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As I understand it, any axiomatisation of a complete linearly ordered field, must be second order in nature. Does that mean that the fact that all complete linearly ordered fields are isomorphic to the reals is a second order theorem and is in fact undecidable in the first order case? My intuition tells me that this is so because there are infinitely many solutions to the continuum hypothesis and each one would give a first order non-isomorphic complete linearly ordered field, could someone tell me if such intuition makes formal logical sense, i.e. is there such a thing as first order non-isomorphic complete linearly ordered fields?

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The usual statement of completeness is inherently a second-order statement: you can't even write it down in first-order logic. And, as it turns out, there is no "clever" way to encode completeness in first-order logic.

So either your question ends there, or you have to figure out what you mean by a "first-order complete linearly ordered field".

There is, incidentally, a very nice way to make sense of that: it's called a real closed field. There is a lot of literature on these, but let me highlight two facts that may be related to your line of thought.

Note that the language of ordered rings consists of the constants $0,1$, the operations $+,-,\cdot$, and the relation $\leq$.

  • Every real-closed field is Dedekind complete with respect to semi-algebraic cuts. That is, any Dedekind cut that can be expressed in the language of ordered rings is, in fact, a principal cut. Similar statements can be made for various other expressions of completeness
  • Let $R$ be a real closed field. For every statement $P$ that can be expressed in the language of ordered rings, it turns out that $P$ holds in the real numbers if and only if $P$ holds in $R$.

There are real closed fields that are not isomorphic to $\mathbb{R}$. The simplest (and smallest) is the field of real algebraic numbers. There are also non-Archimedean real closed fields.

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  • $\begingroup$ Thanks Hurkyl, I was aware of this, but what I am wondering is the relation to the Continuum Hypothesis that I stated. A complete linearly ordered field with the additional axiom of an uncountable cardinal between $\aleph_{0}$ and $2^{\alpeh_{0}$ is not necessarily closed, but could your statements minus the closed bit be modified to still hold for this more general case? $\endgroup$
    – YeatsL
    Commented Jun 9, 2018 at 11:17
  • $\begingroup$ I apologize first because I don't know if I'm allowed to ask this in comments. May I ask for help with a problem of the axiomatization of the field $\mathbb R$, more precisely substituting the axiom of completeness with Cantor's and Archimedean axiom?math.stackexchange.com/questions/3421847/… Sorry for disturbing and possibly breaking the rules. $\endgroup$
    – PinkyWay
    Commented Nov 4, 2019 at 18:02

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