Criterion for a Topological Vector Space to be Hausdorff

Proposition 1: Let $E$ be a topological vector space and let $\mathcal U$ be a collection of neighbourhoods of the origin. Then $E$ is Hausdorff if and only if $\displaystyle{\bigcap_{U \in \mathcal U} U = \{ o \}}$.

• Proof: $\Rightarrow$ Suppose that $E$ is Hausdorff. Then, for each $x \in E$ with $x \neq 0$ there exists a $U_x \in \mathcal U$ such that $x \not \in U_x$. Also, since every $U \in \mathcal U$ is by definition a neighbourhood of the origin we have that $o \in U$ for all $U \in \mathcal U$. Thus:

(1)

\begin{align} \quad \bigcap_{U \in \mathcal U} U \subseteq \bigcap_{x \in E} U_x = \{ o \} \end{align}

• And thus $\displaystyle{\bigcap_{U \in \mathcal U} U = \{ o \}}$.

• $\Leftarrow$ Suppose that $\displaystyle{\bigcap_{U \in \mathcal U} U = \{ o \}}$. Then $x, y \in E$ be such that $x \neq y$. Then $x - y \neq o$. So, there exists a $U \in \mathcal U$ such that $x - y \not \in U$. Since $E$ is a topological vector space, there exists a balanced neighbourhood $V$ of $o$ such that $V + V \subseteq U$, and consequently, $x - y \not \in V + V$.

• Thus, for all $v_1, v_2 \in V$ we have that $x - y \neq v_1 + v_2$, or equivalently, $x - v_1 \neq y + v_2$ for all $v_1, v_2 \in V$. Hence $x - v \neq y + v$ for all $v \in V$. So $(x - V) \cap (y + V) = \emptyset$. But since $V$ is balanced, $-V = V$. Hence:

(2)

\begin{align} \quad (x + V) \cap (y + V) = \emptyset \end{align}

• So for each pair of points $x, y \in E$ with $x \neq y$ there exists neighbourhoods $x + V$ of $x$, and $y + V$ of $y$ such that $(x + V) \cap (y + V) = \emptyset$. Thus $E$ is Hausdorff. $\blacksquare$