Is the principle of transfinite induction (PTI) always not true on a set that is not well ordered?

Question

It seems obvious to me that if for any domino (it is known that all previous dominos fell than this domino will also fell) than it is true that all dominos will fell. It should not matter if there is an infinity of dominos before any domino. Why then the PTI is not true on sets that are not well ordered? It seems to me I understood the proof which indeed shows that PTI is not true when the property in PTI is that x does not belong to B (set that is not well ordered). But is PTI true in my domino example? Is there a way to show that PTI may not be true in domino example? It seems to me that domino example covers all cases.

Answer 1

Every linearly ordered set has a largest well-ordered initial segment (which may as well be empty). Call this segment $S$, and consider the property $x\in S$.

Take any point outside of $S$, if such point exists, $y$, then it holds for every $x<y$ $$\forall u(u<x\to u\in S)\to x\in S,$$ because $S$ is an initial segment, and if $x\notin S$, then there is some $u<x$ such that $u\notin S$ as well. Because $S$ is the largest well-ordered initial segment.

Therefore, if PTI holds, $S$ must be everything, and so well-ordered.

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Of course, this talks about linear orders. We can talk about partial orders, and the same idea works out with a well-founded downwards-closed subset. So we get that well-foundedness to be equivalent to having a recursion property.

Answer 2

The induction principle for well-ordered sets is the following theorem: Let $(X,<)$ be a well-ordered set and let $P(x)$ be a statement pertaining to elements $x$ of $X$. (It can be itentified with the subset $P(X):=\{ x \in X \ | \ P(x)\}$ of $X$). Assume that the following statement holds:

$\forall x(\forall y(y<x\Longrightarrow P(y)) \Longrightarrow P(x))$

Then $\forall z P(z)$ holds. In other words, if $P(x)$ holds for $x$ whenever $P(y)$ holds for all $y<x$, then $P(z)$ holds for all $z$.

Consider a linearly ordered set $(X,<)$ which is not well-ordered. So there is a non-empty subset $Y \subseteq X$ which has no minimum. Consider the statement $P(x): \forall y \in Y,x<y$. Let $x \in X$ such that for all $z\in X$ we have $z<x\Longrightarrow P(z)$. I claim that $P(x)$ holds. Indeed assume for contradiction that there is $y \in Y$ with $y\leq x$. Since $Y$ has no minimum, we may assume that $y<x$. So $P(y)$ holds by our hypothesis. In particular we have $y<y$: a contradiction. We deduce that $P(x)$ holds.

But of course the statement $\forall z P(z)$ is false since $P(y)$ is false when $y \in Y$ and $Y$ is non-empty. So the induction principle fails for $(X,<)$.