This is an embarrassing question, because I learned about this theorem in basic analysis, but haven't realized that I don't really understand its statement until now.

Anyway, it's a famous result that "up to isomorphism, $\mathbb{R}$ is the only Dedekind-complete ordered field". Sources, e.g. (1)(2)(3)(4)

Question: isomorphism in which category?

Isomorphic as fields? Isomorphic as ordered fields? Isomorphic as rings? Isomorphic as sets? (clearly not) Isomorphic as topological spaces? (also clearly not, but you get my point)

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    $\begingroup$ In the category of ordered fields. $\endgroup$ – Berci May 29 '17 at 19:51
  • $\begingroup$ @Berci So in particular we could have a non-orderable field isomorphic to $\mathbb{R}$ as a field? $\endgroup$ – Chill2Macht May 29 '17 at 19:52
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    $\begingroup$ @Chill2Macht No. If $K$ is any ring isomorphic to $\Bbb{R}$, then it can be ordered. The order is the one induced by the isomorphism $f: K \to \Bbb{R}$, i.e. $$x < y \ \ \Longleftrightarrow \ \ f(x) < f(y)$$ $\endgroup$ – Crostul May 29 '17 at 19:54
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    $\begingroup$ I thought $\mathbb R$ was the unique (up to isomorphism of course) complete ordered field. What's the difference between a Dedekind-complete ordered field and a complete ordered field? $\endgroup$ – bof May 29 '17 at 20:05
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    $\begingroup$ @Chill2Macht Note that the order rlatoin in $\Bbb R$ can be exopressed with field properties: $x\le y\iff \exists z\colon x+z^2=y$. $\endgroup$ – Hagen von Eitzen May 29 '17 at 20:10

We are talking about a theory in the language of ordered rings, so a homomorphism should respect the elements of the language:

  • $ \varphi(0) = 0 $
  • $ \varphi(1) = 1 $
  • $ \varphi(-x) = -\varphi(x) $
  • $ \varphi(x+y) = \varphi(x) + \varphi(y) $
  • $ \varphi(xy) = \varphi(x) \varphi(y) $
  • $x < y \implies \varphi(x) < \varphi(y) $

(I assume the language has the $<$ predicate rather than the $\leq$ predicate)

In some contexts, one might require more from a homomorphism: that it preserve something related to completeness.

An isomorphism, of course, is a homomorphism that has an inverse homomorphism.

Incidentally, note that there is only one ring homomorphism between any two complete ordered fields, and everything else follows from the ring structure (e.g. the nonnegative elements are precisely the squares), so the answer to your multiple choice question turns out to be "all of the above".

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    $\begingroup$ So that is kind of like how two Hilbert spaces are isomorphic as Hilbert spaces if and only if they are isomorphic as inner product spaces? (assuming that is even true, which I think it is, but I'm not sure why) $\endgroup$ – Chill2Macht May 29 '17 at 20:22

The claim can be made precise i nthis form:

Let $F$ be a Dedekind-complete ordered field. Then there exists a unique nonzero ring homomorhism $\phi\colon F\to \Bbb R$, and this is a ring (and field and order ...) isomorphism.


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