Problem
Prove that a set $R$ of real numbers is well ordered iff there is no infinite decreasing sequence of numbers $R$. In other words, there is no set of numbers $r_i \in R$ such that
$r_0 > r_1 > r_2 > \cdots$
(Note: the textbook says, verbatim, "iff there is no infinite decreasing sequence of numbers $R$", but I'm assuming there is an "in" missing, and it should be "iff there is no infinite decreasing sequence of numbers in $R$".)
Solution
To prove that a set $R$ of real numbers is well ordered iff (if and only if) there is no infinite decreasing sequence of numbers in $R$, it's necessary to prove both directions of the "if and only if":
1) If a set $R$ of real numbers is well ordered, then there is no infinite decreasing sequence of numbers in $R$:
Proof. If a set $R$ of real numbers is well ordered, then every non-empty subset of $R$ has a minimum element.
Therefore, there is no infinite decreasing sequence of numbers in $R$, because that would be a subset without a minimum element.
2) If there is no infinite decreasing sequence of numbers in a set $R$ of real numbers, then $R$ is well ordered:
Proof. If there is no infinite decreasing sequence of numbers in $R$, then every sequence of numbers in $R$ has a smallest element. Therefore, $R$ is well ordered.
Therefore, $R$ is well ordered iff there is no infinite decreasing
sequence of numbers $R$.
Is this proof complete?
Best Answer
One direction is easy: a (strictly) decreasing sequence $(x_n)$ defines a subset $A=\{x_n: n \in \Bbb N\}$ that has no minimum (because if $x_n \in A$ were a minimum, $x_{n+1} < x_n$ contradicting the minimality).
The other direction needs a bit more care: If we have any non-empty set $A$ without a minimum, then use recursion (plus dependent choice): define $x_0 \in A$ as you want (it's non-empty after all). Note that having defined $x_n$ for some $n$, by assumption it is not $\min(A)$ so there is some $a \in A$ such that $a < x_n$, and then define $x_{n+1} = a$ and we continue the recursion. This defines a strictly decreasing sequence in $A$ (hence in $\Bbb R$) and we have a contradiction. So we have a well-order.
Note that we only use we're in a linear order and nothing specific about $\Bbb R$ e.g. This fact holds in all linear orders: well order iff no decreasing sequence (if you believe in dependent choice).