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Say yn is a bounded sequence of real numbers. If all the convergent subsequences converge to the same limit, say A, then apparently yn is convergent and converges to A.

Can I just show that if bounded sequence yn does not converge to limit A then it has a convergent subsequence whose limit is not equal to A and show that this is a contradiction? Thanks!

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  • $\begingroup$ I'm trying to understand what's the connection between the first two lines and the last two ones...but I can't. $\endgroup$
    – Timbuc
    Feb 17 '15 at 20:49
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Yes you can attempt to take that route for a proof, although you would call it the contrapositive probably instead of proof by contradiction. Note however probably the only easy way to show the contrapositive is to use theorems about convergent subsequences of sequences in compact sets, and define an appropriate sequence and compact set in terms of $A$.

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  • $\begingroup$ Ok cool so I can show a contradiction to the contrapositive and it will work. I know how to do this $\endgroup$ Feb 17 '15 at 21:07

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