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Natural Induction
Needed by:
Natural Order
Recursion Theorem
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Peano Axioms


Historically considered a fountainhead for all of mathematics.


So far we know that $\omega $ is the unique smallest successor set. In other words, we know that $0 \in \omega $, $n \in \omega \implies \ssuc{n} \in \omega $ and that if these two properties hold of some $S \subset \omega $, then $S = \omega $. We can add two important statements to this list. First, that 0 is the successor of no number. In other words, $n^+ \neq 0$ for all $n \in \omega $. Second, that if two numbers have the same successor, then they are the same number In other words, $\ssuc{n} = \ssuc{m} \implies n = m$

These five properties were historically considered the fountainhead of all of mathematics. One by the name of Peano used them to show the elementary properties of arithmetic. They are:

  1. $0 \in \omega $.
  2. $n \in \omega \implies n^+ \in \omega $ for all $n \in \omega $.
  3. If $S$ is a successor set contained in $\omega $, then $S = \omega $.
  4. $\ssuc{n} \neq 0$ for all $n \in \omega $
  5. $\ssuc{n} = \ssuc{m} \implies n = m$ for all $n, m \in \omega $.

These are collectively known as the Peano axioms. Recall that the third statement in this list is the principle of mathematical induction.


Here are the statements.

$0 \in \omega $.
$n \in \omega \Rightarrow n^+ \in \omega $.
Suppose $S \subset \omega $, $0 \in S$, and $(n \in S \Rightarrow n^+ \in S$. Then $S = \omega $.
$n^+ \neq 0$ for all $n \in \omega $.

The last one uses the following two useful facts.

$x \in n \Rightarrow n \not\subset x$.
$(x \in y \land y \in n) \Rightarrow x \in n$

This latter proposition is sometimes described by saying that $n$ is a transitive set. This notion of transitivity is not the same as that described in Relations. Using these one can show:

Suppose $n, m \in \omega $ with $n^+ = m^+$. Then $n = m$.
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