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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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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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    Duh! Right. If $S$ is a well-order then _every_ non-empty set has a minimal element, in particular, $S$ has one.2012-10-24