I want to compare the cardinalities of the following sets:
A = {$f : [0,1] \to \mathbb N$}
B = $\mathbb R$
$\mathbf {What\,I \,know\, so\, far?}$
Both are uncountable sets. We can start by finding an injective function in at least one direction. I started with $g : B \hookrightarrow A, n \mapsto f_n \text{ such that } f_n(y) = \left\{\begin{matrix} 1 & \text{if} \ y=n\\ 0 & \text{if} \ y \neq n. \end{matrix}\right.$
Here $y \in [0,1].$
But I don't think this is the right injective function. I'm also not able to find any injective function in the other direction. My vague guess is that $|B| < |A|$ but don't know if I'm right.
$$|A|=(\aleph_0)^{(2^{\aleph_0})}\le (2^{\aleph_0})^{(2^{\aleph_0})}=2^{((2^{\aleph_0})\cdot\aleph_0)}\le 2^{(2^{\aleph_0}\cdot 2^{\aleph_0})}=2^{(2^{(\aleph_0+\aleph_0)})}=2^{(2^{\aleph_0})}$$ OTOH $$2^{(2^{\aleph_0})}\le (\aleph_0)^{(2^{\aleph_0})}$$ So by Cantor-Bernstein $|A|=2^{(2^{\aleph_0})}$.
As for $B$, $B=2^{\aleph_0}$, and $2^{\aleph_0} < 2^{(2^{\aleph_0})}$ by Cantor's theorem.
None of these steps need the axiom of choice, so this holds in ZF.