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Proposition 8.5.10 (Principle of strong induction). Let $X$ be a well-ordered set with an ordering relation $\le$, and let $P(n)$ be a property pertaining to an element $n \in X$ (i.e., for each $n\in X$, $P(n)$ is either a true statement or a false statement). Suppose that for every $n \in X$, we have the following implication: if $P(m)$ is true for all $m \in X$ with $m <_{X} n$, then $P(n)$ is also true. Prove that $P(n)$ is true for all $n \in X$.

Exercise 8.5.10. Prove Proposition 8.5.10, without using the axiom of choice. (Hint: consider the set $$Y: = \{n \in X : \text{$P(m)$ is false for some $m \in X$ with $m \le_X n$}\},$$ and show that $Y$ being non-empty would lead to a contradiction.)

I know that every non-empty subset of $X$ has a minimum element. I guess that I need to use this to solve this question, but I don't know how.

I appreciate if you give me some help.

shk910
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2 Answers2

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I would just define $Y=\{n: P(n) \text{ is false}\}$. If $Y$ is non-empty, let $m=\min(Y)$, which then exists.

For all $n <_X m$ we know that $P(n)$ cannot be false or $n \in Y$ with $n <_X m$ contradicting the minimality of $m$. So for all $n <_X m$, $P(n)$ holds. The induction hypothesis tells us that $P(m)$ must be true. But then $m \notin Y$, contradiction !

So $P(n)$ is always true and $Y=\emptyset$.

Henno Brandsma
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Suppose that $Y$ is not empty. Since $Y \subseteq X$, a well-ordered set, there is a least $y^{*} \in Y$. Now use the inductive hypothesis to conclude something about $y^{*}$.

Jay
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