Regularity is productclosed
This article gives the statement, and possibly proof, of a topological space property satisfying a topological space metapropertywarning.png"{{{property}}}" cannot be used as a page name in this wiki.warning.png"{{{metaproperty}}}" cannot be used as a page name in this wiki.
View all topological space metaproperty satisfactions  View all topological space metaproperty dissatisfactions

Contents
Statement
Propertytheoretic statement
The property of topological spaces of being a regular space is a productclosed 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
Regular space
Further information: Regular space
A topological space is regular if it is T1 and further, if given a point and an open set containing , there is an open set containing such that .
Product topology
Further information: Product topology
Suppose is an indexing set, and a family of topological spaces, . Then if is the Cartesian product of the s, the product topology on is a topology with subbasis given by all the open cylinders: all sets of the form such that for all but one , , and for the one exceptional , is an open subset of .
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.
Further information: Regularity is box productclosed
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
Proof details
Given: Indexing set , a family of regular spaces. is the product of the s, given the product topology
To prove: is a regular space
Proof: It suffices to show that given any point , and any open subset of containing , there is an open subset such that .
First, there exists a basis element of containing , and inside . Suppose this basis element is taken as . Now, for those that are equal to the whole space, define . For those that are proper open subsets of , use the fact that is regular to find an open set in , such that .
Now consider to be the product of all the s. Clearly by construction, and is a basis element for the topology on . In particular, is open. Further, the closure of is contained in , and hence in . Thus, is as we sought, and the proof is complete.
References
Textbook references
 Topology (2nd edition) by James R. Munkres, ^{More info}, Page 196197, Theorem 31.2(b), Chapter 4, Section 31