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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!

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

3 Answers 3

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.

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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 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 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.

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I've deleted an incorrect comment (I missed "the order topology"). –  Noah S Jun 23 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 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 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.

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