3 added 12 characters in body

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} $\{x\}$ 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} $\{x\}$ 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.

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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 ${x}$ {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 ${x}$ {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.

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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.