\(\newcommand{\T}{\mathscr{T}}\newcommand{\N}{\mathbb{N}}\) Recall the following
If \((P,\leq)\) originates from a lattice,
\begin{align*} (a,b) = \{x\in P: a\lt x \lt b\mid a\in P\cup\{-\infty\}, b\in P\cup\{\infty\} \} \end{align*}forms a base of the associated order topology.
In an attempt to better understand finite topologies, my question is: Does every finite topology originate from this construction? If not, what are the necessary restrictions that must be fulfilled?
Focusing on finite sets, the following notions come in handy.
Atoms don't have to exist, which becomes apparent when considering an infinite descending chain like \(\{(-1/n,1/n)\mid n\in \N \}\) in the usual topology on \(\mathbb R\): The minimum of this chain would be the one-point set \(\{0\}\). It is not hard to see why this cannot happen it in the finite case:
Let \((P, \leq)\) poset, \(a < b \in P\). The following are equivalent:
Let \(a\) be a lower neighbor of \(b\), i.e. maximal in \((-\infty, b)\). By definition, every \(x\in (-\infty, b)\) satisfies \(a\leq x \Leftrightarrow a=x\), or equivalently,
which in turn holds precisely if \((a,b)=\emptyset\). The argument for \((2)\Leftrightarrow (3)\) follows analogously.
Consider \(P:= \{1,2,3,6\}\) ordered by divisibility. It is a lattice, so we have a base consisting of the intervals \((1,6) = \{2,3\}\), \((-\infty,2)=(-\infty,3)= \{1\}\), \((-\infty,6)=\{1,2,3\}\), \((1,\infty)=\{2,3,6\}\), \((2,\infty)=(3,\infty) = \{6\}\), ergo
\begin{align*}\T_P = \left\{ \emptyset, \{1\}, \{6\}, \{2,3\}, \{1,2,3\}, \{2,3,6\}, P \right\}\end{align*}The following observations seem not so bright in retrospect, but originated from an impulse to find open points, i.e. points \(x\) such that \(\{x\}\) is open already: Open points are pretty “isolated”, because the only sequences converging to \(x\) are the ones being eventually constant.
One might fall into the trap of thinking that maximal elements \(m\) with lower neighbors \(l\) (without the latter, they would only be comparable to themselves) have to be open points by representing them as \((l,\infty)\subseteq (m,\infty)\), but that does not have to be the case:
On a similar note, if \(a<b<c\), just because both \((a,b)\) and \((b,c)\) are empty does not mean \((a,c) = \{b\}\), as the first example shows: \((1,6)=\{2,3\}\). Note that again, both points in question are topologically indistinguishable.
Note that this means we have quite some amount of choice: The topology sees the ordering only on a “local” level, that is, the element in relation to some interval \((a,b)\ni x\).
Associate to a poset \((P, \leq)\) the generated equivalence relation of \(\leq\), i.e. the relation where \(a\sim b\) if and only if there is a sequence \(a=x_1, \ldots, x_n = b\) such that at each position, either \(x_i\leq x_{i+1}\) or \(x_{i+1} \leq x_i\) (check this!).
We call the resulting equivalence class “connected component”, and given a specific \(x\), the associated equivalence class is called the “connected component of \(x\)”.
(TODO)