Give a counter-example that shows Bolzano-Weirstrass is unvalid in IR2

[tsuwal]

Homework Statement

Give a counter-example that shows Bolzano-Weirstrass is unvalid in IR2.

Intro:
Bolzano-Weirtrass theorem says that if a sequence (IN->IR) is bounded then there exists a convergent sub-sequence. (this is shown using the Cauchy sequence concept, showing that a Cauchy sequence is bounded and using the lemma of monotonic sub-sequences)

However, this is not valid valid in IR2, if a sequence (IN->IR2) is bounded then we can't assure that the exists a convergent sub-sequence.

Homework Equations

The Attempt at a Solution

It's not easy since you have the tendency of using a pattern. But I guess
Xn=(Cos(n),Sin(n)) might work...

 

[Dick]

Homework Statement

Give a counter-example that shows Bolzano-Weirstrass is unvalid in IR2.

Intro:
Bolzano-Weirtrass theorem says that if a sequence (IN->IR) is bounded then there exists a convergent sub-sequence. (this is shown using the Cauchy sequence concept, showing that a Cauchy sequence is bounded and using the lemma of monotonic sub-sequences)

However, this is not valid valid in IR2, if a sequence (IN->IR2) is bounded then we can't assure that the exists a convergent sub-sequence.

Homework Equations

The Attempt at a Solution

It's not easy since you have the tendency of using a pattern. But I guess
Xn=(Cos(n),Sin(n)) might work...

Can you explain your notation? What are IN, IR and IR2? Bolzano Weierstrass is valid for sequences in $R^2$.

 

[tsuwal]

IR=real number set
IN=natural number set

I'm sorry, I thought Bolzano-Weirtrass was not valid I am IR^2, my book was not clear in that part. I've consulted Wikipedia and confirmed that it is valid in IR^2. Sorry for the mistake.

Thanks.