Proof of sup(A+B) = sup A + sup B | is my thought process correct?

[michonamona]

Homework Statement

Let A and B be subsets of R (real numbers). The vector sum of two sets A and B is written as A+B and is defined to be:

A+B = {a+b : a in A, b in B}

Prove that for all bounded nonempty sets A and B, sup (A+B) = sup A + sup B

The Attempt at a Solution

let A* = sup A, B*=sup B and C*=sup(A+B)

(1) we first prove that C*<= A* + B*
- I understand the proof for this part

(2) Next, we prove that A* + B*<= C*
- This is MY proof:

since C* is the sup (A+B), then for any a+b in set (A+B):

a+b <= C*

Thus, C* is an upper bound for a+b, for any a in A and b in B

This implies that C* is also the upper bound for the sum of the highest possible a in A (namely, A*) and the highest possible b in B (namely, B*). Therefore:

A*+B* <=C*

In conclusion, by (1) and (2):

A*+B*=C*
My question, what is wrong with the logic in proof (2)? The solution has something different but I want to check if my thought process is also correct.

 

[LCKurtz]

Homework Statement

Let A and B be subsets of R (real numbers). The vector sum of two sets A and B is written as A+B and is defined to be:

A+B = {a+b : a in A, b in B}

Prove that for all bounded nonempty sets A and B, sup (A+B) = sup A + sup B

The Attempt at a Solution

let A* = sup A, B*=sup B and C*=sup(A+B)

(1) we first prove that C*<= A* + B*
- I understand the proof for this part

(2) Next, we prove that A* + B*<= C*
- This is MY proof:

since C* is the sup (A+B), then for any a+b in set (A+B):

a+b <= C*

Thus, C* is an upper bound for a+b, for any a in A and b in B

That last sentence isn't very clear. I think you are just trying to restate the fact that C* = sup(A+B).

This implies that C* is also the upper bound for the sum of the highest possible a in A (namely, A*) and the highest possible b in B (namely, B*). Therefore:

A*+B* <=C*

I don't think anyone will be convinced by that "argument".

The sup of A is not the "highest possible" a in A. The sup may not be in the set.

You might have better luck with an indirect argument. Suppose C* < A* + B* and see if you can come up with a contradiction.

 

[michonamona]

That last sentence isn't very clear. I think you are just trying to restate the fact that C* = sup(A+B).



I don't think anyone will be convinced by that "argument".

The sup of A is not the "highest possible" a in A. The sup may not be in the set.

You might have better luck with an indirect argument. Suppose C* < A* + B* and see if you can come up with a contradiction.

I'm trying to prove that C* = A*+B* or sup(A+B) = sup A + sup B. I'm doing this by proving that both C*<=A*+B* and C*>=A*+B* are true. So the question is, what can I conclude from the following hypothesis:

a+b <= C*, for any a in A and b in B

 

[LCKurtz]

I'm trying to prove that C* = A*+B* or sup(A+B) = sup A + sup B. I'm doing this by proving that both C*<=A*+B* and C*>=A*+B* are true. So the question is, what can I conclude from the following hypothesis:

a+b <= C*, for any a in A and b in B

I understand what you are trying to prove. You said you have already shown

C* ≤ A* + B*

and you need to show A* + B* ≤ C* to be finished. I suggested you try to show this by an indirect argument by assuming this is false -- suppose A* + B* > C* and show that can't happen. Give an argument so show that can't happen. If you can do that you will have shown that C* ≤ A* + B* and C* is not less than A*+ B*, so they must be equal.

So use the properties of sup to show that A* + B* can't be greater than C* and you are done.

[edit: corrected typos]

 

[Tedjn]

Let me just clarify why your logic doesn't work. It's true that a + b <= C* for all a in A and b in B. But, as LCKurtz said, A* might not be in A and B* might not be in B, so you cannot immediately conclude that A* + B* <= C* without further argument.