How large can a subset of computable reals, whose comparison function is computable, grow?
For example, rational numbers are computable reals, and its comparison function is computable. As another example, algebraic numbers have the same property. And algebraic numbers include rational numbers.
How far can this subset be extended while keeping the property that "the comparison function is computable"?
Let $r_i$ be the real computed by the $i$th Turing machine. Note that some machines fail to compute reals - "$r_i$" only makes sense for some $i$; let $I$ be the set of indices for machines which do compute reals (so the set of all computable reals is just $\{r_i:i\in I\}$). There are two seemingly-different notions of "computable field of reals" which turn out to be appropriately equivalent:
An injective computable field is a triple $(X,f,g)$ such that
$X$ is a computable subset of $I$ (note that $I$ itself is not computable);
$\{r_i:i\in X\}$ is a field;
distinct $x,y\in X$ have $r_x\not=r_y$ (this is the "injective" bit); and
$f,g$ are computable functions $X^2\rightarrow X$ such that $r_{f(x,y)}=r_{f(x)}r_{f(y)}$ and $r_{g(x,y)}=r_x+r_y$.
A possibly non-injective computable field is a triple $(X,f,g)$ such that
$(X,f,g)$ satisfy each of the conditions above except the third condition; and
the set $\{(x,y)\in X: r_x=r_y\}$ is computable.
The equivalence between these two notions is the following:
If $(X,f,g)$ is a possibly non-injective computable field, there is an injective computable field $(\hat{X},\hat{f},\hat{g})$ with $\{r_x:x\in X\}=\{r_{\hat{x}}:\hat{x}\in\hat{X}\}$. (Moreover the translation between the two is appropriately computable.)
The proof is simple: if $(X,f,g)$ is a possibly non-injective computable field, consider $$X_0=\{x\in X:\forall y\in X(y<x\implies r_x\not=r_y)\}.$$ By assumption, $X_0$ is computable, and $(X_0, f\upharpoonright X_0, g\upharpoonright X_0)$ is the desired injective computable field.
Thinking in terms of injective computable fields generally makes things easier. In particular, there is no largest injective computable field (in particular the field of all computable reals is not "computably presentable"), and every computable real is contained in some injective computable field.
Here's a fun exercise: if $a,b$ are computable reals, must there be an injective computable field containing both $a$ and $b$?
The answer is yes. Think first about the case where $b$ is transcendental over $a$.
