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Let $\{x_n\}$ be a sequence that does not converge, and let $M$ be a real number. I am thinking about how to show that there exists a subsequence of $\{x_n\}$ that converges to $M$. If I could show that there is a rational subsequence, then could I use the fact that every real number is the limit of a convergent sequence of rational numbers?

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    $\begingroup$ $x_n=n\pi$ is a sequence of real numbers that does not converge. It has no convergent subsequences of rational numbers, and no subsequences that converge to anything (except infinity). $\endgroup$ Apr 18 '18 at 13:42
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Not every sequence has a rational subsequence. For example, the sequence given by $$ x_n = \frac{\sqrt{2}}{n} $$ is a convergent sequence of irrrational numbers.

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  • $\begingroup$ How then would I show that if ${x_n}$ be a sequence of reals that does not converge, there exists a subsequence of ${x_n}$ that converges to M? $\endgroup$ Apr 18 '18 at 13:57
  • $\begingroup$ You don’t, because that isn’t a true statement, unless you are given more information (for example, the sequence is bounded) $\endgroup$ Apr 18 '18 at 16:19
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There are sequences of real numbers without a subsequence of rational numbers. Choose $x_{2n-1} = \sqrt{2}$ and $x_{2n} = \sqrt{3}$. You can't show that this has a subsequence converging to $M = 0$ because it does not.

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Sequence:

$x_n =\sqrt {p_n}$, where $p_n$ are the primes enumerated.

Is there a rational subsequence ?

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