Answer to Question #11860 in Real Analysis for sneha tambe
2012-07-13T11:57:20-04:00
let S be a non-empty bounded set in real numbers a) let a>0 , and let aS:={as; s belongs to S} prove that inf(aS)=a inf S, sup(aS)= a sup S
1
2012-07-19T07:36:15-0400
Let s belongs to S, so inf(S) <= s <= sup(S), whence a*inf(S) <= a*s <= a*sup(S), and therefore a*inf(S) <= inf(aS) and sup(aS) <= a*sup(S). We should prove the inverse inequalities: a*inf(S) >= inf(aS) and sup(aS) >= a*sup(S). 1) Suppose a*inf(S) < inf(aS), so inf(S) < inf(aS) / a Therefore there exists s from S such that inf(S) <= s < inf(aS) / a, whence (*) a*s < inf(aS). But a*s belongs to aS, and so the inequality (*) is impossible. Hence a*inf(S) = inf(aS). 2) The proof that sup(aS) >= a*sup(S) is similar. Suppose sup(aS) < a*sup(S), so sup(aS)/a < sup(S) Therefore there exists s from S such that sup(aS)/a < s <= sup(S) whence (**) sup(aS) < a*s. But a*s belongs to aS, and so the inequality (**) is impossible. Hence sup(aS) = a*sup(S).
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