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I tried to prove Well-Ordering Principle by myself, and I finally did it. However, I'm not sure if this proof is correct. Can anyone evaluate my proof?

Proof: Since the set of natural numbers, $\mathbb{N}$, is bounded below (every element of $\mathbb{N}$ is equal or greater than zero), an arbitrary subset (let's name this set $X$) of $\mathbb{N}$ is also bounded below. So, by the greatest lower bound principle, there exists an infimum of $X$. Let $a:=\inf (X)$. Then, every element of $X$ is equal or greater than $a$. Also, for every real number $y$ which is greater than $a$, there exists an element $k$ (which is equal or greater than $a$) of $X$ which is less than $y$. Therefore, $k=a$, and since $a$ is an infimum of $X$, it becomes a minimum element of $X$. (Q.E.D.)

Andrés E. Caicedo
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19017김범준
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    Well-ordering principle over where? There are many theorems that could be named well-ordering principle (the usual meaning of 'well-ordering principle' is 'every set has a well-order.') I guess your 'well-ordering principle' is '$\mathbb{N}$ is well-ordered' and you are working on the theory of ordered field with completeness axiom. Could I ask I am correct? – Hanul Jeon Apr 20 '20 at 13:55
  • Yes, I was trying to prove that N is well-ordered. – 19017김범준 Apr 20 '20 at 14:05
  • Sorry for my ignorance – 19017김범준 Apr 20 '20 at 14:05
  • You are welcome! You do not need to say sorry. – Hanul Jeon Apr 20 '20 at 14:09

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How do you conclude $k=a$? I means consider the set $X:= \{1,2\}$ which has inf of $a:= 1$. Then consider the real number $3$ which has $k:= 2 \in X$ s.t. $2 < 3$, but $k \neq a$. If you are a little more careful, you could use inf's and real numbers to prove the well-ordering principle of the natural numbers, but this is a bit of a vacuous statement, as you certainly need to use the well-orderedness of $\mathbb{N}$ to construct $\mathbb{R}$. Well-orderedness is more or less an axiom of the natural numbers. You can try to prove that it is equivalent to induction too, e.g. as seen here.

Physical Mathematics
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  • Then, it is not appropriate to use properties of real numbers to prove well-orderedness of N? I was trying to prove the well-ordering principle using the least upper bound (or greatest lower bound) property. – 19017김범준 Apr 20 '20 at 14:11
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Actually, the you can prove the well-ordering principle from the least-upper-bound property of $\mathbb{R}$. See the 2nd bullet point from the Wikipedia article here. Steps:

  • Dedekind Cuts method to construct $\mathbb{R}$ and then show that $\mathbb{Q}$ and thus $\mathbb{N}$ are subfields of $\mathbb{R}$ (standard "baby Rudin" textbook stuff).

  • Next, since $\mathbb{R}$ has the least-upper bound property, any subset of $\mathbb{N}$ , say $X$, has an infimum, and thus the arguments provided by the author are correct up to $a=inf(X)$

  • Lastly, we need to show $a$ (which is automatically an element of $\mathbb{R}$) is actually an element of $X$ itself. The author's original argument here needs just a little strengthening or re-wording. Why not use a contradiction proof to show that there cannot be another real number $a^*$ ($\notin X$) which is "in-between" $a$ and any member $n\in X$? For if there was such a gap between $a$ and $n$, then by the "Q is dense in R" property (Rudin theorem 1.20), we could find a rational number $a^*$ such that $a<a^*<n$. This would contradict the fact that $a$ is the infimum of X.

  • Thus $a$ must be an element of $X$, and so any subset of $\mathbb{N}$ must have a least element.

  • Perhaps a logician can comment, but the Axiom of Choice is lurking around somewhere in the 3rd bullet point above. As Jerry Bona says: "who can tell...?"

Mohin Ahmed
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