Equivalence of Well-Ordering Principle and Induction/Proof/WOP implies PFI
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The Well-Ordering Principle implies the Principle of Finite Induction.
- 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.
- 1982: P.M. Cohn: Algebra Volume 1 (2nd ed.) ... (previous) ... (next): $\S 2.1$: Chapter $2$: Integers and natural numbers: The integers