separated uniform space

Let $X$ be a uniform space with uniformity $𝒰$. $X$ is said to be separated or Hausdorff if it satisfies the following separation axiom:

$$\bigcap 𝒰=\mathrm{\Delta },$$

where $\mathrm{\Delta }$ is the diagonal relation on $X$ and $\bigcap 𝒰$ is the intersection of all elements (entourages) in $𝒰$. Since $\mathrm{\Delta }\subseteq \bigcap 𝒰$, the separation axiom says that the only elements that belong to every entourage of $𝒰$ are precisely the diagonal elements $(x,x)$. Equivalently, if $x\ne y$, then there is an entourage $U$ such that $(x,y)\notin U$.

The reason for calling $X$ separated has to do with the following assertion:

$X$ is separated iff $X$ is a Hausdorff space under the topology $T_{𝒰}$ induced by (http://planetmath.org/TopologyInducedByAUniformStructure) $𝒰$.

Recall that $T_{𝒰}=\{A\subseteq X\mid \text{for each }x\in A\text{, there is }U\in 𝒰\text{, such that }U[x]\subseteq A\}$, where $U[x]$ is some uniform neighborhood of $x$ where, under $T_{𝒰}$, $U[x]$ is also a neighborhood of $x$. To say that $X$ is Hausdorff under $T_{𝒰}$ is the same as saying every pair of distinct points in $X$ have disjoint uniform neighborhoods.

Proof.

$(\Rightarrow )$. Suppose $X$ is separated and $x,y\in X$ are distinct. Then $(x,y)\notin U$ for some $U\in 𝒰$. Pick $V\in 𝒰$ with $V\circ V\subseteq U$. Set $W=V\cap V^{-1}$, then $W$ is symmetric and $W\subseteq V$. Furthermore, $W\circ W\subseteq V\circ V\subseteq U$. If $z\in W[x]\cap W[y]$, then $(x,z),(y,z)\in W$. Since $W$ is symmetric, $(z,y)\in W$, so $(x,y)=(x,z)\circ (z,y)\in W\circ W\subseteq U$, which is a contradiction.

$(\Leftarrow )$. Suppose $X$ is Hausdorff under $T_{𝒰}$ and $(x,y)\in U$ for every $U\in 𝒰$ for some $x,y\in X$. If $x\ne y$, then there are $V[x]\cap W[y]=\mathrm{\varnothing }$ for some $V,W\in 𝒰$. Since $(x,y)\in V$ by assumption, $y\in V[x]$. But $y\in W[y]$, contradicting the disjointness of $V[x]$ and $W[y]$. Therefore $x=y$.∎