Give either a proof or a counter example to justify your answer.
It's intuitive to think that if a partial order has exactly one minimal element,then that element must be the smallest element.
However, the back of the book says otherwise.
Can someone provide some counter examples to solidify the idea.
Consider for instance $A := \mathbb{Z} \cup \{ a, b\}$, where we order the elements from $\mathbb{Z}$ in the usual way, and we define $a < b$, thus creating a partial order on $A$. Because there is no $x \in A$ such that $x < a$, $a$ is a minimal element, but not the smallest element because $a < x$ is not satisfied for all $x \in A$. After all, $a < x$ is not defined for any $x \in \mathbb{Z}$. Clearly there is no minimal element in $\mathbb{Z}$, so the condition is satisfied.
One doesn't have to use $\mathbb{Z}$ in particular, but what made it easy is the fact that it is an infinite set (in the sense that I couldn't think of any counterexamples for finite sets). The reason why I used $\mathbb{Z}$ is just because it comes with a default order according to which it has no minimal element (or smallest, for that matter).
It would be true in a finite set.
In fact, suppose $P$ is a partially ordered set with exactly one minimal element $x$, but $x$ is not a smallest element. Then the set $S$ of $s \in P$ which are not comparable to $x$ is nonempty. If $S$ has a minimal element $y$, it can't be a minimal element of $P$, so there is some $z \in P\backslash S$ with $z < y$. But then $x \le z < y$, contradiction. So $S$ has no minimal element, which implies there is an infinite descending chain $s_1 > s_2 > s_3 > \ldots$ in $S$.
