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

1. Nov 7, 2015

### Potatochip911

1. The problem statement, all variables and given/known data
Let $S$ be a nonempty and bounded subset of $\mathbb{R}$, $bS:=\{bs:s\in S\}$ and $b<0$ show that $\inf(bS)=b\sup S$

2. Relevant equations
3. The attempt at a solution

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.

2. Nov 8, 2015

### Samy_A

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 $\forall s \in S$.

3. Nov 8, 2015

### Potatochip911

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.

4. Nov 8, 2015