Is this a valid injection from $\mathbb{R}^{\mathbb{Q} \cap [0,1]} \to \mathbb{R}^\mathbb{N}$?

Question

I am trying to create an injection from $\mathbb{R}^{\mathbb{Q} \cap [0,1]} \to \mathbb{R}^\mathbb{N}$ so that I can claim $|\mathbb{R}^{\mathbb{Q} \cap [0,1]}| \leq |\mathbb{R}^\mathbb{N}|$.

We know that $\mathbb{Q} \cap [0,1]$ is countable, so there exists a bijection $f : \mathbb{N} \to \mathbb{Q} \cap [0,1]$. My proposed map $\Phi : \mathbb{R}^{\mathbb{Q} \cap [0,1]} \to \mathbb{R}^\mathbb{N}$ is defined as $\Phi(g) = g \circ f$ for $g \in \mathbb{R}^{\mathbb{Q} \cap [0,1]}$. This is a one-to-one map since for any $g_1, g_2 \in \mathbb{R}^{\mathbb{Q} \cap [0,1]}$ where $\Phi(g_1) = \Phi(g_2)$ we have

\begin{align*} g_1(f(x)) = \Phi(g_1)(x) = \Phi(g_2)(x) = g_2(f(x)) \end{align*}

for all $x \in \mathbb{N}$.

Edit: Here I am using $A^B$ to represent the set of functions from $B \to A$.

Answer 1

Yes:

Please note that $\Phi$ is in fact bijective, as $h\to h\circ f^{-1}$ is obviously the inverse of $g\to g\circ f$.