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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Further information: Regular space
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.
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
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
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.
- Topology (2nd edition) by James R. Munkres, More info, Page 196-197, Theorem 31.2(b), Chapter 4, Section 31