4
$\begingroup$

Context: This question arises from the unsuccessful search for a proof (that does not use induction) of a result used to prove the equivalency of the Well-Ordering Principle ("WO") and Induction. Assume the Dedekind-Peano Axioms with addition and the following basics:

(A1) $\forall x[x+1=S(x)]$

(A2) $\forall x\forall y[x+S(y)=S(x+y)]$

(O1) $\ \ x<y\ $ means $\ \exists z[x+z=y]$

(O2) $\ \ x\leq y\ $ means $\ x\!=\!y\vee x\!<\!y$

In particular, the question arises on a result (labeled Theorem 1 here) used in the proof of WO implies induction:

Theorem 0: The well-ordering principle implies the principle of induction.

Proof: Let $A\!\subseteq\!\mathbb{N}$ be a set with the properties: (i) $1\!\in\!A$, and (ii) $k\!\in\!A$ implies $S(k)\!\in\!A$. The proof proceeds by contradiction by supposing that $A\!\neq\!\mathbb{N}$ (i.e., $\mathbb{N}\backslash A$ is non-empty). According to WO, if $\mathbb{N}\backslash A$ is non-empty, then it must contain a least element, which can be denoted as $l$. By (i), $l\!\neq\!1$. Therefore, from Theorem 1, there exists some $m\!\in\!\mathbb{N}$ such that $m\!+\!1\!=\!l$, which implies $m\!<\!l$ by (O1). It must be that $m\!\in\!A$ (since assuming otherwise would contradict $l$ being the least element of $\mathbb{N}\backslash A$). By (ii) and (A1), it follows that $S(m)\!=\!m\!+\!1\!=\!l\!\in\!A$. But this contradicts the assumption that $l\!\in\!\mathbb{N}\backslash A$. Hence the set $\mathbb{N}\backslash A$ must be empty, so that $A\!=\!\mathbb{N}$.

Question: My question is not on the above proof (unless there is a way to prove that result without relying on Theorem 1), but on the proof of Theorem 1 itself. I also understand how to prove Theorem 1 using induction. But since Theorem 1 is used above to prove induction (Theorem 0), one is obviously excluded from using induction and limited to using WO to avoid circular reasoning.

Theorem 1: For all $n\!\in\!\mathbb{N}$, $n\!=\!1$ or $n$ is the successor of some $m\!\in\!\mathbb{N}$. In other words: $$\forall x\Big[\ x\!=\!1\ \vee\ \exists y[x\!=\!y\!+\!1]\ \Big]$$

However, I am at a loss as to how to prove this using just WO. I've attempted to use the method of "proof by minimum counterexample" as above, but only come as far as this:

Proof (incomplete): Consider the set $B\!\subseteq\!\mathbb{N}$ defined by: $$B\equiv\{\ n\!\in\!\mathbb{N}\ :\ n\!=\!1\vee\exists m\!\in\!\mathbb{N}[n\!=\!m\!+\!1]\ \}.$$ The proof proceeds by contradiction by supposing that $B\!\neq\!\mathbb{N}$ (i.e., $\mathbb{N}\backslash B$ is non-empty). According to WO, if $\mathbb{N}\backslash B$ is non-empty, then it must contain a least element, which can be denoted as $l$. It follows from the definition of $B$ that $l\!\neq\!1$ and $\neg\exists m\!\in\!\mathbb{N}[l\!=\!m\!+\!1]$.

Am I on the right track?

$\endgroup$
  • 3
    $\begingroup$ The set of polynomials with natural number coefficients ordered the usual way are well ordered, but don't always have a predecessor. $\endgroup$ – DanielV Mar 3 '18 at 4:48
  • 1
    $\begingroup$ You might need an additional axiom: for all $n$, $1 \leq n$. (Does it follow somehow from the other axioms? Or perhaps you define $1$ to be the smallest element of $\mathbb{N}$, from the well-ordering?) In any case, from $1 \leq l$ and $1 \neq l$, you get $1 < l$, so $l = 1+z$ for some $z$. Then $l$ is the successor of $z$. $\endgroup$ – Zach Teitler Mar 3 '18 at 6:18
  • 1
    $\begingroup$ @StephenK. A simple example of a well-ordered set with more than one element lacking a predecessor is given by the natural numbers ordered so that all even numbers come before all odd numbers. Even numbers and odd numbers are then ordered in the usual way. Neither $0$ nor $1$ have predecessors. The set of pairs of natural numbers under lexicographic ordering has infinitely many elements without predecessors. $\endgroup$ – Fabio Somenzi Mar 3 '18 at 17:15
  • 1
    $\begingroup$ @StephenK.$2x^2$ does not have a predecessor, but also $(x \mapsto 2x^2) \ne (x \mapsto 0)$. Also, it is very unusual to start counting at $1$, most examples of peano arithmetic start counting at $0$. $\endgroup$ – DanielV Mar 3 '18 at 19:53
  • 1
    $\begingroup$ @StephenK.The general point is that there are multiple initial elements, that's what it means to not have a unique predecessor. $\endgroup$ – DanielV Mar 3 '18 at 20:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.