Rooted Kripke frames connection to $K$ I was reading the Modal logic book from Chagrov and Zakharyaschev.
I read the following theorem (generation theorem 3.11):
If $N$ is a generated submodel of $M$, then for every point x and every formula $\phi$:
$$M,x \models \phi \iff N,x \models \phi$$
After that I found an interesting corollary:
$$ K = \{ \phi \mid F \models \phi  \text{ for all rooted frames } F \} $$
But I don't get it why?
Since we are talking for induced generated subframe of an ARBITRARY (not necessarily reflexive, transitive ..)  how can we asure, that from $M$ we can get a rooted subframe?
For example F = ({a,b,c}, {(a,b), (b,c), (c,d)}) how can we get an subframe that it's rooted ??
Where does this corollary come from?
 A: I hope you agree that
$$
\mathbf K=\{\phi\mid {"}\mathscr F\vDash \phi{"}\text{ for all frames }\mathscr F\}
$$
That is, $\bf K$ is the theory consisting of all formulas that are true in any Kripke frame.
Suppose $\phi\in\mathbf K$. Clearly if $\mathscr F$ is a rooted frame, then it is a frame, thus $\mathscr F\vDash \phi$.
Conversely, suppose $\phi\notin\mathbf K$. Then there is some frame $\mathscr F=\langle W,R\rangle$, some point $x\in W$, and some model $\mathscr M$ based on $\mathscr F$ such that $(\mathscr M,x)\not\vDash\phi$. Let $\mathscr N\subseteq \mathscr M$ be the submodel generated by $\{x\}$ and $\mathscr G\subseteq \mathscr F$ the (rooted) subframe of $\mathscr F$ that forms the basis of $\mathscr N$. By the theorem (3.11) we see that $(\mathscr N,x)\not\vDash \phi$, and thus $\mathscr G\not\vDash\phi$. Thus we've found a rooted frame that does not satisfy $\phi$.
Therefore $\phi\in \mathbf K$ if and only if $\phi$ holds in all rooted frames.

As for your example, I assume you meant $$\mathscr F=\big \langle\{a,b,c\},\ \{(a,b),(b,c),\underline{(c,a)}\}\big\rangle,$$
instead of $(c,d)$? In this case $\mathscr F$ is already rooted: the subframe generated by any of the three points is equal to $\mathscr F$ itself.
Basically, a frame is rooted whenever there exists some a point $x$, called the root, such that every point $y$ in the frame either is equal to $x$, or can be reached with a path $x\to \dots \to y$. If we take $a$ as the root in the example above, then $a\to b$ and $a\to b \to c$, thus all points in $\mathscr F$ can be reached from $a$ with a path or are equal to $a$ (it happens by chance that $a$ can also reach itself, but that is not necessary for a rooted frame).
