# Answer on Real Analysis Question for Junel

Question #4228

Let S be a nonempty bounded set in R.

a.) let a>0, and let aS := (as:s element S). prove that, inf (aS)= a inf S, and sup (aS)= a sup S

b.) let b<0, and let bS := (bs:s element S). prove that, inf (bS)= b sup S, and sup (bS)= b inf S.

a.) let a>0, and let aS := (as:s element S). prove that, inf (aS)= a inf S, and sup (aS)= a sup S

b.) let b<0, and let bS := (bs:s element S). prove that, inf (bS)= b sup S, and sup (bS)= b inf S.

Expert's answer

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

which contradicts to (*).

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)

(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

which contradicts to (*).

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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