# Analysis Problem for homework, infimum and supremum

• retspool
In summary, this is a problem involving sequences in R where the limit of one sequence is given and the task is to prove two inequalities. The first inequality is fairly straightforward, while the second one involves a more complex approach. Through a hypothetical scenario and the definition of limsup, it can be shown that the second inequality is impossible, thus proving the desired result.
retspool
I have this analysis homework due tomorrow.
This is one of my problems.Let (sn) and (tn) be sequences in R. Assume that lim sn = s ∈ R. Then lim sup(sn +tn) = s+limsup(tn).I don't even know how to approach it. Even though it seems very straight forward.

Last edited:
You will have to prove 2 inequalities, a hard one and an easy one.

The easy inequality follows from

$$\sup_n{a_n+b_n}\leq \sup_n{a_n}+\sup_n{b_n}$$

just take the limit of both sides.

For the other inequality, suppose that

$$\limsup{a_n+b_n}<s+\limsup{b_n}$$

Then there exist an $$\epsilon>0$$ such that

$$\limsup{a_n+b_n}<\limsup{b_n}+s-\epsilon:=L$$

Since $$\limsup{a_n+b_n}<L$$, it follows that

$$\exists n_0:~\forall n>n_0:~a_n+b_n<L$$

But from a certain $$n_1$$ it is true that $$a_n-s>-\epsilon/2$$. Thus from $$\max\{n_0,n_1\}$$, we must have that

$$s-\epsilon/2+b_n<a_n+b_n<L=\limsup{b_n}+s-\epsilon$$

So from a certain moment, we have that $$b_n<\limsup{b_n}-\epsilon/2$$. But the definition of limsup tells us that this is impossible...

[tex]\l

## 1. What is the difference between infimum and supremum?

The infimum of a set is the largest number that is less than or equal to all elements in the set. The supremum, on the other hand, is the smallest number that is greater than or equal to all elements in the set. In other words, the infimum is the greatest lower bound and the supremum is the least upper bound for a set of numbers.

## 2. How do you find the infimum and supremum of a set?

To find the infimum and supremum of a set, you need to first arrange the numbers in the set in increasing order. The infimum will be the first number in the set, and the supremum will be the last number in the set.

## 3. Can the infimum or supremum of a set be a number that is not in the set?

Yes, the infimum or supremum of a set can be a number that is not in the set. This can happen when the set is not bounded or when the set contains irrational numbers.

## 4. What is the significance of infimum and supremum in analysis problems?

Infimum and supremum are important concepts in analysis as they help us determine the boundaries of a set of numbers. They are used to define limit of a sequence and continuity of a function. They also play a crucial role in proving the existence of certain mathematical objects.

## 5. Can a set have more than one infimum or supremum?

No, a set can have at most one infimum and at most one supremum. If the infimum or supremum is not a unique number, then it is not defined for that set. However, a set can have a maximum and a minimum, which are the largest and smallest elements in the set respectively.

• Calculus and Beyond Homework Help
Replies
2
Views
4K
• Calculus and Beyond Homework Help
Replies
6
Views
2K
• Calculus and Beyond Homework Help
Replies
8
Views
7K
• Calculus and Beyond Homework Help
Replies
6
Views
3K
• Calculus and Beyond Homework Help
Replies
3
Views
1K
• Calculus and Beyond Homework Help
Replies
8
Views
2K
• Calculus and Beyond Homework Help
Replies
1
Views
2K
• Calculus and Beyond Homework Help
Replies
2
Views
2K
• Calculus and Beyond Homework Help
Replies
5
Views
2K
• Calculus and Beyond Homework Help
Replies
7
Views
2K