For a three element set ${a, b, c}$, are there any partial orders $R$ that $a$ is the only minimal element under $R$ I personally feel like there is no such partial order. But I am not sure.
By reflexive quality of partial order, ${(a, a), (b, b), (c,c)}$ should be included in the partial order. But by the definition of minimal element, $b$ is minimal if $aRb$ implies $a = b$, so it excludes $(b, b)$ and $(c, c)$
Is that right? Or am I misusing some properties and definitions?
 A: Let $A$ be a set and $R$ a partial order on $A$. The definition of a minimal element is:

$m\in\mathbb A$ is a minimal element if $$\forall x\in A: xRm\implies x=m$$

This also means that

$m\in \mathbb A$ is not the minimal element of $A$ if there exists some $x\in A$ such that $x\neq m$ and $xRm$


Let $A=\{a,b,c\}$, and let $$R=\{(a,a), (b,b), (c,c), (a,b), (a,c), (b,c)\}$$
which you can check is a partial order.
Then, we can prove that $a$ is the only minimal element of $A$:
Part 1: $a$ is a minimal element of $A$:
Let $x\in A$, and assume that $xRa$. Therefore, $(x,a)\in R$, however, the only element of $R$ that has $a$ on its second position is $(a,a)$, therefore, $(x,a)=(a,a)$, meaning $x=a$.
Part 2: $a$ is the only minimal element of $A$:
Setting $x=a$ shows us that (1) $xRb$ is true and (2) $x\neq b$, meaning there exists some $x$ for wehich $x\neq m$ and $xRb$, in other words, $b$ is not the minimal element of $A$.
Similarly, setting $x=a$ shows that $c$ is not the minimal element of $A$, meaning that $a$ is the only minimal element of $A$.
