Let $X = \{(a_n): a_n \in \mathbb{N} \text{ and } a_n \leq a_{n+1}$ for every $n \in \mathbb{N}\}$
Prove $$\vert X\vert = \vert \mathbb R\vert=c$$
without using $2^{\aleph_0} = \aleph_0^{\aleph_0} = c$
My Attempt
I know that $\vert X\vert = \vert \mathbb R\vert=c$, using this $$\forall n \geq 2, n^{\aleph_0} = \aleph_0^{\aleph_0} = c$$ (I still haven't proved that).
I want to show this in another way.
Probably looking for a surjective function between $X$ and $\mathbb{R}$ or the sucessions of natural numbers.
The thing is that the latter demonstration is after the exercise I'm asking about, so I believe that there is a way to prove it without using it, but I haven't found a way.
See that $|X|\leq c$ is easy. For the other inequality, you consider $f: [0,1]\to X$ defined by $f(0.a_1 a_2 a_3 ...) = (a_1, a_1 +a_2, a_1 +a_2 + a_3, ...)$ and you can check that $f$ is one-to-one (finally you use that $|[0,1]|=c$).