In theory, could you use Induction on any countable set?

Question

Let $A$ be a countable set, not necessarily finite. Suppose I had some property P and I wanted to prove $\forall x \in A(P(x))$ By the general algorithm, If the first element of $A$ has such a property, the $n$th element of $A$ has such a property and the next element after it has such a property then every element in $A$ has the property. This seems logical, but I haven't seen it being used outside of $\mathbb{N}$. Why?

Answer 1

In theory, yes. If $S$ is a countable set, then there is a surjective mapping $s$ from the natural numbers (where we can exclude $0$) to $S$. So based on this $s$ we can now use induction to show that all elements of $S$ have some property $P$ (i.e. $\forall x (x \in S \to P(x))$ by doing exactly what you propose: show that the 'first' element has that property (i.e. show that $P(s(1))$), and show that whenever the $n$-th element has property $P$, then the 'next' element has that property (i.e. $\forall n (P(s(n)) \to P(s(n+1)))$). Or, in full, we have:

$$(P(s(1)) \land \forall n (P(s(n)) \to P(s(n+1)))) \to \forall x (x \in S \to P(x))$$

So yes, you can totally do that. In theory. But, here is the likely problem, and why you don’t see much of that in practice. For many sets, whatever linear order you come up with for that set is likely not very conducive to a nice inductive argument.

For example, the set of rational numbers is countable. Now, there are many mapping from the natural numbers to the rational numbers. Here is an easy one: First, let's create all pairs of positive whole numbers that add up to $2$, then to $3$, etc. So you get $$(1,1), (1,2), (2,1), (1,3), (2,2), (3,1), (1,4), (2,3), (3,2), (4,1), ...etc. $$

We can treat these pairs as rational numbers:

$$\frac{1}{1}, \frac{1}{2}, \frac{2}{1}, \frac{1}{3}, \frac{2}{2}, \frac{3}{1}, \frac{1}{4}, \frac{2}{3}, \frac{3}{2}, \frac{4}{1}, ...etc. $$

Notice that every positive rational number is guaranteed to show up in this list at some point, and so we know that the set of all positive rational numbers is countable. And we immediately get a list of all negative rational numbers too:

$$-\frac{1}{1}, -\frac{1}{2}, -\frac{2}{1}, -\frac{1}{3}, -\frac{2}{2}, -\frac{3}{1}, -\frac{1}{4}, -\frac{2}{3}, -\frac{3}{2}, -\frac{4}{1}, ...etc. $$

And finally, by splicing these two lists, and putting a $0$ in front, we get a list of all rational numbers:

$$0, \frac{1}{1}, -\frac{1}{1}, \frac{1}{2}, -\frac{1}{2}, \frac{2}{1}, -\frac{2}{1}, \frac{1}{3}, -\frac{1}{3}, \frac{2}{2}, -\frac{2}{2}, \frac{3}{1}, -\frac{3}{1}, ...etc. $$

(note that the list of rational numbers thus obtained contains duplicate elements, e.g. $\frac{1}{1}=\frac{2}{2}$. Is that a problem? No, not for what you want to do: as long as you can show that all elements in the list have some property $P$, then clearly all rational numbers have property $P$)

OK, so we know that the set of all rational numbers is countable. Easy peasy. Moreover, let $q$ be the function that maps the non-zero natural numbers to the rational numbers. Note that this function is not trivial to define with a closed formula, but with a little effort it can be done. So far so good.

But now here's the real issue: what is it about the $n$-th entry of this list having property $P$ that would imply that the $n+1$-th entry of this list has property $P$? It all depends on what that property $P$ is that you are considering, but likely there is really no clear connection between $P(q(n))$ and $P(q(n+1))$.

And this is something that will generally be true. Yes, for any countable set $S$ there is some function $s$ that maps natural numbers to that set, but it is unlikely that there is a nice logical relationship between $P(s(n))$ and $P(s(n+1))$ that you can exploit to become part of an inductive proof to show that all elements of $S$ have property $P$. So that is why in practice you rarely, if ever, see this kind of method being used.

A final note: All of the above is not to say that there are no practical methods of induction for countable sets other than the natural numbers. As Bill Dubuque in the Comments points out, you can use various descent methods, e.g. Fermat famously used infinite descent to show that there are no whole numbers $a,b,c$ such that $a^n+b^n=c^n$ for $n=3$ or $n=4$. But note that in such cases you are not exploiting any kind of linear or total ordering on the set, but rather exploit a more general well-founded relationship, and thus use a form of the more general well-founded induction.

But my point in this Answer is to argue that the way you were suggesting to use induction, where you have a 'first' element, and always a 'next' element (i.e. by exploiting some kind of well- ordering on the set), is likely not going to work very well.