Nonmetrizable uniformities with metrizable topologies I'm looking for such pathological examples of uniform spaces which are not metrizable, but whose underlying topology is metrizable. Willard in his General Topology text constructs such a uniformity using ordinals. I am asking for examples which do not rely on ordinals.
EDIT: Below is an example by Daniel Tausk using families of pseudometrics. I forgot to mention that, if possible, I would like an example that uses the definition through a diagonal uniformity, not the definition through pseudometrics nor the definition through uniform covers. See Cryptomorphisms for some elaboration on this phenomenon.
 A: The family of all neighbourhoods of the diagonal of $\mathbb{R}$ (with its normal topology) is a uniformity without a countable basis but it generates the normal topology.
The normal topology of $\mathbb{R}$ is the topology generated by the usual distance.
To see that the open sets in the plane that contain the diagonal generate a uniformity let $U$ be such an open set. For each $x$ there is $r_x$ such that $B(x,r_x)^2\subseteq U$. Now let $V$ be the union of the squares $B(x,\frac13r_x)^2$; this is again an open set and it is  not hard to show that $V\circ V\subseteq U$. 
If $U$ is an open set in the plane that contains the diagonal then the vertical section $U[x]$ is open in $\mathbb{R}$, so this uniformity generates at best a subtopology of the normal topology and hence certainly not the discrete topology. To see that it generates the normal topology consider, given $x$ and $r$, the open set $U_{x,r}=B(x,r)^2\cup(\mathbb{R}\setminus\lbrace x\rbrace)^2$; then $U_{x,r}[x]=B(x,r)$.
The uniformity does not have a countable base: if $\langle U_n:n\in\mathbb{N}\rangle$ is a sequence of neighbourhoods of the diagonal then you can use diagonalisation to produce a neighbourhood such that $U_n\not\subseteq U$ for all $n$: make sure that the square $[n,n+1]^2$ contains a point in $U_n$ but not in $U$.
A: Consider the interval $[0,1]$ and for each $x\in[0,1]$ let $d_x$ be the metric in $[0,1]$ defined by $d_x(y,z)=|y-z|$ if $y\ne x$ and $z\ne x$, $d_x(x,y)=d_x(y,x)=1$ if $y\ne x$ and $d_x(x,x)=0$.
Let $\mathcal U$ be the uniform structure defined by the family of metrics $d_x$, $x\in[0,1]$. For each $x\in[0,1]$, the open ball $B_{d_x}(x,1)$ equals $\{x\}$ and thus the topology defined by $\mathcal U$ is discrete (hence metrizable). Assume by contradiction that there exists a metric $d$ in $[0,1]$ defining the uniform structure $\mathcal U$. In particular, $d$ must define the same topology as $\mathcal U$, i.e., $d$ must be discrete. Thus, if $A_n$ denotes the set of those $x\in[0,1]$ such that the open ball $B_d(x,1/n)$ equals $\{x\}$ then $\bigcup_{n=1}^\infty A_n=[0,1]$ and therefore there exists $n\ge1$ such that $A_n$ is infinite. Since the identity map:
$\mathrm{Id}:([0,1],\mathcal U)\to([0,1],d)$
is uniformly continuous, there exists a finite set $F\subset[0,1]$ and $\delta>0$ such that, for all $y,z\in[0,1]$:
$d_x(y,z)<\delta$ for all $x\in F$ implies $d(y,z)<1/n$.
Choose $y\in A_n\setminus F$ and let $z\in[0,1]$ be such that $|z-y|<\delta$, $z\not\in F$ and $z\ne y$. Then $d_x(y,z)=|y-z|<\delta$ for all $x\in F$ and, since $y\in A_n$ and $z\ne y$, we have also $d(y,z)\ge1/n$, a contradiction.
A: Let $X$ be an infinite set.  Take the uniformity $\mathcal U$ generated by those equivalence relations on $X$ that have only finitely many equivalence classes.  The induced topology on $X$ is discrete and thus metrizable.  But the uniformity $\mathcal U$ is not metrizable, because, given any countably many entourages (without loss of generality, equivalence relations with finitely many classes each), it is easy to construct another equivalence relation, with only two classes, that does not include any of the countably many given ones.  (Alternative proof: The completion of this uniform space is the Stone-Cech compactification of the discrete space $X$, which is not metrizable as a topological space.)
