How many injective functions are there in $\mathbb{N}^\mathbb{N}$?

Asked Aug 08 '22 at 16:13

Active Aug 08 '22 at 17:21

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I was wondering what's the cardinality of $\{f \in \mathbb{N}^\mathbb{N} \mid f \text{ is injective}\}$? I know it's uncountable, but what 'type' of uncountable? Is it the same cardinality of all surjective functions over $\mathbb{N}$? What about of all bijective functions over $\mathbb{N}$?

Thanks in advance!

edited Aug 08 '22 at 17:21

emacs drives me nuts

10,778

asked Aug 08 '22 at 16:13

tcb93

4

The set of bijections has cardinality $2^{\aleph_0}$ (same as the power set, same as the set of all functions from $\mathbb{N}$ to itself). As that is contained in the set of injections, and in the set of surjections, the latter sets also have the same cardinality. – Arturo Magidin Aug 08 '22 at 16:27
See here for the cardinality of the set of all functions. – Arturo Magidin Aug 08 '22 at 16:29
see my answer here for the set of bijections https://math.stackexchange.com/a/4418195/399263 – zwim Aug 08 '22 at 18:22
You can find some posts about the set of bijections on this site, such as "Is symmetric group on natural numbers countable?" and various questions linked there. And, naturally, you have $|\operatorname{Bij}(\mathbb N,\mathbb N)|\le |\operatorname{Inj}(\mathbb N,\mathbb N)| \le \aleph_0^{\aleph_0}$. – Martin Sleziak Aug 09 '22 at 08:53

How many injective functions are there in $\mathbb{N}^\mathbb{N}$?

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