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I thought the set of natural number functions would be of the same cardinality as the countably infinite product of $\mathbb{N}$, which is countable.

Each natural number function can be identified with an infinite-tuple of $\mathbb{N}$ by letting the $i$th entry be the image of the number $i$ under the function.

Asaf Karagila
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user125093
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5 Answers5

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Cantors diagonalisation method: If it is countable then it is $F=\{f_1,f_2,\ldots\}$. But the function $g(i):=f_i(i)+1$ is not in $F$. A contradiction to the initial assumption.

Proof:

For every $i $ we have $$g(i)=f_i (i)+1 \implies g (i)\ne f_i (i) \implies g \ne f_i $$ and so $g \notin F$.

miracle173
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Every real number in $(0,1)$ has either one or two decimal representations. (Some have two, like $0.1000\ldots$ which is equal to $0.0999\ldots$.) But any way, each real number in $(0,1)$ gives at least one decimal representation. And that decimal representation can be viewed as a function from $\mathbb{N}$ to $\mathbb{N}$ by taking the $n$th digit after the decimal. (Actually, it's more restrictive: from $\mathbb{N}$ to $\{0,1,2,3,4,5,6,7,8,9\}$.)

For example, $0.72465\ldots$ can be used to define a function $$\begin{align}1&\mapsto7\\ 2&\mapsto2\\ 3&\mapsto4\\ 4&\mapsto6\\ 5&\mapsto5\\ \vdots&\phantom{\mapsto{}}\vdots\end{align}$$

So the set of functions from $\mathbb{N}$ to $\mathbb{N}$ contains a subset that is as large as the real interval $(0,1)$. (And seemingly even stronger, so does the set of functions from $\mathbb{N}$ to $\{0,1,2,3,4,5,6,7,8,9\}$. And seemingly even stronger still, if we used binary instead of decimal, so does the set of functions from $\mathbb{N}$ to $\{0,1\}$.)

2'5 9'2
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The cardinality of the natural number functions is $\mathbb{N}^{\mathbb{N}} \geq 2^{\mathbb{N}}$ and $2^{\mathbb{N}}$ is uncountable

WLOG
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The countably infinite product of $\mathbb{N}$ is not countable, I believe, by Cantor's diagonal argument.

Eric Auld
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  • I thought it's the case that a countable product of countable sets is countable. – user125093 Feb 22 '14 at 06:38
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    Countable union of countable sets is countable. Perhaps that is what you had in mind. – André Nicolas Feb 22 '14 at 06:40
  • Andre, yeah that must be the source of the confusion. Thank you! – user125093 Feb 22 '14 at 06:41
  • I have a question. Is the union of the set of natural number functions countable? – Valentino Feb 19 '15 at 18:15
  • @valentino That is actually the same thing as the countably infinite product of $\mathbb{N}$. Both can be abbreviated $\mathbb{N}^\mathbb{N}$. This is the same as how $\mathbb{R}^3$ can be seen as the space of functions from ${1,2,3} \to \mathbb{R}$ – Eric Auld Feb 19 '15 at 18:25
  • I'm sorry! That was somewhat vague I meant the set of functions ${f\mid f:\mathbb{N}\to\mathbb{N}}$ is the union of this set countable? – Valentino Feb 19 '15 at 18:29
  • @valentino That's what I mean as well. :) So no, by the diagonalization argument. – Eric Auld Feb 20 '15 at 13:41
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Let's consider the following set: $F = \{ f:D \in 2^\mathbb{N} \rightarrow \mathbb{N} \ | \ f(x) = x \}$;

Notice how there is a function $f$ for each possible set of natural numbers residing inside $2^\mathbb{N}$, which will allow us to draw the following conclusion: $|F| = |2^\mathbb{N}|$.

Now, observe how $F$ is a much smaller set than $\mathbb{N}^\mathbb{N}$(because it contains only functions of the form $f(x)=x$), but even so, it's uncountable. Therefore $\mathbb{N}^\mathbb{N}$ is evidently uncountable since it's a superset of $F$.

$F \subseteq \mathbb{N}^\mathbb{N}, |F| = |2^\mathbb{N}| \implies |\mathbb{N}^\mathbb{N}|=|2^\mathbb{N}| \ \ \ \square $

Alex B.
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