Topological Vector Space is Hausdorff iff T1

Theorem
Let $K$ be a topological field.

Let $\struct {X, \tau}$ be a topological vector space over $K$.


 * $(1) \quad$ $\struct {X, \tau}$ is Hausdorff
 * $(2) \quad$ $\struct {X, \tau}$ is a $T_1$ space
 * $(3) \quad$ for each $x \in X \setminus \set { {\mathbf 0}_X}$, there exists an open neighborhood $U_x$ of ${\mathbf 0}_X$ such that $x \not \in U_x$.

$(1)$ implies $(2)$
This is precisely $T_2$ Space is $T_1$ Space.

$(2)$ implies $(3)$
Suppose that $\struct {X, \tau}$ is a $T_1$ space.

Then:
 * for each $x, y \in X$ with $x \ne y$ there exists an open neighborhood $U_{x, y}$ such that $x \in U_{x, y}$ and $y \not \in U_{x, y}$.

Letting $x \in X \setminus \set { {\mathbf 0}_X}$ and $y = {\mathbf 0}_X$ we get the result.

$(3)$ implies $(1)$
Suppose that:
 * for each $x \in X \setminus \set { {\mathbf 0}_X}$, there exists an open neighborhood $U_x$ of ${\mathbf 0}_X$ such that $x \not \in U_x$.

Let $x, y \in X$ be such that $x \ne y$.

Then $x - y \ne {\mathbf 0}_X$.

Consider $U_{x - y}$.

From Open Neighborhood of Point in Topological Vector Space contains Sum of Open Neighborhoods: Corollary, there exists a symmetric set $V_{x, y}$ such that:
 * $V_{x, y} + V_{x, y} \subseteq U_{x - y}$

Since $V_{x, y} = -V_{x, y}$, we have:
 * $V_{x, y} - V_{x, y} \subseteq U_{x - y}$

For brevity, let $V = V_{x, y}$.

We show that $\paren {x + V} \cap \paren {y + V} = \O$.

that $\paren {x + V} \cap \paren {y + V} \ne \O$.

Let $z \in \paren {x + V} \cap \paren {y + V}$.

Then there exists $u, v \in V$ such that $x + u = z$ and $y + v = z$.

Then we have $x - y = \paren {z - u} - \paren {z - v} = v - u$.

We have $v - u \in V - V$, and so $v - u \in U_{x - y}$.

But $x - y \not \in U_{x - y}$, so this is a contradiction.

Hence $\paren {x + V} \cap \paren {y + V} = \O$.

From Translation of Open Set in Topological Vector Space is Open, $x + V$ is an open neighborhood of $x$ and $y + V$ is an open neighborhood of $y$.

Hence $\struct {X, \tau}$ is Hausdorff.