Is there a way to prove that a set is bounded using calculus techniques?

In summary, the conversation discusses proving that a subset S of R is bounded, with the set being defined as S=(n^2-1)/(n^3-1) where n is a natural number. The conversation also touches on using proof techniques to show that the set is bounded both above and below, and introduces the concepts of bounded subsets and the operations of negation and addition on bounded sets.
  • #1
Anne5632
23
2
Homework Statement
Prove that S which is a subset of R is bounded
Relevant Equations
S=(n^2-1)/(n^3-1)
S is an element of natural numbers
I know that for a set to be bounded it is bounded above and below, for the bound below is it 0 and n cannot equal 1 and u paper bound is inf but how do I prove that it is bounded?
 
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  • #2
Anne5632 said:
Homework Statement:: Prove that S which is a subset of R is bounded
Relevant Equations:: S=(n^2-1)/(n^3-1)
S is an element of natural numbers

I know that for a set to be bounded it is bounded above and below, for the bound below is it 0 and n cannot equal 1 and u paper bound is inf but how do I prove that it is bounded?
Did you try writing down the first few elements in the set to get an idea? I assume we are taking about ##n > 1## here?
 
  • #3
PeroK said:
Did you try writing down the first few elements in the set to get an idea? I assume we are taking about ##n > 1## here?
the inf of the set is 0 so does that count in the interval?
 
  • #4
Anne5632 said:
the inf of the set is 0 so does that count in the interval?
Yes, it should be clear that for all ##n > 1## we have ##s_n > 0##. ##S## is, therefore, bounded below by ##0##. What about an upper bound?
 
  • #5
PeroK said:
Yes, it should be clear that for all ##n > 1## we have ##s_n > 0##. ##S## is, therefore, bounded below by ##0##. What about an upper bound?
1
 
  • #6
Anne5632 said:
1
How do you prove that is the question?

Have you done many proofs?
 
  • #7
PeroK said:
How do you prove that is the question?

Have you done many proofs?
No I haven't done proofs
I simplified the fraction in the equation given and factorised out (x-1) then got a polynomial to degree one in top and a polynomial to degree 2 on the bottom.
If i want to show it's smaller than 1 should I rewrite 1 into a polynomial with degree 2÷polynomial to degree 2
 
  • #8
Anne5632 said:
No I haven't done proofs
I simplified the fraction in the equation given and factorised out (x-1) then got a polynomial to degree one in top and a polynomial to degree 2 on the bottom.
If i want to show it's smaller than 1 should I rewrite 1 into a polynomial with degree 2÷polynomial to degree 2
You can either do a direct proof, which is to show directly that ##\frac{n^2 -1}{n^3 - 1} \le 1##. Or, you can do a proof by contradiction by assuming that for some ##n## we have ##\frac{n^2 -1}{n^3 - 1} > 1## and reaching a contradiction.

Cancelling the common factor of ##n - 1## is an option.
 
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  • #9
PeroK said:
You can either do a direct proof, which is to show directly that ##\frac{n^2 -1}{n^3 - 1} \le 1##. Or, you can do a proof by contradiction by assuming that for some ##n## we have ##\frac{n^2 -1}{n^3 - 1} > 1## and reaching a contradiction.

Cancelling the common factor of ##n - 1## is an option.
Got the bounds I think , thanks.
Part 4 of that q asked:
Let B be a bounded subset of R. Prove -B + S is bounded from below.

How would I know the bounds of B? Does it have no lower bound?
 
Last edited:
  • #10
Anne5632 said:
Got the bounds I think , thanks.
Part 4 of that q asked:
Let B be a bounded subset of R. Prove -B + A is bounded from below.

How would I know the bounds of B? Does it have no lower bound?
What's A?
 
  • #11
PeroK said:
What's A?
Sorry meant to write S,
A is the set S described in the question above
 
  • #12
Anne5632 said:
Sorry meant to write S,
A is the set S described in the question above
First, if ##X## is a bounded subset of ##\mathbb R##, then the set ##-X \equiv \{-x: x \in X\}## is also a bounded subset of ##\mathbb R##.

And, more generally ##X## is bounded above iff ##-X## is bounded below.

Also, if ##X## and ##Y## are bounded above, then the set ##X + Y \equiv \{x+y: x \in X, y \in Y\}## is bounded above. Similarly, if ##X## and ##Y## are bounded below, then so is ##X + Y##.

These are things that I assume you will be shown in your course or asked to prove as an exercise. You need to learn some of the techniques that are used to prove things like this. The proofs are straighforward, but not always easy to see for someone new to formal proofs.

Have you seen anything like this on your course? It seems to me that you are working in the dark here.
 
  • #13
PeroK said:
First, if ##X## is a bounded subset of ##\mathbb R##, then the set ##-X \equiv \{-x: x \in X\}## is also a bounded subset of ##\mathbb R##.

And, more generally ##X## is bounded above iff ##-X## is bounded below.

Also, if ##X## and ##Y## are bounded above, then the set ##X + Y \equiv \{x+y: x \in X, y \in Y\}## is bounded above. Similarly, if ##X## and ##Y## are bounded below, then so is ##X + Y##.

These are things that I assume you will be shown in your course or asked to prove as an exercise. You need to learn some of the techniques that are used to prove things like this. The proofs are straighforward, but not always easy to see for someone new to formal proofs.

Have you seen anything like this on your course? It seems to me that you are working in the dark here.
No I haven't, and none of the other exercises are similar.
But I'll looks throught the theorems in that topic
 
  • #14
Anne5632 said:
No I haven't, and none of the other exercises are similar.
But I'll looks throught the theorems in that topic
If we take the first one.
PeroK said:
First, if ##X## is a bounded subset of ##\mathbb R##, then the set ##-X \equiv \{-x: x \in X\}## is also a bounded subset of ##\mathbb R##.
A simple proof of that would be;

Let ##U## be an upper bound for ##X##.

Let ##y \in -X##. Then ##y = -x## for some ##x \in X##. Now ##x \le U \ \Rightarrow \ -x \ge -U##. Hence ##y \ge -U##.

As ##y## was any member of ##-X## we see that ##-U## is a lower bound for ##-X## and so ##-X## is bounded below.

Then you would do something similar for a lower bound ##L##.

That's the style of proof that I assume is required for these problems.
 
  • #15
PeroK said:
If we take the first one.

A simple proof of that would be;

Let ##U## be an upper bound for ##X##.

Let ##y \in -X##. Then ##y = -x## for some ##x \in X##. Now ##x \le U \ \Rightarrow \ -x \ge -U##. Hence ##y \ge -U##.

As ##y## was any member of ##-X## we see that ##-U## is a lower bound for ##-X## and so ##-X## is bounded below.

Then you would do something similar for a lower bound ##L##.

That's the style of proof that I assume is required for these problems.
Would the formula
Inf(N)=-sup(-N)
be useful?
 
  • #16
Anne5632 said:
Would the formula
Inf(N)=-sup(-N)
be useful?
That result is similar and has the same sort of proof as the one I gave. But, ultimately, pure mathematics isn't about plugging numbers into formulas; it's about being able to construct logically valid proofs.
 
  • #17
PeroK said:
That result is similar and has the same sort of proof as the one I gave. But, ultimately, pure mathematics isn't about plugging numbers into formulas; it's about being able to construct logically valid proofs.
True,
To find the lower bound of -B I thought I could use that theorem
 
  • #18
Anne5632 said:
True,
To find the lower bound of -B I thought I could use that theorem
Well, let's see you attempt to prove it. You haven't actually posted anything yet that gives me any indication of what you can do!
 
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  • #19
@Anne5632 :
If you've done a bit of Calculus or advanced pre-Calculus, you can consider ##f(x)=\frac {x^2-1}{x^3-1} ## and use standard optimization techniques: Take derivative f'(x), set to 0, etc.
 

Related to Is there a way to prove that a set is bounded using calculus techniques?

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

A set is considered bounded if there exists a finite number that serves as an upper and lower limit for all of the elements in the set. This means that all of the elements in the set fall within a certain range, and there is no element that is infinitely large or small.

2. How do you prove that a set is bounded?

To prove that a set is bounded, you must show that there exists a finite number that serves as an upper and lower limit for all of the elements in the set. This can be done by finding the maximum and minimum values of the set, or by showing that the set is contained within a larger bounded set.

3. What is the difference between a bounded set and an unbounded set?

A bounded set has a finite upper and lower limit, while an unbounded set does not have any limits and can contain infinitely large or small elements. In other words, an unbounded set extends infinitely in one or both directions, while a bounded set is contained within a finite range.

4. Can a set be bounded in one dimension but unbounded in another?

Yes, a set can be bounded in one dimension but unbounded in another. For example, a set of points on a line may have a finite range, but if those same points are plotted on a two-dimensional graph, they may extend infinitely in one direction.

5. Why is it important to prove that a set is bounded?

Proving that a set is bounded is important because it allows us to determine the behavior and characteristics of the set. Bounded sets have specific properties and can be analyzed and manipulated in certain ways, while unbounded sets have different properties and may require different methods of analysis.

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