Is the set of all functions $\mathbb N$ $\to$ $\mathbb Q$ uncountable?

Question

Is the set of all functions from $\mathbb N$ to $\mathbb Q$ uncountable?

Answer 1

Hint For any real number $x$ you can define a function $f_x : \mathbb N \to \mathbb Q$ by

$$f_x(n) =\frac{\lfloor nx \rfloor}{n} \,.$$

Now, the mapping $x \to f_x$ is a one to one function from $\mathbb R$ to your set.

Answer 2

The cardinality of the set of all functions from $\mathbb N$ to $\mathbb Q$ is $$|\mathbb Q^{\mathbb N}|=|\mathbb Q|^{|\mathbb N|}=\aleph_0^{\aleph_0} = 2^{\aleph_0} = \mathfrak c > \aleph_0.$$ So this set is uncountable.

See also:

• Is $\aleph_0^{\aleph_0}$ smaller than or equal to $2^{\aleph_0}$?
• Is the sets of all maps from $\mathbb{N}$ to $\mathbb{N}$ countable?
• Comparing the two cardinals $\aleph_0^{\aleph_0}$ and $2^{\aleph_0}$
• What is $\aleph_0$ powered to $\aleph_0$?
• What's the cardinality of all sequences with coefficients in an infinite set?

Some of the above links are taken from: Overview of basic results on cardinal arithmetic

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If you do not want to compute the cardinality of this set, just show that it is not countable, you can use diagonal argument. See, for example, some answers in these questions:

• The set of natural number functions is uncountable
• The set of all functions from $\mathbb{N} \to \{0, 1\}$ is uncountable?
• Show that the set of functions $\mathbb{N}\to\{0,1\}$ is not countable
• Infinite Cartesian product of countable sets is uncountable

(You should think a little bit about why showing this for functions $\mathbb N\to\mathbb N$ and $\mathbb N\to\mathbb Q$ is equivalent.)

Answer 3

Yes, in fact you can see it in this way: split the set $A=\{f\colon \mathbb N\to \mathbb N\}$ into $B=\{f\colon \mathbb N\to \mathbb N\colon f(n)=9 \mbox{ eventually}\}$ and $C=A\setminus B$. Then you can map bijectively $C$ to $[0,1)$ by sending $f$ to $0,f(1)f(2)f(3)\ldots$ On the other hand, $B$ is countable because it is a countable union of countable sets.