- #1

jgens

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Proof 1: Let [itex]n \in \mathbb{N}[/itex]. Clearly [itex]n+1 \in \mathbb{N}[/itex] and [itex]n < n+1[/itex]. Since, given any natural number, it's possible to explicitly construct a larger natural number, [itex]\mathbb{N}[/itex] contains no greatest element.

Proof 2: Clearly [itex]1 \in \mathbb{N}[/itex]. Now suppose that [itex]\mathbb{N}[/itex] is bounded above, in which case [itex]\mathbb{N}[/itex] is a bounded, non-empty subset of the Real numbers. Since [itex]\mathbb{N}[/itex] satisfies the necessary conditions, [itex]\sup\{\mathbb{N}\}[/itex] exists. Because [itex]\sup\{\mathbb{N}\}[/itex] is an upper bound for [itex]\mathbb{N}[/itex], it follows that for any natural number [itex]n[/itex] we have that [itex]n < \sup\{\mathbb{N}\}[/itex]. Since [itex]n+1[/itex] is also a natural number, [itex]n+1 < \sup\{\mathbb{N}\}[/itex] which implies that [itex]n < \sup\{\mathbb{N}\} - 1[/itex]. This contradicts the fact that [itex]\sup\{\mathbb{N}\}[/itex] is a least upper bound and consequently, the assumption that [itex]\mathbb{N}[/itex] is bounded above must have been incorrect. Therefore, [itex]\mathbb{N}[/itex] is unbouded above, completing the proof.

Now, my question is this: What is the difference between the two proofs? From what I can gather, the first

*only*demonstrates that [itex]\mathbb{N}[/itex] contains no greatest element; while the second demonstrates that [itex]\mathbb{N}[/itex] contains no greatest element

*and*is in fact, unbounded above. Is this sort of thinking correct or am I just fundamentally confused about something? Any clarifications or advice are appreciated. Thanks!