How are Uniform Spaces Hausdorff?

Question

Definition: Let $X$ be a set. A set $U\subseteq X\times X$ is called entourage of the diagonal if $\Delta=\{(x,x):x\in X\}\subseteq U$ and $U=U^{-1}.$

Let $\Phi$ be a family of entourages of the diagonal. The pair $(X,\Phi)$ is called uniform space if for all entourages $U,V$ the following conditions are satisfied:

$1.U\in \Phi,U\subseteq V \implies V\in \Phi.$

$2.U,V\in \Phi\implies U\cap V\in \Phi.$

$3.U\in \Phi \implies (\exists V\in \Phi)V\circ V=\{(x,z):(\exists y\in X),(x,y),(y,z)\in V\}\subseteq U.$

$4.\bigcap\Phi=\Delta.$

The sets of the form $$U[x]=\{y\in X:(x,y)\in U\}$$ gives a local base at $x$ for the topology induced by the uniformity $\Phi.$

Now in this paper $I^K$-cauchy Functions by Sleziak,Toma,Das it says The last condition of the Definition implies that the induced topology is Hausdorff.

Now by last condition it means the $4$th one $\bigcap \Phi = \Delta.$

Let $x,y\in X;x\neq y.$ We need to find at least two $U\cap V=\phi,U,V\in \Phi$ such that $x\in U,y\in V.$Now according to the definitions wlg $$U=U[x]\\ V=V[y]$$. Now if they have a point say $(a,b)$ in common then to be in $U[x]$, $a=x$ and to be in $V[y]$ $a=y$ and thus making $x=y$ which contradicts our initial assumption that $x\neq y.$ So they are Hausdorff. But what has condition $4$ to do with it $?$

Also , from the proof , did we just find out that whenever two points are distinct in an Uniform Space all their nbd's are disjoint?

Thank You.

Answer 1

Let $x,y\in X$ and assume that every neighborhood of $x$ intersects every neighborhood of $y$. In particular we'll have: $$\forall U\in\Phi,\ \exists z_U\in X,\ z_U\in U[x]\cap U[y].$$ Which can be translated as $$\forall U\in\Phi,\ \exists z_U\in X,\ (x,z_U),(y,z_U)\in U.$$ Now, since $U$ is an entourage of $\Delta$, $U=U^{-1}$, hence we conclude: $$\forall U\in\Phi,\ \exists z_U\in X,\ (x,z_U),(z_U,y)\in U.$$

We now proceed by contradiction: assume that $x\neq y$, then, by Condition 4, there exists $U\in\Phi$ such that $(x,y)\not\in U$. By Condition 3, there exists $V\in\Phi$ such that $V\circ V\subset U$. But we have a contradiction with the fact that $(x,z_V),(z_V,y)\in V$ hence $(x,y)\in V\circ V\subset U$.

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You went wrong there:

Now if they have a point say $(a,b)$ in common

This is nonsense: $U[x]$ is a subset of $X$, not of $X\times X$.

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did we just find out that whenever two points are distinct in an Uniform Space all their nbd's are disjoint?

Certainly not!

Answer 2

If $x \neq y$, then $(x,y) \notin \Delta$, so if (4) holds, there is some $U \in \Phi$ such that $(x,y) \notin U$. Now let $V \in \Phi$ be such that $V \circ V \subseteq U$.

Now suppose $z \in V[x] \cap V[y]$. Then $(x,z) \in V, (z,y) \in V$ and symmetry yields that $(x,y) \in V \circ V$ so $(x,y) \in U$, which is a contradiction with how $U$ was chosen. So $V[x]$ and $V[y]$ are disjoint neighbourhoods for $x$ and $y$.

Note the abstract similarity with the proof for metrics: $d(x,y) = r > 0$ and we use a two balls of radius $\frac{r}{2}$ (Which corresponds to $V[x], V[y]$) and the triangle inequality then does the rest. Uniformities are inspired on metrics, so this often helps.