3
$\begingroup$

For reference, this question is taken from page 47 of 'Classic Set Theory' by Derek Goldrei.

The chapter is called 'the construction of the natural numbers', so not everything we 'typically' know about natural numbers is known at this point. We have so far that: $\mathbb{N}$ with element $0$ (the empty set) and the successor function $S : \textbf{n} \longrightarrow \textbf{n}^+$ satisfies Peano's axioms (http://mathworld.wolfram.com/PeanosAxioms.html).

We know that $\mathbb{N}$ is a well-ordered set, well ordered by $\in$ but we do not know the usual properties for addition, multiplication and exponentiation hold (associativity, commutativity etc).

Any help would be greatly appreciated. Thanks.

$\endgroup$
1
$\begingroup$

HINT:

Assuming that you define $n^+$ as the set $n\cup\{n\}$, then by induction you can show that if $n$ is a natural number then all its elements are natural numbers.

$\endgroup$
  • $\begingroup$ I'm not entirely sure how to do that. For the base case do I use $x = 0$? $\endgroup$ – Math Student 91 Apr 26 '14 at 14:57
  • 1
    $\begingroup$ Well, if $0=\varnothing$, then certainly there is no element of $\varnothing$ which is not a natural number. $\endgroup$ – Asaf Karagila Apr 26 '14 at 14:57
  • $\begingroup$ So for the base case, we know $0 \in \mathbb{N}$? Then the induction hypothesis is that for $x \in \mathbb{N}$, we have $x \in \mathbb{N}$. Then we want to show that for $x \in n^+$, we have $x \in \mathbb{N}$. We have that $x \in n^+$ implies that $x \in n \cup \{n\}$, so $x \in n$ or $x \in \{n\}$. Is this right so far? $\endgroup$ – Math Student 91 Apr 26 '14 at 15:03
  • 1
    $\begingroup$ Yes, only that you mean to use $\subseteq$ for some of these $\in$ symbols. In particular $0\subseteq\Bbb N$ and the second one in the induction hypothesis. $\endgroup$ – Asaf Karagila Apr 26 '14 at 15:07
  • 1
    $\begingroup$ The part after "implies" is plain wrong. Here you definitely switched $\subseteq$ and $\in$. $\endgroup$ – Asaf Karagila Apr 26 '14 at 15:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.