products of compact pavings are compact
Suppose that is a paved space for each in index set![]()
. The product
(http://planetmath.org/GeneralizedCartesianProduct) is the set of all functions
![]()
such that for each , and the product of subsets is
Then, the product paving is defined by
Theorem 1.
Let be compact paved spaces for . Then, is a compact paving on .
Note that this result is a version of Tychonoffβs theorem applying to paved spaces and, together with the fact that all compact pavings are closed subsets of a compact space, is easily seen to be equivalent
![]()
to Tychonoffβs theorem.
Theorem 1 is simple to prove directly. Suppose that is a subset of satisfying the finite intersection property. Writing for gives
| (1) |
for any . By the finite intersection property, this is nonempty whenever is finite, so . Consequently, satisfies the finite intersection property and, by compactness of , the intersection
![]()
is nonempty. So equation (1) shows that is nonempty.
| Title | products of compact pavings are compact |
|---|---|
| Canonical name | ProductsOfCompactPavingsAreCompact |
| Date of creation | 2013-03-22 18:45:09 |
| Last modified on | 2013-03-22 18:45:09 |
| Owner | gel (22282) |
| Last modified by | gel (22282) |
| Numerical id | 5 |
| Author | gel (22282) |
| Entry type | Theorem |
| Classification | msc 28A05 |
| Related topic | SumsOfCompactPavingsAreCompact |