Without LEM or the axiom of choice, we can prove that the [Eudoxus reals](https://ncatlab.org/nlab/show/Eudoxus+real+number) are equivalent to the Cauchy reals but can't prove either of those equivalent to the Dedekind reals.

However, we can prove that the [*multi-valued* Cauchy reals](https://ncatlab.org/nlab/show/Cauchy+real+number#generalisations) are equivalent to the Dedekind reals. My question is if we can say the same about a kind of multi-valued Eudoxus real?

In particular, a multi-valued Eudoxus real will be represented by a set $R \subseteq \mathbb Z \times \mathbb Z$ (thought of as a relation) such that $\forall x \in \mathbb Z. \exists a \in \mathbb Z. xRa$ and the set $\{a + b - c: \exists x, y \in \mathbb Z. xRa \land yRb \land (x+y)Rc\}$ is bounded. Relations $R_1$ and $R_2$ represent the same multi-valued Eudoxus real when the set $\{a - b : \exists x \in \mathbb Z. xR_1a \land xR_2b\}$ is bounded.

We turn this into an ordered field in the same way as we do the (single-valued) Eudoxus reals. The Eudoxus reals are subfield of the multi-valued Eudoxus reals, and they are equal if we assume the axiom of countable choice.

We can embed a multi-valued Eudoxus real $e$ into the Dedekind reals via the Dedekind cut $(\{p \in \mathbb Q: p < e\}, \{q \in \mathbb Q: e < q\})$, as usual. My question: **is this embedding surjective?**