For any topological space $X$ one can consider the so called frame of all open subsets of $X$ to be denoted by $\mathcal{O}(X)$. If $f:X \to Y$ is continuous taking the inverse image we get the morphism between frames $\mathcal{O}(f): \mathcal{O}(Y) \to \mathcal{O}(X)$. In this manner we obtain a contravariant functor. On the other side, for a frame $F$ one can define the space $Pt(F)$ of points of $F$ to be $Hom(F,\{0,1\})$ declaring all the sets of the form $\Phi(U):=\{p \in Pt(F): p(U)=1\}$ as opens. If $g:F \to G$ is a frame morphism we get a continuous map in the opposite direction $Pt(g):Pt(G) \to Pt(F)$ by $Pt(g)(q)=q \circ g$.
A topological space $X$ is called sober if any irreducible subset $A$ of $X$ is a closure of the unique point. This is equivalent to the condition that the map $\delta:X \to Pt(\mathcal{O}(X))$ is a homeomorphism where $\delta_x(U):=1$ if $x \in U$ and $0$ otherwise.
Take $X$ to be any topological space (not necessarilly sober). Then even the map $\delta$ need not to be homeomorphism, still $\mathcal{O}(\delta):\mathcal{O}(Pt(\mathcal{O}(X))) \to \mathcal{O}(X)$ is an isomorphism of frames. This isomorphism turns out to be equal to $\Phi^{-1}$. In particular $\mathcal{O}(X)$ is always spatial (see below).
Now we would like to repeat the argument but for the other direction: we call a frame $F$ to be spatial if the following condition holds: if it is not true that $a \leq b$ in $F$ then there exists $p \in Pt(F)$ such that $p(a)=1$ and $p(b)=0$. Now it is not true that for any frame $F$ the map $Pt(\Phi)$ is always a homeomorphism: it is $\iff$ $Pt(\Phi)$ is surjective $\iff \Phi$ is injective $\iff F$ is spatial. In this case $Pt(\Phi)=\delta^{-1}$.
Question 1: Is the above reasoning concerning spatiality of $F$ correct?
In particular it is not true that always (i.e. for any frame $F$-not necessarily spatial) $Pt(F)$ is isomorphic to $Pt(\mathcal{O}(Pt(F)))$: it happens precisely when $F$ is spatial which holds iff $Pt(F)$ is sober. I'm pretty sure that I am missing something here: so let me ask
Question 2: Is the space $Pt(F)$ always sober? Is it true that always $Pt(F) \cong Pt(\mathcal{O}(Pt(F)))$ or it is true precisely when $F$ is spatial?
I would be grateful if someone can clarify these issues.
