MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

During my work with order preserving homeomorphisms, I got interested in the double arrow space and, subsequently, in the lexicographic square. I would really like to find examples of spaces like these two. Specifically, I am looking for a space $X$ with the following properties:

  1. $X$ is a compact linearly ordered topological space (compact LOTS)
  2. $X$ is densely ordered, that is for all $x,y\in X$ if $x <y$ then there exists $z\in X$ such that $x<z<y$. (I think one can say that in this case $X$ "has no jumps".)

Besides the spaces mentioned above, closed real intervals also satisfy these conditions. I think that maybe the pseudo-arc is another example, although I am still trying to understand its construction.

Any other examples or references would be highly appreciated!

Thank you!

share|cite|improve this question
A related question: for what posets $P$ is it the case that any linearization of $P$ is compact in the order topology? – Noah Schweber Jun 23 '14 at 20:19
Q: How many topological spaces are linearly ordered? $\tag*{}$ A: LOTS! $\tag*{}$ – Asaf Karagila Jun 23 '14 at 23:01
. . . Thank you for that. :P But I think my question is actually not that silly. – Noah Schweber Jun 23 '14 at 23:24
up vote 5 down vote accepted

One of my favorite spaces has this property: the extended long ray.

First, the long ray itself: this is just the space $L$ gotten by pasting together $\omega_1$-many copies of $[0, 1)$ in the natural way. Formally, $L$ is the lexicographic order on $\omega_1\times[0, 1)$, with both viewed as linear spaces in the natural way.

Now, $L$ is densely ordered, but obviously $L$ is not compact (although any countable open cover admits a finite subcover, and every sequence contains a convergent subsequence; and, as an unrelated nice property, every proper initial segment of $L$ is homeomorphic to $[0, 1)$). However, if we add a point at the end, we get the extended long ray $L^*$.

$L^*$ is still densely ordered, and unlike $L$ is compact; in fact, if I remember correctly, it is both the Stone-Cech compactification and the one-point compactification of $L$. $L^*$ is a nice counterexample to a bunch of things, as is $L$ itself; in particular, the differentiable structures admitted by $L$ are many and nasty.

I suspect the wonderful book "Counterexamples in Topology" (Steen & Seebach) has many more examples of such spaces.

share|cite|improve this answer
The reason why the Stone-Cech compactification of $L$ conincides with the one-point one is the fact that scalar-valued continuous functions on $L$ are eventually constant (the proof is verbatim the same as for $\omega_1$, where the same phenomenon occurs). – Tomek Kania Jun 24 '14 at 7:42
Thanks a lot! I like this example! Just one question: I understand why $L$ is densely ordered, and I guess that adding a point does not change it. But is there a general theory that describes what happens to an order on a certain space after a compactification? – Ludolila Sep 11 '14 at 13:49

The linear orders which are compact in the order topology are precisely the complete totally ordered sets $X$. By complete I mean that every subset of $X$ has a least upper bound including $\emptyset$ and $X$. If $R\subseteq X$ has no least upper bound, then let $S$ be the set of all upper bounds of $R$. Then $\bigcup_{r\in R}\{x\in X|x<r\}\cup\bigcup_{s\in S}\{x\in X|x>s\}$ is an open cover with no finite subcover.

For the converse, assume $X$ is a complete totally ordered set. Let $\mathcal{U}$ be a non-principal ultrafilter on $X$. Then for each $a\in X$ either $\{x\in X|a\leq x\}\in\mathcal{U}$ or $\{x\in X|a\geq x\}\in\mathcal{U}$. Let $A=\{x\in X|a\geq x\}\in\mathcal{U}$ and let $B=\{x\in X|a\leq x\}\in\mathcal{U}$. Then $A$ is downwards closed and $B$ is upwards closed and $\{A,B\}$ is a partition of $X$. Therefore there is some $x\in X$ where if $a\leq x$, then $a\in A$ and if $a\geq x$, then $a\in B$. Let $(a,b)$ be an interval around $x$. Then $a\in A,b\in B$, so $(a,1],[0,b)\in\mathcal{U}$, so $(a,b)=(a,1]\cap[0,b)\in\mathcal{U}$. Therefore $X$ is compact since every non-principal ultrafilter converges.

share|cite|improve this answer
I've deleted an incorrect comment (I missed "the order topology"). – Noah Schweber Jun 23 '14 at 20:18
Thanks a lot for this theorem, I wasn't aware of it and it looks helpful. But I'm not sure that I understand the converse direction of the proof. On the third line you write that $A=...\in U$ and $B=... \in U$, but they can't both be in $U$, I think. Also, $\{A, B\}$ is not a partition, since $a\in A\cap B$. Then you take the interval $(a,b)$ which is kind of abuse of notation, because it's not the same $a$, right? – Ludolila Sep 11 '14 at 13:42
And I think we need to prove that there exists an $x$ such that every (basic) nbhd of $x$ is in $U$, right? But I don't understand who is this $x$ and how you reach the conclusion in the last two lines. I would really appreciate if you can elaborate (or even suggest an online source I can use). – Ludolila Sep 11 '14 at 13:44

A compact Suslin line would be another important example although they only exist in some models of set theory. Dedekind completions of suitable Aronszajn lines give you more ZFC examples.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.