Locally Convex Space is Hausdorff iff induces Hausdorff Topology

Theorem

Let $\struct {X, \mathcal P}$ be a locally convex space.

Let $\tau$ be the standard topology on $\struct {X, \mathcal P}$.

Then $\struct {X, \mathcal P}$ is Hausdorff if and only if $\struct {X, \tau}$ is a Hausdorff topological space.

Proof

Sufficient Condition

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

Let $x \in X$ have $x \ne {\mathbf 0}_X$.

If there exists no such $x$, then we have the claim by vacuous truth.

Otherwise, since $\struct {X, \tau}$ is Hausdorff, there exists $U, V \in \tau$ with $x \in U$ and ${\mathbf 0}_X \in V$ and $U \cap V = \O$.

Then $x \not \in V$ and ${\mathbf 0}_X \not \in V$.

From Open Sets in Standard Topology of Locally Convex Space, there exists $\epsilon > 0$ and seminorms $p_1, p_2, \ldots, p_n$ such that:

$\set {z \in X : \map {p_k} z < \epsilon, \text { for all } k \in \set {1, 2, \ldots, n} } \subseteq V$

Then we have, since $x \not \in V$:

$x \not \in \set {z \in X : \map {p_k} z < \epsilon, \text { for all } k \in \set {1, 2, \ldots, n} }$

Then there exists $k \in \set {1, 2, \ldots, n}$ such that:

$\map {p_k} x \ge \epsilon > 0$

In particular:

$\map {p_k} x \ne 0$

with $p_k \in \mathcal P$.

So $\struct {X, \mathcal P}$ is Hausdorff.

$\Box$

Necessary Condition

Suppose that $\struct {X, \mathcal P}$ is Hausdorff.

Let $x, y \in X$ have $x \ne y$.

We aim to find $U, V \in \tau$ such that $x \in U$ and $y \in V$ with $U \cap V = \O$.

Since $x \ne y$, we have $y - x \ne {\mathbf 0}_X$.

So from the definition of a Hausdorff locally convex space, there exists $p \in \mathcal P$ such that:

$\map p {y - x} \ne 0$

That is:

$\map p {y - x} > 0$

since seminorms map onto the non-negative real numbers.

Let:

$\epsilon = \map p {y - x}$

Then consider:

$U = \set {z \in X : \map p {z - x} < \epsilon/2}$

and:

$V = \set {z \in X : \map p {z - y} < \epsilon/2}$

From Open Sets in Standard Topology of Locally Convex Space, we have that $U, V \in \tau$.

From Seminorm Maps Zero Vector to Zero, we have:

$\map p { {\mathbf 0}_X} = 0 < \epsilon/2$

So we have $x \in U$ and $y \in V$.

We now show that $U \cap V = \O$.

Suppose that $U \cap V \ne \O$.

Let $z \in U \cap V$.

Then:

$\map p {z - x} < \epsilon/2$

and:

$\map p {z - y} < \epsilon/2$

Then, we have:

\(\ds \map p {y - x}\)

\(=\)

\(\ds \map p {\paren {y - z} + \paren {z - x} }\)

\(\ds \)

\(\le\)

\(\ds \map p {y - z} + \map p {z - x}\)

property $(\text N 3)$ of a seminorm

\(\ds \)

\(=\)

\(\ds \cmod {-1} \map p {z - y} + \map p {z - x}\)

property $(\text N 2)$ of a seminorm

\(\ds \)

\(=\)

\(\ds \map p {z - y} + \map p {z - x}\)

\(\ds \)

\(<\)

\(\ds \epsilon\)

\(\ds \)

\(=\)

\(\ds \map p {y - x}\)

a contradiction.

So there exists no such $z \in U \cap V$, so $U \cap V = \O$.

Since $x, y \in X$ were arbitrary, we have that $\struct {X, \tau}$ is Hausdorff.

$\blacksquare$