How do I prove that the inf(S) = -sup(-S) where -S={-s l s in S}?

I thought the inf is the greatest lower bound and hence it should be sup (-S) ?

First, I note you have posted several questions on the "Number Theory" forum which have nothing to do with "Number Theory". Questions involving limits, inf, sup, etc. should be under "Calculus" or "Analysis".

Second, suppose S= {x| x> 1}. Obviously, its inf is 1. Now, -S= {-x| x> 1}= {y| y< -1} (I multiplied both sides of x> -1 by -1 and then let y= -x). It should be obvious that sup -S= -1. That is, inf S= 1= - sup -S, not "sup -S".

Now, as to the proof. Let a= inf(S). Then a is the "greatest lower bound" on S. Since a is a lower bound, if x is in S then \(\displaystyle x\ge a\) and, since a is the

**greatest** lower bound, if b is any lower bound on S, \(\displaystyle a\ge b\).

Let y be any member of -S. Then x= -y is in S and so \(\displaystyle x= -y\ge a\). Multiplying both sides of that by -1, \(\displaystyle -x= y\le -a\) so -a is an upper bound on S.

To show it is the

**least** upper bound, suppose b is any upper bound on -S. That is, if y is in -S, \(\displaystyle b\ge y\). By definition of -S, For any x in S, x= -y for some y in -S. Since \(\displaystyle b\ge y\), \(\displaystyle -b\le -y= x\). Since x could be any member of S, -b is a lower bound on S. Since a is the

**greatest** lower bound on S, \(\displaystyle a\ge -b\). Now, multiply both sides of that by -1 to get \(\displaystyle -a\le b\). That shows that -a is the

**least** upper bound on -S, so inf(S)= -sup(-S).