Direct Product of Topological Vector Spaces is Hausdorff iff Hausdorff Factor Spaces

Theorem

Let $K$ be a topological field.

Let $I$ be a set.

Let $\family {X_i}_{i \in I}$ be an $I$-indexed family of topological vector spaces over $K$.

Let:

$\ds X = \prod_{i \mathop \in I} X_i$

be the direct product of $\family {X_i}_{i \in I}$.

Equip $X$ with the product topology.

Then $X$ is Hausdorff if and only if $X_i$ is Hausdorff for each $i \in I$.

Proof

For each $i \in I$, let $\FF_i$ be the set of open neighborhoods of ${\mathbf 0}_{X_i}$, where ${\mathbf 0}_{X_i}$ is the zero vector of $X_i$.

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

Lemma

Then:

$\ds \prod_{i \mathop \in I} \bigcap \FF_i = \bigcap \FF$

$\Box$

Sufficient Condition

Suppose that $X_i$ is Hausdorff for each $i \in I$.

From Characterization of Hausdorff Topological Vector Space, we have:

$\bigcap \FF_i = \set { {\mathbf 0}_{X_i} }$

Then we have:

$\ds \bigcap \FF = \prod_{i \mathop \in I} \set { {\mathbf 0}_{X_i} } = {\mathbf 0}_X$

Then from Characterization of Hausdorff Topological Vector Space, $X$ is Hausdorff.

$\Box$

Necessary Condition

Suppose that $X$ is Hausdorff.

Then:

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

Hence, we have:

$\ds \prod_{i \mathop \in I} \bigcap \FF_i = \bigcap \FF = \set { {\mathbf 0}_X} = \prod_{i \mathop \in I} \set { {\mathbf 0}_{X_i} }$

Hence we obtain:

$\ds \bigcap \FF_i = \set { {\mathbf 0}_{X_i} }$

for each $i \in I$.

Hence from Characterization of Hausdorff Topological Vector Space, $X_i$ is Hausdorff.

$\blacksquare$