I'm essentially reopening [this old question](https://mathoverflow.net/questions/37849/closedness-of-finite-dimensional-subspaces) of Ricky Demer which was never fully answered.


Essentially the original question: Suppose we have a topological field $F$ which is complete, Hausdorff, and non-discrete, and we put a Hausdorff topology on $F^n$ so as to make it a topological vector space over $F$; is is this topology necessarily the product topology (and hence complete, and hence closed in anything it embeds in)?


As the link shows, the answer is yes if $\tau$ comes from an absolute value on $F$, and it's easy to see the same argument works if it comes from a field ordering on $F$.


Note that the argument there shows that this question reduces to the following lemma:


Suppose we have a topological field $(F,\tau)$ which is Hausdorff and non-discrete, and we give $F$ a second topology $\tau'$, which is also Hausdorff, such that $(F,\tau')$ is a topological vector space over $(F,\tau)$. Does this force $\tau=\tau'$? What if we assume that $(F,\tau)$ is complete?


(I'm isolating completeness as a separate, possibly-unnecessary condition because the argument there only uses completeness in the reduction to the lemma, not in proving the lemma for valued fields.)


Related to [this question](https://mathoverflow.net/questions/74418/does-the-weak-approximation-theorem-hold-for-general-topological-fields) 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.



Finite dimensional vector spaces over a complete but not-necessarily-valued field
Well, now I feel silly -- on looking through Wieslaw again, I see he does give examples of non-discrete, non-straight fields, just not in that section. For instance, take two absolute values on the rationals; the topology they generate together still make the rationals a topological field, is not discrete, and is obviously finer than either of the ones you started with. Since it isn't even minimal, it can't be straight.


I'm not going to accept my own answer on this since this example presumably isn't complete, and I'd like a complete example. Which might well be in here too, if I keep looking... (Is the completion of this again a field? If so that should work, but I'd need to check if that's true.)


**Edit:** Nope, the completion of this isn't a field, so I'm still lacking for a complete example.


**Edit again**: OK, I'm now pretty sure Wieslaw gives one, so the answer is **no**. Wieslaw gives an example of a complete normed field which is not given by any absolute value (here "normed" means instead of |ab|=|a||b|, we only require |ab|≤|a||b| and |-a|=|a|). Furthermore, he shows given a power-multiplicative norm, you can find an absolute value that generates a coarser topology (so if a power-multiplicative norm wasn't equivalent to an absolute value, it isn't minimal). (Here power-multiplicative only means we require |an|=|an| for positive n, not for negative n.) And after a bit of staring at his example, I'm pretty sure it's power-multiplicative. So unless anyone can show that I've missed something, I'm going to consider this one closed.


