Counterexample for an Alternate Definition for Uniform Spaces.

Let $\mathcal{D}$ be a filter on $X^2$ such that:

1. $(\forall D\in\mathcal{D})(\Delta(X)\subseteq D)$.
2. $(\forall D\in\mathcal{D})(D\circ D\in \mathcal{D})$.
3. $(\forall D\in\mathcal{D})(D^{-1}\in \mathcal{D})$.

Show that $(X,\mathcal{D})$ is not necessarily a uniform space.

Just 2 has been altered compared with the original definition of uniform spaces.

• Note that condition 1. $\Delta(X) \subseteq D$ implies $D \subseteq D \circ D$, so condition 2. $D \circ D \in \mathcal{D}$ is automatic since $\mathcal{D}$ is a filter. – Martin Feb 4 '13 at 15:45
• you're right. But is there an example that satisfies both 1 and 3 but is not a uniformity? – user59671 Feb 4 '13 at 15:52

1 Answer

Condition 2. is void since for $D \in \mathcal{D}$ we have $\Delta(X) \subseteq D$ so that $D \subseteq D \circ D$ and hence $D \circ D \in \mathcal{D}$ since $\mathcal{D}$ is a filter.

Take a three-point space $X = \{x,y,z\}$ and consider the filter $\mathcal{D}$ on $X \times X$ generated by $D = (X \times X) \setminus \{(x,z), (z,x)\}$, so $\mathcal{D} = \left\{D, D \cup \{(x,z)\}, D \cup \{(z,x)\}, X \times X\right\}$ and 1. and 3. are clearly satisfied.

Observe that $D \circ D = X \times X$ so that there is no $E \in \mathcal{D}$ such that $E \circ E \subseteq D$, and $\mathcal{D}$ is not a uniformity.