most recent 30 from http://mathoverflow.net 2013-05-19T03:23:04Z http://mathoverflow.net/feeds/question/30746 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/30746/a-version-of-the-zorn-lemma Alexandre Borovik 2010-07-06T09:01:30Z 2010-07-06T09:28:49Z

<p>Let $S$ be a linearly ordered set. A pair $(X,Y)$ of subsets $X$, $Y$ of $S$ is called a pre-cut if $x &lt; y$ (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 $Y \subseteq V$. </p> <p>The following property easily follows from the Zorn Lemma:</p> <p>(*) Every pre-cut is contained in a maximal pre-cut.</p> <p>Is the reverse true, that the validity of property (*) for all linearly ordered sets implies (in ZF) the Zorn Lemma?</p>

http://mathoverflow.net/questions/30746/a-version-of-the-zorn-lemma/30747#30747 Michael Greinecker 2010-07-06T09:23:07Z 2010-07-06T09:28:49Z

<p>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 &lt; x &lt; 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.</p>