Euclidean frame validity

I have trouble understanding why $$\Diamond P \to \square\Diamond P$$ is valid in Euclidean frames.

I found a proof online which is detailed as follows:

Proof. Suppose $$F$$ is a Euclidean frame, and $$M$$ a model based on $$F.$$ Suppose $$\models_w \Diamond A$$. Then there is a $$v$$ such that $$w\mathrel{R}v$$ and $$\models_v A.$$ Now, for any $$u$$ with $$w\mathrel{R}u$$, we have $$u\mathrel{R}v$$ since $$R$$ is Euclidean. So $$\models_u \Diamond$$A. Since $$u$$ is arbitrary, $$\models_w \Box\Diamond A,$$ and therefore $$\models_w \Diamond A \to \Box\Diamond A.$$

My question is what if the arbitrary $$u$$ is chosen as world $$v.$$ My understanding of necessarily true is that from every world from the current world should satisfy the condition. However, at world $$v,$$ there is no path to another world where $$p$$ is true since $$p$$ is only true at $$v$$ itself. Doesn't that imply that $$\Diamond A$$ is false at $$v$$ and hence $$\Box\Diamond A$$ is false as well? If that's the case, why is the axiom valid?

Thank you in advance for any explanation!

• It's not just any arbitrary $u$, it's an arbitrary $u$ with $wRu$. It's OK if $R$ also happens to be reflexive as well as euclidean. – Ryan A Jan 9 '18 at 4:00
• In expressions like $$\models_w \diamond A \rightarrow \diamond A,$$ you shouldn't keep alternating in and out of MathJax. The whole thing should be between a single pair of dollar signs or double dollar signs. See my edits to the question. $\qquad$ – Michael Hardy Jan 9 '18 at 18:45

Maybe it is worth addressing precisely the question asked: what if the arbitrary $$u$$ is chosen as world $$v$$.

One thing to note here is this: Euclideanness implies that the endpoint of any path has a loop. That is, $$wRv$$ implies $$vRv$$ (indeed $$wRu$$ and $$wRv$$ together imply $$uRv$$; now take $$u=v$$).

Thus although there might be no path from $$v$$ to another world, there is a path from $$v$$ to itself, $$v$$ being an endpoint of a path (the one starting in $$w$$). Thus also $$v$$ possesses a path to a point where $$p$$ is true - namely a path to itself.

$u$ is an arbitrary world such that $w R u$, not just any old arbitrary world. If $v$ happens to be one of these worlds, then we know that $w R v$ must hold.

Suppose that $\models_w \Diamond P$. Then we get a world $v$ with $w R v$ and $\models_v P$. We then consider an arbitrary world that can be reached from $w$ (as required by the semantics of $\Box$) and call this $u$. This means that $w R u$, and since the frame was assumed to be Euclidean, $w R u$ and $w R v$ implies that $u R v$, from which we conclude that $\models_u \Diamond P$. Since the $u$ represents any world that can be reached from $w$, we have thus shown that $\models_w \Box \Diamond P$.

My question is what if the arbitrary $$u$$ is chosen as world $$v.$$ My understanding of necessarily true is that from every world from the current world should satisfy the condition. However, at world $$v,$$ there is no path to another world where $$p$$ is true since $$p$$ is only true at $$v$$ itself.

Why do you say that? $$v$$ need not be the only accessible world which witnesses the truth of $$\vDash_w\Diamond A$$.

In any case, in an Euclidean frame, by definition, any world $$v$$ accessible from $$w$$ is also accessible from any world $$u$$ that is itself accessible from $$w$$ . $$(wRv~\& ~wRu)\implies uRv$$. This inlcudes includes $$v$$ itself. $$(wRv~\& ~wRv)\implies vRv$$ . Since $$v$$ is the witness to $$\vDash_w\Diamond A$$, it is accessible from $$w$$, so therefore $$vRv$$ and hence $$\vDash_v \Diamond A$$ is okay.

But more, $$\vDash_u\Diamond A$$ is true for all arbitrary worlds, $$u$$, accessible from $$w$$ when $$\vDash_w\Diamond A$$, since they can each access some world $$v$$ where $$\vDash_v A$$ when $$\vDash_w \Diamond A$$ and the frame is Euclidean.

Hence $$\vDash_w\Box\Diamond A$$ if $$\vDash_w \Diamond A$$, and so therefore $$\Diamond A\to \Box\Diamond A$$ is a theorem in an Euclidean frame