# How to prove a set is a bounded set?

• Cyn
In summary, the set S1 = {x ∈ R2 : x1 ≥ 0,x2 ≥ 0,x1 + x2 = 2} is a bounded set, as every value between 0 and 2 that satisfies x1 + x2 = 2 is smaller than the upper bound of 2.
Cyn
1. I have to show that
S1 = {x ∈ R2 : x1 ≥ 0,x2 ≥ 0,x1 + x2 = 2}
is a bounded set.
2. So I have to show that sqrt(x1^2+x2^2)<M for all (x1,x2) in S1.3. I have said that M>0 and we have 0<=x1<=2 and 0<=x2<=2.
And x2 = 2-x1
We can fill in sqrt(x1^2 + (2-x1)^2) = sqrt (0^2 + (2-0)^2) = 2 < M = 3.
And we can fill in sqrt (x1^2 + (2-x1)^2) = sqrt (2^2 + (2-2)^2) = 2 < M = 3.
Every value between the 0 and the 2 that satisfy x1+x2 = 2 is smaller than this M. So the set is bounded.
Is this correct?

Cyn said:
Every value between the 0 and the 2 that satisfy x1+x2 = 2 is smaller than this M.
Can you prove this?

The approach works, although I would make it easier: Just use 0<=x1<=2 and 0<=x2<=2 to find a maximal value the square root can get, and use that as limit. You don't have to find the smallest possible M.

Cyn said:
1. I have to show that
S1 = {x ∈ R2 : x1 ≥ 0,x2 ≥ 0,x1 + x2 = 2}
is a bounded set.
2. So I have to show that sqrt(x1^2+x2^2)<M for all (x1,x2) in S1.3. I have said that M>0 and we have 0<=x1<=2 and 0<=x2<=2.
And x2 = 2-x1
We can fill in sqrt(x1^2 + (2-x1)^2) = sqrt (0^2 + (2-0)^2) = 2 < M = 3.
And we can fill in sqrt (x1^2 + (2-x1)^2) = sqrt (2^2 + (2-2)^2) = 2 < M = 3.
Every value between the 0 and the 2 that satisfy x1+x2 = 2 is smaller than this M. So the set is bounded.
Is this correct?

Bounded usually means that ##L_1 \leq x_1 \leq U_1## and ##l_2 \leq x_2 \leq U_2## for some finite ##L_1,L_2,U_1,U_2## with ##U_i \geq L_i## for all ##i##. Looking at ##\sqrt{x_1^2 + x_2^2}## is not necessary (but is harmless); one version of "bounded" is true if and only if the other is true as well. That is, the two versions are equivalent.

In your case you have ##0 \leq x_i \leq 2## for ##i = 1,2##, and that is all you need. You can always enclose the square with vertices (0,0), (),2), (2,0), (2,2) inside a larger circle if you want to, and that will bound the norm ##\sqrt{x_1^2 + x_2^2}##.

## 1. What does it mean for a set to be bounded?

A set is considered bounded if it has a finite upper and lower limit. This means that there exists a number M such that all elements in the set are less than or equal to M, and a number m such that all elements in the set are greater than or equal to m.

## 2. How can I prove that a set is bounded?

To prove that a set is bounded, you need to show that there exists an upper and lower limit for all elements in the set. This can be done by finding the maximum and minimum values in the set, or by using the definition of a bounded set mentioned above.

## 3. Can a set be bounded in one direction?

Yes, a set can be bounded in one direction. This means that the set has either an upper or lower limit, but not both. For example, the set [0, ∞) is bounded in one direction, with 0 being the lower limit.

## 4. Is a bounded set always a finite set?

Not necessarily. A bounded set can be either finite or infinite. As long as there exists an upper and lower limit for all elements in the set, it can be considered a bounded set.

## 5. Can a set be unbounded?

Yes, a set can be unbounded if it does not have an upper or lower limit. This means that the elements in the set can increase or decrease without bound. An example of an unbounded set is the set of all positive integers.

• Calculus and Beyond Homework Help
Replies
1
Views
1K
• Calculus and Beyond Homework Help
Replies
3
Views
565
• Calculus and Beyond Homework Help
Replies
26
Views
2K
• Calculus and Beyond Homework Help
Replies
5
Views
2K
• Calculus and Beyond Homework Help
Replies
1
Views
1K
• Calculus and Beyond Homework Help
Replies
14
Views
3K
• Calculus and Beyond Homework Help
Replies
3
Views
1K
• MATLAB, Maple, Mathematica, LaTeX
Replies
5
Views
1K
• Set Theory, Logic, Probability, Statistics
Replies
54
Views
4K
• Calculus and Beyond Homework Help
Replies
11
Views
1K