Question #11860

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

Expert's answer

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).

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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