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"A bounded sequence of real numbers converges to x if every convergent subsequence of the sequence converges to x." I require a counterexample to prove that the theorem fails if the hypothesis that the sequence is bounded is dropped. Also, am I correctly interpreting the question?

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    $\begingroup$ Welcome to Math.SE! It is not clear what you mean with your last sentence: what is the question you are referring to? $\endgroup$
    – Hrodelbert
    Nov 3, 2015 at 9:16
  • $\begingroup$ @Hrodelbert My guess is that the OP is asked to show that the theorem fails if the hypothesis that the sequence is bounded is dropped. So in the last sequence they are probably asking whether this is the same as finding a sequence with the above properties.(Which is, as far as can say, correct. Perhaps it would be worth adding that unbounded sequence cannot be convergent, to make clear that it is indeed a counterexample.) $\endgroup$ Nov 3, 2015 at 13:29
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    $\begingroup$ @MartinSleziak I agree that this is the most likely explanation, but without the OP telling us so we cannot be sure. $\endgroup$
    – Hrodelbert
    Nov 3, 2015 at 13:52
  • $\begingroup$ Yes. That is what I mean $\endgroup$
    – Non-Being
    Nov 3, 2015 at 21:26

3 Answers 3

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Consider the sequence $(x_n) \in {}^{\def\N{\mathbf N}\N}\def\R{\mathbf R}\R$ given by $$ x_n = \begin{cases} n & \text{if $n$ is even}\\ 0 & \text{if $n$ is odd} \end{cases} $$ Then every convergent subsequence of $(x_n)$ converges to 0, but $(x_n)$ does not converge.

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Consider the trivial example $(x_n)_{n\in\mathbb{N}}$ given by $x_n=n$, which obviously does not converge. This sequence has no convergent subsequence, so the condition from the theorem holds vacuously for any $x$ (i.e. for any $x\in\mathbb{R}$, any convergent subsequence of $(x_n)$ converges to $x$).

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  • $\begingroup$ But does an example which is vacuously true suffice? $\endgroup$
    – Non-Being
    Nov 3, 2015 at 21:33
  • $\begingroup$ Technically yes, since the theorem does not require a converging subsequence to actually exist. At least not explicitly - their existence follows from the boundedness condition, which we are throwing away here. $\endgroup$ Nov 3, 2015 at 22:37
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By definition, every unbounded sequence contains a sub-sequence that diverges to plus or minus infinity.

Thus not all sub-sequences can converge to the same value.

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  • $\begingroup$ Yes. I know that. But I am talking about a particular case where the subsequences which are convergent converge to dance limit. $\endgroup$
    – Non-Being
    Nov 3, 2015 at 21:30

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