# Homework Help: Supremum / Infimum

1. Aug 11, 2008

### danago

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

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

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:

$$\begin{array}{l} a \le \sup A \\ b \le \sup B \\ \end{array}$$

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 im a bit stuck. I cant really see why sup(A+B) isnt 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,

2. Aug 11, 2008

### 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.

3. Aug 11, 2008

### danago

I guess thats what im 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 :(

4. Aug 11, 2008

### Defennder

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

5. Aug 11, 2008

### Werg22

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.

6. Aug 11, 2008

### 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?

7. Aug 11, 2008

### 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.

8. Aug 11, 2008

### danago

Ok fair enough. So what ive done is sufficient to answer the question then?

9. Aug 11, 2008

### Defennder

Yes.

10. Aug 11, 2008

### danago

Alright thanks everyone