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?

Re: Proof involving convergent sequences.

Quote:

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 converges to .

Well is a subsequence of itself.

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?

Re: Proof involving convergent sequences.

Quote:

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 .

You do know how to write the negation of convergence?

Construct a subsequence of that does not converge to .

Re: Proof involving convergent sequences.

Quote:

Originally Posted by

**Plato** Well I did not read it carefully enough.

You can suppose that

.

You do know how to write the negation of convergence?

Construct a subsequence of

that does not converge to

.

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?

Re: Proof involving convergent sequences.

Re: Proof involving convergent sequences.

Quote:

Originally Posted by

**Plato** Can

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.

Re: Proof involving convergent sequences.

Quote:

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?

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.

Re: Proof involving convergent sequences.