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If we consider Euclidean space based on its use in geometry (i.e. only as a place to explore geometric theorems, like Euclid did in the "Elements"), can it be defined as "space where Euclid's axioms (or their more rigorous formalisations, e.g. Hilbert's) apply"? This seems very simple and intuitive - are there caveats to it?

Is there more to definitions of Euclidean space than formalisations of spaces whose properties Euclid was trying to describe by his five axioms?

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  • I also have this question. As far as I have been able to tell, the answer depends on what axioms one uses for Euclidean geometry, but if one wants to restrict to the model of Euclidean geometry which can be represented by $\mathbb{R}^n$, then I think it suffices (based on the fact that up to isomorphism the real numbers are the only Dedekind complete ordered field and that Dedekind completeness of $\mathbb{R}$ is equivalent to its Cauchy completeness) to ask for (1) Tarski's /Hilbert's axioms + (2) complete metric space. – Chill2Macht Jul 02 '17 at 19:55
  • (1)https://math.stackexchange.com/questions/348587/whats-a-non-standard-model-of-tarskian-euclidean-geometry (2)https://math.stackexchange.com/questions/2178037/does-euclidean-geometry-require-a-complete-metric-space (3)https://math.stackexchange.com/questions/80930/what-is-the-modern-axiomatization-of-euclidean-plane-geometry/1991372#1991372 (4) https://math.stackexchange.com/questions/2180838/is-metric-cauchy-completeness-outside-the-realm-of-first-order-logic (5) https://math.stackexchange.com/questions/2185555/do-the-real-algebraic-numbers-satisfy-a-type-of-completeness-axiom-excluding-fre – Chill2Macht Jul 02 '17 at 19:59
  • The key result (using Tarski's axioms) is the representation theorem (all of the key results along these lines seem to be called representation theorem) which says that any model (whatever that means, I don't know formal logic) of Euclidean geometry using Tarski's axioms is isomorphic as a model (I don't know specifically what this means either) to $R^2$ for $R$ some real closed field ($\mathbb{R}$ is a real closed field, but so are the algebraic numbers). I imagine similar representations hold for $R^n$ if one changes the upper and lower dimension axioms to make the dimension $n$ – Chill2Macht Jul 02 '17 at 20:02
  • Take what I say with a grain of salt, since otherwise this might be a case of the blind leading the blind, but anyway I found this page especially helpful: https://ncatlab.org/nlab/show/Euclidean+geometry#TarskiAxioms Wikipedia's page on Tarski's axioms also has helpful analysis and discussion which the ncatlab page lacks, but on the other hand it (1) doesn't mention the representation theorem, which for me is crucial to understand all of this and (2) is somewhat less organized. – Chill2Macht Jul 02 '17 at 20:04
  • For what we intuitively understand to be Euclidean geometry (e.g. the definition of $\pi$ and stuff like that) we definitely need a little more than Tarski's axioms (and Hilbert's axioms might not actually be more powerful, even though they are written using second-order, rather than first-order, logic), hence "complete metric space", see e.g. here: https://books.google.de/books?id=IMgg0Uc00I4C&pg=PA593&lpg=PA593&dq=affine+geometry+tarski%27s+axioms&source=bl&ots=jfYHmjrX8Q&sig=IxkfZ98uG6chdJas7VchlY_ADbs&hl=en&sa=X&redir_esc=y#v=onepage&q=affine%20geometry%20tarski's%20axioms&f=false – Chill2Macht Jul 02 '17 at 20:15
  • The relevant quote being on p. 593 (Logic from Russell to Church): "There are notions of textbook geometry that cannot be expressed in G [Tarski's axioms] -- for example, the notions of the circumference and area of a circle, and the notion of a polygon with arbitrarily many vertices". See also this question: https://math.stackexchange.com/questions/689658/which-kinds-of-geometry-have-an-angle-measure The idea is simple enough: $\pi$ is transcendental, the algebraic numbers can't express $\pi$, so an arbitrary real closed field can't express $\pi$. – Chill2Macht Jul 02 '17 at 20:25

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