Cardinality of a space of continuous functions Let $C(\lbrace 0,1 \rbrace^{\mathbb{N}},[0,1])$ be the space of continuous functions from $(\lbrace 0,1 \rbrace^{\mathbb{N}},\tau_p)$ to $[0,1]$ and let $\mathcal{M} = \lbrace f \in  C(\lbrace 0,1 \rbrace^{\mathbb{N}},[0,1]) \: | \: f$ is injective$\rbrace$. I want to prove that $card(\mathcal{M}) = 2^{\aleph_0}$ (or $card(C(\lbrace 0,1 \rbrace^{\mathbb{N}},[0,1])) = 2^{\aleph_0}$). Actualy, I only need $card(\mathcal{M}) \leq 2^{\aleph_0}$.   
Observations: $\tau_p$ is the topology of pointwise convergence and [0,1] have the usual topology.
I will be very grateful with any suggestions.
 A: There's an easy upper bound here, which is the same as for $C(\mathbb{R},\mathbb{R})$: $2^\mathbb{N}$, with pointwise convergence (equivalently, product topology) is a separable space, this means it has a countable dense subset (eventually $0$ sequences for instance). Let $S$ be such a subset. 
Then $C(2^\mathbb{N},[0,1]) \to [0,1]^S$ defined by restriction ($f\mapsto f_{\mid S}$) is injective, because $[0,1]$ is Hausdorff and $S$ is dense. Therefore $C(2^\mathbb{N},[0,1])$ has cardinality at most $|[0,1]|^{|S|} = (2^{\aleph_0})^{\aleph_0} = 2^{\aleph_0}$. Thus the cardinality of $\mathcal{M}$ is also bounded by this. 
But also, $|C(2^\mathbb{N},[0,1])| \geq 2^{\aleph_0}$ (think constant functions). 
Also, $[0,1]$ contains a homeomorphic copy of $2^\mathbb{N}$ (the Cantor set), and up to rescaling by a $>0$ scalar this yields continuum many injective continuous functions $2^\mathbb{N} \to [0,1]$ (they're all different because of their maximum for instance), hence $|\mathcal{M}| \geq 2^{\aleph_0}$
Thus by Cantor-Bernstein, $|C(2^\mathbb{N},[0,1])| = |\mathcal{M}| = 2^{\aleph_0}$
