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- Thread starter dwsmith
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- Mar 10, 2012

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What is asked?$S = \{x : (x - a)(x - b)(x - c)(x - d) < 0\}$, where $a < b < c < d$

This questioned shouldn't be to difficult but would it be best to multiply out?

And how is the $a < b < c < d$ going to affect it?

- Feb 5, 2012

- 1,621

Hi dwsmith,$S = \{x : (x - a)(x - b)(x - c)(x - d) < 0\}$, where $a < b < c < d$

This questioned shouldn't be to difficult but would it be best to multiply out?

And how is the $a < b < c < d$ going to affect it?

It's clear that the set \(S\) contains elements \(a<x<b\) or \(c<x<d\). Otherwise, \((x - a)(x - b)(x - c)(x - d) >0\). That is,

\[S=\{x : a<x<b \mbox{ or }c<x<d\}=(a,b)\cup(c,d)\]

Now I suppose it is obvious as to what is the supremum and what is the infimum. Isn't?

Kind Regards,

Sudharaka.

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$\text{inf} \ S = a + c$ and $\text{sup} \ S = b + d$Hi dwsmith,

It's clear that the set \(S\) contains elements \(a<x<b\) or \(c<x<d\). Otherwise, \((x - a)(x - b)(x - c)(x - d) >0\). That is,

\[S=\{x : a<x<b \mbox{ or }c<x<d\}=(a,b)\cup(c,d)\]

Now I suppose it is obvious as to what is the supremum and what is the infimum. Isn't?

Kind Regards,

Sudharaka.

- Feb 5, 2012

- 1,621

\(a+c\) may not be a lower bound and \(b+d\) may not be an upper bound. A simple example to contradict your supremum and infimum would be, \(a=1,b=2,c=3,d=4\). Then,$\text{inf} \ S = a + c$ and $\text{sup} \ S = b + d$

\[S=(1,2)\cup(3,4)\]

Now it is clear that, \(1+3=4\) is not a lower bound of \(S\). \(2+4=6\) although an upper bound for this example is not the least upper bound.

The simplest way to think about this would be to draw the two intervals \((a,c)\) and \((b,d)\) on a real line(Note that, \(a<b<c<d\)) and see what are the upper bounds and lower bounds of \(S\).

Kind Regards,

Sudharaka.