Why restrict the principle of transfinite induction to well-ordered sets?
For an arbitrary poset, there remains only one necessary and sufficient condition to get the transfinite induction working. I suggest to call such posets trans-inductive.
I use the non-reflexive definition of posets, and the interval notations ("*" means, there's no lower bound.)
[*, y] = { x | x<y or x=y } and
[*, y) = { x | x<y }
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Thm. Let (L, <) be a poset. The following are equivalent.
(i) Every totally ordered, nonempty subset S \subseteq L has a minimum.
(ii) If A \subseteq L, such that
for any x in L, if [*, x) \subseteq A, then x in A, Then A = L.
Def. A poset with the above property (i) is called trans-inductive.
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Proof.
(i) => (ii): Assume A != L, but [*, x) in A always implies that x in A. Have a look at B = L-A, and a chain C \subseteq B that's maximal in B.
[*, min(C)) = \emptyset \subseteq A => min(C) in A.
=> contradiction :)
!(i) => !(ii): Assume there's a chain C \subseteq L with no minimum. Let f: (\NN, >) -> (C, <) be injective and order-preserving. Set A = {f(2k) | k in \NN}.
A breaks the rule of (ii).
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See also the discussion on Tree (set theory).
Most topics in PlanetMath are taken from books or papers. Possibly the above is not in any book. I don't know if PlanetMath is the right place to introduce a new concept to mathematics, but I don't know another way. Write a paper?? |
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