Regularity is product-closed

This article gives the statement, and possibly proof, of a topological space property satisfying a topological space metaproperty
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Statement

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.

Definitions used

Hausdorff space

Further information: Hausdorff 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 $\bar{V} \subset U$ .

Product topology

Further information: 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.

Proof

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