Why is this epistemic logic situation valid

Question

So this is epistemic logic and as far as I know this is similar to predicate logic so that is the reason for tagging this this way.

So I can see that $p$ is indeed true in worlds $w,s$, but I don't see this for the other statements. Like I agree partly with "$\square p$ is true precisely in worlds $w,u,s$" But I don't see why $v$ is different to $u$. They both have a knowledge relation to $s$ and they both have $\neg p$ but $u$ has knowledge of $p$ and $u$ does not?

This is example 5.26 in this pdf. (page 158)

Answer 1

This diagram doesn't depict an epistemic Kripke frame, which are always required to be reflexive. Note that in the document you take this from, the following comes immediately before this diagram:

Now let us also look at a much more abstract example, that does not admit an epistemic interpretation. This will allow you to see how our mechanism of interpretation works in the more general case.

So this is clearly intended not to be an instance of an epistemic Kripke frame.

In the depicted frame, neither $u$ nor $v$ are accessible to themselves, so the $\bar{p}$'s occurring in $u$ or $v$ respectively don't affect the truth of any modal statements at those worlds. In particular, the only world accessible to $u$ is $s$, where $p$ holds, so $\square p$ holds in $u$. However, $v$ can access $u$, where $\bar{p}$ does hold, so clearly $p$ does not hold at every world accessible from $v$, and $\square p$ thus doesn't hold in $v$.