Can the Sum of Supremums be Greater than the Supremum of the Sum?

[danago]

Suppose that $$A,B \subseteq \Re^+$$ are non empty and bounded sets. Define + and . as the following set opetations:

$$\begin{array}{l}<br /> A + B = \{ a + b|a \in A,b \in B\} \\ <br /> A.B = \{ ab|a \in A,b \in B\} \\ <br /> \end{array}$$

Prove that $$\sup (A + B) \le \sup A + \sup B$$


I started by letting $$a \in A,b \in B$$. From the definition of supremum:

$$\begin{array}{l}<br /> a \le \sup A \\ <br /> b \le \sup B \\ <br /> \end{array}$$

I then added the two inequalities to give:

$$a + b \le \sup A + \sup B$$

Since this holds true for any a and b, sup A+sup B is an upper bound on the set A+B. This is where I am a bit stuck. I can't really see why sup(A+B) isn't strictly equal to sup A + sup B. I tried coming up with a few example sets A and B but that didnt really help anything.

Any suggestions?

Thanks,
Dan,

 

[Werg22]

You've already established that sup A + sub B is an upper bound for the A + B. All that remains is to show that it's the least upper bound.

 

[danago]

You've already established that sup A + sub B is an upper bound for the A + B. All that remains is to show that it's the least upper bound.

I guess that's what I am having trouble with. I tried assuming that there exists some other upper bound 'r' which is less than sup A + sup B, and then showing that this leads to a contradiction. No luck though :(

 

[Defennder]

He isn't required to show that it is the least upper bound, only that it is an upper bound.

 

[Werg22]

He isn't required to show that it is the least upper bound, only that it is an upper bound.

Yes, but he already showed this in the first post, my understanding was that he was curious to see why it's the least upper bound.

danago, if k is the supremum of A + B, then for any epsilon > 0, there's an element x in A + B satisfying |k - x| < epsilon. All you have to show is that sup A + sup B has this property.

 

[Defennder]

If you want to prove that anyway, just note that $$\sup A - \frac{\epsilon}{2} < a \ \mbox{for} \ \exists a \ \mbox{and} \ \forall \epsilon > 0$$. The same applies for b. See how to continue from here?

 

[HallsofIvy]

But his question was, "Why is it not the least upper bound?"

danago, the fact that you are asked to prove that $sup A+B\le \sup A+ \sup B$ does not necessarily imply they are not equal. Only that you are not asked to prove that.

 

[danago]

But his question was, "Why is it not the least upper bound?"

danago, the fact that you are asked to prove that $sup A+B\le \sup A+ \sup B$ does not necessarily imply they are not equal. Only that you are not asked to prove that.

Ok fair enough. So what I've done is sufficient to answer the question then?

 

[Defennder]

Yes.

 

[danago]

Alright thanks everyone