# Non-Empty Set of Natural Numbers with no Greatest Element is Denumerable

Jump to navigation
Jump to search

## Theorem

Let $A$ be a non-empty set of natural numbers.

Let $A$ have no greatest element.

Then $A$ is a countably infinite set.

## Proof

Aiming for a contradiction, suppose $A$ is finite.

From Subset of Naturals is Finite iff Bounded, it follows that $A$ is bounded.

Hence $A$ has a greatest element.

This contradicts the fact that $A$ has no greatest element.

Hence by Proof by Contradiction it follows that $A$ is not finite.

From Subset of Natural Numbers is either Finite or Denumerable, $A$ is a countably infinite set.

$\blacksquare$

## Sources

- 2010: Raymond M. Smullyan and Melvin Fitting:
*Set Theory and the Continuum Problem*(revised ed.) ... (previous) ... (next): Chapter $3$: The Natural Numbers: $\S 8$ Definition by finite recursion: Exercise $8.1$