What does isomorphism mean in "$\mathbb{R}$ is the Dedekind-complete ordered field up to isomorphism"? 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)
 A: 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".
A: 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.

