Generated subframe of a frame

Question

Let's define that $F=(W,R) $ is a Kripke frame where $W=\mathbb{Z}$ and $R=\{(u,v)\mid v=u+1 \}$.

Then we can have generated subframe $F_{0}=(W_{0},R_{0})$, where $W_{0}=\mathbb{N}$ and $R_{0}=\{ (u,v)\in\mathbb{N}^{2}\mid v=u+1 \}$.

Why is this actually a generated subframe?

Why $W_{0}=\mathbb{N}$?

Why $-1\not\in\ W_{0} $?

The $F_{0}$ is the smallest subframe generated by $\{0 \}$.

But isn't $F'=(\{0,1\},\{(0,1)\})$ also a subframe of $F$?

Also why $F''=(\{-1,0 \},\{(-1,0) \})$ is not a subframe of $F$?

If we take the intersection of all the $X\subseteq W$ such that $\{ 0\}\subseteq X$ and $(X,R\cap X)$ (generated subframe of $F$), then shouldn't we get $X=0$ instead of $\mathbb{N}$?

I am so lost here.

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Edit: Let's say that $F'$ is a generated subframe of $F$ defined above.

Now, $1\in \{0,1\}$ and $2\in W$ and $1R2$, but $2\not\in \{0,1\}$. Contradiction.

Is this the reason why $W_{0}$ must be the whole set of natural numbers?

So, if we start with a set with just $\{0,1\}$ we end up adding a $2$ in it. Then $3$ and so on. So, the smallest possible subframe generated by $0$ in this case must have set of natural numbers is it's universe.

Answer 1

See generated subframe:

A frame $\mathbf G=\langle G,S,W\rangle$ is a generated subframe of a frame $\mathbf F=\langle F,R,V\rangle$, if the Kripke frame $\langle G,S\rangle$ is a generated subframe of the Kripke frame $\langle F,R\rangle$ (i.e., $G$ is a subset of $F$ closed upwards under $R$, and $S$ is the restriction of $R$ to $G$), and

$W = \{ A\cap G, \ A\in V \}$.

Note. $V$ is a set of subsets of $F$.

In your example, we have that the original Kripke frame is $F= (W,R)$, where $W= \mathbb Z$, and $V= \{ \{ 0 \} \}$ .

The generated subframe $F_0 = (W_0,R_0)$, where $W_0 = \mathbb N$, is the smallest subframe generated by $V$.

Why ? Consider the above definition: we start from $\{ 0 \}$ and we have to consider the relation $R_0 = \{ (u,v) \in \mathbb N \times \mathbb N \mid v=u+1 \}$, i.e. the restriction of $R$ to $\mathbb N$.

We want that $W_0$ is closed upward under $R$, and this means that we have to start from $0$ and "apply" the relation $xRy$.

Obviously: $0R1$, and thus we have to "throw in" $1$. Now we have to repeat the "procedure" with $1$, adding $2$; and so on.

In this way, the subset of $W = \mathbb Z$ generated starting from $\{ 0 \}$ and closed upward under $R$ is $\mathbb N$.