# Math Help - Proof involving convergent sequences.

1. ## Proof involving convergent sequences.

Prove that a sequence xn converges to x if and only if every subsequence of xn has a subsequence that converges to x.

I'm pretty stuck on this one. I know the "=>" is very easy, because every subsequence of a convergent sequence is convergent.

But what about the "<=" way? I have spent 4 hours trying this now and nothing I do seems to make it work. Anyone have any tips?

2. ## Re: Proof involving convergent sequences.

Originally Posted by nauticaricky
Prove that a sequence xn converges to x if and only if every subsequence of xn has a subsequence that converges to x.
I'm pretty stuck on this one. I know the "=>" is very easy, because every subsequence of a convergent sequence is convergent.
Suppose that every subsequence of $x_n$ converges to $x$.

Well $x_n$ is a subsequence of itself.

3. ## Re: Proof involving convergent sequences.

The second half of the statement is if every subsequence of xn has a subsequence that converges to x. Its talking about subsequences of the subsequences. Does that make a difference?

4. ## Re: Proof involving convergent sequences.

Originally Posted by nauticaricky
The second half of the statement is if every subsequence of xn has a subsequence that converges to x. Its talking about subsequences of the subsequences. Does that make a difference?
Well I did not read it carefully enough.
You can suppose that $(x_n)\not\to x$.
You do know how to write the negation of convergence?

Construct a subsequence of $(x_n)$ that does not converge to $x$.

5. ## Re: Proof involving convergent sequences.

Originally Posted by Plato
Well I did not read it carefully enough.
You can suppose that $(x_n)\not\to x$.
You do know how to write the negation of convergence?

Construct a subsequence of $(x_n)$ that does not converge to $x$.
So if I suppose what you said, I would be using the contrapositive. So I would want to find a subsequence of xn that itself has no convergent subsequence?

6. ## Re: Proof involving convergent sequences.

Originally Posted by nauticaricky
So if I suppose what you said, I would be using the contrapositive. So I would want to find a subsequence of xn that itself has no convergent subsequence?
If you understand the negation of convergence correct then you can say:
$\exists c>0$ and there is an increasing sequence of integers $N_1 such that $\left|x-x_{N_n}\right|\ge c$.
Can $\left(x_{N_n}\right)$ have subsequence that converges ?

7. ## Re: Proof involving convergent sequences.

Originally Posted by Plato
Can $\left(x_{N_n}\right)$ have subsequence that converges ?
I don't think so, because any subsequence will eventually go into the range of those N's, and then will diverge.

8. ## Re: Proof involving convergent sequences.

Originally Posted by nauticaricky
I don't think so, because any subsequence will eventually go into the range of those N's, and then will diverge.
Yes, that is the whole point! But can you construct all the details?

9. ## Re: Proof involving convergent sequences.

I am trying. XNn can have no convergent subsequence. Suppose it does. Then there is XNnk a subsequence and For any e<0, there is A such that |x-XNnk|<e if Nnk>A. But any subsequence will eventually contain Nn>A such that |x-xNn|>e, as you said. Is this correct? I am not sure how to use math symbols on here, but I hope you get the idea of what I'm saying.

10. ## Re: Proof involving convergent sequences.

Originally Posted by nauticaricky
I am trying. XNn can have no convergent subsequence. Suppose it does. Then there is XNnk a subsequence and For any e<0, there is A such that |x-XNnk|<e if Nnk>A. But any subsequence will eventually contain Nn>A such that |x-xNn|>e, as you said. Is this correct? I am not sure how to use math symbols on here, but I hope you get the idea of what I'm saying.
You do not have an intuitive notion of convergence.
To say that $(x_n)\to x$ it means that almost all of the terms of $x_n$ are 'close' to $x$.
But all the terms of $x_{N_n}$ are all at least $c$ away from $x$.
So if 'close to' means $\frac{c}{2}$ that is impossible.