1. ## Convergent subsequence

Prove or disprove:

Let $\displaystyle x_n \in R$ and suppose the sequence $\displaystyle \{\sin(x_n)\}$ converges. The sequence $\displaystyle \{{x_n}\}$ must contain a convergent subsequence.

I am not being able to figure out on how to prove or disprove this. Any help will be appreciated.

2. Originally Posted by serious331
Prove or disprove:

Let $\displaystyle x_n \in R$ and suppose the sequence $\displaystyle \{\sin(x_n)\}$ converges. The sequence $\displaystyle \{{x_n}\}$ must contain a convergent subsequence.

I am not being able to figure out on how to prove or disprove this. Any help will be appreciated.
False:

Let $\displaystyle x_k = 2\pi k$

3. Originally Posted by southprkfan1
False:

Let $\displaystyle x_k = 2\pi k$
Thanks southparkfan.

does the sequence $\displaystyle \{sin(2\pi k)\}$ converge to 0? because it is 0 for all values of k?

and how can we show that the subsequence of $\displaystyle \{(2\pi k)\}$ doesn't converge?

4. Originally Posted by serious331
how can we show that the subsequence of $\displaystyle \{(2\pi k)\}$ doesn't converge?
You are joking aren’t you? You cannot be serious.

5. A convergent sequence is always bounded!

6. Given any $\displaystyle M>0$, let $\displaystyle N=\frac{M}{2\pi}$. Then $\displaystyle k>N$ implies $\displaystyle 2\pi k > M$. And so $\displaystyle \lim_{k\to \infty} 2\pi k = +\infty$. It's pretty clear why any subsequence cannot converge since the tail of any subsequence will still tend to $\displaystyle +\infty$.

7. Originally Posted by Plato
You are joking aren’t you? You cannot be serious.
Okay, my question was retarded. Make as much fun as you can!

8. Originally Posted by serious331
Okay, my question was retarded. Make as much fun as you can!
I have to say Plato, I did think that was a tiny bit arrogant to ask. It should be obvious, but better to ask than to just assume!

If you wanna formalize why there can be no convergent subsequence, we can assume to the contrary. Let [tex]s_{n} = 2\pi n[/math[ and assume that there is a subsequence $\displaystyle (s_{n_{k}})$ such that given $\displaystyle \epsilon > 0$, there exists $\displaystyle N$ such that $\displaystyle n_{k} > N$ implies $\displaystyle |s_{n_{k}} - a| < \epsilon$, where $\displaystyle a$ is the finite number that we suppose the subsequence will converge to. And so this means for all $\displaystyle n_{k} > N$, $\displaystyle s_{n_{k}} < \epsilon + a$ (assume that $\displaystyle a > 0$ because if it wasn't, then clearly after some point the sequence would have to be negative). But since $\displaystyle \epsilon + a > 0$, we know for all $\displaystyle n > \frac{\epsilon + a}{2\pi}$, $\displaystyle 2\pi n > \epsilon + a$, and so rises the contradiction.

9. Originally Posted by Pinkk
I have to say Plato, I did think that was a tiny bit arrogant to ask. It should be obvious, but better to ask than to just assume!

If you wanna formalize why there can be no convergent subsequence, we can assume to the contrary. Let [tex]s_{n} = 2\pi n[/math[ and assume that there is a subsequence $\displaystyle (s_{n_{k}})$ such that given $\displaystyle \epsilon > 0$, there exists $\displaystyle N$ such that $\displaystyle n_{k} > N$ implies $\displaystyle |s_{n_{k}} - a| < \epsilon$, where $\displaystyle a$ is the finite number that we suppose the subsequence will converge to. And so this means for all $\displaystyle n_{k} > N$, $\displaystyle s_{n_{k}} < \epsilon + a$ (assume that $\displaystyle a > 0$ because if it wasn't, then clearly after some point the sequence would have to be negative). But since $\displaystyle \epsilon + a > 0$, we know for all $\displaystyle n > \frac{\epsilon + a}{2\pi}$, $\displaystyle 2\pi n > \epsilon + a$, and so rises the contradiction.
Thank you for the detailed explanation Pinkk. Not everyone is as smart as Mr. Plato.

I have another question. how can we prove that if the sequence $\displaystyle \{|x_n|\}$ converges, then the sequence $\displaystyle \{x_n\}$ has a convergent subsequence.

10. Since $\displaystyle |x_{n}|$ converges to let's say $\displaystyle a \ge 0$ (obviously the limit must be nonnegative since $\displaystyle |x_{n}| \ge 0$ by virtue of the absolute value), then we have some $\displaystyle N$ such that $\displaystyle n > N$ implies $\displaystyle ||x_{n}| - a| < 1$, so $\displaystyle |x_{n}|$ is bounded by $\displaystyle 1+a$ for all $\displaystyle n > N$. So let $\displaystyle M = \max\{|x_{1}|, |x_{2}|, ... , |x_{N}|, 1+a\}$. Then we have $\displaystyle |x_{n}|\le M$ for all $\displaystyle n$, so by definition, $\displaystyle (x_{n})$ is a bounded sequence. Thus, by the Bolzano-Weierstrass Theorem, $\displaystyle (x_{n})$ has a convergent subsequence.