Consider the following statement:

If $X$ is partially ordered set such that every chain in $X$ has un upper bound, then for every $x \in X$ there is a maximal element $m$ in $X$ such that $x \le m$.

What is the relation, if any, to Zorn's lemma? Is it weaker, stronger or maybe just nonsense?

  • $\begingroup$ Do you mean that there is a larger/equal element to every element in $x$, or that there is a maximal element in $X$? $\endgroup$ – naslundx Mar 1 '14 at 8:41
  • $\begingroup$ For every $x$ in $X$ there is larger/equal element which is maximal in $X$. $\endgroup$ – martin Mar 1 '14 at 8:43
  • $\begingroup$ By "Zorn's lemma" I guess you mean the assertion that every partially ordered set has a maximal chain? $\endgroup$ – bof Mar 1 '14 at 8:45
  • $\begingroup$ Zorn's Lemma: If $X$ is partially ordered set such that every chain in $X$ has an upper bound, then there is a maximal element in $X$. $\endgroup$ – martin Mar 1 '14 at 8:47

The statement is equivalent to Zorn's lemma.

It implies Zorn's lemma quite easily, because they have the same requirements from the partial order, and if there is a maximal element above each point, then there's certainly a maximal element.

On the other hand, assuming Zorn's lemma, and given a partial order $(P,\leq)$ satisfying these requirement, consider for $x$ in the partial order the set $P_x=\{m\in P\mid x\leq m\}$, then the restriction of $\leq$ to $P_x$ satisfies Zorn's lemma again, and therefore it has a maximal element, $m$ which is maximal in $P$, and so $x\leq m$.

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  • $\begingroup$ Very interesting, thank you. Small typo: in the last line should be $P_x$. $\endgroup$ – martin Mar 1 '14 at 8:57
  • $\begingroup$ No, it should be $P$. We want a maximal element in $P$; Zorn's lemma gives us $m$ as a maximal element of $P_x$, but it's also maximal in $P$. $\endgroup$ – Asaf Karagila Mar 1 '14 at 8:58
  • $\begingroup$ So in the definition of $P_x$, $m$ is also maximal in $P$ and not just larger than $x$. Am I right? $\endgroup$ – martin Mar 1 '14 at 9:01
  • $\begingroup$ That is correct. $\endgroup$ – Asaf Karagila Mar 1 '14 at 9:11

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