# 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

### 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 product-closed

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