Let $X$ be a nonempty set and $\mathcal U$ be a uniformity on $X.$ I know that $\mathcal U$ is said to be metrizable if $\exists$ a metrix $d$ on $X$ such that the uniformity $\mathcal U(d)$ induced by the metric $d$ is $\mathcal U$ itself.
Now I know that every metric $d$ on $X$ induces a $\tau(d)$ topology and also a unifomity $\mathcal U$ on $X$ induces a topology $\tau(\mathcal U)$. Now my question is: If $\mathcal U$ be a uniformity on $X$ induced by the metric $d$ can we conclude $\tau(d)=\tau(\mathcal U)?$
or more generally can we say,
a uniformity $\mathcal U$ on $X$ is metrizable $\iff\exists$ a metric $d$ on $X$ such that $\tau(d)=\tau(\mathcal U)?$
Be careful: a metric $d$ is the most "detailed" structure, it induces a uniformity $\mathcal{U}(d)$ in the standard way, and a uniformity $\mathcal{U}$ in turn induces a topology $\tau(\mathcal{U})$ in another standard way, but such that $\tau(d)$ (the metric topology, directly induced by the open balls) is exactly $\tau(\mathcal{U}(d))$.
But both steps are irreversible in general: if we have any uniformisable topology $\tau$ on $X$, there can be two different (i.e. not isomorphic as uniform spaces) uniformities $\mathcal{U}_1$, $\mathcal{U}_2$ on $X$ such that $\tau(\mathcal{U}_1) =\tau(\mathcal{U}_2)= \tau$. There can also be different metrics on a set that give the same uniformities (like the $d_p$, $p>1$ metrics on $\mathbb{R}^n$),so $\mathcal{U}(d_1) = \mathcal{U}_2$, while $d_1 \neq d_2$. Such metrics are called "uniformly equivalent" (and metrics $d_1, d_2$ that give $\tau(d_1) =\tau(d_2)$ are called "equivalent topologies".
But if we know a uniformity $\mathcal{U}$ is metrisable in your sense, $\exists d$ a metric on $X$, such that $\mathcal{U} = \mathcal{U}(d)$ and so $\tau(\mathcal{U}) = \tau(\mathcal{U}(d)) = \tau(d)$ as well.
If we have a metric space $(X,d)$ like $(\mathbb{R},|.|)$ and we consider $(X,\mathcal{U}_f)$, the fine uniformity w.r.t. the standard topology on $\mathbb{R}$. Then $\tau(\mathcal{U}_f)$ is just $\tau(d)$ but $\mathcal{U}_f$ is not a metrisable uniformity (see the overview paper here (end of paragraph 2.13). So the condition is not an equivalence.