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Let $S$ be a linearly ordered set. A pair $(X,Y)$ of subsets $X$, $Y$ of $S$ is called a pre-cut if $x < y$ (strict inequality) for all $x \in X$ and $y \in Y$. Pre-cuts are naturally ordered: $(X,Y) \le (U,V)$ if $X \subseteq U$ and $Y \subseteq V$.

The following property easily follows from the Zorn Lemma:

(*) Every pre-cut is contained in a maximal pre-cut.

Is the reverse true, that the validity of property (*) for all linearly ordered sets implies (in ZF) the Zorn Lemma?

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Let $(X,\leq)$ be a linearly ordered set and (U,V) a precut in X. Define (U*,V*) as the precut you get by closing U under smaller elements and V under larger elements. If (U*,V*) cover X, we are done. Otherwise, there exists x in X such that u < x < v for all u in U* and v in V*. Then (U*',V*'), given by U*'={u:u$\leq$ x} and V*'={v: v> x} is a maximal precut containing (U,V). We don't need the axiom of choice for this argument, so no, this is not equivalent to Zorns lemma.

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  • $\begingroup$ Thanks! As an algebraist, I have a habit to apply the Zorn Lemma automatically, as soon as it formally applicable ... A good lesson. $\endgroup$
    – Alexandre Borovik
    Commented Jul 8, 2010 at 7:02

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