# Correctness of proof that an ordered field S that has the supremum property also has the infimum property

First question I have is how would you describe the relationship between an ordered field and an ordered set and continue the proof by treating the field as a set? I want to say that right in the beginning to make things clear but I don't know how to say it.

Here's the rest of the proof:

Let A be a nonempty subset of S s.t. it is bounded above and D be the set of all lower bounds for A.
Fix a to be elements of A. Then for each d in subset D d is a lower bound for A so d $\leq$ a for all a in A.

Since d is bounded above by A it has the supremum $\gamma$ . Since every a is an upper bound for d, $\gamma \leq a$ for each element a in subset A. So $\gamma$ is a lower bound for A and is clearly an infimum since it is an upper bound for the set of lower bounds.

• So far no progress towards a proof. Use the fact that $x\le y$ if and only if $-y\le -x$. – André Nicolas Sep 20 '14 at 17:56
• @AndréNicolas: This looks like progress to me, if you change "bounded above" in the third paragraph to "bonded below". matsmv is saying, the set $D$ of lower bounds of $A$ is bounded above, so it has a supremum; and this supremum of $D$ is the infimum of $A$. – TonyK Sep 20 '14 at 17:59
• I was wrong in my comment above, which I am tempted to erase but won't. – André Nicolas Sep 20 '14 at 18:31

This argument is on the right track but the assumption that $A$ is bounded above is irrelevant. You need to assume $A$ is bounded below, so that the set $D$ of all lower bounds of $A$ is non-empty. Then you can apply the supremum property to the set $D$.