# Rational Numbers and Sequences

Can the rational numbers be arranged in a sequence? If so, consider any such sequence of all the rational numbers. Show that every real number is a subsequential limit of this sequence.

Since rational number is countably infinite, I see that rational numbers can be arranged in a sequence. But I'm lost how to proceed

Let $$q_1, q_2, \ldots$$ be an enumeration of the rational numbers.

Consider some real number $$r$$.

• Suppose $$r$$ has an infinite decimal expansion. By considering all possible truncations of the decimal expansion, you obtain a sequence of rational numbers $$q'_1, q'_2 ,\ldots$$ that converges to $$r$$. You can use this to get a subsequence of your original sequence $$(q_i)$$ that converges to $$r$$. (Start your new sequence with $$q'_1$$, which must appear in the original sequence somewhere. If $$q'_2$$ appears before $$q'_1$$ in the original sequence, skip it; else append $$q'_2$$ to your sequence. Do the same for $$q'_3$$, and so on.)

• Suppose $$r$$ has a finite decimal expansion. You can do a silly trick to give it an infinite decimal expansion (e.g. $$1.2 = 1.1999\cdots$$, or $$-2.348 = -2.3479999\cdots$$) and then perform the above procedure.

• Your second bullet point starts with "If $r$ is rational, it has a finite decimal expansion." However, most rational numbers have an infinite, periodic decimal expansion instead (e.g., $\frac{1}{3} = 0.3333\cdots$). Perhaps this is what you meant, but the rest of the bullet point indicates otherwise. – John Omielan Nov 12 at 3:12
• @JohnOmielan Sorry, that was a silly mistake on my part. I've edited my answer. – angryavian Nov 12 at 5:04

Hint for second part:

For a real $$r$$ and an $$\epsilon \gt 0$$, is $$Q \cap (r-\epsilon, r + \epsilon)$$ finite?

Let $$(q_n) = q_1, q_2, \ldots$$ be an enumeration of the rational numbers.

Consider some real number $$a$$. We can recursively define a strictly increasing function

$$\tag 1 \alpha: \Bbb N \to \Bbb N$$

as follows:

$$\quad \alpha(1) = \text{the smallest natural number } k \text{ such that } \big [ q_k \lt a \big ] \land \big [ a - q_k \lt 1 \big ]$$

Assume $$\alpha$$ is defined on $$\{1,2,\dots,n\}$$. Define

$$\quad \alpha(n+1) = \text{the smallest natural number } k \text{ such that }$$
$$\quad \quad \big [k \gt \alpha(n) \big ] \land \big [q_k \gt q_{\alpha(n)} \big ] \land \big [q_k \lt a \big ]\land \big [a - q_k \lt \frac{1}{n+1}\big ]$$

It is easy to see that the subsequence $${\big ( q_{\alpha(n)}\big )}_{ n \in \Bbb N}$$ is strictly increasing and converges to $$a$$.