I'm trying to calculate the cardinality of B:
$$ B = \{f \in \mathbb R \to \mathbb R | \forall x \in \mathbb R.f ([x]) = f(x)\} $$
- $[x]$ is the floor value of $x$.
I think the cardinality is $\aleph$ , but I'm not sure. I found an injective function from $\mathbb Z \to \mathbb R$ to $B$ but can't find an injective function for the other direction. Ideas?
Note $$B^\prime=\mathbb{R}^{\mathbb Z}$$ the set of functions from $\mathbb Z$ to $\mathbb R$.
$$ \begin{array}{l|rcl} \varphi : & B & \longrightarrow & B^\prime\\ & f & \longmapsto & \begin{array}{l|rcl} f^\prime : & \mathbb Z & \longrightarrow & \mathbb R \\ & z & \longmapsto & f(z) \end{array} \end{array}$$ is an injection. As you mentioned in your OP, there also exists an injection from $B^\prime$ to $B$.
Hence $B$ and $B^\prime$ have the same cardinality. As the cardinality of $B^\prime$ is the one of the continuum $\mathfrak c=2^{\aleph_0}$, $B$ also has $\mathfrak c$ for cardinality.