# Help with real analysis/advanced calculus

• Oct 2nd 2012, 08:45 AM
salmaster
Hey can anyone help me prove this problem?

Let A and B be a subsets of R and bounded above.
Let InfA=a and InfB=b.
Let C= {xy ; x belongs to A, and y belongs to B}
Prove that InfC=ab
• Oct 2nd 2012, 08:54 AM
Plato
Re: Help with real analysis/advanced calculus
Quote:

Originally Posted by salmaster
Let A and B be a subsets of R and bounded above.
Let InfA=a and InfB=b.
Let C= {xy ; x belongs to A, and y belongs to B}
Prove that InfC=ab

Did you mean that to read below?
• Oct 2nd 2012, 09:02 AM
johnsomeone
Re: Help with real analysis/advanced calculus
This problem isn't true as stated.

1) I think you're intending to assume that A and B are bounded below.

2) Let A = {-1, 1}, B = {-1}. Then a = -1, b = -1, so ab = 1. Also C = {-1, 1}, so inf(C) = -1.
Thus inf(C) is NOT equal to ab.
• Oct 2nd 2012, 09:04 AM
salmaster
Re: Help with real analysis/advanced calculus
Yes I'm sorry I meant bounded below.
• Oct 2nd 2012, 09:24 AM
Plato
Re: Help with real analysis/advanced calculus
Quote:

Originally Posted by salmaster
Yes I'm sorry I meant bounded below.

Let $A=\{-1,2\}~\&~B=\{2,-3\}$ then $C=\{-2,3,4,-6\}$
Does the statement hold in that case?
• Oct 2nd 2012, 09:35 AM
salmaster
Re: Help with real analysis/advanced calculus
I don't think that we are supposed to prove it with specific examples of ordered pairs for A, B, and C..
• Oct 2nd 2012, 09:45 AM
Plato
Re: Help with real analysis/advanced calculus
Quote:

Originally Posted by salmaster
I don't think that we are supposed to prove it with specific examples of ordered pairs for A, B, and C..

The point is that in reply #5 $\inf(A)=-1,~\inf(B)=-3~\&~\inf(C)=-6$ so that is a counterexample to what you are trying to show.
So the statement you have is false.