# Set Theory: Fully ordered but not well ordered set proof

If a nonempty set $A$ is fully ordered but not well ordered, prove there is a sequence of elements such that $a_1 \gt a_2 \gt a_3 \gt ...$

I know the intuition behind this that since $A$ is not well ordered, there does not exist a least element in $A$. And since $A$ is fully ordered, any two elements in $A$ are comparable. But is this enough to prove that there is such a sequence?

How would I write this out mathematically?

Consider a finite set $\{a_i\}$ that is fully ordered. Than, any two $a_1,a_2 \in \{a_i\}$ are comparable. $\implies a_1 \le a_2$ or $a_2 \le a_1$.

$\implies$(?)every subset of $\{a_i\}$ will have a least element. Thus it is well ordered. This shows that if A is fully ordered but not well ordered, then there exists an infinite sequence of the elements.

• Prove that finite sets must be well-ordered if they are fully-ordered, complete the proof. – DHMO Apr 19 '17 at 18:24
• @DHMO I have tried to complete the proof but is the implication (where the question mark is) correct? I don't know if I can deduce that statement from the two elements being comparable. – Suhy Suh Apr 19 '17 at 18:38

$A$ not being well-ordered does not imply $A$ has no least element. For example $[0,1]$ is not well-ordered but does have a least element.
However, since $A$ is not well-ordered, but is linearly-ordered (fully-ordered, totally-ordered, completely-ordered, terminology-de-jour-ordered), there has to be some $B\subset A$ such that $B$ has no least element (if no such set existed, then $A$ would satisfy all requirements to be well-ordered).
So choose an arbitrary $a_1 \in B$. But $a_1$ is not lowest, so pick an arbitrary $a_2 \in \{ b \in B\mid b < a_1\}$. Repeat inductively.