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It is known that

A $T_1$-topological space is metrizable $\iff$ it has a compatible uniformity with a countable base.

Let $(X,\tau')$ be a $T_1$-topological space and $\mathcal U'$ be a compatible uniformity on $X$ (i.e. $\tau_{\mathcal U'}=\tau')$ with countable base. Then by the above result $X$ is metrizable.

Now my question is

Is it always possible to find a metric $d$ on $X$ such that $\mathcal U'$ is compatible with $d$ i.e. $$\mathcal U(d)=\mathcal U'?$$

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Once you have $\mathcal{U}(d) =\mathcal{U}'$, the topology induced by that metric is the same as the topology induced by $\mathcal{U}'$, so $\tau'$.

This is because the intermediate step is superfluous in :

$d$ metric induces a uniformity, uniformity induces a topology.

is the same as metric induces that topology in one go, using the open balls.

This is clear when you look at the definitions of how we induce these topolgoies resp. uniformities:

Recall that a base for $\mathcal{U}(d)$ (as entourages) is all sets $\{(x,y) \in X^2: d(x,y) < \varepsilon\}$ and when we have a base for the entourage uniformity $\mathcal{U}$ its induced topology on $X$ has as a base all sets $B[x]$ with $B$ in that base and $x \in X$, and for the standard $d$-uniformity base these $B[x]$ are just the open balls around $x$ with radius $\varepsilon$.

Also, in the proof of the uniform metrisability theorem (Birkhoff-Kakutani?) we explicitly construct a compatible metric for the uniformity with a countable base, so in any scenario the answer is yes, you can.

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  • $\begingroup$ Yes I have edited it. But even can we guarantee the metrisability of the uniformity? $\endgroup$ – Jave Nov 21 '18 at 4:53
  • $\begingroup$ @Jave yes, if it has a countable uniform base. That's what the theorem says, right? $\endgroup$ – Henno Brandsma Nov 21 '18 at 5:54
  • $\begingroup$ Why the downvote? $\endgroup$ – Henno Brandsma Nov 21 '18 at 5:54
  • $\begingroup$ The theorem doesn’t say whether the obtained metric will be compatible with the given uniformity $\endgroup$ – Jave Nov 21 '18 at 5:57
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    $\begingroup$ @Jave yes, that’s what I wrote in my answer. $\endgroup$ – Henno Brandsma Nov 21 '18 at 6:16

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