I found a statement said that the proposition "Mathematical induction can be applied on any total order set" is False.
(The place I found the statement might not be believable.)
This means that there is a counterexample where cannot the math induction be applied.
I didn't find out much information about the relation of Mathematical induction and total-ordered set. So I guess somewhere I thought wrong.
I have no robust concept about them though, however I thought that any total-ordered set can be applied math induction, stated with my naive intuition below:
Because Mathematical induction is based on well-ordering principle, which is applied on $\mathbb{Z^+}$, I'd consider that any total-ordered set is isomorphic to $\mathbb{Z^+}$. By applying a topological sort on the total-ordered set, we can get a chain. The chain is actually the longest path of the total ordered set, so it preserves the order property. The isomorphism $f$ can be defined by: For all elements $x$ in the total-ordered set, $f(x)=$ " $x$'s order number in the chain".
Hence, I "guess" that math induction can be applied on the total-ordered set,
based on the assumption that "If any set is isomorphic to $\mathbb{Z^+}$, and that isomorphism preserves the order information, then math induction can be applied on it", which I cannot tell is correct or not.
Thanks for any hint and correction!
Best Answer
One very general form of induction is well-founded induction. Suppose $\le$ well-founds $S$. Since any non-empty subset of $S$ has a $\lt$-minimal element, contrapositively $$(\forall x\in S(x\lt y\to\phi(x))\to\phi(y))\to\forall y\in S(\phi(y)).$$
One can't generalise this to total ordering, which doesn't guarantee an analogous property of $S$'s non-empty subsets.
However, one can sometimes induct without knowing how to well-found a set. For example, real induction relies on the fact that subsets of $\Bbb R$ have infima and suprema.