Find cardinality of a set {$f \in \mathbb{N}^{\mathbb{N}}|f\le h$} where $h(n)=n+1$

Question

On a set $\mathbb{N}$ is defined a partial order relation $f \le g \iff \forall{n\in\mathbb{N}} f(n) \le g(n) $.

Also let $h: \mathbb{N}\to\mathbb{N}$ given by a formula $h(n)=n+1$.

Find cardinality of a set {$f \in \mathbb{N}^{\mathbb{N}} f\le h$}

I can't figure out how to find cardinality here and also what is $f$ in the end, is it just an arbitrary function?

Answer 1

Notice that if $f:\mathbb N\to\{0,1\}$, then $f\leq h$.
So $|\{0,1\}^{\mathbb N}| \leq |\{f\in \mathbb N^{\mathbb N} : f \leq h\}|$.

Answer 2

Note that the cardinality of the set of sequences $$ S=\big\{(a_n): a_n\in\{0,1\}\big\} $$ is equal to the cardinality of $\mathbb R$ and so is the cardinality of $\mathbb N^{\mathbb N}$.

The set $A$ in the OP clearly satisfies $$ |S|\le |A|\le |\mathbb N^{\mathbb N}| $$ Hence $$ |A|=2^{\aleph_0}. $$