Technical aside to the previous post.

This is not a part 2 of the previous post. In this post I will clarify a remark I made offhandedly in the previous post.

In the last post I informally introduced topologies by constructing them as a natural lattice of observations on an abstract space. For each $x \in X$ a filter $\mathscr{N}(x)$ was constructed with the interpretation that $U \in \mathscr{N}(x) \iff x$ was observed to be in $U$. Recall that we have assumed that the space $X$ is an abstract mental construction where it may be the case that idealized mental points $x \in X$ may not always be empirically observed, that is it is possible that we do not have $\{x\} \in \mathscr{N}(x)$.  Here I am giving the possibility of $\{x\} \in X$ the interpretation of empirically determining the precise position of $x$. (I later gave the space where it is always possible to determine the precise position of abstract points empirically a name, classical space, and I showed that its lattice of observations is $\mathscr{P}(X)$ and that its logic is classical.)

Moreover I claimed that since we had such a filter $\mathscr{N}(x)$ for each $x \in X$ then we could make a topology $\Omega X$ from these filters which I called "neighborhood filters". However that is imprecise and seemingly not quite right since neighborhood filters has a technical definition. We must put more conditions on the original filters to ensure that the resulting topology induced by those filters has neighborhood filters that are the same as the originals. The point of this post is to make more precise that remark. Since the sufficient condition is rather technical I chose to gloss over this point in the earlier post since I did not feel it was essential to introducing topologies through a "lattice of observations" interpretation. However, we shall see that even this extra technical requirement is naturally satisfied by the filter system constructed through the "lattice of observations" construction. So we were still quite right to call it a neighborhood system.

First let us define some terms: Let $X$ be a space and $\mathscr{F} \subset \mathscr{P}(X)$. Then $\mathscr{F}$ is a filter (of sets) on $X \iff$ (definition) the following conditions hold:

\[1.\qquad \forall A \in \mathscr{F} \forall B \in \mathscr{P}(X)\:[A \subset B \implies B\in \mathscr{F}]\]
\[2. \qquad \forall A,B \in \mathscr{F}\:[A \cap B \in \mathscr{F}]\]

Now suppose for each $x \in X$ we let $\mathscr{F}(x)$ be a filter such that $\forall A \in \mathscr{F}(x)$ we have $x \in A$. Then we shall call the resulting collection of filters $(\mathscr{F}(x))_{x\in X}$ a filter system for $X$. Now let us use a filter system to construct a topology.

Let \[\Omega_{\mathscr{F}} X = \Omega X := \{U \subset X\:|\: \forall x \in U \exists V \in \mathscr{F}(x)\:[V \subset U]\}\]

Then the $\forall x$ term in the conditions on the set ensures that $\emptyset \in \Omega X$ vacuously. Also clearly $X \in \Omega X$ because each $\mathscr{F}(x)$, as a filter, must contain $X$. Also it is not difficult to see that $\Omega X$ is closed under arbitrary unions. Lastly to show $\Omega X$ is closed under finite intersections, let $A, B \in \Omega X$. Then we may suppose $A \cap B$ is nonempty, because we already know that $\Omega X$ contains the empty set. So let $x \in A \cap B$. Then, by definition of $\Omega X$, $\exists V_1,V_2 \in \mathscr{F}(x)$ such that $V_1 \subset A$ and $V_2 \subset B$. But $\mathscr{F}(x)$ is a filter so $V:=V_1 \cap V_2 \in \mathscr{F}(x)$ and clearly $V \subset A \cap B$. So it follows $A \cap B \in \Omega X$. Hence $\Omega X$ is a topology. We shall call it the topology induced by the filter system $(\mathscr{F}(x))_{x\in X}$.

Now let $\Omega X$ be defined as above (the topology induced by the filter system $(\mathscr{F}(x))_{x\in X}$). Let us make a new filter system, which we shall call the neighborhood filter system or n-hood system, from $\Omega X$. To do this define
\[\mathscr{N}(x) := \{N\subset X\:|\:\exists U \in \Omega X \:[x\in U \subset N]\}\]
Then it is not difficult to check that $x$ is in every $N \in \mathscr{N}(x)$ and that each $\mathscr{N}(x)$ is a filter. We shall call each $\mathscr{N}(x)$ the neighborhood filter (or n-hood filter) at $x$. Then $(\mathscr{N}(x))_{x\in X}$ is a filter system for $X$ called the n-hood system for $X$ induced by $\Omega X$.

So the question naturally arises: When do we have that $\forall x \in X \: [\mathscr{F}(x) = \mathscr{N}(x)]$? To begin to answer this question let us note that we already have $\mathscr{N}(x) \subset \mathscr{F}(x)$ for each $x$. To see this let $x \in X$ and $N \in \mathscr{N}(x)$. Then $\exists U \in \Omega X$ such that $x \in U$ and $U \subset N$. But $x \in U \in \Omega X \implies \exists V \in \mathscr{F}(x) \:[V\subset U]$. So $V \subset N$. But $\mathscr{F}(x)$ is a filter and $V \in \mathscr{F}(x)$. So $N \in \mathscr{F}(x)$. Hence $\mathscr{N}(x) \subset \mathscr{F}(x)$, as desired.

However to show the other containment, that is $\mathscr{F}(x) \subset \mathscr{N}(x)$ for each $x \in X$, we need an extra condition on the filter system $\mathscr{F}(x)$. This is the technical condition I mentioned earlier. The condition is as follows:

\[(*) \qquad \forall x \in X \forall U \in \mathscr{F}(x) \exists V \in \mathscr{F}(x) \forall y \in  V \exists W \in \mathscr{F}(y) \:[W \subset U]\]

Let us show that this condition is sufficient. Let $x \in X$ and $U \in \mathscr{F}(x)$ Now let $A:= \{z \in U\:|\: \exists W \in \mathscr{F}(z)\:[W\subset U]\}$. Clearly $A \subset U$. Also, since $z \in A \implies \exists W \in \mathscr{F}(z)\:[W \subset U]$, we have that $A \in \Omega X$ (A is open). Moreover,  by ($*$) and the definition of $A$, $\exists V \in \mathscr{F}(x) \forall y \in V [y \in A]$. That is $V \subset A$. So in particular $x \in A \in \Omega X$. So $A \in \mathscr{N}(x)$. But $A \subset U$. So $U \in \mathscr{N}(x)$. Therefore $\mathscr{F}(x) \subset \mathscr{N}(x)$, as desired.

The condition $*$ is actually necessary. This is because $(\mathscr{N}(x))_{x\in X}$  satisfies $*$. Indeed if $x \in X$ and $U \in \mathscr{N}(x)$ then by definition of $\mathscr{N}(x)$, there exists a $V \in \Omega X$ with $x \in V \subset U$. Then with such a $V$, and using that open sets are neighborhoods of each of their points, it is not difficult to verify $*$. For additional reference see General Topology by Stephen Willard.

Hence we have shown that a filter system induces a topology whose n-hood system is the original filter system $\iff$ the original filter system satisfies $*$.

Now let us show that we were correct all along to call our filter system from the last post induced by observations on the space $X$ a neighborhood system. That is it naturally satisfies $*$ so it induces a topology (lattice of observations) whose n-hood system is the original filter system. To see this let $x$ be an abstract point in space and $U \in \mathscr{N}(x)$ (where $\mathscr{N}(x)$ is as defined in the previous post). Then by definition $x$ has been empirically observed to be in $U$. Then we may take $U = V$ and still satisfy $*$. Indeed if $y \in U = V$, then we may interpret this containment as knowing empirically that $y \in U$. Therefore $U \in \mathscr{N}(y)$ because $\mathscr{N}(y)$ is defined to be the list of areas of $X$ where we have empirically observed $x$. That is areas $A \subset X$ where we know that $y \in A$. Hence $U \in \mathscr{N}(y)$. So take $W = U \in \mathscr{N}(y)$ and clearly $*$ is satisfied.

Note to the reader: Some of the reasoning from the previous paragraph may seem strange. In the previous paragraph we were not considering $X$ as a set containing points but rather as an abstract space, as I mentioned in my previous post, where containment is only defined through the lattice of observations (topology). Formally we have switched to an alternative logic (intuitionism induced by the lattice of observations) which ensures that points are only defined when observed (and they can only be observed in neighborhoods by definition of the lattice of observations). This is a confusing point and it is in my hope that in later posts I will be able to clarify it. We may eventually be able to put this on more rigorous ground through topos theory and pointless topology which I hope to introduce in later posts.