Is this lemma equivalent to the axiom of choice?

Question

Given any pre-ordering $\preceq$ of an arbitrary set $X$ is the following lemma:

$$\text{There exists an inclusion minimal set }S\text{ satisfying }\{a\preceq b:b\in S\}=X\\\iff \text{ Every chain in }(X,\preceq)\text{ has a lower bound}$$

Equivalent to the axiom of choice? Also assuming the axiom of choice if every chain in $(X,\preceq)$ does have a lower bound, then is it true that for every set $Q\subseteq X$ we have:

$$Q\text{ is an inclusion minimal set satisfying }\{a\preceq b:b\in Q\}=X\\\iff Q\text{ is a maximal anti-chain of minimal elements in }(X,\preceq)$$

Sorry if this is elementary, I found several mistakes in some stuff I typed while writing out an analog of Dilworth's theorem for pre-orders and just want to be sure its correct before I write it again formally.

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To clarify what I mean by chains and anti-chains in $(X,\preceq)$, if we define an undirected graph $G$ such that $V(G)=X$ and $E(G)=\{\{u,v\}\subseteq X:u\leq v\}$ then for any sets $C,A\subseteq X$ we can write:

$$C\text{ is a chain of }(X,\preceq)\iff C\text{ is a clique of }G\\\iff \text{Every pair of elements in }C\text{ is adjacent in }G$$ $$A\text{ is an anti-chain of }(X,\preceq)\iff A\text{ is an independent set of }G\\\iff \text{No pair of elements in }C\text{ is adjacent in }G$$

Also by a minimal element in $(X,\preceq)$ I mean any $m\in X$ satisfying $\forall x\in X(x\leq m\implies m\leq x)$.

Answer 1

You seem to flip some directions of the order, take $\Bbb N$, every chain has a lower bound, but there is no set of maximal elements. You also want to quantify over all $X$, in the sense that choice is equivalent to that for any preordered $X$ etc.

Now, the answer is simple, yes. This is equivalent to choice. Simply note that every partially ordered set is preordered, and every partially ordered set with a minimal set of "generators" must have that all those "generators" are maximal elements. So we get the following version of Zorn's lemma, which is of course equivalent to the standard one:

(Zorn$^+$) Suppose $X$ is a partially ordered set where every chain has an upper bound. Every $x\in X$ lies beneath a maximal element.

So your lemma implies Zorn's pretty obviously. The other direction is also simple: move from the preordered set to its antisymmetric quotient, then use choice to choose representatives for the maximal equivalences classes in the quotient order.

To your second question, which we used in the above proof, yes. If $Q$ is an inclusion minimal set, then no two elements in it can be comparable, otherwise you could omit one of them. And they have to be maximal (or minimal, if you reverse your comparison), otherwise you won't span everything.