Equivalence of Well-Ordering Principle and Induction

Theorem
The Well-Ordering Principle, the Principle of Mathematical Induction and the Principle of Complete Induction are logically equivalent.

That is:


 * Principle of Mathematical Induction: Given a subset $$S \subseteq \N$$ of the natural numbers which has these properties:
 * $$0 \in S$$;
 * $$n \in S \implies n+1 \in S$$
 * then $$S = \N$$.

iff:


 * Principle of Complete Induction: Given a subset $$S \subseteq \N$$ of the natural numbers which has these properties:
 * $$0 \in S$$;
 * $$\left\{{0, 1, \ldots, n}\right\} \subseteq S \implies n+1 \in S$$
 * then $$S = \N$$.

iff:


 * Well-Ordering Principle: Every non-empty subset of $$\N$$ has a minimal element.

Proof
To save space, we will refer to:
 * The Well-Ordering Principle as WOP;
 * The Principle of Mathematical Induction as PMI;
 * The Principle of Complete Induction as PCI.

PMI implies PCI
Let us assume that the PMI is true.

Let $$S \subseteq \N$$ which satisfy:
 * $$(A)$$: $$0 \in S$$;
 * $$(B)$$: $$\left\{{0, 1, \ldots, n}\right\} \subseteq S \implies n+1 \in S$$.

We want to show that $$S = \N$$, that is, the PCI is true.

Let $$P \left({n}\right)$$ be the propositional function:
 * $$P \left({n}\right) \iff \left\{{0, 1, \ldots, n}\right\} \subseteq S$$

We define the set $$S'$$ as:
 * $$S' = \left\{{n \in \N: P \left({n}\right) \text { is true}}\right\}$$

$$P \left({0}\right)$$ is true by $$(A)$$, so $$0 \in S'$$.

Assume $$P \left({k}\right)$$ is true where $$k > 0$$.

So $$k \in S'$$, and by hypothesis, $$\left\{{0, 1, \ldots, k}\right\} \subseteq S$$

So by $$(B)$$, $$k + 1 \in S$$.

Thus $$\left\{{0, 1, \ldots, k, k + 1}\right\} \subseteq S$$.

That last statement means $$P \left({k + 1}\right)$$ is true.

This means $$k + 1 \in S'$$.

Thus we have satisfied the conditions:
 * $$0 \in S'$$;
 * $$n \in S' \implies n + 1 \in S'$$.

So PMI gives that $$S' = \N$$, which means $$S = \N$$.

Thus PMI implies PCI.

PCI implies WOP
Let us assume that the PCI is true.

Let $$\varnothing \subset S \subseteq \N$$.

We need to show that $$S$$ has a minimal element, and so demonstrate that the WOP holds.

With a view to obtaining a contradiction, assume that:
 * $$(C)$$: $$S$$ has no minimal element.

Let $$P \left({n}\right)$$ be the propositional function:
 * $$n \notin S$$

Suppose $$0 \in S$$.

We have that $$0$$ is a lower bound for $$\N$$.

Hence by Lower Bound for Subset, $$0$$ is also a lower bound for $$S$$.

$$0 \notin S$$, otherwise $$0$$ would be the minimal element of $$S$$.

This contradicts our supposition $$(C)$$, namely, that $$S$$ does not have a minimal element.

So $$0 \notin S$$ and so $$P \left({0}\right)$$ holds.

Suppose $$P \left({j}\right)$$ for $$0 \le j \le k$$.

That is:
 * $$\forall j \in \left[{0 \, . \, . \, k}\right]: j \notin S$$

where $$\left[{0 \,. \, . \, k}\right]$$ denotes the closed interval between $0$ and $k$.

Now if $$k + 1 \in S$$ it follows that $$k + 1$$ would then be the minimal element of $$S$$.

So then $$k + 1 \notin S$$ and so $$P \left({k+1}\right)$$.

Thus we have proved that:
 * 1) $$P \left({0}\right)$$ holds;
 * 2) $$\left({\forall j \in \left[{0 \, . \, . \, k}\right]: P \left({j}\right)}\right) \implies P \left({k+1}\right)$$.

So we see that PCI implies that $$P \left({n}\right)$$ holds for all $$n \in \N$$.

But this means that $$S = \varnothing$$, which is a contradiction of the fact that $$S$$ is non-empty.

So, by proof by contradiction, $$S$$ must have a minimal element.

That is, $$\N$$ satisfies the Well-Ordering Principle.

Thus PCI implies WOP.

WOP implies PMI
We assume the truth of the Well-Ordering Principle.

Let $$S \subseteq \N$$ which satisfy:
 * $$(D)$$: $$0 \in S$$;
 * $$(E)$$: $$n \in S \implies n+1 \in S$$.

We want to show that $$S = \N$$, that is, the PMI is true.

With a view to obtaining a contradiction, assume that:
 * $$S \ne \N$$

Consider $$S' = \N \setminus S$$, where $$\setminus$$ denotes set difference.

From Set Difference Subset, $$S' \subseteq \N$$ and so from WOP, $$S'$$ has a minimal element.

A lower bound of $$\N$$ is $$0$$.

By Lower Bound for Subset, $$0$$ is also a lower bound for $$S'$$.

By hypothesis, $$0 \in S$$.

From the definition of set difference, $$0 \notin S'$$.

So this minimal element of $$S'$$ has the form $$k + 1$$ where $$k \in \N$$.

Thus $$k \in S$$ but $$k + 1 \notin S$$.

From $$(E)$$, this contradicts the definition of $$S$$.

Thus if $$S' \ne \varnothing$$, it has no minimal element.

This contradicts the Well-Ordering Principle, and so $$S' = \varnothing$$.

So $$S = N$$.

Thus we have proved that WOP implies PMI.

Final assembly
So, we have that:
 * PMI implies PCI: The Principle of Mathematical Induction implies the Principle of Complete Induction;
 * PCI implies WOP: The Principle of Complete Induction implies the Well-Ordering Principle;
 * WOP implies PMI: The Well-Ordering Principle implies the Principle of Mathematical Induction.

This completes the result.