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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.

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    $\begingroup$ So far no progress towards a proof. Use the fact that $x\le y$ if and only if $-y\le -x$. $\endgroup$ Commented Sep 20, 2014 at 17:56
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    $\begingroup$ @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 $A$. $\endgroup$
    – TonyK
    Commented Sep 20, 2014 at 17:59
  • $\begingroup$ I was wrong in my comment above, which I am tempted to erase but won't. $\endgroup$ Commented Sep 20, 2014 at 18:31

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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 $D$.

(André Nicolas is usually right, but he is too pessimistic this time.)

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  • $\begingroup$ I think the idea was right; "bounded above" looks like a typo for "bounded below". $\endgroup$
    – TonyK
    Commented Sep 20, 2014 at 18:01
  • $\begingroup$ So how would you connect the relationship between the ordered set and the ordered field? And how will the supremum property be assumed if we assume A is bounded below? $\endgroup$
    – cambelot
    Commented Sep 20, 2014 at 18:23
  • $\begingroup$ Oh. To echo André's comment: I was wrong in my comment above, which I am tempted to erase but won't. $\endgroup$
    – TonyK
    Commented Sep 20, 2014 at 18:34
  • $\begingroup$ Is the relationship as simple as: F is an ordered set, which is also a field? If so, how would you edit this proof? $\endgroup$
    – cambelot
    Commented Sep 20, 2014 at 18:49
  • $\begingroup$ @matsmv: What you are trying to prove is that every non-empty set that is bounded below has a greatest lower bound, given that every non-empty set that is bounded above has a smallest upper bound. So you need to assume a set is non-empty and bounded below, and then prove that it has a greatest lower bound. The way you do that involves looking at the smallest upper bound of the set of all lower bounds. $\endgroup$ Commented Sep 20, 2014 at 22:37

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