Principle of Finite Induction/One-Based

It has been suggested that this page or section be merged into Equivalence of Well-Ordering Principle and Induction/Proof/WOP implies PFI. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Mergeto}} from the code.

This page has been identified as a candidate for refactoring of advanced complexity. In particular: As part of the merge, set this result up as a contributing result to the above result. Sources need careful treatment. Until this has been finished, please leave {{Refactor}} in the code. New contributors: Refactoring is a task which is expected to be undertaken by experienced editors only. Because of the underlying complexity of the work needed, it is recommended that you do not embark on a refactoring task until you have become familiar with the structural nature of pages of $\mathsf{Pr} \infty \mathsf{fWiki}$. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Refactor}} from the code.

Theorem

Let $S \subseteq \N_{>0}$ be a subset of the $1$-based natural numbers.

Suppose that:

$(1): \quad 1 \in S$

$(2): \quad \forall n \in \N_{>0} : n \in S \implies n + 1 \in S$

Then:

$S = \N_{>0}$

Proof 1

Consider $\N$ defined as a naturally ordered semigroup.

The result follows directly from Principle of Mathematical Induction for Naturally Ordered Semigroup: General Result.

$\blacksquare$

Proof 2

Let $T$ be the set of all $1$-based natural numbers not in $S$:

$T = \N_{>0} \setminus S$

Aiming for a contradiction, suppose $T$ is non-empty.

From the Well-Ordering Principle, $T$ has a smallest element.

Let this smallest element be denoted $a$.

We have been given that $1 \in S$.

So:

$a > 1$

and so:

$0 < a - 1 < a$

As $a$ is the smallest element of $T$, it follows that:

$a - 1 \notin T$

That means $a - 1 \in S$.

But then by hypothesis:

$\paren {a - 1} + 1 \in S$

But:

$\paren {a - 1} + 1 = a$

and so $a \notin T$.

This contradicts our assumption that $a \in T$.

It follows by Proof by Contradiction that $T$ has no such smallest element.

Hence it follows that $T$ can have no elements at all.

That is:

$\N_{>0} \setminus S = \O$

That is:

$S = \N_{>0}$

Sources

• 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $1$: Properties of the Natural Numbers: $\S 20$
• 1971: Robert H. Kasriel: Undergraduate Topology ... (previous) ... (next): $\S 1.17$: Finite Induction and Well-Ordering for Positive Integers: $17.1$
• 1982: P.M. Cohn: Algebra Volume 1 (2nd ed.) ... (previous) ... (next): Chapter $2$: Integers and natural numbers: $\S 2.1$: The integers: $\mathbf{I}$
• 2008: Paul Halmos and Steven Givant: Introduction to Boolean Algebras ... (previous) ... (next): Appendix $\text{A}$: Set Theory: Induction