MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question

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.

share|cite|improve this answer
Thanks! As an algebraist, I have a habit to apply the Zorn Lemma automatically, as soon as it formally applicable ... A good lesson. – Alexandre Borovik Jul 8 '10 at 7:02

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.