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In my text book the proof of well ordering principle goes like this:

Let A be an arbitrarily given nonempty set which is to be well-ordered. Consider the family A* of all well-ordered sets ($A_0$, $\le _0$), where $A_0 \subseteq A$. We partially order A* by writing ($A_0, \subseteq _0) \subseteq ^*(A_1, \subseteq _1)$ if and only if

(i) $A_0 \subseteq A_1$

(ii) x, y $\in A_0$ and x $\le _0y$ imply x $\subseteq _1 y$

(iii) x $\in A_1 - A_0$ implies y $\le _1 x$ for all y $\in A_0$


The proof goes like this but they say nothing about whether A* can be empty or not. Shouldn't they first proof that A* is nonempty(well-defined?)? If it's okay not to show that, why is it so?

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up vote 1 down vote accepted

Technically it should be mentioned that $A^*\ne\varnothing$, but this is obvious, since $\langle\varnothing,\varnothing\rangle\in A^*$. For that matter, $\big\langle\{a\},\{\langle a,a\rangle\}\big\rangle\in A^*$ for all $a\in A$.

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can an empty set also be well-ordered? – mingyu Feb 16 '13 at 1:43
@mingyu: The empty relation is a partial order of the empty set; and clearly it is a well-ordering as well. Every non-empty set has a minimal element! – Asaf Karagila Feb 16 '13 at 1:46

Singletons, and finite sets, can always be well-ordered. So $A^*$ is never empty.

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Yes, it is required that $A^*$ be non-empty, so this indeed needs to be verified, which in this case is trivial. Quite often proofs with Zorn's Lemma tend to follow a very routine trajectory of lots of details to check, so sometimes the first (usually most trivial step) of checking non-emptiness is forgotten.

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