Dedekind completion of ordered fields

The existence of additive inverses fails in $\mathbb{S}^*.$

A member of $\mathbb{S}^*$ (a Dedekind cut in $\mathbb{S})$ is a subset A of $\mathbb{S}$ such that $A$ has an upper bound, $A$ has no least upper bound, and $(\forall x\in A )(\forall y \lt x )(y\in A).$ If $A$ and $B$ are Dedekind cuts, then $A+B$ is defined to be $\{x \mid (\exists a\in A)(\exists b\in B)(x\lt a+b)\},$ which you can see is a Dedekind cut itself.

You can check that $\{x \mid x \lt 0\}$ is a Dedekind cut, and that it is the additive identity in $\mathbb{S}^*.$

If $\mathbb{R}$ is a proper subset of $\mathbb{S},$ then there is some member of $\mathbb{S}$ greater than all members of $\mathbb{R}$ (proving this uses the ordered field axioms in $\mathbb{S}).$ It follows that $A=\{x\in\mathbb{S}\mid (\exists r\in\mathbb{R})(x\lt r)\}$ is a Dedekind cut.

The element $A$ has no additive inverse in $\mathbb{S}^*,$ which you can see as follows:

Assume $A$ had an additive inverse $B.$ Then there exist $a_0\in A$ and $b_0\in B$ such that $-1 \lt a_0+b_0.$ By the definition of $A,$ there is some $r_0\in \mathbb{R}$ such that $a_0\lt r_0,$ and it follows that $-1\lt r_0+b_0.$ But then we have $r_0+1\in A,$ $b_0$ in $B,$ and the sum of the two is non-negative, which is a contradiction.

You may be interested in knowing that there is a kind of completion of $\mathbb{S}$ akin to Dedekind completion:

Instead of considering every cut $(A,B)$, take only good cuts: cuts $(A,B)$ where for any $0<\varepsilon \in \mathbb{S}$, there is $(a,b) \in A \times B$ such that $b-a < \varepsilon$.

Then you really get a dense ordered field extension $\widetilde{\mathbb{S}}$ which is complete in the sense that it has no proper dense ordered field extension. This is the same as completing $\mathbb{S}$ using Cauchy $\lambda$-sequences where $\lambda$ is the least cardinal of a subset of $\mathbb{S}$ without bounds. This is also the same as completing $\mathbb{S}$ seen as an ordered field with its canonical uniform structure.