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I know the WOP is treated like axiom of a natural number, but I was curious if I can prove WOP defined for the set of natural numbers N by following:

Suppose A is a subset of N, which then obeys all axioms from Peano postulates, and let us assume that A does not have a least element. Then, 1 in A must be a successor to another natural number, which is 0. However, 0 is neither in A nor N. (If 0 is in N, then we argue that 0 must be a successor to -1, which is not part of N). Also, 1 in A being a successor contradicts with the Peano axiom (1 cannot be a successor). Hence, our assumption that A having no least element is not true. Therefore, A has a least element. Without out loss o generality, every subset of the N has a least element.

I understand this is silly proof, but I wanted to make sure that I have reasoning for WOP in N before moving on.

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    $\begingroup$ Possible duplicate of Well-Ordering and Mathematical Induction $\endgroup$ – Andres Mejia Jun 23 '16 at 16:31
  • $\begingroup$ WOP is equivalent to the principle of induction. There are answers to this question abundant on the internet, as well as on SE $\endgroup$ – Andres Mejia Jun 23 '16 at 16:32
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Well ordering principle is a direct consequence of the Principle of Mathematical Induction and here we need the second version of Principle of Mathematical Induction stated below:

Principle of Mathematical Induction Second Version: If $A$ is a subset of $\mathbb{N}$ such that

  • $1 \in A$
  • If $1, 2, \ldots, n \in A$ then $(n + 1) \in A$

then $A = \mathbb{N}$.

Let $A$ be a non-empty subset of $\mathbb{N}$. If $1 \in A$ then obviously $A$ has a least member and hence let $1 \notin A$. Consider $B = \mathbb{N} - A$. Then we can see that $B$ is also a subset of $\mathbb{N}$ and $1 \in B$.

Let's assume that $A$ does not have a least member. Using this assumption I will prove that if $1, 2,\ldots, n \in B$ then $(n + 1) \in B$. If $1, 2,\ldots, n \in B$ then $1, 2, \ldots, n \notin A$ and hence any member in $A$ is greater than $n$. If $(n + 1) \in A$ then it would become the least member of $A$ and this is not allowed by our assumption. Hence $(n + 1) \in B$. It now follows by Principle of Mathematical Induction that $B = \mathbb{N}$ and hence $A = \mathbb{N} - B = \emptyset$ which is contrary to the assumption that $A$ is non-empty.

It follows from this contradiction that $A$ must have a least member. It is possible to prove the Principle of Mathematical Induction by assuming the truth of Well Ordering Principle. So in a sense both these principles are equivalent, but it appears that the principle of induction captures the essence of natural numbers in a more intuitive and obvious manner and perhaps that is the reason it was selected as one of the Peano's axioms.

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  • $\begingroup$ Thank you very much for your answer! Does that mean predicate P(n): {n in N} rather than what I specified? What is wrong with my predicate from above, though? $\endgroup$ – MathWanderer Jun 23 '16 at 18:36
  • $\begingroup$ @MathWanderer: Your predicate looks itself like an inductive step. It should look more like an statement about some property of $n$ which can be true or false. And then you need to show that if $P(n)$ is true for $n = 1, 2, \ldots, k - 1$ then it is true for $n = k$ also. $\endgroup$ – Paramanand Singh Jun 24 '16 at 3:48
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In my opinion, you have essentially restated the axiom by making the same assumption in a more subtle fashion (assumptions about the naturals). Generally, axioms are axioms because they cannot be proven independently using the other axioms. For instance, if you are familiar with analysis, the existence of suprema in the reals is typically axiomatic, but the existence of infima is not, since it can be proven from the suprema axiom.

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  • $\begingroup$ Thank you for your answer. I thought that the axioms like WOP do have proofs as I recall that axioms usually born from some theorems. Looking back at my work, I believe I used circular logic when I claimed that the proof contradicts with the Peano axiom about 1 cannot be a successor (another word that 1 is least element), but I thought I made right logic by claiming the contradiction that 0 is not a member of N. $\endgroup$ – MathWanderer Jun 22 '16 at 17:32
  • $\begingroup$ @MathWanderer - well, I'm sure you could come up with a situation where WOP is provable but it would have to be a system with axioms that are fairly unconventional. $\endgroup$ – The Count Jun 22 '16 at 17:34
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    $\begingroup$ I am not sure what you mean by Peano axioms, there are different versions (the original second order, or the first order) and some additional differences of detail. But WOP is provable using the induction part of the Peano axioms. $\endgroup$ – André Nicolas Jun 22 '16 at 17:57
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I came up with proof now:

Let A be a non-empty subset of the set of natural numbers N, we let as assume that A does not contain a least element. Set B=N-A, meaning B is a subset of N. Let predicate P(n) be defined as follow: P(n): for all elements n of B, n in B implies that n+1 is in B. Let T be a truth set of P(n):

Basis Step: P(1) is true as 1 in N indicates that 1 cannot be in A.

Inductive Step: Let as assume that P(1)^P(2)^...^P(n) is true, where n is a natural number. That means {1,2,...,n} is not a subset of A. That means P(n+1) is true since A does not have a least element. Hence, truth set T for P(n) is equal to B. This means B=N.

Since B=N, we deduce that A is empty, which contradicts with the fact that A is non-empty. Hence, our assumption is wrong and A does contains a least element. Without loss of generality, every subset of natural numbers does contain least. element.

*From "Let A be a non-empty subset...", is this considered as fact? I am worried since I use word 'Let", which usually indicates the assumption.

I am very sorry for not using symbols...I just started to learn Latex.

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  • $\begingroup$ I think you have the right idea in your mind but you have not expressed it properly. The standard proof of Well Ordering Principle is based on this idea and I have given an answer containing this proof. The statement $P(n)$ should just be $n \in B$. $\endgroup$ – Paramanand Singh Jun 23 '16 at 16:31

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