Cartesian Product of Natural Numbers with Itself is Countable
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Theorem
The Cartesian product $\N \times \N$ of the set of natural numbers $\N$ with itself is countable.
Proof
This is simply a special case of Cartesian Product of Countable Sets is Countable.
$\blacksquare$
Sources
- 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $1$: Mappings: $\S 15$
- 2000: James R. Munkres: Topology (2nd ed.) ... (previous) ... (next): $1$: Set Theory and Logic: $\S 7$: Countable and Uncountable Sets: Example $2$
- 2000: James R. Munkres: Topology (2nd ed.) ... (previous) ... (next): $1$: Set Theory and Logic: $\S 7$: Countable and Uncountable Sets: Corollary $7.4$
- 2005: René L. Schilling: Measures, Integrals and Martingales ... (previous) ... (next): $2.5 \ \text{(iv)}$
- 2008: Paul Halmos and Steven Givant: Introduction to Boolean Algebras ... (previous) ... (next): Appendix $\text{A}$: Set Theory: Countable Sets
- 2010: Raymond M. Smullyan and Melvin Fitting: Set Theory and the Continuum Problem (revised ed.) ... (previous) ... (next): Chapter $1$: General Background: $\S 2$ Countable or uncountable?