Talk:Zorn's Lemma

Zorn's lemma applies to any partially ordered set, not just ordered ones. I haven't looked at this proof to see if it's being used anywhere. Qedetc 15:02, 7 June 2011 (CDT)


 * The way "ordered set" is used here, it includes "partially ordered set" in its concept. Click on the link to see. For a non-partially ordered set, "totally ordered" is generally used. The clue is in the discussion of "chains" which are less relevant in a totally ordered set because there is only one chain in such. So there is no problem here.


 * The vexed question of "some people understand a term to mean this, some understand it to mean that" is always going to be a problem here, and always going to raise questions which can not be answered to the satisfaction of all. --prime mover 15:09, 7 June 2011 (CDT)

Oh whoops. For some reason I was reading "ordered" as "totally ordered", which isn't even how I'd normally read it.

Qedetc 17:56, 7 June 2011 (CDT)

The last edit on this page either introduced circularity by invoking Hausdorff Maximality (the page for which has 3 proofs, two invoke Zorn's Lemma) or obscured the relation to the axiom of choice (invoked in the 3rd proof of Hausdorff Maximality). The edit also removed a large amount of discussion. I'd agree that the proof currently offered for AoC implies ZL could be shorter / cleaner, but it seems like the edit overreached a bit.

Qedetc 16:14, 17 February 2012 (EST)


 * What he said. Axiom of Choice Implies Zorn's Lemma is straight out of, with a bit of expansion of the points which aren't immediately obvious - so that in itself is a cogent reason enough to keep it. I for one think it is particularly elegant (despite being so long). --prime mover 16:23, 17 February 2012 (EST)