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I was talking with a friend of mine and we wonder how many sequences of rational numbers on $[0,1]$ there exists. My first attempt was to consider that every sequence like that must be a subset of $\mathbb{Q} \times \mathbb{N}$ and because this is an infinite countable set, then all their subsets (functions) must be an infinite countable.

But he said to me that this is wrong, because all the functions from $\mathbb{N}$ to $\mathbb{Q}$ must be |$\mathbb{Q}^{\mathbb{N}}|$

After a while, I thought that my attempt is really wrong because all the sequences like that lives in the power set of $\mathbb{Q} \times \mathbb{N}$.

But right now I'm thinking that if the number of all sequences of rational numbers on $[0,1]$ has the cardinality of real numbers, then we can choose (can we, right?) an element of each sequence and label it with a real number, then we can construct a sequence of rational labelled with real numbers which is a contradiction, then the number of all sequences of rational numbers on $[0,1]$ must be the same of the natural numbers.

Do anyone know the right answer? (if it exists)

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  • $\begingroup$ Note that the same rational can appear in multiple sequences . . . $\endgroup$ – Noah Schweber Feb 8 '16 at 7:30
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Using your notation, $|\mathbb{Q}^{\mathbb N}| \ge |\mathbb{R}|$ because any real is the limit of a sequence of rationals. The other way around, $|\mathbb{Q}^{\mathbb N}| \le |\mathbb{R}|$ because any sequence of rationals can be mapped to a real, for example by building one off the concatened expansion in some arbitrary base of the numerators and denominators. Thus $|\mathbb{Q}^{\mathbb N}| = |\mathbb{R}|$ so the number of rational sequences has the same cardinality as $\mathbb R$.

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  • $\begingroup$ You won't be able to find a number suitable for "in some arbitrary base" as the numerators and denominators can be arbitrarily large. $\endgroup$ – mathcounterexamples.net Feb 8 '16 at 7:46
  • $\begingroup$ @mathcounterexamples.net Of course they can be arbitrarily large, and I don't see why that would be a problem. Just to frame your objection, are you disputing that $|\mathbb{Q}^{\mathbb N}| \le \mathbb{R}$ or just the possible construction that I hinted? $\endgroup$ – dxiv Feb 8 '16 at 7:54
  • $\begingroup$ Just the possible construction that you hinted. $\endgroup$ – mathcounterexamples.net Feb 8 '16 at 10:14
  • $\begingroup$ @mathcounterexamples.net Let $\{p_n / q_n\}$ be a finite sequence of rationals written as irreducible fractions. Let $r_n \in [0, 1]$ be the real defined by concatenating, after the decimal point, the digits of $p_1, q_1, p_2, q_2 ... p_n, q_n$. The sequence $r_n$ is increasing with $n$ and is bounded above by $1$, so it has a limit $r \in [0, 1]$ as $n \to \infty$. Moreover, two different rational sequences $\{p_n / q_n\}, \{p'_n / q'_n\}, $ will map to reals $r_n, r'_n$ that differ starting at a digit, say the $k^{th}$, so $|r_n - r'_n| \ge 10^{-(k+1)}$, thus their limits will be different. $\endgroup$ – dxiv Feb 8 '16 at 16:18
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It has to be uncountable because you can represent any real number with a sequence of rationals (in particular only 0, 1 (base 2)).

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