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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

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  • $\begingroup$ Wait. Are $q$ and $k$ fixed? $\endgroup$ – fleablood May 14 '20 at 22:22
  • $\begingroup$ no; it means that $x \sim y$ iff $\exists q \in \mathbb{Z}, k \in \mathbb{N}:x - y= d(q,k)$. $\endgroup$ – TomS May 14 '20 at 22:28
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    $\begingroup$ 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. $\endgroup$ – fleablood May 14 '20 at 22:29
  • $\begingroup$ thanks for clarification; that was my idea as well $\endgroup$ – TomS May 14 '20 at 22:35
  • $\begingroup$ Nonsense. There are no d's mentioned in the problem. $\endgroup$ – William Elliot May 15 '20 at 11:22
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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.

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

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