Why locally convex topological vector spaces Hausdorff?

Question

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!

Answer 1

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.

Answer 2

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 $p(a-b)>0$.