# Convergence

• May 11th 2010, 01:25 AM
alexandrabel90
Convergence
How do you prove that if a sequence is bounded then it has only one convegent subsequence?
• May 11th 2010, 04:20 AM
Failure
Quote:

Originally Posted by alexandrabel90
How do you prove that if a sequence is bounded then it has only one convegent subsequence?

You can't because that statement is false. For a trivial example consider $x_n := (-1)^n$. The subsequence $x_{2n}$ converges to 1 and the subsequence $x_{2n+1}=-1$ converges to -1.
• May 11th 2010, 06:27 AM
ogajawasungu
I think you must have missed something;I assume your sequence is bounded and monotone,but still this has infinite convergent sub-sequences converging to one limit.
• May 11th 2010, 07:37 AM
Failure
Quote:

Originally Posted by ogajawasungu
I think you must have missed something;I assume your sequence is bounded and monotone,but still this has infinite convergent sub-sequences converging to one limit.

Monotonicity was not part of the original problem statement, however. Also, if a sequence converges (as is the case if it is bounded and monotone), then every of its subsequences converges to the same limit (I don't even consider finite subsequences here, since talk of convergence makes hardly any sense for finite sequences).
And just to mention another possible criticism of the original question: if a sequence has a convergent subsequence, it necessarily has infinitely many (but, of course, it might be that they all converge to the same limit). And of course, if a sequence converges, all of its subsequences converge, and converge to the same limit at that.
• May 13th 2010, 07:29 AM
hollywood
I think we're looking for this:

Bolzano-Weierstrass theorem - Wikipedia, the free encyclopedia

Any bounded sequence has a convergent subsequence. But not necessarily only one, as Failure's counter-example shows.

- Hollywood