Let $X$ be a linearly ordered topological space with a countable dense subset. Does it necessarily follow that $X$ is metrizable?

EDIT: Apollo's comment int he answers implies the answer is negative. Let $X$ be the open unit interval $(0,1)$ and adjoin to every real number $x$ a "ghost number" $x'$ such that $x'$ is the immediate successor of $x$. The "real rationals" are dense in this space. Simply note that sets of the form $(y, x]$ with $x$ and $y$ real and $[x',y)$ with $x'$ ghost and $y$ real form a basis, and these sets all contain a real rational. This space cannot be metrizable, because the subspace topology on the set of all ghost reals is exactly that of the Sorgenfrey line.

  • $\begingroup$ Not an answer but a fun comment: Here is a simple criterion of when trees are metriable: google.com/… $\endgroup$ – Peter Arndt Mar 1 '11 at 20:29
  • $\begingroup$ Yes, if $X$ is connected, then $X$ has no jumps (points $x < y$ with no points inbetween), and then $d(X) = w(X)$. Urysohn does the rest. $\endgroup$ – Henno Brandsma May 25 '18 at 10:09

You already found a (classical) counterexample: the double arrow ($[0,1] \times \{0,1\}$, ordered lexicographically), which is even compact and separable. There is however a nice metrization theorem for linearly ordered spaces (due to Lutzer): a linearly ordered space $X$ is metrizable (in the order topology) iff the diagonal $D = \{(x,x) : x \in X\}$ is a countable intersection of open subsets of $X \times X$ (a $G_\delta$). This condition is also necessary and sufficient for countably compact regular spaces as well, not just the ordered ones.


No. Take $[0,1]\times\{0,1\}$ with the lexicographic order. This gives a counterexample --- it is separable (for example $\mathbb{Q}\times\{1\}$ is a countable dense set), yet it is not metrizable. One way to see this is to notice that the subspace $[0,1]\times\{1\}$ (homeomorphic to the Sorgenfrey line) is not second-countable, hence not metrizable. The counter-example can also be viewed as an example of an Alexandroff "double-point" construction, which is an example of the general construction of "(special) resolution" (which is a nice technique for generating counterexamples).

(Edited to incorporate comments --- original answer was incorrect, citing Sorgenfrey line as a counterexample.)

  • $\begingroup$ The Sorgenfrey line is not linearly ordered. $\endgroup$ – mathahada Mar 1 '11 at 0:04
  • $\begingroup$ The Sorgenfrey line is a subspace of a linear order with the order topology, but yes, it is not an order topology $\endgroup$ – Apollo Mar 1 '11 at 0:19
  • $\begingroup$ What is the linear order the Sorgenfrey line is a subspace of? Is this linear order separable, too? $\endgroup$ – mathahada Mar 1 '11 at 0:20
  • 1
    $\begingroup$ Actually, now that I think about it, yes it is separable: $[0,1]\times\{0,1\}$ ordered lexicographically. $\endgroup$ – Apollo Mar 1 '11 at 0:24

I believe this follows from the Urysohn metrization theorem. Order topologies are regular (even completely normal, according to Wikipedia), and your hypothesis of a dense countable subset should imply second countability, unless I'm missing something obvious. Namely open intervals with endpoints in your dense set will form a basis for the topology. So the two hypotheses of Urysohn's metrization theorem are satisfied.

Edit: This works if you have a countable dense subset $D$ in the sense that for every $x$ and $y$ there exists a $d\in D$ with $x < d < y$, but that's not the sense of density being used!

  • $\begingroup$ You should also ensure the maximal and minimal elements of the order, if any exist, are in your countable dense set in order for the corresponding set of intervals to be a basis. $\endgroup$ – Joel David Hamkins Mar 1 '11 at 0:51
  • $\begingroup$ Indeed. And of course, that's easily arranged! $\endgroup$ – Jim Conant Mar 1 '11 at 1:16
  • $\begingroup$ In my example the space is not second countable. For every real number $x$ the open interval $U=(x-1,x')=(x-1,x]$ has maximal element $x$. A basis element contained in $U$ and containing $x$ must also have maximal element $x$ so there is a surjection from the set of all basis element to the reals $\endgroup$ – mathahada Mar 1 '11 at 9:37
  • $\begingroup$ I mean Apollo's example, not mine:) $\endgroup$ – mathahada Mar 1 '11 at 9:37
  • $\begingroup$ I think I was using a different definition of "dense." I was taking dense to mean that between any two elements there exists an element from my dense subset, but I guess it should mean that the closure is the whole space in this context. (Which is how you define separable.) This is one of those situations where overlapping terminology causes confusion. :) $\endgroup$ – Jim Conant Mar 1 '11 at 11:04

If you use the definition of dense used by Jim Conant, there is a fairly explicit way (apparently going back to Cantor) to metrize the space by embedding it in the unit interval. So let $(L,\leq)$ be a linearly ordered set and $C=(c_ 1,c_2,\ldots)$ a countable subset such that for all $x_1< x_2$ there is a $c\in C$ with $x_1< c < x_2$. Then it is easily seen that $u:L\to [0,1]$ given by $u(x)=\sum_{n\in L_x} 1/2^n$ with $L_x=(n\in\mathbb{N}:c_n< x)$ is an order-preserving function. So one can identify $L$ with a subset of $[0,1]$ and metrize it that way.

A similar construction can be used if $C$ just satisfies that there is a $c\in C$ with $x_1\leq c\leq x_2$ whenever $x_1 < x_2$ with a little more effort.

These results are used in mathematical economics to represent preference relations by utility functions. The book "Representations of Preference Orderings" by Bridges and Mehta contains a lot of related results.


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