# Connectedness of a linear ordered topological space

From a short search in the internet I was unable to find whether there are easy conditions for connectedness in Linearly ordered topological spaces. It looks to me like connectdness in the linear order case should be simpler, but was wondering if perhaps there is no such simple characterizations.

For example, I think that the following should be true:

Given a Linearly ordered topological space $$(X,<)$$:

1. $$X$$ is connected if and only if it is path connected.
2. $$X$$ is connected if and only if every bounded subset is order-complete.
3. $$X$$ is connected if and only if every bounded set is connected.

I saw a result stating that $$X$$ connected is orderable if and only if $$A \subseteq X\times X$$ is disconnected in $$X\times X$$ w.r.t the product topology, where $$A=X\times X \setminus \Delta$$ and $$\Delta$$ is the diagonal set in $$X\times X$$.

So my questions are whether any of these proposed statements true, and is there perhaps another simple characterization which is true?

• An order topology is connected iff it is a linear continuum. One direction is even proved on wikipedia, the other is proved in the first google result Apr 5, 2021 at 8:52
• @HennoBrandsma I corrected the statement to the phrasing I saw in a paper by Samuel Eilenberg. Apr 5, 2021 at 9:49
• @HennoBrandsma I wrote it incorrectly even though I thought I corrected it. I will edit to the correct (hopefully) formulation. The paper is "Ordered topological spaces" from 1940. Apr 5, 2021 at 12:38
• But that's another question: $X$ is given to be connected and you want a criterion for orderability, not $X$ is ordered and you want a criterion for connectedness.. The general orderability question (for non-connected spaces too) was later solved by several authors independently. Apr 5, 2021 at 12:46
• @HennoBrandsma But doesn't it give a necessary condition for connectedness? Also, that was among the results which I was able to find regarding connectedness and LOTS before asking this question. Apr 5, 2021 at 13:09

Let $$X$$ be an ordered topological space $$(X,<)$$.

A cut $$(A,B)$$ of $$X$$ (by which I mean $$A,B \subseteq X$$, both non-empty, $$A \cap B = \emptyset$$, $$A \cup B = X$$, and also for all $$a \in B$$ and all $$b \in B$$ we have $$a < b$$) is called a jump if $$A$$ has a maximum and $$B$$ has a minimum, and a gap if neither is the case.

Theorems: $$X$$ is connected iff $$X$$ has no gaps or jumps. $$X$$ is compact iff it has no gaps, and a minimum and a maximum.

This was the way we used to discuss it in university (90's). Now Munkres has popularised the term "linear continuum" and in his text book shows that $$X$$ is connected iff it is a linear continuum which means that

• $$X$$ is order complete (every subset that is bounded above has a supremum).
• $$X$$ is order-dense: for all $$x < y$$ in $$X$$ there exists some $$z \in X$$ with $$x < z < y$$.

Note that the second implies $$X$$ has no jumps and the former that $$X$$ has no gaps (so there is a clear connection with my "classical" terminology). But Munkres' equivalent conditions are probably easier to verify in practice.. First application is that $$\Bbb R$$ and all its subsegments and intervals are connected.

Your conditons do not quite work: in $$\Bbb Z$$ all bounded subsets are order complete (even finite) but $$\Bbb Z$$ is not connected. Your condition 3. is somewhat circular and not really useful as a criterion. It quite easily follows from both above characterisations though: gaps or jumps manifest on a bounded subset already e.g.

Finally, 1. is certainly false: $$[0,1]^2$$ in the lexicographic order topology is connected but not path-connected.