Example

The space $\mathbf{R}^n$ with its usual topology and the order structure that is the product of the order structure of its factors is an ordered topological vector space. On the other hand, for $n \geq 2$, when $\mathbf{R}^n$ carries the lexicographical order (S, III, § 2.6), the usual topology is not compatible with the ordered vector space structure of $\mathbf{R}^n$.

Let A be a set; the vector space $\mathscr{B}(A ; \mathbf{R})$ of real valued bounded functions defined on A, with the topology defined by the norm $\|x\| = \sup_{t \in A} |x(t)|$ and the order structure induced by the product order structure of $\mathbf{R}^A$, is an ordered topological vector space.

In an ordered topological vector space E, the set of elements $x \leq 0$ is closed; since translations are homeomorphisms, we deduce that, for all $a \in E$, the set of elements $x \geq a$ (resp. $x \leq a$) is closed. Since $\{0\}$ is the intersection of the sets $x \geq 0$ and $x \leq 0$, it follows that $\{0\}$ is closed and that E is Hausdorff.