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The following are the Peano Postulates:

There exists a set $\mathbb{N}$ with an element $1 \in \mathbb{N}$ and a function $s:\mathbb{N} \to \mathbb{N}$ that satisfy the following three properties.

a. There is no $n \in \mathbb{N}$ such that $s(n)=1$.

b. The function $s$ is injective

c. Let $G \subseteq \mathbb{N}$ be a set. Suppose that $1 \in G$, and that if $g \in G$ then $s(g) \in G$. Then $G=\mathbb{N}$.

We then need to prove this lemma:

Let $a \in \mathbb{N}$. Suppose that $a \neq 1$. Then there is a unique $b \in N$ such that $a=s(b)$.

So, the proof begins by proving that if there were such a $b$, then $b$ is unique.

Then, to show that $b$ exists, we just need to define $G=\{1\} \cup \{c \in \mathbb{N} |$there is some $b \in \mathbb{N}$ such that $s(b)=c \}$ and prove that $G=\mathbb{N}$ because this will immediately imply the existence part of the lemma.

I don't understand how proving that $G=\mathbb{N}$ will imply that $b$ exists. What is the idea that this proof is using to prove that $b$ exists and how is it dependent on $G=\mathbb{N}$?

mauna
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Actually it is not so good idea to use the same letter $b$ for these different numbers. $b$ (replace it with $d$, for instance) in the definition of $G$ is not the same $b$ from your lemma. $G=\mathbb{N}$ means that each number except $1$ is a successor of some number. Therefore $a$ is also a successor for some number $b$.

njguliyev
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  • at this point, $s$ is not defined as the successor function yet. We just know that is is a unary operation and injective. How can we still conclude that $a$ is also a successor for sume number $b$? – mauna Aug 22 '13 at 17:16
  • @mauna, look at the definition of $G$: it is the union of ${1}$ and the set of those numbers $c \in \mathbb{N}$ which can be obtained from some number (denoted $b$) by applying $s$. If $G=\mathbb{N}$ then every (!) number $\ne 1$ can be obtained by applying $s$. – njguliyev Aug 22 '13 at 17:47