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

  • #1

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.
 

Answers and Replies

  • #2
Samy_A
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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##.
 
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  • #3
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.
 
  • #4
Samy_A
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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 :smile:.)
 
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