Zorn's Lemma: Why maximal and not minimal?

Zorn's Lemma ensures that every inductively ordered set has a maximal element. There is an obvious duality between the notion of a maximal and minimal element.

Question 1: Given this duality, could we also formulate a dual Lemma that gives the existence of a minimal element for any inductively ordered set?

Question 2: I suppose this fails, since i have never heard or seen of such a Lemma and there might be some good counterexamples. However, how does this failure can be traced back to the equivalence of Zorn's Lemma with the well-ordering principle of the integers? Why does the latter imply that we can say more about maximal elements rather than for minimal elements?

Solution 1:

From every order $\leq$ you can create a dual order: define $a \leq^* b \iff b \leq a$. Then, $a$ is maximal in $\leq^*$ if and only if it is minimal in $\leq$. Thus, everybody formulates the Kuratowski-Zorn lemma for maximal elements, because it can obviously be applied when searching for minimal elements: just consider a dual order.

Solution 2:

Note that the set $( 0 , 1 ]$ is inductively ordered by the usual ordering $\leq$ (inductively ordered taken, as per Planet Math, to mean that every chain has an upper bound). However there is clearly no $\leq$-minimal element in $(0,1]$.

However there is a duality present. Call a partially ordered set $( X , \leq )$ dually-inductively ordered if every chain has a lower bound. Then $( X , \leq )$ is dually-inductively ordered set iff its reverse $( X , \geq )$ is inductively ordered, and $\leq$-minimal elements correspond exactly to $\geq$-maximal elements. Therefore, Zorn's Lemma is equivalent to the statement that every dually-inductively ordered set has a minimal element.