Proving well ordering is total relation

Question

Assuming R is well-ordered relation on a set A. Wikipedia claims that every well ordered relation is also a total one.Even tough in first sight it seems trivial, how can I prove it?

Answer 1

For the sake of contradiction, take an antichain $C \subseteq A$ in the well-ordered set $(A, \leq)$, that is a set of pairwise non-comparable elements ($\forall x,y \in C, x \not \leq y \text{ and } y \not \leq x$). Thus it contains many minimal but no smallest element. On the other hand $C$ contains a smallest element as $\leq$ is a well-order. This is a contradiction.

Definition (minimal, smallest element)

If $(A, \leq)$ is a partially ordered set, $X$ is a nonempty subset of $A$, and $a \in A$, then

• $a$ is a minimal element of $X$ if $a \in X$ and $\forall x \in X, x \not \leq a$;
• $a$ is the smallest element of $X$ if $a \in X$ and $\forall x \in X, a \leq x \lor a = x$.

Answer 2

It's simple, you only need to take a pair $\{a,b\}$ of a well-ordered set such that $a \neq b$ , now this pair has a minimum element, so either $a < b$ or $b < a$.