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I found a post concerning this question, but I cannot understand the proof given there of the fact I'm talking about.

The link is : Verification of proof that the empty set is well ordered

I think I can prove indirectly that the empty set is well ordered in the following way :

(1) suppose the empty set is not well ordered

(2) that is, suppose it is false that every non empty subset of the empty set has a first element

(3) it means there exists at least some set S such that

  • S is a non empty subset of the empty set

  • and S has no first element

(4) which requires the first conjunct " S is a non empty subset of the empty set" to be true

(5) but this is impossible, for the empty set has only one subset, which is empty.

However , I cannot manage to give a direct proof of the same fact.

I cannot go further than this : (1) Let S be an arbitrary set (2) Assume it is true that : S is a non empty subset of the empty set (3) Derive from this that : S has a first element. ... but how?

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Apply the definition :

$\emptyset$ is well-ordered if it has a total order and every non-empty subset of $\emptyset$ has a least element in this ordering.

It means :

$\forall S (S \subseteq \emptyset \land S \ne \emptyset \to \ldots)$.

But the only subset of $\emptyset$ is $\emptyset$ itself.

Thus, $S \subseteq \emptyset \land S \ne \emptyset$ is False, and $(S \subseteq \emptyset \land S \ne \emptyset \to \ldots)$ is True, for every $S$.

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You can prove it by exhaustion:

  • for every pair of distinct elements of the empty set, the elements are comparable. This is vacuously true, so that you have a total order (whatever the comparison rule);

  • for every non-empty subset of the empty set, there is a least element. This is vacuously true.

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