Verified answer

Recall that if x = sup T, then (1) x ≥ t for every t in T and (2) if t_1 < x, there exists t in T such that t ≥ t_1.

This says that (1) x is an upper bound and (2) it is the least of all possible upper bounds.

Similarly, x = inf T provided (1) x ≤ t for every t in T and (2) if t_1 > x, then there exists t in T such that t ≤ t_1.

This says that (1) x is a lower bound and (2) it is the greatest of all lower bounds.

Okay, on to your statement.

Proof: Let x = -sup{-s| s ∈ S}. Then note that

(1) - x ≥ -s for all s ∈ S

hence x ≤ s for all s ∈ S showing that x is a lower bound of S.

Second, suppose that t > x. To see that t is not a lower bound of S note that -t < -x. Hence -t is not an upper bound of {-s | s ∈ S}. It follows that there exists s ∈ S such that -t < -s. Then

(2) t > s

so that t is not a lower bound of S. It follows that x is the greatest lower bound of S--i.e.

inf(S) = x = - sup{-s| s ∈ S}.

info: For any x in A we've that inf A <= x <= sup A. info: If a is an bigger certain of A and b is a decrease certain of B, then sup A <= a and b <= inf A. permit -S = {-s | s is in S}. word that inf S <= x for all x in S. Then -inf S => -x. Then -inf S is an bigger certain for -S. consequently, sup -S <= -inf S. Multiply the two facets by technique of -a million yields inf S <= -sup -S. word that sup -S => -x for all x in S. Then -sup -S <= x. Then -sup -S is a decrease certain of S. Then -sup -S <= inf S. when you consider that inf S <= -sup -S and -sup -S <= inf S, we end that inf S = -sup -S.