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I need to find the cardinality of the following set:

$$ A = \{f \in \mathbb N \to \mathbb N|\forall n, m \in \mathbb N.(n \geq m) \to (f(n) \leq f (m))\} $$

I'm not sure if $A$ is countable or not. I tried to find an injective function from $P(N)$ to $A$ but couldn't find one.

Any ideas?

Henrik supports the community
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user401516
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1 Answers1

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Yes, A is countable. A function $f \in A$ is unique identified by $f(1)$, and its values on a (finite) set of arguments where it changes (the size of that set can be at most $f(1)$, because is it non-increasing). Call this set $V$.

Let $p_i$ denote the i-th prime. Each $f \in A$ maps to a unique natural number $x$ which can be constructed as follows: start with $x=2^{f(1)}$, and for each $v \in V$, multiply $x$ by $p_v^{f(v)}$.

This provides an injective function from $A$ to the natural numbers; the other direction is trivial. So $A$ is a countable set.

Glorfindel
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  • thanks! brilliant. Im still wondering about the other direction though, do I need another function or is there an easier way? – user401516 Dec 30 '16 at 14:52
  • The other direction is trivial; for $n \in \mathbb N$, take $f$ to be the constant function with value $n$. – Glorfindel Dec 30 '16 at 14:54
  • Are you sure it is an injective function? it sounds right but how can I prove that? – user401516 Dec 30 '16 at 16:27
  • It uses the same idea as for proving that the set of finite subsets of $\mathbb N$ is countable, which is explained here. These proofs depend on the fact that every natural numbers has a unique prime factorization (and that there are infinitely many primes). – Glorfindel Dec 30 '16 at 16:57