Convergence

Aug 2009
639
2
How do you prove that if a sequence is bounded then it has only one convegent subsequence?
 
Jul 2009
555
298
Zürich
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 \(\displaystyle x_n := (-1)^n\). The subsequence \(\displaystyle x_{2n}\) converges to 1 and the subsequence \(\displaystyle x_{2n+1}=-1\) converges to -1.
 
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May 2010
6
0
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.
 
Jul 2009
555
298
Zürich
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.