# reference request: writing intervals of totally-ordered sets in a “nice” way

Let $$\Omega$$ be a totally-ordered set.

We need to introduce some symbolism and terminology, which is all very natural and intuitive. If $$A$$ and $$B$$ are subsets of $$\Omega$$ we write $$A whenever $$a for all $$a\in A$$ and $$b\in B$$. By an interval of $$\Omega$$, we mean any subset $$I$$ such that there exist sets $$A$$ and $$B$$ with $$A and $$\Omega=A\cup I\cup B$$. In the special case where $$A=\emptyset$$, the interval $$I$$ is called an initial segment of $$\Omega$$.

It is desirable to write intervals of $$\Omega$$ as we might write intervals of $$\mathbb{R}$$. Thus, we say that an interval $$I$$ of $$\Omega$$ is of real form if it can be written in one of the following ways for some $$a,b\in\Omega$$: $$(-\infty,b):=\{x\in\Omega:x $$(-\infty,b]:=\{x\in\Omega:x\leq b\}$$ $$(a,b):=\{x\in\Omega:a and so on, also for $$[a,b)$$, $$(a,b]$$, $$[a,b]$$, $$(a,\infty)$$, and $$[a,\infty)$$, defined in the obvious ways.

Unfortunately, intervals of $$\Omega$$ need not always have real form. Consider for instance the case $$\Omega=[0,1)\cup(1,2]$$ endowed with the usual ordering. Then $$[0,1)$$ is an interval, but $$1\notin\Omega$$ and so it can't be written in real form. On the other hand, due to the fact that $$\Omega\subseteq[0,2]$$ in this case, we can still write the interval $$[0,1)$$ very nicely.

Fact. For any totally-ordered set $$\Omega$$, there exists a superset $$\Gamma\supseteq\Omega$$ which is endowed with a "compatible" total order (in the sense that $$x in $$\Omega$$ implies $$x in $$\Gamma$$), and such that every interval of $$\Gamma$$ has real form.

This is not hard to show. If $$b\in\Omega$$ we define $$I_b:=(-\infty,b].$$ Let $$\Gamma$$ denote the set of all initial segments of $$\Omega$$. Then $$\subseteq$$ is a total order on $$\Gamma$$ compatible with the order on $$\Omega$$, and the map $$b\mapsto I_b$$ is an injection from $$\Omega$$ into $$\Gamma$$ which means we can identify $$\Omega$$ with a subset of $$\Gamma$$. It is straightforward to see that every interval of $$\Gamma$$ is of real form.

Question 1. Has this been discussed in the literature? If so, does anyone have a reference?

Question 2. The term "real form" sounds unsatisfying, but I can't think of anything better. Is there already a term for this in the literature? Or just a better suggestion from you guys?

Thanks!

• A general term for interval is order convex which can also be used with partial orders. – William Elliot Feb 21 at 23:54
• As a matter of fact, a set in a connected linear order is order convex iff it is an interval. So the real forms are the connected orders. – William Elliot Feb 22 at 10:18
• @WilliamElliot connected in the order topology? – Ben W Feb 22 at 14:03
• Yes............ – William Elliot Feb 22 at 22:06