Regularity is product-closed

Property-theoretic statement
The property of topological spaces of being a regular space is a product-closed property of topological spaces.

Verbal statement
An arbitrary (finite or infinite) product of regular spaces, when endowed with the product topology, is also a regular space.

Regular space
A topological space $$X$$ is regular if it is T1 and further, if given a point $$x \in X$$ and an open set $$U$$ containing $$x$$, there is an open set $$V$$ containing $$x$$ such that $$\overline{V} \subset U$$.

Product topology
Suppose $$I$$ is an indexing set, and $$X_i$$ a family of topological spaces, $$i \in I$$. Then if $$X$$ is the Cartesian product of the $$X_i$$s, the product topology on $$X$$ is a topology with subbasis given by all the open cylinders: all sets of the form $$\prod_i A_i$$ such that for all but one $$i$$, $$A_i = X_i$$, and for the one exceptional $$i$$, $$A_i$$ is an open subset of $$X_i$$.

A basis for this topology is given by finite intersections of open cylinders: these are products where finitely many coordinates are proper open subsets, and the remaining are whole spaces.

Related facts
An analogous proof to this one shows that the property of being a regular space is also closed under taking arbitrary box products. Note that this statement is independent of the statement about arbitrary products; neither can be deduced directly from the other, because the property of being regular is not closed either under passing to a coarser topology or under passing to a finer topology.

Proof outline
The proof proceeds as follows:


 * Start with the point in the product space, and the open set containing it
 * Find a basis open set containing the point, which lies inside this open set
 * For each coordinate on which the projection of the basis open set is a proper subset, find a smaller open subset whose closure is contained inside the given projection
 * Reconstruct from these a smaller basis open set whose closure lies in the given basis open set

Proof details
Given: Indexing set $$I$$, a family $$\{ X_i \}_{i \in I}$$ of regular spaces. $$X$$ is the product of the $$X_i$$s, given the product topology

To prove: $$X$$ is a regular space

Proof: It suffices to show that given any point $$x = (x_i) \in X$$, and any open subset $$U$$ of $$X$$ containing $$x$$, there is an open subset $$V \ni x$$ such that $$\overline{V} \subset U$$.

First, there exists a basis element of $$X$$ containing $$x$$, and inside $$U$$. Suppose this basis element is taken as $$W = \times_{i \in I} W_i$$. Now, for those $$W_i$$ that are equal to the whole space, define $$V_i = W_i$$. For those $$W_i$$ that are proper open subsets of $$X_i$$, use the fact that $$X_i$$ is regular to find an open set $$V_i \ni x_i$$ in $$X_i$$, such that $$\overline{V_i} \subset W_i$$.

Now consider $$V$$ to be the product of all the $$V_i$$s. Clearly $$x \in V$$ by construction, and $$V$$ is a basis element for the topology on $$X$$. In particular, $$V$$ is open. Further, the closure of $$V$$ is contained in $$W$$, and hence in $$U$$. Thus, $$V$$ is as we sought, and the proof is complete.

Textbook references

 * , Page 196-197, Theorem 31.2(b), Chapter 4, Section 31