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Let $J$ be an arbitrary (non-empty) index set, and let $\left\{ X_\alpha \right\}_{\alpha \in J}$ be an indexed family (or collection) of topological spaces, and let $\Pi_{\alpha \in J} X_\alpha$ denote their Cartesian product.

Then here's Theorem 19.4 in the book Topology by James R. Munkres, 2nd edition:

If each space $X_\alpha$ is a Hausdorff space, then $\Pi_{\alpha \in J} X_\alpha$ is a Hausdorff space in both the box and product topologies.

So far so good!

Now my question is, does the converse also hold? That is, if $\Pi_{\alpha \in J} X_\alpha$ is a Hausdorff space in either the box or the product topologies, then does it follow that each space $X_\alpha$ is Hausdorff also?

Of course, if $\Pi_{\alpha \in J} X_\alpha$ is Hausdorff in the product topology, then it is Hausdorff in the box topology also.

Saaqib Mahmood
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1 Answers1

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Yes, the converse is true. If $X=\prod X_\alpha$ is Hausdorff in the box topology, then every factor $X_\alpha$ is Hausdorff, provided that no factor is empty. To prove this, take two distinct points $x_\beta,y_\beta$ in $X_\beta$, and choose $x_\alpha=y_\alpha\in X_\alpha$ for any $\alpha\ne\beta$. Now consider the points $x=(x_\alpha)_\alpha$ and $y=(y_\alpha)_\alpha$ in $X$. By Hausdorffness there exist disjoint open boxes around $x$ and $y$. Can you show that their projections to $X_\beta$ are disjoint as well?

One may also show that every factor is homeomorphic to a subspace of $X$, implying at once that if the box product is $T_0$ ($T_1$, Hausdorff, (completely) regular), then the factor is also $T_0$ ($T_1$, Hausdorff, (completely) regular).

Stefan Hamcke
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