# Thread: infinium and supremum question

1. ## infinium and supremum question

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.

2. 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