Proving if ##b<0\Rightarrow \inf(bS)=b\sup S##

[Potatochip911]

Homework Statement

Let $S$ be a nonempty and bounded subset of $\mathbb{R}$, $bS:=\{bs:s\in S\}$ and $b<0$ show that $\inf(bS)=b\sup S$

Homework Equations

3. The Attempt at a Solution [/B]
I have actually figured out how to do it one way (I will post that way below) but I am trying to figure out how to show this using limits since I can see a lot of similarities between the two methods but I'm struggling with some of the math for the limit version.

1st method: Let $u$ be the supremum of $S$ so $u=\sup S$ this implies that $\forall s\in S$ we have that $u\geq s$, multiplying by a $b<0$ we obtain $bu\leq bs$ so we know that for the set $bS$ that $bu$ is a lower bound which implies $bu\leq \inf(bS)$, i.e. $1) \hspace{3mm} b\sup S\leq \inf(bS)$

Going the other way, let $v$ be the infimum of the set $bS$ so $\forall s\in bS$ we have $v\leq bs$ so $\frac{v}{b}\geq s$ which implies that $\frac{v}{b}$ is an upper bound for the set $S$ i.e. $\forall s\in S\Rightarrow \frac{v}{b}\geq \sup S$ so we obtain $v\leq b\sup S$
$2) \hspace{3mm} \inf(bS)\leq b\sup S$

From $1) \mbox{ and } 2)$ we have that $\inf(bS)=b\sup S$

Now I also wanted to show this using limits, so instead of going the other way I would just use $1)$ and show that $\forall \varepsilon >0$ we have that $\left |bS-b\sup S\right|<\varepsilon$. The problem I run into when trying to show this is that $b<0$ so multiplying by $b$ flips the inequality so I had some questions about this. Since $\left |b\right |=-b$ does this mean that if I start with: For all $\varepsilon_0 >0$ we have that for $s\in S\Rightarrow \left |s-\sup S\right |<\varepsilon_0$. This is where I start to get confused since I want $\left |bS-b\sup S\right |<b\varepsilon_0=\varepsilon$. In order to put $b$ inside the absolute value I think I have to multiply by $-b$ which is in fact $>0$ so the inequality doesn't flip however this suggests an odd (in my opinion) choice of $\varepsilon_0=\frac{\varepsilon}{-b}$ since then $\varepsilon$ must be negative in order for $\varepsilon_0$ to be positive. Mathematically what I'm saying is: $$\left |s-\sup S\right |<\varepsilon_0=\frac{\varepsilon}{-b}\Longrightarrow -b\left |s-\sup S \right |<-b\frac{\varepsilon}{-b}\Longrightarrow \left |bs-b\sup S\right |<\varepsilon$$

Edit: I see now that $\varepsilon$ doesn't have to be negative in order for $\varepsilon_0>0$ since $-b>0$. If someone could just double check that this is indeed correct though I would appreciate it.

 

[Samy_A]

Now I also wanted to show this using limits, so instead of going the other way I would just use $1)$ and show that $\forall \varepsilon >0$ we have that $\left |bS-b\sup S\right|<\varepsilon$.

This is not correct. Maybe it is a typo and you meant $|\inf(bS)-b\sup S|<\varepsilon$.

The problem I run into when trying to show this is that $b<0$ so multiplying by $b$ flips the inequality so I had some questions about this. Since $\left |b\right |=-b$ does this mean that if I start with: For all $\varepsilon_0 >0$ we have that for $s\in S\Rightarrow \left |s-\sup S\right |<\varepsilon_0$.

This is wrong if you mean that this should hold for $\forall s \in S$.

 

[Potatochip911]

This is not correct. Maybe it is a typo and you meant $|\inf(bS)-b\sup S|<\varepsilon$.
This is wrong if you mean that this should hold for $\forall s \in S$.

I actually just tried to prove the wrong thing using limits. I'm too tired right now though to figure out how to fix this so I'll try again in the morning.

 

[Samy_A]

Your first method looks fine.

While it may be nice/fun to prove something in two different ways, here I fear that a proof with limits will be little more than a contrived rewording of your first proof. (Of course that only means that I don't immediately see an elegant proof using limits .)