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](https://mathoverflow.net/questions/71110/finite-dimensional-vector-spaces-over-a-complete-but-not-necessarily-valued-field) 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.



Does the weak approximation theorem hold for general topological fields?
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...


