Characterization of Hausdorff Topological Vector Space

Theorem

Let $K$ be a topological field.

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

The following statements are equivalent:

$(1) \quad$ $\struct {X, \tau}$ is a Hausdorff topological vector space

$(2) \quad$ the intersection of all open neighborhoods of ${\mathbf 0}_X$ in $\struct {X, \tau}$ is $\set { {\mathbf 0}_X}$

$(3) \quad$ $\set { {\mathbf 0}_X}$ is closed in $\struct {X, \tau}$.

Proof

$(1)$ implies $(2)$

Suppose that $\struct {X, \tau}$ is a Hausdorff topological vector space.

Let $\FF$ be the set of open neighborhoods of ${\mathbf 0}_X$.

From Topological Vector Space is Hausdorff iff T1, $\struct {X, \tau}$ is $T_1$.

Clearly:

$\set { {\mathbf 0}_X} \subseteq \bigcap \FF$

Now let $x \in X \setminus \set { {\mathbf 0}_X}$.

Then there exists an open neighborhood $U \in \FF$ of ${\mathbf 0}_X$ such that $x \not \in U$.

In particular, $x \not \in \bigcap \FF$.

So we obtain $\bigcap \FF = \set { {\mathbf 0}_X}$.

$\Box$

$(2)$ implies $(3)$

Let $\FF$ be the set of open neighborhoods of ${\mathbf 0}_X$.

Suppose that $\bigcap \FF = \set { {\mathbf 0}_X}$.

Let $x \in \map \cl {\set { {\mathbf 0}_X} }$.

From Condition for Point being in Closure, for each open neighborhood $U_x$ of $x$ such that ${\mathbf 0}_X \in U_x$.

From Classification of Open Neighborhoods in Topological Vector Space, we equivalently have:

for each open neighborhood $U$ of ${\mathbf 0}_X$ we have ${\mathbf 0}_X \in U + x$.

That is:

$x \in -U$ for each open neighborhood $U$ of ${\mathbf 0}_X$.

From Dilation of Open Set in Topological Vector Space is Open, $U$ is an open neighborhood of ${\mathbf 0}_X$ if and only if $-U$ is.

Hence, we have $x \in U$ for each open neighborhood $U$ of ${\mathbf 0}_X$.

So $x \in \bigcap \FF = \set { {\mathbf 0}_X}$.

Hence we have $x = {\mathbf 0}_X$, so $\map \cl {\set { {\mathbf 0}_X} } \subseteq \set { {\mathbf 0}_X}$.

Since $\set { {\mathbf 0}_X} \subseteq \map \cl {\set { {\mathbf 0}_X} }$, we have:

$\map \cl {\set { {\mathbf 0}_X} } = \set { {\mathbf 0}_X}$

From Set is Closed iff Equals Topological Closure:

$\set { {\mathbf 0}_X}$ is closed in $\struct {X, \tau}$.

$\Box$

$(3)$ implies $(1)$

Suppose that $\set { {\mathbf 0}_X}$ is closed in $\struct {X, \tau}$.

Let $x \in X$.

We have $\set x = x + \set { {\mathbf 0}_X}$.

From Translation of Closed Set in Topological Vector Space is Closed Set, $\set x$ is closed.

So for all $x \in X$, $\set x$ is closed.

Hence from Definition 2 of a $T_1$ space, $\struct {X, \tau}$ is a $T_1$ space.

From Topological Vector Space is Hausdorff iff T1, $\struct {X, \tau}$ is Hausdorff.

$\blacksquare$