equivalence classes and cardinality

Question

I need to prove that every equivalence class created by the equivalnce relation $\sim$ on $\mathbb{R}$, that is defined by: $a\sim b \Leftrightarrow (a-b) \in \mathbb{Q}$, is $\aleph_0$. Furthermore, I need to prove that the group of all equivalence classes is uncountable. I have an idea how to move from step one (proving it's $\aleph_0$) to step two (proving that the group of all equivalence classes is uncountable), by using the laws of addition in cardinality. However, I have no idea how to even approach the first step. Any help or idea will be appreciated!

Thanks a bunch!

Let $B$ be any equivalence class, and let $b$ be a fixed element of $B$. Then $B$ consists of all $a$ of the form $b+r$, where $r$ ranges over the rationals. This gives us a one to one correspondence between $B$ and the rationals. — André Nicolas, Jan 31 '14 at 00:03

score 3 · Accepted Answer · answered Jan 31 '14 at 00:00

3

Hint: Take any $b\in\Bbb R$. Then $a\sim b$ if and only if $a=b+p$ for some $p$ such that...what?

answered Jan 31 '14 at 00:00

Cameron Buie

102,994

Can't believe I failed to see that. Thanks a lot!! – Studentmath Jan 31 '14 at 00:09

score 2 · Answer 2 · answered Jan 31 '14 at 00:09

2

HINT: The rationals are countable. If $x-y\in\Bbb Q$ then there is a unique $q\in\Bbb Q$ such that $x+q=y$.

answered Jan 31 '14 at 00:09

Asaf Karagila

393,674

equivalence classes and cardinality

2 Answers2