Hello, This is my problem: let $\displaystyle d:X\times X\rightarrow R$ be a distance on X. We know d is continous with the prodcut topology when the topology given to X is the metric topology. Show tha if d is continous with the product topology when the topology given to X is $\displaystyle \mathcal{F}$ then this topology is finner than the metric toplogy.

Attempts: we know that there is a coarsest topology for which d is continous is $\displaystyle \{d^{-1}(A):A$ open set in R$\displaystyle \}$ I tried to prove that this topology is the product topology of the metric but this doesn't seem to be the case.

I also tried to prove directly that if U is open in the metric topology then U is in $\displaystyle \mathcal{F}$

Any comments will be appreciated.