Set representation of natural numbers In Wolfgang Wechler's Universal Algebra for Computer Scientists,
he says that natural numbers can be represented as finite sets:
$0$ stands for $\emptyset$, $1$ for {$\emptyset$}, $2$ for {$\emptyset$,{$\emptyset$}}, $3$ for {$\emptyset$,{$\emptyset$,{$\emptyset$}}} and so on.
Then he says that a more readable way is that $0 = \emptyset$ and $ n = \{0,1,\dots\,n-1\}$ for $n \ge 1$.
But to me this seems different: in my opinion $3$ would be coded as $ \{\emptyset,\{\emptyset\},\{\emptyset,\{\emptyset\}\}\} $.
Isn't this a mistake in the book? If it is, which construction is the "right" one?
 A: The book has a typo in its description of $3$, and you are correct.
To expand a bit.
Given a set $a$, we define $s(a)$, the successor of $a$, to be the set $s(a) = a\cup \{a\}$. 
A set $X$ is said to be inductive if and only if:


*

*$\emptyset \in X$; and

*If $a\in X$, then $s(a)\in X$.


The Axiom of Infinity in ZF Set Theory states that there exists at least one inductive set. Given an inductive set $X$, we define
$$\mathbb{N}_X = \bigcap\{ S\subseteq X\mid S\text{ is inductive.}\}.$$
One can then prove that if $X$ and $Y$ are inductive sets, then $\mathbb{N}_X=\mathbb{N}_Y$; we define "the natural numbers", $\mathbb{N}$, to be this (now shown to be well-defined) set.
So what is $\mathbb{N}$? It contains $\emptyset$; we call it $0$. It contains $s(0) = s(\emptyset) = \emptyset\cup\{\emptyset\} = \{\emptyset\}$, which we call $1$.
It contains $s(1)$, which we call $2$. What is $2$?
$$2 = s(1) = s(\{\emptyset\}) = \{\emptyset\} \cup \{\{\emptyset\}\} = \{\emptyset,\{\emptyset\}\}.$$
Then we have $s(2)$, called (unsurprisingly), "$3$"; and we have:
$$3 = s(2) = s(\{\emptyset,\{\emptyset\}\}) = \{\emptyset,\{\emptyset\}\}\cup\Bigl\{\{\emptyset,\{\emptyset\}\}\Bigr\} = \Bigl\{\emptyset,\{\emptyset\},\{\emptyset,\{\emptyset\}\}\Bigr\} = \{0,1,2\}.$$
And so on. The set contains exactly $\emptyset$ and its successors, and it corresponds to the naive idea of the "natural numbers:". The fact that it is the "smallest inductive set" implies that induction holds: if $S\subseteq \mathbb{N}$ is such that $0\in S$ and, whenever $n\in S$ we have $s(n)\in S$ (that is, $S$ is inductive), then $S=\mathbb{N}$. 
