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Let $(a_n)$ be a sequence of rational numbers, where all rational numbers are terms. (i.e. enumeration of rational numbers)

Then, is there any convergent sub-sequence of $(a_n)$?

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    $\begingroup$ Hint: It has a bounded subsequence. $\endgroup$ – André Nicolas Jun 3 '16 at 18:43
  • $\begingroup$ Even stronger, for any rational number $q$ there is a subsequence of the sequence $(a_n)$ converging to $q$! $\endgroup$ – Johannes Huisman Jun 3 '16 at 18:51
  • $\begingroup$ Since it may be interpreted both ways (see some comments below the answers), do you mean convergence to a rational, or to any real? $\endgroup$ – Clement C. Jun 3 '16 at 19:10
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The argument given by Elio Joseph can be refined in order to prove that for any given real number, there always exists a subsequence which converges to it.

For that, let $x \in \mathbb{R}$ be a real number. Consider the interval $[x-1,x+1]$. It contains infinitely many rationals, but in particular it contains one rational, which is some $a_{n_1}$ of our enumeration.

Now, having chosen $a_{n_i}$, we consider the interval $[x-\frac{1}{i+1}, x+\frac{1}{i+1}]$. This contains infinitely many rationals. In particular, we have that it contains one rational which is not one of the first $n_i$ of our enumeration. We take such rational to be our $a_{n_{i+1}}$.

It is clear that $a_{n_i} \to x$.

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Even better - for every real $x$, there is a subsequence of $(a_n)$ converging to $x$.

Indeed, start by letting $a_{n_1}$ be any element of the sequence in $(x-1,x+1)$. Then, given $a_{n_1},\ldots,a_{n_k}$, since there are infinitely many rationals in each interval, there must exist $n_{k+1}>n_k$ such that $a_{n_{k+1}}\in(x-\frac1{k+1},x+\frac1{k+1})$. Inductively this creates a subsequence $(a_{n_k})$ which converges to $x$.

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Yes there is.

Consider the interval $I:=[0,1]$. Since it contains infinitely many rationals, your sequence will have value in $I$ forever.

So you can extract the sub-sequece such that all the values $(a_{\varphi(n)})$ of the new sequence are in $I$ :

$$\forall n \quad a_{\varphi(n)}\in I.$$

Since $I$ is a compact and $(a_{\varphi(n)})$ is a sequence with values in this compact, you can extract of convergent sub-sequence of $(a_{\varphi(n)})$.

Which gives you the convergent sub-sequence of $(a_n)$ you wanted.

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  • $\begingroup$ Could you elaborate a bit, for example $[0,1]\cap \mathbb Q$ is not compact so you might want to add some of your assumptions $\endgroup$ – user190080 Jun 3 '16 at 18:49
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    $\begingroup$ @user190080 It's still a sequence in $[0,1]$, so there is a subsequence converging in $[0,1]$. The limit needs not be in $\mathbb{Q}$. $\endgroup$ – Clement C. Jun 3 '16 at 18:55
  • $\begingroup$ @ClementC. I think that user190080 assumed that the limit needed to be in $\mathbb{Q}$. Both interpretations are somewhat valid, since OP does not clarify this. However, I think that in the way the question is worded, the most plausible interpretation is that of user190080, since no mentioning of real numbers was made. $\endgroup$ – Aloizio Macedo Jun 3 '16 at 19:01
  • $\begingroup$ @ClementC. yes, that's true. I wasn't exactly criticizing the actual argument via a bounded sub-sequence but the used wording. $\endgroup$ – user190080 Jun 3 '16 at 19:01

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