In a higher algebra book that I'm working through, the natural numbers are constructed in the following manner:-
Consider the class $S$ of all finite sets. Now, $S$ is partitioned into equivalence classes based on the equivalence relation that two finite sets are equivalent if there exists a one-to-one correspondence between them, i.e. if they are equipotent. And each of these equivalence classes are given a label, corresponding to the number of one-to-one correspondences.
So, $S= S_1 \bigcup S_2 \bigcup S_3 \bigcup....$where $S_1, S_2, S_3,$ etc are disjoint equivalence classes, and to $S_n$, we give the label of the $n$th natural number. This is how the natural numbers are constructed.
Now, as I understand it, the number of elements of $S_n$ for any $n$, has to be infinite. For instance, the number $5$ is the label given to $S_5$. But $5$ can be represented in an infinite number of ways: five chairs, tables, coins, pencils, pens, etc. So, this means that $S_5$ is an infinite class, and so is any $S_n$.
The only way I see this possible is, the class $S$ that we started out with, has to be an infinite class. Is this true? Basically, what I'm asking is: Is the class of all finite sets infinite? How do you prove this?
Proof that $S$ in indeed infinite:
The class of all sets is obviously infinite. Now for every set $M$ there exists a set $\{M\}$ containing $M$ as only element. But $\{M\}$ is a one-element set, that is, element of $S_1$. Therefore $S_1$ has as many elements as the class of all sets. Obviosly, then $S$ must also have as many elements as the class of sets, since all elements of $S_1$ are also elements of $S$.
Edit:
Andres Caicedo requested that I add also the proof that the class of all sets is infinite, so there it is:
In the proof above, I've given a bijection between the class of all sets and the class $S_1$ of one-element sets, $M\mapsto\{M\}$. Now $S_1$ is a proper subclass of the class of all sets (all its members are sets, but there exist sets — for example the empty set — which are not in $S_1$). However a finite class cannot have a bijection to a proper subclass, therefore the class of all sets cannot be finite.