## Return to Answer

2 typo corrected

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 samller 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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Let $(X,\leq)$ be a linearly ordered set and (U,V) a precut. Define (U*,V*) as the precut you get by closing U under samller 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.