Relations on Sets

Question

Let $A \neq \emptyset$ and $O$ an antisymmetric relation on $A$. Suppose that for every $B \subseteq A$, $B \neq \emptyset$, $\exists$ least $B$ i.e. $\exists b \in B$ s.t $\forall c \in B$ $bOc$. Now I want to proof that $O$ is a total order in $A$. I haven't really made any progress in this question, I know that for being a total order, a relation $R$ has to be reflexive, transitive, antisymmetric and total. I'm given $O$ is antisymmetric, but from the least element existence I don't know how to proceed.

Answer 1

• Take $x,y\in A$. Then by considering $B=\{x,y\}$ and applying the hypothetis, we have that $x$ or $y$ is a minimum. Let's suppose it's $x$. Then $xOy$. And if it's $y$, then $yOx$. So the relation is total.
• Take $x\in A$. Then by considering $B=\{x\}$ we have that $x$ is the miminum : $xOx$. So the relation is reflexive.
• Take $x,y,z\in A$ such that $xOy$ and $yOz$. By considering $B=\{x,y,z\}$, we have that the minimum is $x,y$ or $z$. It can't be $z$ since $yOz$, and it can't be $y$ since $xOy$. So it's $x$. So $xOz$.