Can we prove that every ordered space is normal without choice?

Question

In ZFC, every linear ordered space with 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 have seen every proof of this statement uses choice. Even if (as I know) the proof of "Every linear continuum is normal" uses the axiom of choice.

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

score 10 · Accepted Answer · edited Nov 10 '13 at 08:57

10

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.

edited Nov 10 '13 at 08:57

user642796

53,641

answered Nov 10 '13 at 08:14

Asaf Karagila

405,794

@Arthur: Thanks! – Asaf Karagila Nov 10 '13 at 09:07

Can we prove that every ordered space is normal without choice?

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