1. ## Sequences/Subsequences

I need to show that
if every subsequence of X = (xn) has a subsequence that converges to 0, then the lim X = 0.

Any help would be greatly appreciated!

2. Originally Posted by CrazyCat87
I need to show that
if every subsequence of X = (xn) has a subsequence that converges to 0, then the lim X = 0.

Any help would be greatly appreciated!
All this is saying is that every subsequence of your sequence is convergent. What does that mean?

3. Originally Posted by CrazyCat87
I need to show that
if every subsequence of X = (xn) has a subsequence that converges to 0, then the lim X = 0.
Is a sequence a subsequene of itself?

4. It's saying that if the subsequence converges to 0, then it must mean that the sequence also converges to 0.

5. Originally Posted by CrazyCat87
It's saying that if the subsequence converges to 0, then it must mean that the sequence also converges to 0.
No that is not what you first posted.
You posted: If each subsequence of $x_n$ converges to 0 then the sequence $x_n$ converges to 0
Well $x_n$ is a subsequence of $x_n$.

6. hintrove that limsup x_n=0, and liminf x_n=0.

7. I would prove it by showing that a sequence $\{x_n\}$ which does not converge to $0$ has a subsequence which has no subsequence converging to $0$. This is easy. The sequence $\{x_n\}$ must be frequently outside of some $\epsilon$-neighbourhood of $0$; take the subsequence which consists of those points.

8. Originally Posted by Bruno J.
I would prove it by showing that a sequence $\{x_n\}$ which does not converge to $0$ has a subsequence which has no subsequence converging to $0$. This is easy. The sequence $\{x_n\}$ must be frequently outside of some $\epsilon$-neighbourhood of $0$; take the subsequence which consists of those points.
So you're saying to prove by contradiction?
If so then wouldn't you have to prove that if there exists a subsequence of X that has no subsequence that converges to 0, then X does not converge to 0?

9. Originally Posted by CrazyCat87
So you're saying to prove by contradiction?
If so then wouldn't you have to prove that if there exists a subsequence of X that has no subsequence that converges to 0, then X does not converge to 0?
no, what BrunoJ suggested was a proof by contrapostitive. he pretty much gave you the proof. you can formalize it if you want