# Answer to Question #4228 in Real Analysis for Junel

Question #4228
Let S be a nonempty bounded set in R. a.) let a&gt;0, and let aS := (as:s element S). prove that, inf (aS)= a inf S, and sup (aS)= a sup S b.) let b&lt;0, and let bS := (bs:s element S). prove that, inf (bS)= b sup S, and sup (bS)= b inf S.
First we show that

(1)& sup(aS)= a sup(S)

(2)& inf(-S)=
-sup(S).

and then deduce from these statement all
others.

Proof of (1).
By definition x = sup T
if x>=t
for any& t from T,
and for any p>0 there exists& t'& from T such that
t' > x-t

First we show that sup(aS) <= a sup(S)
Indeed, since&

s <= sup(S)
for all s from S, then
& as <= a
sup(S),
whence
& sup (aS) <= a sup(S),

To prove the converse
inqeuality& sup(aS) >= a sup(S) suppose that&
sup(aS) < a
sup(S)
Then there exists b such that
(*) sup(aS) < ab < a
sup(S)
But this means that s < b for all s from S, and so sup(S) <=
b.
Hence
& a sup(S) <= ab
Hence
sup(aS) = a sup(S).

Proof of (2).
Since
& s <= sup(S)
for
all s from S, it follows that
& -s >= -sup(S),
whence

inf(-S)>=-sup(S).

Conversely, suppose that inf(-S) > -sup(S), so
there exists x such that
(**) inf(-S) > x > -sup(S).

Then the
first part of this inequality inf(-S) > x
implies that&
-s
> x
for all s from S, so
s < -x,
whence sup(S)<= -x, and
therefore -sup(S)>= x, which contradicts to
(**).

================
Now we can prove that

(3)& inf(aS) = a
inf(S).

Indeed:
& inf(aS) = inf( - (-aS) )& = by (2) =
=
-sup(-aS)
= -sup(a*(-S)) = by (1) =
= - a *
sup(-S) = by (2) =
= a * inf(S)

Let b<0. Let us
prove that
(4) inf (bS)= b sup S
(5) sup (bS)= b inf S

Proof
of (4).
Since -b>0 we obtain
& inf(bS) = inf (-b * (-S) ) = by
(3)
= -b * inf(-S) = = by (2)
= -b * (-sup(S))
=
= b * sup(S).

Proof of (5).
Again
&
sup(bS) = sup (-b * (-S) ) = by (1)
= -b * sup(-S) =
= by (4)
= -b * inf(-S) = = by (2)
= -b *
(-sup(S)) =
= b * sup(S)

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