# supremum and infimum

• Dec 4th 2010, 06:11 AM
the prince
supremum and infimum
Hello!
can someone help me to prove this

Let A*B={a*b;a,b>0 a is element of A,b element of B.Prove that:
inf(AB)=inf(A)inf(B)
sup(AB)=sup(A)sup(B)

i've done the proof with inf(A+B)= inf (A) + inf(B) and inf(-A)= -inf(A) but i can't prove the problem above
• Dec 4th 2010, 06:32 PM
alexandros
Quote:

Originally Posted by the prince
Hello!
can someone help me to prove this

Let A*B={a*b;a,b>0 a is element of A,b element of B.Prove that:
inf(AB)=inf(A)inf(B)
sup(AB)=sup(A)sup(B)

i've done the proof with inf(A+B)= inf (A) + inf(B) and inf(-A)= -inf(A) but i can't prove the problem above

We have $\displaystyle ab\leq \sup AB$ for all $\displaystyle a\in A, b\in B$
Fix $\displaystyle a$. Now $\displaystyle b\leq \frac{1}{a} \cdot \sup AB$ for all $\displaystyle b\in B$.
But then $\displaystyle \sup B\leq \frac{1}{a} \cdot \sup AB$.
i.e. $\displaystyle a\leq \frac{1}{\sup B} \cdot \sup AB$.
Now this is true for all $\displaystyle a\in A$.
Hence $\displaystyle \sup A\leq \frac{1}{\sup B} \cdot \sup AB$

.
i.e. $\displaystyle \sup A\cdot \sup B\le \sup AB$

We also have : for all a,b,belonging to A ,B $\displaystyle ab\leq SupASupB$

HENCE Sup(AB)$\displaystyle \leq SupASupB$

Thus Sup(AB)=SupASupB