\alpha,w)=1$,
$V(\Box(\alpha\rightarrow\Box\alpha),w)=1$ and $V(\Box \alpha,w)=0$,
so we need $V(\alpha,w_0)=0$ for some $w_0$ in $S(w)$.
For the remaining $w_i\in S(w)$ (if any), $C(\mathrm{Z})$ implies $0<L_{w_0}(w_i)<\infty$. Then:


*

*For any $w_1\in S(w)$ with $L_{w_0}(w_1)=1$, if $V(\alpha,w_1)=1$ then $V(\alpha\rightarrow\Box\alpha,w_1)=0$
(because $w_1$ sees $w_0$ where $V(\alpha,w_0)=0$), hence $V(\Box(\alpha\rightarrow\Box\alpha),w)=0$
and $\mathrm{Z}$ is valid at $w$.
So we need to take $V(\alpha,w_1)= 0$ at all such $w_1$.

*For any $w_2\in S(w)$ with $L_{w_0}(w_2)=2$, if $V(\alpha,w_2)=1$ then
$V(\alpha\rightarrow\Box\alpha,w_2)=0$
(because $w_2$ sees some $w_1\in S(w)$ with $L_{w_0}(w_1)=1$ where we took $V(\alpha,w_1)=0)$,
hence $V(\Box(\alpha\rightarrow\Box\alpha),w)=0$ and $\mathrm{Z}$ is valid at $w$.
So we need to take $V(\alpha,w_2)=0$ at all such $w_2$.
We use induction to conclude that to invalidate $\mathrm{Z}$ at $w$ we need to take
$V(\alpha,w_i)=0$ at all $w_i\in S(w)$ with $L_{w_0}(w_i)<\infty$,
otherwise $V(\Box(\alpha\rightarrow\Box\alpha),w)=0$ and $\mathrm{Z}$ is valid at $w$.
But since there are no other worlds in $S(w)$ we have $V(\Diamond\alpha,w)=0$,
therefore $\mathrm{Z}$ is still valid at $w$.
Necessity: Assume $w$ sees some $w_0$ and there are also worlds $w_x\in S(w)$ with $L_{w_0}(w_x)=\infty$.
Then we take:


*

*$V(\alpha,w_i)=0$ for all $w_i\in S(w)$ with $L_{w_0}(w_i)<\infty$,

*$V(\alpha,w_x)=1$ for all $w_x\in S(w)$ with $L_{w_0}(w_x)=\infty$,
as well as $V(\alpha,w_y)=1$ for any $w_y$ seen by such $w_x$.
This is possible because no such $w_x$ sees a $w_i\in S(w)$ with $L_{w_0}(w_i)<\infty$
(otherwise it would itself have $L_{w_0}(w_x)<\infty$).
For this valuation $V(\alpha\rightarrow\Box\alpha,w_x)=1$,
hence $V(\Box(\alpha\rightarrow\Box\alpha),w)=1$ continues to hold, as well as $V(\Box\alpha,w)=0$.
But this time $V(\Diamond\alpha,w)=1$ and $\mathrm{Z}$ is invalid at $w$.
