Does there exist a totally disconnected topological space T whose topology is the order topology of a linearly ordered set and whose (small inductive) dimension is equal to 1? There exist topological spaces which satisfy all the other conditions, but I do not know of any whose topology is an order topology.


No. According to this paper zero-dimensionality (in small, or large inductive, or covering definition) is equivalent to being totally disconnected, and all non-such ordered spaces are one-dimensional (in all three definitions). Dimension theory is pretty dull in ordered spaces.

  • $\begingroup$ Thanks for the reference; I’m surprised that the result is so recent. $\endgroup$ – Brian M. Scott Oct 12 '14 at 21:13
  • $\begingroup$ @BrianM.Scott Maybe nobody cared enough, both dimension theory and ordered spaces are somewhat out of fashion? It might have been folklore in some circles, and never written down. $\endgroup$ – Henno Brandsma Oct 13 '14 at 3:39
  • $\begingroup$ Or perhaps most of the people who actually were interested in ordered spaces weren’t all that much interested in dimension theory. (Mind you, I could well be over-generalizing from a single instance — me!) $\endgroup$ – Brian M. Scott Oct 13 '14 at 3:48
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    $\begingroup$ That $\operatorname{ind}(X) \le 1$ (trivial) and even $\dim(X) \le 1$ (slightly less so) for ordered spaces was well-known, I'm sure, so that sort of trivialises the dimension theory before you even start thinking about it. $\endgroup$ – Henno Brandsma Oct 13 '14 at 3:51
  • $\begingroup$ Many thanks for such a complete answer to my question $\endgroup$ – Garabed Gulbenkian Oct 13 '14 at 20:30

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