In ZFC, every linear ordered space respect to the order topology is completely normal. I saw the this proof and proof of this statement in the book "Counterexamples of topology" (Example 39). But as I seen every proof of this statement uses choice. Even if (as I know) the proof of "Every linear continuum is normal" uses axiom of choice.

So I think that choice is essential to prove this statement. That is true? Thanks for any help.


You cannot do that without the axiom of choice. See the following paper by van Douwen:

Eric K. van Douwen, "Horrors of Topology Without AC: A Nonnormal Orderable Space". Proceedings of the American Mathematical Society Vol. 95, No. 1 (Sep., 1985), pp. 101-105

And also related is this paper by Krom:

Melven Krom , "A linearly ordered topological space that is not normal". Notre Dame J. Formal Logic Volume 27, Number 1 (1986), 12-13.

  • $\begingroup$ @Arthur: Thanks! $\endgroup$ – Asaf Karagila Nov 10 '13 at 9:07

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