Locales as spaces of ideal/imaginary points I recently saw a video of a presentation of Andrej Bauer here about constructive mathematics; and there are two examples of locales he mentions that strike me : he explains quickly what the space of random reals might be, by saying that it's the locale of reals that are in every measure $1$ subset of $[0,1]$ (for instance): as he says, of course there are no such reals, but that should not stop us from considering the space of these reals, which may have interesting topological properties even if it has no points.
Similarly in a constructive setting (or classical setting without AC) some rings may have no prime/maximal ideals, and so their spectrum as usually conceived is uninteresting. But that should not stop us from studying the space of prime/maximal ideals with the Zariski topology, even if it has no points.
My questions are related to these examples specifically and to generalizations:
Is the first example of random reals in any way connected to the random reals one mentions in forcing ? e.g. is forcing to add some random reals in any way connected to considering the topos of sheaves on the locale of random reals ?
Has the second example been extensively studied ? What sort of properties can we get from the study of this "Zariski locale" ?
Is there some form of general theory of locales as spaces of imaginary points ? For instance is this how one usually sees locales intuitively; or better is there some actual theory (more than a heuristic) of constructing pointless (or with few points) spaces of objects that we'd like to exist but don't actually exist ? This is very vague so I'll give a further example of what one might envision: if two first-order structures $A$ and $B$ aren't isomorphic but $A\cong_p B$, we might want to study the space of isomorphisms of $A$ and $B$, which would ideally be a pointless locale. One could say something similar about generic filters of a poset when one is trying to do some forcing : from the point of view of the small model, these generic filters don't exist: we could envision a space of generic filters. In these four cases we have some objects that don't exist (random reals, maximal ideals, isomorphisms) but that we can define and that in some very vague sense ought to exist, and so we construct the space of these objects; but it turns out that this space can have no points at all: is there a general theory of this sort of thing ?
These questions are very vague so I hope they're appropriate. I'll appreciate answers with references, but I'd also very much like answers that themselves provide some intuition (though a bit more technical than what I've expressed in the question), and some thoughts.
 A: Yes. There is a theory of locales considered as spaces whose points live in a forcing extension which I call Boolean-valued pointfree topology. Let me briefly outline how the theory works.
Motivating Example 1: Suppose that $B$ is a complete Boolean algebra. Then $B$ is a frame. The points in the frame $B$ are precisely the generic ultrafilters on $B$. If $B$ is atomless, then there are no generic ultrafilters on $B$, but forcing adds generic ultrafilters to $B$. Therefore, forcing is a process that adds points to the frame $B$. One should consider process that not only adds points to the frames which happen to be complete Boolean algebras, but forcing should be a process that adds points to all frames.
Motivation 2: Suppose that $L$ is a frame. Then the points in the frame $L$ are precisely the frame homomorphisms $\phi:L\rightarrow 2$. The most natural way to generalize the notion of a point to a $B$-valued point would be to define a $B$-valued point to be a frame homomorphism $\phi:L\rightarrow B$. It is not to hard to show that every frame embeds into a complete Boolean algebra, so for each frame $L$ one can find a complete Boolean algebra for which there are plenty of frame homomorphisms $\phi:L\rightarrow B.$
Pitfall: It is tempting to simply consider the collection of points in a frame $L$ as a topological space in a forcing extension $V[G]$, but this approach is deficient since $L$ may still have pointfree aspects when interpreted in $V[G]$. One should therefore take care to interpret frames as frames instead of topological spaces in forcing extensions (if you interpret frames as simply topological spaces in forcing extensions, then nothing works and you end up with a substandard theory).
Method 1: Suppose that $L$ is a frame in a model $M$. The model $M$ need not be a ground model, well-founded, nor a model of ZFC (or much of anything for that matter). Then $L$ is a distributive lattice in the universe $V$. Let $\mathcal{A}$ be the collection of all subsets $R\subseteq L$ with $R\in M$. Then each
$R\in\mathcal{A}$ has a least upper bound in $V$. Let $C_{\mathcal{A}}$ be the collection of all ideals $I\subseteq L,I\in V$ such that if $R\in\mathcal{A}$, then $\bigvee^{V}R\in I$. Then $C_{\mathcal{A}}$ is a frame. The frame $C_{\mathcal{A}}$ is the most natural structure that one would obtain simply by interpreting the frame $L$ as a frame in $V$ as an object in $V$.
This method is very general. However, I prefer the Boolean-valued model approach to interpreting frames in forcing extensions since complete Boolean algebras are some of the most important examples of frames.
Method 2: 
Suppose that $L$ is a frame that contains a complete Boolean algebra $B$ as a subframe. Then $L$ is a $B$-valued structure closed under mixing where we define Boolean valued equality and inequality by
$b\leq\|x=y\|$ if and only if $b\wedge x=b\wedge y$ and $b\leq\|x\leq y\|$ if and only if $b\wedge x\leq b\wedge y$. Since $L$ is a $B$-valued structure closed under complete mixing, one should interpret $L$ as an object in the Boolean-valued universe $V^{B}$.
Theorem: 


*

*Suppose that $L$ is a frame that contains a complete Boolean algebra $B$ as a subframe. Then $$V^{B}\models\text{$L$ is a frame.}$$

*Suppose that $B$ is a complete Boolean algebra. If $$V^{B}\models\text{$\dot{X}$ is a frame},$$
then there is a frame $L$ that contains $B$ as a subframe and where
$$V^{B}\models\text{The frames $L$ and $\dot{X}$ are isomorphic.}$$
Theorem: Suppose that $L$ is a frame that contains a complete Boolean algebra $B$ as a subframe. Suppose that $P$ is one of the following properties. Then $L$ satisfies $P$ is and only if $V^{B}\models\text{$L$ satisfies $P$.}$


*

*regularity

*complete regularity

*paracompactness

*ultraparacompactness

*zero-dimensionality
Now, suppose that $L$ is a frame and $B$ is a complete Boolean algebra. Then the coproduct $L\oplus B$ contains copies $L$ and $B$ as subframes. Therefore, since $B$ is a subframe of $L\oplus B$, we should consider $L\oplus B$ as a frame inside $V^{B}$. 
Theorem: Suppose that $L$ is a frame and $P$ is one of the following properties. Then $L$ satisfies $P$ if and only if $V^{B}\models\text{$L\oplus B$ satisfies $P$.}$


*

*regularity

*compactness

*connected local connectedness
Suppose now that $L$ is a frame and $B$ is a complete Boolean algebra. Suppose that $P$ is one of the following properties. Then if $L$ satisfies $P$, then
$V^{B}\models\text{$L\oplus B$ satisfies $P$}$


*

*paracompactness

*ultraparacompactness

*complete regularity

*second countability
Theorem: Suppose that $L$ is a regular frame. Then there is a complete Boolean algebra $B$ such that 
$V^{B}\models\text{$L\oplus B$ is a Polish space.}$
Hooray, any regular frame can be given points and made spatial by sufficient forcing. Non-regular frames can also be given points by forcing, then in $V^{B}$, the points in the $B$-valued frame are precisely the frame homomorphisms $\phi:L\rightarrow B.$
I kindly request that you do not upvote this answer.
