Every ordered abelian group $G$ can be completed to give a larger ordered abelian group $\bar{G}$. The original abelian group $G$ embeds into $\bar{G}$ as a dense subset, and every non-empty subset of $G$ that is bounded from above has a supremum in $\bar{G}$. The completion can be constructed using Dedekind cuts, or by Cauchy sequences or Cauchy nets.
I'm interested in the ring $\mathbb{R}(x)$ of rational functions in $x$. We can give this ring an order by saying that for any $f \in \mathbb{R}(x)$, $f > 0$ if and only if $f(x) > 0$ for large enough $x \in \mathbb{R}$.
I've tried, but failed to visualize the completion of this ring. I see how we can make the completion of $\mathbb{R}(x)$ into a ring as well. However, it can not be a field because all complete orderd fields are isomorphic to $\mathbb{R}$.
We can even embed all power series into this monstrosity: if $a_0, a_1, a_2 \dots \in \mathbb{R}$ and $a_i \geq 0$, then let $f_n(x) = a_0 + \frac{a_1}{x} + \dots + \frac{a_n}{x^n}$. The supremum of all $f_i$ is an element of $\bar{\mathbb{R}(x)}$, which we can identify with $\sum_{i=0}^\infty \frac{a_i} {x^i}$.
How can I visualize this space? Is it too large to be visualized?