Easy convergent subsequence question.

[pantin]

Homework Statement

Consider the sequence {x_k} = {(arctan(k^2+1),sink)} in R^2. Is there a convergent
subsequence? Justify your answer.

Homework Equations

Every bounded sequence in R^n has a convergent subsequence.

The Attempt at a Solution

To show {x_k} is bounded: The range of arctan(k^2+1) is (-pi/2, pi/2) and the range of sin(k) is [-1,1].

A sequence is bded from above and below if there exist an M and an m both belong to R, such that a_k<=M and a_k>=m for all a_k in the sequence. Here, I want to state let M=1 and m=-1. because they are max and min y coordinate value for {x_k}, is this correct?

 

[morphism]

No, it's not. But you're close. What you've stated is not the correct definition of a bounded sequence in R^2.

 

[HallsofIvy]

Actually, I think that depends on the precise definition of "bounded" in R² in your text and, in turn, upon the norm used.


Morphism's point (I think) is that you have shown that the x and y of (x,y) are each bounded. For R² you must show that |(x,y)|= $\sqrt{x^2+ y^2}$ has an upper bound. That, of course, is easy.

It is also possible to define |(x,y)| as max |x|,|y| as a norm on R² in which case you are done! Of course, that is not the standard norm.

 

[pantin]

we just have 2 similar concepts on the text, one is

every bded sequence in R has a convergent subsequence

the second is what I stated before..

every bded sequence in R^n has a convergent subsequence

or should I use the definition of upper and lower bd?