# Homework Help: Having a difficult time with supremum proof

1. Sep 3, 2011

### SMG75

1. The problem statement, all variables and given/known data

Suppose a = sup(A) and b = sup(B). Let A + B = $\left\{$x + y;x$\in$A; y$\in$B$\right\}$. Show that a + b = sup(A + B).

2. Relevant equations

3. The attempt at a solution

I'm honestly not sure where to start. Any help guys?

2. Sep 3, 2011

### micromass

Well, a=sup(A), and b=sup(B). What does that mean for you??

You need to show a+b=sup(A+B). What does that mean?? What do you need to prove?

3. Sep 3, 2011

### SMG75

I suppose saying I have no clue where to start was a bit of an overstatement.

a is not only the upperbound of A but the smallest possible upperbound. Same for b/B. This is the definition of a supremum.

When the sets are added together, the supremum of that new set, call it C, is the same as the supremum's of sets A and B individually added together. That is, c=sup(A+B)=sup(C)=a+b.

Here is my intuition: if you take the largest values from set A and set B and add them together, this is the greatest number in (A+B). Call this value z. The supremum of (A+B) lies at the next incremental value greater than z. Does this make sense?

4. Sep 3, 2011

### gb7nash

Yes, this is what we want to show.

Ah, but you don't know A or B has a largest value! This is only true if max(A) = sup(A), and max(B) = sup(B).

Go through your notes/book. What is the mathematical definition of a supremum? Least upper bound is not enough.

5. Sep 3, 2011

### SMG75

If S is a nonempty subset of numbers, then M=sup(S) if and only if M is an upperbound for S AND for any z>0, M-z is not an upperbound for S.

Is this a solid definition?

6. Sep 3, 2011

### gb7nash

Almost. You need to say mathematically what it means for M-z to not be an upperbound for S:

M=sup(S) if and only if M is an upperbound for S AND for any z>0, there exists an s in S such that s > M-z.

I wish I could draw a number line to illustrate this, but this basically says that if M is a supremum, we can choose any z>0 and we'll always find a value between M-z and M that's in S.

7. Sep 3, 2011

### SMG75

That makes perfect sense, actually. Where should I go from here?

8. Sep 3, 2011

### gb7nash

Well, we need to prove two things:

1) a+b is an upperbound.
2) For a fixed z>0, there exists s in A+B such that s > a+b-z

For 1), How do you show a value is an upperbound of a set?

9. Sep 3, 2011

### SMG75

Let A$_{w}$=set of whole parts of elements of A+B
A$_{w}$ is bounded from above because A+B is bounded. Let a$_{w}$=max(A$_{w}$. And let A$_{1}$ $\subseteq$ A+B be the subset of A+B containing all elements that begin a$_{w}$. ...(decimal expansion). Assume a$_{w}$ $\geq$ 0. Let a$_{1}$=max (1/10)'s digit of all elements in A+B. Let A$_{2}$ $\subseteq$ A$_{1}$ be the subset of all elements in A$_{1}$ beginning a$_{w}$.a$_{1}$...
Continue in this manner...
Let C=a$_{w}$.a$_{1}$a$_{2}$a$_{3}$...
Claim: C=sup(A+B)
given any x$\in$A+B, a$_{w}$ $\geq$ the whole part of x. If they're equal, a$_{1}$ $\geq$ the (1/10)'s digit of x. If these are equal, a$_{2}$ $\geq$ (1/100)'s digit of x...
For any n$\geq$1 there is an element of A$\ni$X$_{n}$ such that X$_{n}$=a$_{w}$.a$_{1}$a$_{2}$a$_{3}$...a$_{n}$
$\left|$C-X$_{n}$$\right|$$\leq$$\frac{1}{10^{n}}$, so any number less than s-$\frac{1}{10^{n}}$ is not an upperbound. Since this is true for all n, no number below C is an upperbound. Thus, C=sup(A+B)=a+b.

Am I way off here?

10. Sep 3, 2011

### SMG75

The part that says \left and \right is supposed to be absolute value. I'm not sure how to input that.

11. Sep 3, 2011

### SMG75

Can anyone tell me if I'm on the right track here?

12. Sep 3, 2011

### micromass

I don't see what your post 9 actually has to do with proving that sup(A+B)=a+b...
You're making it way more complicated than it needs to be.

First, you need to prove that a+b is an upper bound of A+B. For that, take x+y in A+B. Can you prove that $x+y\leq a+b$?

13. Sep 3, 2011

### SMG75

Well, since x$\in$A, it is by definition less than a, which is the upperbound of A. Similarly, y is less than b. Therefore, x+y$\leq$a+b. I'm not sure how to formalize this intuition, though.

14. Sep 3, 2011

### micromass

Yes, that's ok. This proves that a+b is an upper bound of A+B. Now prove that it is the least upper bound...

15. Sep 3, 2011

### SMG75

I'm a little stuck in thinking about this. Intuitively, I know a is the least upperbound of A by definition, and b is the least upperbound of B, similarly. Having a least upperbound of A+B greater than a+b implies that either a or b was not a least upperbound to begin with. I'm not sure where to go from here...

16. Sep 3, 2011

### micromass

You could prove that there exists a sequence in A+B that converges to a+b.

In general:
If z is an upper bound of a set Z, and if there exists a sequence $(z_n)_n$ in Z that converges to z, then z is the supremum of Z.

Now, try to prove that "in general"-statement, and try to see why such a sequence should exist.

17. Sep 3, 2011

### SMG75

I think I was following well until this point, but now I'm totally lost. Would you mind explaining this? By the way, I really appreciate the help.

18. Sep 3, 2011

### micromass

You did see sequences yet, did you??

Well, assume that z is an upper bound of Z, and let's say that there are element $z_n\in Z$ such that $z_n\rightarrow z$. Then I claim that z is the least upper bound.

Let's try to prove this.
Let's assume that w is another upper bound of Z. We will want to prove that $z\leq w$. Do you follow until here?? Can you prove this?

19. Sep 3, 2011

### SMG75

Yes, I follow until there. I'm not sure how I prove that, though...

20. Sep 3, 2011

### micromass

$z_n\leq w$. Now take the limit...