Showing that S is unbounded by contradiction

[cbarker1]

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

 

[fresh_42]

Start with the definition. What does '$S$ is bounded' mean?

 

[cbarker1]

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".

 

[fresh_42]

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.

 

[cbarker1]

I am confused...are we still using proof by contradiction?

 

[fresh_42]

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.

 

[PeroK]

I suggest that the triangle inequality is your friend in this case.

 

[PeroK]

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.

 

[cbarker1]

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...

 

[pasmith]

If $S$ is bounded, then there exists $R > 0$ such that $n \leq \|(x_n,y_n)\| \leq R$ for every $n \geq 1$. Is that possible?

 

[cbarker1]

no. I think that your condition is not possible I can pick a N bigger than R...