Consider the set $(0,1)$ and denote every $a \in (0,1)$ by it's decimal expansion $$ a=0.a_1a_2a_3\ldots $$ Now, define the equivalence relation $a \sim b$ if and only if $a_p = b_p$ for every prime number $p$; and let $A$ denote the set of equivalence classes.
Is it possible to establish an injection $f:[0,1]\rightarrow A$?
What I actually want to prove is that is that $|A|=2^{\aleph_0}$. So far, I've been able to prove that $A$ is not countable but that is not enough if one wants to avoid the use of the continuum hypothesis.
Is it actually possible to prove this without using the continuum hypothesis?
First of all, note that some numbers have more than one decimal expansion. $0.5=0.4\bar 9$. So for sake of concreteness let us assume that when such situation occurs we choose the expansion ending with an infinite string of $0$'s.
HINT: It is in fact much much much easier to define an injection from $2^\Bbb N$ into $A$. Given any $f\colon\Bbb N\to\{0,1\}$, map $f$ to the real number $a$ such that for the $n$-th prime number $a_{p_n}=f(n)$, and otherwise $a_k=0$.
Now prove that if $f\neq g$ then they are mapped to distinct real numbers which are not equivalent, and therefore the map is in fact injective into $A$.