Missing Points for Topology on Locale I am just beginning my study of locales and frames (point-free topologies), and in reading Johnstones `Stone Spaces', I have come across the theorem that the points of a locale have a natural topological structure. However, I have explicitly constructed a very simple example, and it raised a question for me 
(EDIT: As I wrote this, I answered my own question, but would still appreciate input on the question as it is now written). 
Are there points that get discarded when constructing a topology on the set of points of a locale?
Further, why are there so many points when a topology that generates that frame needs only two elements?
Consider the frame $2=\{0, 1\}$, and the frame $A=\{0, a, b, 1\}$, where $a$ and $b$ can't be compared. 
I have found 9 points for $A$, by considering all possible right Galois adjoints $2\to A$ that preserve meets (here, that just means preserving the ordering), as each point $A\to 2$ has a uniquely determined right adjoint.
Johnstone then defines $\phi:A\to \mathcal{P}(\text{pt}A)$ by setting $\phi (x)=\{p:A\to 2|p(x)=1\}$. This is supposed to induce a topology on the set of all points of $A$. 
My specific question is that the constant point $p_0(x)=0$ is (I believe) a legitimate point, but clearly is not included in the image of $\phi$.
Is this point $p_0$ supposed to be discarded when constructing the topology on $\text{pt}A$? 
Further, the discrete topology on two elements generates the frame $A$ given above, but the set of points of $A$ has 8 (or 9?) elements - are there supposed to be this many> Is there an intuitive reason why so many arise?
 A: Actually, your frame $A$ only has two points. Recall that a frame morphism $p\colon A\to 2$ must preserve all meets and joins. In particular, it must preserve the empty meet, $1$, and the the empty join, $0$. So $p(0) = 0$ and $p(1) = 1$, and it remains to determine where $a$ and $b$ go. But we can't send $a$ and $b$ both to $1$ (since then $a\land b = 0$, but $p(a) \land p(a) = 1$) or both to $0$ (since then $a\lor b = 1$, but $p(a)\lor p(b) = 0$). So the only options are $p(a) = 0$ and $p(b) = 1$, or $p(a) = 1$ and $p(b) = 0$.
Another way to look at it is that a point of a frame is determined by a principal prime ideal, or equivalently a completely prime filter (see II 1.3 in Johnstone). Prime ideals are proper, so they don't contain $1$, and $\{0\}$ is not a prime idea (since $a\land b = 0$). So the only two prime ideals in $A$ are $\{0,a\}$ and $\{0,b\}$. 
You tried to count the points $A\to 2$ by counting their right adjoints $2\to A$. But you made two mistakes here. First, such a right adjoint must preserve meets, and this is a stronger condition than just preserving the order (for example, a meet-preserving map preserves $1$). This still leaves you with 4 maps, sending $1$ to $1$ and $0$ to any of the four elements of $A$. Second, counting the meet-preserving maps $2\to A$ will count the join-preserving maps $A\to 2$ (their left adjoints), but a frame morphism must also preserve finite meets. Only two of the four maps (the ones which send $0$ to $a$ and to $b$) have finite-meet-preserving left adjoints.
