real analysis

**flower3** · August 17th 2009, 12:37 PM

Give an example to show that it is not true ,ingeneral ,that :
sup(AB)=sup(A)sup(B) and inf (AB)=inf(A)inf(B)

**Bruno J.** · August 17th 2009, 01:23 PM

Hint for the first one:

$A: \: 1,-1,1,-1,...$
$B: \: -1,1,-1,1,...$

**putnam120** · August 17th 2009, 02:17 PM

Also consider $a_n=\cos\left(\frac{n\pi}{2}\right)$ and $b_n=\sin\left(\frac{n\pi}{2}\right)$ .

**Plato** · August 17th 2009, 02:57 PM

Originally Posted by Bruno J.

hint
$A: \: 1,-1,1,-1,...$
$B: \: -1,1,-1,1,...$

But that depends upon how $AB$ is defined.
If it is defined by $AB=\left\{ {xy:x \in A\;\& \,y \in B} \right\}$ then $\sup(A)=\sup(AB)=\sup(B)$

How is $AB$ defined?

**Opalg** · August 18th 2009, 12:32 AM

Originally Posted by flower3

Give an example to show that it is not true ,ingeneral ,that :
sup(AB)=sup(A)sup(B) and inf (AB)=inf(A)inf(B)

How about $A = \{1,2\},\ B=\{-1,-2\}$ ?

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