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Can we prove or disprove that every set can be well-ordered to have a maximal element in ZF or ZFC?

Or is it the area where different models have different answers to this?

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The empty set cannot "be well-ordered to have a maximal element". $\;$ –  Ricky Demer Mar 18 at 23:23
    
By taking the dual order, this is just the WOP and the answers all well known. –  Git Gud Mar 18 at 23:25
    
@GitGud A well-ordered set whose dual is also well-ordered is finite. Surely you don't mean to say all sets are finite in ZFC? –  Zhen Lin Mar 18 at 23:26
    
@ZhenLin: I'm sure that if you search hard enough, you'll come across someone who would argue that ZFC proves that infinite sets don't exist. (For example, because appealing to the "existence" of an infinite set is the same as appealing to Leprechauns and unicorns...) –  Asaf Karagila Mar 18 at 23:28
    
@ZhenLin I meant you take a well-ordering and you reverse the order, i.e., you take the dual order. Minima become maxima in the dual order. –  Git Gud Mar 18 at 23:31

2 Answers 2

up vote 14 down vote accepted

Take a nonempty set $S$ and pick $x\in S$. Well order $S\backslash\{x\}$. Extend the resulting well-order to $S$ by making $x$ larger than every other element. You get a well-order with a maximum.

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If the set is empty, then it doesn't have any elements, in particular maximal elements. But that's the trivial, and uninteresting case. So let's casually chuck it aside, and assume we're talking about non-empty sets.

Of course that without the axiom of choice we cannot prove that every set can be well-ordered to begin with. So the axiom of choice is necessary here.

Assuming it, though, we have two cases:

  1. The set is finite, in which case it will definitely have a maximal element in any well-order.

  2. The set is infinite, in which case we can put it in bijection with many different ordinals. Many of which have maximal elements.

So the answer is indeed positive.

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Case 0: The set is empty, in which case there is no well-order in which the set will have a maximal element. $\;$ –  Ricky Demer Mar 18 at 23:28
    
True, I'll add that. Thanks. –  Asaf Karagila Mar 18 at 23:29

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