※ 引述《poacherking (阿堡)》之銘言：
: 想請問大大
: SUP(-A)=-INF(A)
: 好像很值觀
: 但是卻不知證明如何下手
: 請鞭小力一點
You don't tell us what is the set "A."
So, let me guess the set A is a nonempty set of real numbers which is
bounded below. (Because R is a complete Archimedean ordered field, we
can discuss this problem by using some good properties.)
Let's begin...
Let -A = {-x: x in A}.
A is bounded below => there exists m in R s.t. m ≦ x, for all x in A.
=> -m ≧ -x, for all x in A.
=> -A is bounded above.
=> sup(-A) exists. (Let -s = sup(-A))
Then,
(1): -x ≦ -s => x ≧ s, for all x in A.
=> s is a lower bound of A.
(2): n ≦ x, for all x in A. (let n be any lower bound of A)
=> -n ≧ -x, for all x in A
=> -n is an upper bound of -A.
=> -n ≧ -s, since -s = sup(-A).
=> n ≦ s.
By (1) and (2), s = inf(A).
Hence, inf(A) = - sup(-A) <=> -inf(A) = sup (-A).
Q.E.D.
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