Equivalence of Well-Ordering Principle and Induction/Proof/WOP implies PFI

From ProofWiki
Jump to navigation Jump to search


The Well-Ordering Principle implies the Principle of Finite Induction.

That is:

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


Principle of Finite 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 $S = \N$.


To save space, we will refer to:

The Well-Ordering Principle as WOP
The Principle of Finite Induction as PFI.

We assume the truth of WOP.

Let $S \subseteq \N$ which satisfy:

$(D): \quad 0 \in S$
$(E): \quad n \in S \implies n+1 \in S$.

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

Aiming for a contradiction, suppose that:

$S \ne \N$

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

From Set Difference is Subset, $S' \subseteq \N$.

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 $k \in \N$.

We can consider the von Neumann construction of the natural numbers.

By definition of natural number addition, it is noted that $k + 1 \in \N$ is the immediate successor element of $k \in \N$.

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

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

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

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

So $S = N$.

Thus we have proved that WOP implies PFI.