Proofs of the Existence of No Greatest Natural Number

Click For Summary

Discussion Overview

The discussion revolves around the proofs of the statement "there exists no greatest natural number," exploring two distinct proofs provided by a participant. The focus includes the nature of these proofs, their implications regarding the boundedness of natural numbers, and the underlying concepts of supremum and upper bounds.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant presents two proofs: the first shows that for any natural number n, n+1 is also a natural number, implying no greatest element exists in the natural numbers.
  • The second proof argues that if the natural numbers were bounded above, it would lead to a contradiction regarding the supremum of the set of natural numbers.
  • Another participant introduces the concept of ω as an infinite limit ordinal, suggesting a connection to the discussion.
  • A later reply points out a minor error in the second proof regarding the use of < and ≤ in relation to the supremum.
  • This same reply also notes that the proof fails to demonstrate that sup(N)-1 is an upper bound for N, questioning the validity of the contradiction sought by the original poster.
  • The original poster expresses gratitude for the feedback and acknowledges the errors pointed out.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the validity of the proofs, as one participant identifies errors in the second proof while another suggests a different perspective involving limit ordinals. The discussion remains unresolved regarding the completeness and correctness of the proofs.

Contextual Notes

There are unresolved issues regarding the assumptions made in the proofs, particularly concerning the treatment of the supremum and the implications of boundedness. The discussion also highlights the need for clarity in mathematical notation and definitions.

jgens
Gold Member
Messages
1,575
Reaction score
50
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!
 
Physics news on Phys.org
i thought it would have had something to do with [itex]\omega[/itex] being an infinite limit ordinal
 
Last edited:
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.
 
Thanks Hurkyl! I'm glad that you pointed out those errors, I really should know better.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 3 ·
Replies
3
Views
5K
  • · Replies 21 ·
Replies
21
Views
4K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 17 ·
Replies
17
Views
6K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K