# Why locally convex topological vector spaces Hausdorff?

In A Course in Functional Analysis, John B. Conway, 100p, it is written that

Definition. A locally convex topological vector space (LCTVS) is a TVS whose topology is defined by a family of seminorms $$\mathcal{P}$$ such that $$\cap_{p\in\mathcal{P}}\{x:p(x)=0\}=\{0\}$$.

This attitude that has been adopted in this book is that all topological spaces are Hausdorff. The condition in Definition above that $$\cap_{p\in\mathcal{P}}\{x:p(x)=0\}=\{0\}$$ is imposed precisely so that the topology defined by $$\mathcal{P}$$ be Hausdorff. In fact, suppose that $$x\neq y$$. Then there is a $$p\in \mathcal{P}$$ such that $$p(x-y)\neq 0$$; let $$p(x-y)>\epsilon>0$$. If $$U=\{z:p(x-z)<\epsilon/2\}$$ and $$V=\{z:p(y-z)<\epsilon/2\}$$, then $$U\cap V=\emptyset$$ and $$U$$ and $$V$$ are neighborhoods of $$x$$ and $$y$$, respectively.

Here I cannot agree with "$$x\neq y\Rightarrow \exists p\in\mathcal{P},\,p(x-y)>0$$". Isn't $$\mathcal{P}$$ the set of seminorms that is a priori given? Thank you in advance!

• Seminorms have to be non-negative, so if $x\neq y$ then there has to be some seminorm such that $p(x-y) > 0$ else $x-y \in \cap_{p\in\mathcal{P}}\{x:p(x)=0\}$. Mar 5, 2021 at 17:39
• $\cap_{p\in\mathcal{P}}\{x:p(x)=0\}=\{0\}$. Mar 5, 2021 at 18:23

The condition on $$\mathcal{P}$$ says that the only vector $$x$$ all of whose seminorms (coming from $$\mathcal{P}$$) are zero is the zero vector itself. So if $$x\neq y$$, and hence $$x - y \neq 0$$, there must be some $$p \in \mathcal{P}$$ witnessing that.
Given two distinct vectors $$a,b$$, consider the difference vector $$a-b$$. This is nonzero, so the condition $$\bigcap_{p\in \mathcal{P}}\{x: p(x)=0\}=\{0\}$$ implies that for some $$p\in\mathcal{P}$$ we have $$p(a-b)\not=0$$.
Since seminorms take on only nonnegative values, this means $$p(a-b)>0$$.