Cardinality of the set of bounded sequences in $\Bbb N$

Question

This is from "Mathematical logic, A course with exercises" Chapter 7 question 7. The question is to determine the cardinality of $\{f\in {\mathbb N}^{\mathbb N}: (\exists p\in \mathbb N)(\forall n\in \Bbb N)(f(n)\le p)\}$

I know this is the set of bounded sequences but my intuition is that it should be denumerable, since it is determined by the upper bound and the number of repetition of each sequence member.

However, the solution manual says the cardinality is $2^{\aleph_0}$ since it includes $2^\Bbb N$. How can I see this? What is an inclusion map?

Thanks for any help!

Answer 1

Note that any sequence consisting of just $0$ and $1$ is bounded. Thus, any sequence in $2^{\mathbb{N}}$ would be inside the set of bounded sequences.