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Natural Numbers

  1. Mar 9, 2010 #1


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    Earlier today, I was thinking about the statement that "there exists no greatest natural number" and immediately, two proofs sprang to my mind. Since my question depends on these, I'll write them out below . . .

    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!
  2. jcsd
  3. Mar 9, 2010 #2
    i thought it would have had something to do with [itex]\omega[/itex] being an infinite limit ordinal
    Last edited: Mar 9, 2010
  4. Mar 9, 2010 #3


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    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].[/quote]
    Minor error: the thing that immediately follows has [itex]\leq[/itex], not <.

    As for your second proof, you have neglected somewhere along the line to show that sup(N)-1 is an upper bound on N, so you haven't yet shown the contradiction you were seeking.

    Aside from those errors, your proofs look fine, and prove what you think they prove.
  5. Mar 10, 2010 #4


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    Thanks Hurkyl! I'm glad that you pointed out those errors, I really should know better.
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