# Cardinality of a quotient set of [0,1]

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

• Wait. Are $q$ and $k$ fixed? – fleablood May 14 '20 at 22:22
• no; it means that $x \sim y$ iff $\exists q \in \mathbb{Z}, k \in \mathbb{N}:x - y= d(q,k)$. – TomS May 14 '20 at 22:28
• Okay, so $[x]=\{y|x\sim y\} =\{y| x-y=\frac q{2^k}$ for some integer $q$ and some positive integer $k\} = \{x \pm \frac q{2^k}| q\in \mathbb Z; k\in \mathbb Z\}$. So $[x]$ is countable. And as $[0,1]$ is uncountable and $[0,1]=\cup_{[x]\in [0,1]/\sim} [x]$ that is a union of countable sets whose cardinality is uncountable. As the union of countably many sets is countable this must be an uncountable union. So $[0,1]$ must be uncountable. – fleablood May 14 '20 at 22:29
• thanks for clarification; that was my idea as well – TomS May 14 '20 at 22:35
• Nonsense. There are no d's mentioned in the problem. – William Elliot May 15 '20 at 11:22

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.

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