Background: The transfer principle in nonstandard analysis implies that any nonstandard model of the reals is a commutative (for additively and multiplicatively). It is also well-known that the set $\beta(\mathbb R)$ of all ultrafilters on $\mathbb R$ is not a nonstandard model of $\mathbb R$, for instance since addition and multiplication do not extend commutatively.
Among the standard results in nonstandard analysis, is that a set is compact iff every point in an enlargement of the set is near standard.
Now, in the context of $\beta(\mathbb R)$, call an ultrafilter on a subset $S\subseteq \mathbb R$ near standard if it contains the neighborhood filter $N_x$ for some $x\in \mathbb R$. The classical result that a set is compact iff every ultrafilter on it converges can now be restated as "$S$ is compact iff every element in $\beta(S)$ is near standard".
Is this analogy just conincidental, or can $\beta(\mathbb R)$ be considered as a non-commutative nonstandrard model of $\mathbb R$ in a useful way?
