Every real closed field $R$ has in integer part $I$. That is, $I$ is a discrete ordered subring of $R$ such that for each $x \in R$ there is $z \in I$ such that $z \leq x < z + 1$.
If $R$ is Archimedean, then $\mathbb{Z}$ is the unique integer part and if $R$ is not Archimedean, my understanding is that $R$ may have non-isomorphic integer parts. My question is:
Is it always possible to find an integer part with cardinality strictly less than the cardinality of $R$?
Is the cardinality of any integer part always strictly less than the cardinality of $R$? Do different integer parts have same cardinality?
Well, for instance, if $R$ is countable, then clearly there cannot be an integer part of strictly smaller cardinality. More generally, for any infinite cardinal $\kappa$ there is a real-closed field of cardinality $\kappa$ with integer part of cardinality $\kappa$. For instance, you could start with a field of rational functions over $\mathbb{Q}$ in variables $(x_\alpha)_{\alpha<\kappa}$ ordered so that $x_\alpha$ is infinitely larger than $x_\beta$ for $\beta<\alpha$. The real closure of this ordered field will then have cardinality $\kappa$ and an integer part must have cardinality at least $\kappa$ since the "floors" of the $x_\alpha$s must all be distinct.
(Alternatively, this can be proved by a simple compactness argument. Consider the theory of real-closed fields, augmented to have a unary relation symbol which defines an integer part and $\kappa$ constant symbols that are distinct elements of the integer part. This theory is finitely satisfiable, since any finite subset involves only finitely many of the constant symbols. So it is consistent, and by Löwenheim-Skolem it has a model of cardinality $\kappa$.)
Any two integer parts $I$ and $J$ of an ordered field $R$ have the same cardinality. Indeed, there is a bijection $I\to J$ given by taking $i\in I$ to the unique $j\in J$ such that $j\leq i<j+1$.