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 and, since a is thegreatestlower bound, if b is any lower bound on S, .

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

To show it is theleastupper bound, suppose b is any upper bound on -S. That is, if y is in -S, . By definition of -S, For any x in S, x= -y for some y in -S. Since , . Since x could be any member of S, -b is a lower bound on S. Since a is thegreatestlower bound on S, . Now, multiply both sides of that by -1 to get . That shows that -a is theleastupper bound on -S, so inf(S)= -sup(-S).