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It is well known that:

A sequence of real numbers converges $\{a_n\}$ to p, if and only if every subsequence $\{a_{n_k}\}$ converges to p.

I am wondering if this similar statement holds.:

A sequence of real numbers $\{a_n\}$ converges to p if and only if every subsequence that does converge, converges to p.

The $\rightarrow$ part follows from the first statement, it is the $\leftarrow$ part I have a problem with. I need to prove that:

All subsequences that does converge, converges to p $\rightarrow$ the main sequence converges to p.

Is this true?, also, does the answer depend on wheter we allow convergence to $-\infty,\infty$ or not?(I think I need to assume this, because it seems the statement is false if not, considering 1,2,1,3,1,4,1,5,1,6,1,7......)

The problem I have with proving this, is that in the first statement we knew that all subsequences converges, and the main sequence is also a subsequence, so we are done, but now we do not know if the main sequence converges.

PS: I am assuming that atleast one subsequence converges, but I guess that if we include convergence to $-\infty,\infty$, this does not matter?

You do not need to give me the entire proof if you don't want, maybe only a hint if you know?(if the statement is true)

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I don't think this is true. Consider the sequence $0,1,0,2,0,3,0,4,0,5,...$ Any sub-sequence that converges must converge to $0$ but the sequence itself does not converge.

Edit: If we count $-\infty,\infty$ as possible limit points then I think this result is true. Suppose every subsequence that converges, converge to a point $x$. Look a subsequence that doesn't converge. If this subsequence goes to $-\infty,\infty$ then we have violated the assumption. Therefore, this subsequence must be bounded. Then by Bolzano Weierstrass Theorem, there must exist a convergent subsequence of this subsequence. This subsequence must converge to $x$. Next, recall that if a sequence in a bounded interval has exactly one limit point then that sequence must converge to that limit point. Therefore, the result follows.

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  • $\begingroup$ If you include $+\infty$ then you have one subsequence that converges to $0$ and another that converges to $+\infty$, which arguably violates the OP's hypothesis. $\endgroup$ – Ian Dec 23 '14 at 2:27
  • $\begingroup$ Opps, I didn't see OP's edits. $\endgroup$ – 1-___- Dec 23 '14 at 2:32
  • $\begingroup$ Thank you, but there is one thing with the proof I have a question about. "If this subsequence goes to −∞,∞ then we have violated the assumption. Therefore, this subsequence must be bounded. ", even if the subsequence does not go to infinity or minus infinity, it may still be unbounded? $\endgroup$ – user119615 Dec 23 '14 at 2:59
  • $\begingroup$ A sequence is either bounded or not bounded. If it is not bounded then it must go to $\infty$ or $-\infty$. $\endgroup$ – 1-___- Dec 23 '14 at 3:00
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    $\begingroup$ Thank you very much for all your help and time. It was very smart and creative for you to come up with this strategy! Have a merry christmas! $\endgroup$ – user119615 Dec 23 '14 at 3:57

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