Having a difficult time with supremum proof

[SMG75]

Homework Statement

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

Homework Equations

The Attempt at a Solution

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

 

[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?

 

[SMG75]

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?

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?

 

[gb7nash]

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.

Yes, this is what we want to show.

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?

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.

 

[SMG75]

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.

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?

 

[gb7nash]

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?

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.

 

[SMG75]

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.

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

 

[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?

 

[SMG75]

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?

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\inA+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\geq1 there is an element of A\niX_{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?

 

[SMG75]

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

 

[SMG75]

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

 

[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?

 

[SMG75]

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?

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

 

[micromass]

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

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...

 

[SMG75]

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...

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...

 

[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.

 

[SMG75]

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.

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.

 

[micromass]

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.

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?

 

[SMG75]

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?

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

 

[micromass]

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

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

 

[HallsofIvy]

Personally, I wouldn't use "sequences". Certainly if x is any member of A, then x< a and if y is any member of B, then y< b. It follows that x+ y< a+ b so a+ b certainly if an upper bound for A+ B.

Now, suppose z< a+ b and let x be any member of A. then b> z- x and, since b is the least upper bound for B, there exist y in B such that z-x< y< b. What can you say about x+ y?

 

[SMG75]

Personally, I wouldn't use "sequences". Certainly if x is any member of A, then x< a and if y is any member of B, then y< b. It follows that x+ y< a+ b so a+ b certainly if an upper bound for A+ B.

Now, suppose z< a+ b and let x be any member of A. then b> z- x and, since b is the least upper bound for B, there exist y in B such that z-x< y< b. What can you say about x+ y?

I'm mostly confused by b>z-x in your post. Can you explain why that is true?