Why does the order topology of some set have to contain **all** subsets on that set in the form of [a_0, b) and (a, b_0]?

Question

I'm currently reading Munkres' topology book, and when he defines the order topology on a set, he states that the order topology contains all subsets of that set in the form [a_0, b) and (a, b_0], along with all sets (a, b), where a_0 and b_0 are the smallest and largest elements in that set respectively. Why is it that we need to define all of those sets, as opposed to just one for each? For example, if I had [a_0, c) and (b, d) within my basis, and open sets in a topology are simply unions of basis elements, I can construct the subset [a_0, d) by taking the union of those two elements, so why is it necessary that I include [a_0, d) in my basis?

Answer 1

The order topology on $(X,<)$ is the smallest topology such that all sets of the form $L(x) = \{y \in X: y < x\}$, where $x \in X$ (the left sets) and all sets of the form $R(x) = \{y \in X: y > x\}$ where $x \in X$ (the right sets) are both open.

All these sets together form a so-called subbase for the order topology.

The base generated by the subbase is the set of all finite intersections of sets from the subbase (including intersections of one set, which ensures all subbase elements are also in the base).

As $L(x_1) \cap L(x_2) = L(\min(x_1,x_2))$ and $R(x_1)\cap R(x_2) = R(\max(x_1,x_2))$, we only get new base elements by intersecting a set of the form $L(x_1)$ and one of the form $R(x_2)$, which gives us open intervals $(x_2,x_1)$ with endpoints in $X$. These are truly new types of open sets.

If $X$ has a minimum $m$, then clearly $L(x) = [m,x)$, and so these sets, being in the generating subbase are in the base too. Munkres wants to be explicit in including these in the form $[m,x)$, instead of $L(x)$. Similar considerations hold if $X$ has a maximum $M$ and then $R(x) = (x,M]$. Munkres wants to be explicit in the fact that basic neighbourhoods of $m$ and $M$ (if they exist) look a bit different from the basic neighbourhoods of other points, that are open intervals with endpoints on both sides.

The subbase and resulting base are not meant to be minimal, just convenient. E.g. for $\mathbb{R}$ we can limit ourselves to intervals and left/right sets (aka open segments) with endpoints in $\mathbb{Q}$, because of the order density of the rationals. In $[0,1]$ we can for basic neighbourhoods of $0$ suffice with sets $[0,\frac{1}{n})$, instead of all $[0,x)$ etc. But we want to describe the order topology in full generality, so in that we need to include the full base, not some thinned out subset of it.