Cardinality of a quotient set of [0,1]

Question

Let $[0,1] \subset \mathbb{R}$. Let $x,y \in [0,1]$ and $q \in \mathbb{Z}$, $k \in \mathbb{N}$.

Define the equivalence relation

$$x \sim y \iff x-y = \frac{q}{2^k}$$

for some $q,k$.

How do I find the cardinalities of

1. the equivalence classes $[x]$
2. the quotient set $[0,1] \; /_\sim$

My guess is

1. $|[x]| = \aleph_0$ because the d’s are countable
2. $|[0,1] \; /_\sim| = \mathfrak{c}$ because $|[0,1]| = \mathfrak{c}$

but I don‘t see a rigorous proof.

Thanks

Answer 1

Question has been clarified and answered by fleablood:

Let $[0,1] \subset \mathbb{R}$. Let $x,y \in [0,1]$ and $q \in \mathbb{Z}$, $k \in \mathbb{N}$.

Define the equivalence relation

$$x \sim y \iff \exists (q,k) \in \mathbb{Z} \times \mathbb{N}: x-y = \frac{q}{2^k}$$

The cardinalities of

1. the equivalence classes $[x]$ and
2. the quotient set $[0,1] \; /_\sim$

are

1. $|[x]| = \aleph_0$ because $|\mathbb{Z} \times \mathbb{N}| = \aleph_0$
2. $|[0,1] \; /_\sim| = \mathfrak{c}$ because $|[0,1]| = \mathfrak{c}$ and $[0,1]$ is the union of its equivalence classes, each of which is countable; if the quotient set itself would be countable, then $[0,1]$ would be countable as well; therefore the quotient set is uncountable.

$$[0,1] = \bigcup_{[x] \in [0,1]/_\sim} [x]$$