Showing that S is unbounded by contradiction

In summary: Yes, you can. But what's the point?The point is that the sequence is moving away from the closed ball.
  • #1
cbarker1
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Homework Statement
Let ##S## be a subset of ##\mathbb{R}^2## with the standard metric. Show that if there exists a sequence ##(x_n, y_n)## in ##S## s.t. ##|(x_n,y_n)|\ge n## for all ##n \ge 1##, then ##S## is unbounded
Relevant Equations
A set ##S## is bounded if there is a closed ball B(r,p)=\{(x,y)\in \mathbb{R}^2: |(x,y)-p|\le R}\ such that ##B(r,p)## is a subset of ##S##.
Dear Everyone,

I am attempting a proof of contradiction for this problem. I am stuck on next step.

My attempt:
Assume that ##S## is bounded. Choose a ##N=\text{greatest integer function of} R+1##...

Here is where I am stuck. I want to show that the sequence is moving away from the closed ball as N is getting larger.

Thanks,
Cbarker1
 
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  • #2
Start with the definition. What does '##S## is bounded' mean?
 
  • #3
A set S is bounded if there is a closed ball B(r,p)=\{(x,y)\in \mathbb{R}^2: |(x,y)-p|\le R}\ such that B(r,p) is a subset of S. Based by the assumption, I am "given a R" but i can't know the "R".
 
  • #4
O.k., so we have: ##S## bounded ##\Longleftrightarrow (\exists \,p\in S)\, (\exists \,R>0)\, : \,S\subseteq B(p,R)##

We want to show that ##S## in unbounded, which means: ##(\forall \,p\in S) \,(\forall R>0)\, : \,S\nsubseteq B(R,p).## Now we choose any ##p\in S## and any ##R>0## and we will have to find a point ##q\in S## with ##q\not\in B(p,R)##.

Finally translate ##q\not\in B(p,R)## and think about what that point ##q## could be.
 
  • #5
I am confused...are we still using proof by contradiction?
 
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  • #6
cbarker1 said:
I am confused...are we still using proof by contradiction?
No. We prove the statement directly. I have only derived what unbounded means, i.e. not bounded with the definition you gave for boundedness.

You can write it as an indirect proof, but there is no need to.
 
  • #7
I suggest that the triangle inequality is your friend in this case.
 
  • #8
cbarker1 said:
A set S is bounded if there is a closed ball B(r,p)=\{(x,y)\in \mathbb{R}^2: |(x,y)-p|\le R}\ such that B(r,p) is a subset of S. Based by the assumption, I am "given a R" but i can't know the "R".
It's the other way round. If ##S## is bounded, then ##S## is a subset of some closed ball.
 
  • #9
I talked to my professor about this question. He recommended me to do proof by contradiction on this question. I need to practice this method. It is good practice for me to do so. I know that the sequence needs move away from the closed ball...
 
  • #10
If [itex]S[/itex] is bounded, then there exists [itex]R > 0[/itex] such that [itex]n \leq \|(x_n,y_n)\| \leq R[/itex] for every [itex]n \geq 1[/itex]. Is that possible?
 
  • #11
no. I think that your condition is not possible I can pick a N bigger than R...
 

Related to Showing that S is unbounded by contradiction

1. How do you show that S is unbounded by contradiction?

To show that S is unbounded by contradiction, we assume that S is bounded and then use logical reasoning to arrive at a contradiction. This contradiction proves that our initial assumption of S being bounded is false, thus showing that S is unbounded.

2. Why is it important to show that S is unbounded by contradiction?

Showing that S is unbounded by contradiction is important because it helps us understand the limitations of a given system or problem. It also allows us to identify any flaws in our assumptions or logic, leading to a deeper understanding of the problem at hand.

3. Can you give an example of showing that S is unbounded by contradiction?

One example of showing that S is unbounded by contradiction is the proof that the set of prime numbers is unbounded. We assume that there is a largest prime number, but then we can always find a larger prime number by adding 1 to it, leading to a contradiction.

4. Are there any alternative methods for showing that S is unbounded?

Yes, there are other methods for showing that S is unbounded, such as using the definition of a limit or using mathematical induction. However, using contradiction is a commonly used and effective method for proving unboundedness.

5. What are the potential limitations of using contradiction to show that S is unbounded?

One limitation of using contradiction to show that S is unbounded is that it may not always be applicable or straightforward to use. It also requires careful and rigorous logical reasoning, which can be challenging for some individuals. Additionally, it may not provide any insights into how or why S is unbounded, only that it is indeed unbounded.

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