Let S be a non-empty subset of the real numbers that is bounded below. Define −S
by −S = {− x | x ∈S }. Prove: (i) −S is bounded above. (ii) sup(−S) = − inf S .
To prove that −S is bounded above, we need to show that there exists a real number M such that −x ≤ M for all x in −S.
Since S is a non-empty subset of the real numbers, it is bounded below. Therefore, there exists a real number L such that L ≤ x for all x in S.
Now, let's consider any element y in −S. By the definition of −S, y is of the form −x for some x in S.
Since L ≤ x for all x in S, it follows that −L ≥ −x for all x in S. In other words, −L is an upper bound for −S.
Therefore, we have shown that −S is bounded above.
Now, let's prove that sup(−S) = −inf(S).
To show that the supremum of −S is −inf(S), we need to verify two properties:
1. For any y in −S, −inf(S) ≤ y:
Let y ∈ −S. By definition, y is of the form −x for some x in S. Since inf(S) is the greatest lower bound of S, we have inf(S) ≤ x for all x in S. Therefore, it follows that −inf(S) ≥ −x for all x in S, which implies −inf(S) is less than or equal to y. Thus, −inf(S) is a lower bound for −S.
2. For any ε > 0, there exists y in −S such that −inf(S) + ε > y:
Let ε > 0 be given. Since inf(S) is the greatest lower bound of S, there must exist x0 in S such that inf(S) ≤ x0 < inf(S) + ε. Since −x0 is in −S, it follows that −inf(S) + ε > −x0, satisfying the condition.
By satisfying these two properties, we have shown that −inf(S) ≤ y for all y in −S, making it the least upper bound (supremum) of −S.
Therefore, we can conclude that sup(−S) = −inf(S).