micromass said: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 said: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.
SMG75 said: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 said: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.
SMG75 said: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 said: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.
gb7nash said: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?
micromass said: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 [itex]x+y\leq a+b[/itex]?
SMG75 said:Well, since x[itex]\in[/itex]A, it is by definition less than a, which is the upperbound of A. Similarly, y is less than b. Therefore, x+y[itex]\leq[/itex]a+b. I'm not sure how to formalize this intuition, though.
micromass said: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...
micromass said: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 [itex](z_n)_n[/itex] 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 said: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 said: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 [itex]z_n\in Z[/itex] such that [itex]z_n\rightarrow z[/itex]. 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 [itex]z\leq w[/itex]. Do you follow until here?? Can you prove this?
SMG75 said:Yes, I follow until there. I'm not sure how I prove that, though...
HallsofIvy said: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?
A supremum is the least upper bound of a set of numbers. It is the smallest number that is greater than or equal to all the numbers in the set.
Supremum is important because it helps to establish the existence and uniqueness of certain mathematical objects, such as limits and infimums. It also plays a crucial role in proving the convergence of sequences and series.
In order to prove a supremum, you must first show that the number in question is an upper bound for the set. Then, you must show that it is the smallest upper bound, meaning that no smaller number can satisfy the conditions of being an upper bound.
One common challenge is determining the existence of a supremum, as it is not always clear if there is a smallest upper bound for a given set. Another challenge is finding a way to prove the uniqueness of the supremum, as there may be multiple numbers that satisfy the conditions of being an upper bound.
Understanding supremum proofs can benefit other areas of science by helping to analyze and understand data. It can also aid in making predictions and drawing conclusions based on mathematical models. Additionally, understanding supremum proofs can improve problem-solving skills and critical thinking abilities, which are valuable in any scientific field.