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It there any relation between the axiom of choice and Euclidean Geometry ??

I mean what are the known statements, theorems or results in euclidean geometry that are dependent on AC ?? (this question has been asked by someone else as a comment to my last question. See HERE)

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  • $\begingroup$ Both things are called "axioms"... $\endgroup$
    – Asaf Karagila ♦
    Commented Jan 15, 2013 at 19:21
  • $\begingroup$ @Asaf: As a set theorist you mean no relation exists. No statement or theorem or result in euclidean geometry is dependent on AC ? Is it provable ?? $\endgroup$
    – user30669
    Commented Jan 15, 2013 at 19:27
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    $\begingroup$ I never said anything about that. It also depends on what you define as Euclidean geometry. One could argue that the Banach-Tarski theorem applies to Euclidean geometry; while another could argue that the classical Euclidean geometry is effectively a constructive theory and there is no appeal to infinitary arguments which would require the axiom of choice. I only said that both the axioms of Euclidean geometry and the axiom of choice are named "axioms". $\endgroup$
    – Asaf Karagila ♦
    Commented Jan 15, 2013 at 19:35
  • $\begingroup$ @Asaf: I see :) $\endgroup$
    – user30669
    Commented Jan 15, 2013 at 19:46
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    $\begingroup$ What do you mean by a relation? As it stands, this is not a real question. $\endgroup$
    – Emil Jeřábek
    Commented Jan 15, 2013 at 20:01

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Tarski gave a first-order formulation of Euclidean geometry and proved it complete (without using the axiom of choice). In particular, that means that Euclidean geometry is the same in every model of ZF, with or without choice. One might question whether Tarski's formulation actually captures all of Euclidean geometry, but it certainly includes everything in Euclid.

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