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Does every non-empty set have a well ordering with greatest element?

It is well known that every set has a well ordering. But can we also assume that this well ordering has greatest element?

[Edited to remove an ambiguity revealed by the answers and comments]

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up vote 9 down vote accepted

Yes. Let $\langle S,\le\rangle$ be a well-order. If $S$ already has a greatest element with respect to $\le$, we’re done. If not, let $s_0$ be the $\le$-least element of $S$, and define a new well-ordering $\preceq$ of $S$ as follows:

  • if $s,t\in S\setminus\{s_0\}$, then $s\preceq t$ iff $s\le t$;
  • if $s\in S$, then $s\preceq s_0$.

This produces a well-ordering of $S$ with $s_0$ as the largest element: it simply moves $s_0$ from the beginning of the order to the end.

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@Pete: That $S\ne\varnothing$ is implicit in the question. – Brian M. Scott Jul 13 '12 at 21:06
If $S = \mathbb Z$, how do you pick $s_0$? – Rudy the Reindeer Oct 24 '12 at 6:57
@MattN.: If $S=\Bbb Z$, you don’t have a well-order to begin with. – Brian M. Scott Oct 24 '12 at 11:13
Duh! Right. If $S$ is a well-order then every non-empty set has a minimal element, in particular, $S$ has one. – Rudy the Reindeer Oct 24 '12 at 11:19

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