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 $X_i$ 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 ${\mathbf 0}_X$.

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.

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.