Topological field without the Stone-Weierstrass property

68 Views Asked by At

Say that a (Hausdorff) topological field $\mathbb{F}$ (more generally, a topological ring) has the Stone-Weierstrass property if for all Hausdorff spaces $X$ every unital separating subalgebra $\mathcal{A}\subseteq C(X, \mathbb{F})$ with the property below is dense in $C(X, \mathbb{F})$ with compact-open topology.

We have $\sigma\circ g\in \mathcal{A}$ for every $g\in\mathcal{A}$ and every field automorphism $\sigma:\mathbb{F}\to\mathbb{F}$ that is also a homeomorphism (automorphism in the category of topological fields).

Examples of fields with the SW property are $\mathbb{R}$ and $\mathbb{C}$, where in the latter case consideration of conjugation $z\mapsto \overline{z}$ is required for the Stone-Weierstrass theorem to be true (which justifies usage of field automorphisms in definition above).

More generally, any valuable field other than $\mathbb{C}$ satisfies the SW property with no field automorphisms necessary, see The Stone-Weierstrass theorem for valuable fields by Chernoff, Rasala and Waterhouse.

What's an example of a topological field without the SW property?