the Cartesian product of a finite number of countable sets is countable
Theorem 1
The Cartesian product of a finite number of countable sets is countable![]()
.
Proof: Let be countable sets and let . Since each is countable, there exists an injective function . The function defined by
where is the th prime
is, by the fundamental theorem of arithmetic![]()
, a bijection between
and a subset of and therefore is also countable.
Note that this result does not (in general) extend to the
Cartesian product of a countably infinite![]()
collection
![]()
of countable
sets. If such a collection contains more than a finite number of sets
with at least two elements, then Cantor’s diagonal argument can be
used to show that the product
is not countable.
For example, given , the set consists of all countably infinite sequences![]()
of zeros and ones.
Each element of can be viewed as a binary fraction and can
therefore be mapped to a unique
real number in and is, of course, not countable.
| Title | the Cartesian product of a finite number of countable sets is countable |
|---|---|
| Canonical name | TheCartesianProductOfAFiniteNumberOfCountableSetsIsCountable |
| Date of creation | 2013-03-22 15:19:45 |
| Last modified on | 2013-03-22 15:19:45 |
| Owner | BenB (9643) |
| Last modified by | BenB (9643) |
| Numerical id | 13 |
| Author | BenB (9643) |
| Entry type | Theorem |
| Classification | msc 03E10 |
| Synonym | The product of a finite number of countable sets is countable |
| Related topic | CardinalityOfACountableUnion |
| Related topic | AlgebraicNumbersAreCountable |
| Related topic | CardinalityOfTheRationals |