Find a sequence $\{x_n\}$ such that for any $y \in \Bbb R$, there exists a subsequence $\{x_{n_i} \}$ converging to $y$. Now, I wish to prove the very same thing, but what comes to my mind, am I even able to define $$a_n=n\text{-th rational number}$$ because there is not "smallest" number from which to start. Actually the subsequence must be something like $a_{n_i}$ s.t. $n_i$ is increasing, how can i achieve such thing. When i enumerate $\mathbb{Q}$, how am i able to construct such sequence?
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An enumeration of $\mathbb Q$ is a bijection $b\colon\mathbb{N}\longrightarrow\mathbb Q$. Having that bijection, you can say that $b(1)$ is the first rational number, that $b(2)$ is the second rational number, and so on. So, yes, now you can talk about the $n$th rational number.
Note that in order to find a sequence such that, for each $n\in\mathbb N$, there is a subsequence converging to it, you don't need all this. Just consider the sequence$$1+1,1+\frac12,2+\frac12,1+\frac13,2+\frac13,3+\frac13,1+\frac14,2+\frac14,3+\frac14,4+\frac14,\ldots$$
José Carlos Santos
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So then, do i just define a new ordering on $\mathbb{Q}$? – Michal Dvořák Oct 09 '18 at 13:24
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1@MichalDvořák Yes. – José Carlos Santos Oct 09 '18 at 13:24
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1I've added another answer. – José Carlos Santos Oct 09 '18 at 13:33
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That actually makes things clear, thanks for the solution! – Michal Dvořák Oct 09 '18 at 13:38
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Actually, is there a way to write that in a closed form? (the sequence you gave) – Michal Dvořák Oct 10 '18 at 11:27
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@MichalDvořák Right now, I don't see how to do that. Perhaps that you might post it as another question. – José Carlos Santos Oct 10 '18 at 12:47