4
$\begingroup$

The weak approximation theorem states that given a field $F$ and nontrivial inequivalent absolute values $|\cdot|_1,\ldots,|\cdot|_n,$ and letting $F_i$ denote $F$ with the topology from $|\cdot|_i$, then the diagonal in $F_1 \times \ldots \times F_n$ is dense.

So suppose now we have the same setup, except now the topologies don't necessarily come from absolute values (but do make $F$ a topological field, and are all still distinct and nondiscrete and of course Hausdorff). Does the result still hold?

Related to this older question in that one way to come up with a counterexample for both simultaneously would be to find a field with two (nondiscrete, Hausdorff) topologies with one strictly finer than the other.

$\endgroup$
2
$\begingroup$

Well, I feel silly -- this is answered early on in Wiesław's "Topological Fields" now that I look. The answer is no, distinct (non-discrete) topologies on a field need not be independent, they can be comparable, or even incomparable but still dependent.

For a simple example, take two values on the rationals; the topology generated by both together will still make the rationals a topological field, and is not discrete, and is strictly finer than either of the ones you started with.

I think you can recover it if you add some conditions weaker than coming from an absolute value, but, well, now you know where to look for this sort of thing...

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.