# Thread: Trouble with a product topology problem

1. ## Trouble with a product topology problem

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 $\displaystyle \mathcal{F}$

2. The open balls $\displaystyle B(x_{0},\epsilon) := \{ y \in X : d(x_{0},y)<\epsilon \}$ are a basis for the topology on X, which is induced by the metric d.
It is therefor sufficient to show that every open ball is also in $\displaystyle \mathcal{F}$.

For every $\displaystyle x_{0} \in X$ consider the map

$\displaystyle h: X \rightarrow \mathbb{R}$
$\displaystyle h(x):=d(x,x_0)$
, which is the composition of
$\displaystyle f:X \rightarrow X \times X$
,$\displaystyle f(x):=(x,x_{0})$
and
$\displaystyle d:X\times X \rightarrow \mathbb{R}$

both are continious in the topology $\displaystyle \mathcal{F}$, so the composition is also continous and we have
$\displaystyle B(x_{0},\epsilon)=h^{-1}((-\infty , \epsilon)) \in \mathcal{F}$

3. I think you solved my problem, thanks