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I have been working on this problem which I stumbled upon:

Consider the enumeration $(q_n)_{n\in\mathbb{N}}$ of the set $\mathbb{Q}\cap[0,1].$ Does this sequence have convergent subsequences? If yes, then what is the limit to which these subsequences converge?

My attempt summarized:

Every limit point of a sequence is a limit of a subsequence thereof, which implies that $\mathbb{Q}\cap[0,1]$ has infinitely many convergent subsequences. I still can't think of what these limits might be. Since our enumeration contains rational numbers only, between $0$ and $1$ and these rational numbers lie closely in R, we could say that our limit points are all $x \in [0,1]$.

If I am correct to some degree, I can't really construct a more formal and easy to understand proof. I'd appreciate any help.

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    $\begingroup$ Yes, you can construct a sequence of rational numbers converging to any real number. See for example this: math.stackexchange.com/questions/209001/… $\endgroup$
    – wormram
    Jun 23, 2021 at 23:19
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    $\begingroup$ Bolzano-Weierstrass says that every bounded sequence has a convergent sub-sequence. $\endgroup$
    – Doug M
    Jun 23, 2021 at 23:30
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    $\begingroup$ The key point is that if if $x \in [0, 1]$, then any open interval $(x - \delta, x + \delta)$ contains infinitely many rationals, and so the $q_n$ must visit each such interval infinitely often. So you can find an increasing sequence $n_1 < n_2 < \ldots$, such that $q_{n_k} \in (x - 1/k, x + 1/k)$ for each $k = 1, 2, \ldots$ This gives you a subsequence of the $q_n$ that tends to $x$. $\endgroup$
    – Rob Arthan
    Jun 23, 2021 at 23:47

1 Answer 1

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Since, by virtue of the \color{magenta{Bolzano-Weierstrass theorem}, every bounded sequence has a convergent subsequence, this sequence too has a convergent subsequence. Since several of them can be obtained and the actual limit will depend upon the subsequence chosen, a couple of examples should suffice.

Consider $1, \frac{1}{2}, \frac{1}{3}, ...$. This one converges to $0$.

Consider $1,1,1,...$. This one converges to $1$.

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  • $\begingroup$ But the second sequence does not arise as a subsequence of an enumeration of the rationals, nor is the first very likely to do either. So your "examples " are spoiling an otherwise OK answer. $\endgroup$ Sep 14, 2021 at 18:25

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