Exercise 2.13 in Introduction to Set Theory, 3rd ed. by Hrbacek & Jech

Question

Exercise 2.13 on page 46 reads

(Double Induction) Let $P(x,y)$ be a property. Assume

(**) If $P(k,l)$ holds for all $k,l \in \mathbb{N}$ such that $k < m$ or ($k=m$ and $l<n$), then >$P(m,n)$ holds.

Conclude that $P(m,n)$ holds for all $m,n \in \mathbb{N}$.

Since the authors state on page 44

(The Induction Principle, Second Version) Let $P(x)$ be a property (possibly with parameters). Assume that, for all $n\in\mathbb{N}$,

(*) If $P(k)$ holds for all $k<n$, then $P(n)$ holds.

Then $P(n)$ holds for all $n \in \mathbb{N}$.

I was trying to solve the exercise applying the second version of induction (answers here use a double application of regular induction, which is stated on page 42 of the book)

My argument is as follows: Assume (**) is true. Let $n=n_{0}$ be fixed and consider the property $Q(m):P(m,n_{0})$. Then (**) implies the condition (*) for $Q$ and by the second version of induction we have $\forall m\, P(m,n_{0})$.

Question: Since $n_{0}$ was arbitrary, can I conclude that $P(m,n)$ holds for all $m,n\in\mathbb{N}$?

If not, I guess I can perform an additional symmetric application of the second version of induction. This time fixing $m=m_{0}$ in order to conclude $\forall n\, P(m_{0},n)$.

Hence, do I need to apply the second version of induction above once or twice?

Answer 1

Yes, your first argument works. However, I think this is a good place for thinking about the alternative way of framing induction as using well ordering to show that the set of counterexamples must be empty.

First, note that $\Bbb N \times \Bbb N$ is well ordered (meaning every non-empty subset has a $\lt$-least element) by the lexicographic ordering $(a, b) \lt (c, d) \iff a \lt c \lor (a=c \land b \lt d)$. This is easy to see.

Let $S \subseteq \Bbb N \times \Bbb N$ Since $\Bbb N$ is well-ordered, any non-empty subset has a smallest first coordinate $x$. Consider $S_x = \{b \mid (x, b) \in S \}$. Then $S_x$ is a non-empty subset of $\Bbb N$, so it has a least element $y$. By definition, we know that $(x, y) \in S$. If also $(a, b) \in S$ with $(x, y) \neq (a, b)$, then by definition of $x$, we know that $x \leq a$. If $x=a$, then by definition of $y$, we know that $y \lt b$. Thus, we know that $(x, y) \lt (a, b)$ and $(x, y)$ is the $\lt$-least element of $S$, as required.

Now assume $(\ast \ast)$ and suppose $P(m, n)$ were false for some $(m, n)$. Then there is a $\lt$-least such $(m, n)$ for which $P$ fails. What can it be?

That least counterexample of $P$ can't be $(0, 0)$ because for $(0, 0)$, the hypothesis of $(\ast \ast)$ is vacuously true.

The least counterexample also can't have the form $(0, l+1)$ because the only smaller elements have the form $(0, b), b \leq l$, and since we are assuming that $(0, l+1)$ is the least counterexample, we know that $P(0, b)$ holds for $b \leq l$, which means (via $(\ast \ast)$) that $P(0, l+1)$ holds after all.

Now we show the least counterexample of $P$ can't have the form $(k+1, 0)$. Since $(k+1, 0)$ is, by assumption, the least pair for which $P$ fails, we know that $P(a, b)$ holds whenever $a \lt k+1$. We don't have to worry about $(k+1, n+1)$ because it's never the case that $n+1 \lt 0$. Thus, the hypothesis of $(\ast \ast)$ holds and we know that $P(k+1, 0)$ holds, contrary to our assumption that $(k+1, 0)$ is a counterexample.

Finally, similar reasoning shows that the least counterexample can't have the form $(k+1, l+1)$.

In short, the set of counterexamples can't have a least element. But every non-empty subset of $\Bbb N \times \Bbb N$ has a least element, so the set of counterexamples must be empty. That means $P$ holds for all $(m, n)$.