infinium and supremum question

**lllll** · September 19th 2008, 09:23 PM

I'm a little stuck on this question:

Let S be a nonempty bounded subset of $\mathbb{R}$ and let $b<0$ . Show that $inf(bS) = bsup(S)$

so far I have:

$\beta = inf(S) \ \therefore \ b\beta = inf(bS) \ \forall \ s \in S \ , \ \beta \leq s$

$b\beta \leq bS \ \forall \ s \in S \ \rightarrow -n\beta \geq -nS$

therefore $b\beta$ is a upper bound of $bS$ .

That's what I have so far.

**bkarpuz** · September 20th 2008, 07:56 AM

Originally Posted by lllll

I'm a little stuck on this question:

Let S be a nonempty bounded subset of $\mathbb{R}$ and let $b<0$ . Show that $inf(bS) = bsup(S)$

so far I have:

$\beta = inf(S) \ \therefore \ b\beta = inf(bS) \ \forall \ s \in S \ , \ \beta \leq s$

$b\beta \leq bS \ \forall \ s \in S \ \rightarrow -n\beta \geq -nS$

therefore $b\beta$ is a upper bound of $bS$ .

That's what I have so far.

Let $S\subset\mathbb{R}$ be bounded, and set $\beta :=\inf\{S\}$ .
Then $\beta$ is finite, therefore, we know that
$t\geq\beta \text{ for all }t\in S.\qquad(*)$
To complete the proof, we have to show that $\beta b=\sup\{bS\}$ holds, where $b<0$ .
Note that $bS=\{bt:t\in S\}$ .
Multiplying both sides of (1) with $b$ , we get
$bt\leq b\beta \text{ for all }t\in S,$
which is equivalent to
$bt\leq b\beta \text{ for all }bt\in bS.$
This proves that $b\beta=\sup\{bS\}$ , which is the desired identity.

You may wish to check the following book:
Introduction to Calculus and ... - Google Book Search

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