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Let $\{x_n\}$ be a sequence such that $\forall k > 1 \in \Bbb N$ its subsequence $\{x_{pk}\}$ converges to $1$.

Does $\{x_n\}$ converge to $1$?

I'm a bit confused by this problem since no constraints on $p$ are given. The only reasonable way seems to consider $p$ a natural number. Intuitively the statement seems to be true. Let's try to consider different values for $k$: $$ k = 2:\{x_{pk}\} = \{x_2, x_4, \dots, x_{2p} \}\\ k = 3:\{x_{pk}\} = \{x_3, x_4, \dots, x_{3p} \}\\ \cdots\\ k = n:\{x_{pk}\} = \{x_n, x_{2n}, \dots, x_{np} \}\\ $$

All of the sequences above converge to $1$, namely: $$ \lim_{p\to\infty}x_{2p} = 1\\ \lim_{p\to\infty}x_{3p} = 1\\ \cdots\\ \lim_{p\to\infty}x_{np} = 1\\ $$

At this point, I was thinking about the union of all the sequences above. All of them are subsequences of $x_n$, and since $k$ is an arbitrary natural number we may consider the following set: $$ X = \bigcup\limits_{k=1}^{\infty} x_{pk} $$

Moreover, since every sequence in the union converges to $1$ then it must follow that $X$ also converges to one. But $X$ is essentially equivalent to $\{x_n\}$ which would imply that: $$ \lim_{n\to\infty}x_n = 1 $$

I'm not sure the reasoning above might be applied to the problem. So I would like to ask for verification or a valid solution in case mine is wrong. Also, in case the idea in mine is fine then is it rigorous enough to consider the proof complete? Thank you!

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Of course not (in general). Take $x_n$ to be $0$ if $n$ is prime and $1$ otherwise.

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  • $\begingroup$ Thank you for your answer. Sorry for the confusion but I've missed adding an important notice about $x_{pk}$. Which is $\{x_{pk}\}$ is a subsequence of $\{x_n\}$. $\endgroup$ – roman Mar 15 at 17:30
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    $\begingroup$ I don't get the relevance $\endgroup$ – mathworker21 Mar 15 at 17:47
  • $\begingroup$ You are right, after thinking more carefully on this $x_n$ is indeed not necessarily convergent, even given the fact it has convergent subsequences. $\endgroup$ – roman Mar 15 at 17:51
  • $\begingroup$ $x_n$ is not convergent $\endgroup$ – mathworker21 Mar 15 at 18:23

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