EDIT on 5/23/12 adding discussion about $T_0$ separation.
The inspiration for this post (first in a series) comes from the most excellent book Topology Via Logic by Steven Vickers.
Let $X$ be a space. I am saying space instead of set because I don't want to bring up in the readers mind all of the baggage that comes with calling it a set. Now suppose we want to explore this space. We do not know a priori where specific points are nor do we have the ability to distinguish points in all cases. Indeed let us even go so far to say that the idea of individual points is a comforting abstraction and a mental construct, just as someone would do if they tried to imagine actual space itself as consisting of identifiable points in a vector space. Now suppose in our minds we choose one of these hypothetical and abstract "points" in $X$ and ask questions about it. We first can say "I definitely know (at least) that $x \in X$." Then looking closer, possibly after much energy expenditure, we might identify a smaller region $U$ of $X$ and say "aha, $x \in U$. Then we start recording the regions known to contain $x$, placing them in a list called $\mathscr{N} (x)$. After much time we are able to generate quite a long list, however we may never get $\{x\} \in \mathscr{N}(x)$. Suppose this process is allowed to continue indefinitely. What are some obvious rules of this list. Well, if $U \in \mathscr{N}(x)$ and $U \subset V$ then we should have $V \in \mathscr{N}(x)$ because if $V$ contains $U$ and $x$ is known to be in $U$, then it certainly is known to be in $V$. Also if $U, V \in \mathscr{N}(x)$ then $x$ is known to be in both $U$ and $V$. So $x$ is known to be in $U \cap V$. Hence $\mathscr{N}(x)$ is closed under finite intersections. As some readers might recognize, $\mathscr{N}(x)$ is a filter. If we repeat this process for each $x \in X$ we can use the resulting system of neighborhood filters to get a topology $\Omega X$ on $X$. Hence $\Omega X$ represents a lattice of observations on $X$. That is if we take a $U \in \Omega X$ then for some abstract $x \in X$ we know (observation) that $x \in U$ (except for the empty set which is defined by the property that we know for sure that no x can be observed there).
It is interesting to note that the only space for which you can actually observe all points precisely, that is for all, originally hypothetical $x \in X$, we actually have $\{x\} \in \mathscr{N}(x)$, is for $\Omega X = \mathscr{P}(X)$. Let us call this space classical space and let us call $\mathscr{P}(X)$ the lattice of classical observations. Now $\mathscr{P}(X)$ is well known to be a boolean algebra and a model for classical logic. That is the logic of observations of points (questions about whether certain $x$ are in certain elements of $\Omega X$) on classical space is classical. So if the proposition $p$ is "$x \in U"$ then $\sim p$ (read not $p$) is "$x \in U^c$", which makes perfect sense because in classical space $\Omega X = \mathscr{P}(X)$, that is "$x \in U^c$" is a always a possible observation for any $U$. Then since $X = U \cup U^c \in \Omega X$ ($U^c$ is the complement of $U$), we certainly have that "$x \in U$" ($p$) or "$x \in U^c$" ($\sim p$) . So $p \vee \sim p$ is true for all observations in the lattice of classical observations about points in classical space. That is the law of excluded middle holds on observations in classical space. Let us rephrase noting that the law of excluded middle holding on a lattice of observations on a space $X$ is equivalent to $\forall U \in \Omega X \: [U^c \in \Omega X]$. That is it is equivalent to the lattice of observations (topology) on $X$ being closed under complementation. It is not hard to see that for a $T_0$ space, the "excluded middle property" actually ensures that the lattice of observations is classical ($T_0$ means that for any two distinct points there exists a neighborhood of one of the points not containing the other). This is because if $x \in X$ and $y \in \{x\}^c$ then $\exists U \in \Omega X$ such that $x \in U$ and $y \notin U$ or $y \in U$ and $x \notin U$. However since complements are open by assumption that $\Omega X$ is boolean, $U^c$ is open in either case. So either way $y$ is in an open set not containing $x$. Hence $\{x\}^c$ is open. So $\{x\}$ is open by the assumption that $\Omega X$ is boolean. So $\Omega X$ contains all singletons. Hence it is discreet. This ultimately will not be good enough for us because the definition of a $T_0$ space requires that every time some $x$ is "objectively" (in the Platonic sense of classical mathematics) not equal to some $y$ we must have that we can observe one without observing the other. We would like to rephrase and state that if $x$ is known (observed) to be distinct from $y$ then this means literally that we can produce a neighborhood in the lattice of observations containing one point and not the other and $T_0$ separation for us becomes a necessity since it literally means that the points are known to be distinct (known by means of the lattice of observations), otherwise we cannot know objectively. From this we can now characterize the excluded middle property on the lattice of observations on a space: A space $X$ with lattice of observations $\Omega X$ is classical $\iff$ every abstract $x \in X$ can be observed objectively ($\{x\}\in \Omega X$) $\iff$ the lattice of observations satisfies the excluded middle property and it can be observed that any two hypothetical distinct points (mental abstraction) are observably distinct ($T_0$ separation).
Let us formalize some of these ideas. First some vocabulary.
Let $(P,\leq)$ be a partially ordered set. We will call $(P,\leq)$ a frame if
\[1. \qquad \forall A \subset P \: [\bigvee A \in P]\]
\[2. \qquad \forall A \subset P, |A| < \infty \:[\bigwedge A \in P]\]
\[3. \qquad \forall A \subset P\:\forall x\in P\:[x\wedge \bigvee A = \bigvee \{x\wedge a\:|\:a\in A\}]\]
Now note, as Vickers does, that 2. is wholly unnecessary and in fact unduly restrictive. We can get a property stronger that 2. using only 1. and the fact that $P$ is a poset as follows. First note that 1. guarantees that we have a least element, call it $0$, in P. This is because $0 = \bigvee \emptyset$. So let $A \subset P$. Then let $L$ be the set of lower bounds of $A$, which is non-empty because $0 \in A$. Then by 1., $s:=\bigvee L \in P$. Let $a \in A$. Then $a$ is a u.b. for $L$. So $s \leq a$. Hence $s$ is a l.b. for $A$. Let $t$ be another l.b. for $A$. Then $t \in L$. So $t \leq s$. Hence $s = \bigwedge A \in P$. So we automatically get that $P$ contains arbitrary infimums.
So why are we even considering these frames? Well, let $X$ be a space and $\Omega X$ a topology on $X$. Then $\Omega X$ is a frame under set inclusion. Indeed properties 1-3 follow immediately because $\Omega X$ is closed under arbitrary unions (joins) and finite intersections (meets). But wait, we have just shown that $\Omega X$ must then contain arbitrary meets. Doesn't that imply that it contains arbitrary intersections? No. Let $A \subset \Omega X$. Then $\bigwedge A$ is the largest set in $\Omega X$ that is $\leq$ (contained in) every $a \in A$. So it is the union (join) of all open sets contained in the intersection of $A$. It is the interior of $\bigcap A$. Hence $\bigwedge A = (\bigcap A)^\circ$.
I will stop here for now. In the next post I will introduce (complete) Heyting Algebras and develop a logic of observations on an arbitrary topology. Just as the classical logic of propositions on $X$ has lattice of observations $\mathscr{P}(X)$ (discrete topology) and satisfies the excluded middle, the logic of propositions on an arbitrary topology $\Omega X$ (we shall call it an intuitionist logic of propositions on $X$) has lattice of observations $\Omega X$ and the excluded middle property need not hold. I may also discuss some intuition (if I haven't hinted at it already) as to why these notions are useful, particularly the abandonment of the excluded middle property.
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