Does this sequence have convergent subsequences?

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.

• Yes, you can construct a sequence of rational numbers converging to any real number. See for example this: math.stackexchange.com/questions/209001/… Jun 23, 2021 at 23:19
• Bolzano-Weierstrass says that every bounded sequence has a convergent sub-sequence. Jun 23, 2021 at 23:30
• 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 $x$. Jun 23, 2021 at 23:47

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