Talk:Principle of Mathematical Induction/Well-Ordered Set

Should this be in Category:Proofs by Induction along with the various principles of induction? --Cynic (talk) 22:13, 9 June 2009 (UTC)

Shrug. If you like. I was never a big fan of categorising results according to the style of proof used to verify it, myself. There might be more than one way to prove something, one way doing it directly, one way indirectly, one way by induction, etc. But feel free to add it there if you want to. --Matt Westwood 05:32, 10 June 2009 (UTC)

Theorem statement
Isn't:
 * $\forall s \in S: \left({\forall t \in S: t \preceq s \implies t \in T}\right) \implies s \in T$

true for any $T \subseteq S$, since $\forall s \in S: s \preceq s$? Did I make a mistake somewhere? --abcxyz 06:33, 28 June 2012 (UTC)


 * I've noticed from going back to the initial source that it should in fact be
 * $\forall s \in S: \left({\forall t \in S: t \prec s \implies t \in T}\right) \implies s \in T$
 * ... I'm going to review this carefully and make sure of it. --prime mover 06:45, 28 June 2012 (UTC)


 * The second statement appears correct; it is similar to other 'order induction' variants I have seen (eg. for induction on subsets of the reals). --Lord_Farin 06:57, 28 June 2012 (UTC)


 * This is taking more time to think about than I was expecting. The initial exposition of this theorem from Devlin has that the smallest element of $S$ is an element of $T$, and uses that in the proof (which is minimal). He then states that "condition 1 is superfluous, since it is a consequence of condition (ii)" (the condition given here). My work in the original page was a (flawed) attempt to prove that. However, yours glosses over the detail by saying "By the rule of transposition (and the definition of a smallest element) $\forall t \in S: t \prec s \implies t \in T$" which I'm not sure I can follow.


 * But it's hot and muggy and I can't think straight (plus I'm a bit fick) so I'm probably talking utter shite as usual. --prime mover 07:20, 28 June 2012 (UTC)