Dual of Zorn's Lemma? [closed]

It seems to me that the dual of Zorn's Lemma should be true: if $S$ is a non-empty partially ordered set and every chain of $S$ has a lower bound in $S$, then $S$ has at least one minimal element.

Zorn's Lemma comes up quite often in Commutative Algebra, but I don't recall ever seeing a dual of it or any applications of a dual. Are there any interesting applications of the dual, assuming it is true?

P.S. I am not a logician.

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closed as not a real question by Bill Johnson, Joel David Hamkins, Pete L. Clark, Yemon Choi, fedjaSep 15 '10 at 3:42

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

The principle is equivalent to Zorn's lemma, if you just turn the order upside-down. –  Joel David Hamkins Sep 15 '10 at 1:23
This is too easy to be a homework problem. I vote to close. –  Bill Johnson Sep 15 '10 at 1:46
Hannay, in my opinion this sort of question belongs on math.stackexchange.com and not here. –  Asaf Karagila Sep 15 '10 at 1:47
Maybe a more interesting version of the question is: If we take the axiom of choice to mean "Every epi has a right inverse," what can be said of the "axiom", "Every monic has a left inverse"? Of course I haven't thought about this at all, but my guess is that this would fail in a lot of topoi? –  Dylan Wilson Sep 15 '10 at 2:22
@Dylan Wilson: I agree that that is a more interesting "dual to axiom of choice". But I disagree that it is a "version of the question" that OP asked. –  Theo Johnson-Freyd Sep 15 '10 at 5:45

If your poset is $\langle S,<\rangle$ and it has the property that every chain has a lower bound, define the order $R$ on $S$ which is: $aRb \iff b< a$.
It is easy to see that every chain has an upper bound, satisfying Zorn's lemma and that the maximal element in $R$ is the minimal element in the original order.