Let $A=\mathbb R,\ aSb \iff a-b\in \mathbb Q $

  • What is the cardinality of $[\pi]_S$ ?
  • Prove that the quotient group $\mathbb R/S$ is uncountable.
  • Well I think that cardinality is zero because for all $a-b=\pi\notin\mathbb Q$ so this equivalence class is empty.

  • I find it strange that this quotient group is uncountable since it consists of elements only from the rational numbers and they are countable. Even with a union of all the equivalence classes we will have only $\mathbb Q$ and not $\mathbb R$.

Please share your thoughts on how to solve this.


Note: This is from set theory intro course so I probably won't understand solutions that utilize knowledge from abstract algebra, rings or group theory.

  • 1
    $\begingroup$ Hm. I'm fairly certain that there is at least one other student from your course that is asking questions here. The first question was asked just the other night, and the second question was asked more than several times before. $\endgroup$
    – Asaf Karagila
    Dec 20, 2013 at 20:15
  • $\begingroup$ @AsafKaragila you probably mean math.stackexchange.com/users/76802/oria-gruber, I checked, he didn't ask this. $\endgroup$
    – GinKin
    Dec 20, 2013 at 20:18
  • $\begingroup$ Fairly close to your first question: math.stackexchange.com/questions/612340/… - I answered anyway, because it seems to me that you failed to understand the basics of the question. My recommendation is that you review the definitions of an equivalence relation again. $\endgroup$
    – Asaf Karagila
    Dec 20, 2013 at 20:19
  • $\begingroup$ @SalechAlhasov you mean in this case $\pi$ is an unknown so we can write $x$ instead ? $\endgroup$
    – GinKin
    Dec 20, 2013 at 20:26

1 Answer 1


Note that equivalence classes ARE NEVER EMPTY. If anything $\pi\in[\pi]_S$. You can show that $[x]_S$ is always countable by construction a bijection between $[x]_S$ and $\Bbb Q$.

The quotient set (also group, but you mentioned that you don't want algebraic arguments) is not made of rational numbers, but rather of an element from each equivalence class. By knowing that the equivalence classes are countable, you can easily prove that there must be uncountably of them, because their union is $\Bbb R$.

  • $\begingroup$ Let me rephrase, $[\pi]$ is the eqivalence class of $\pi$ with $S$, so for any $a-b=\pi$ but it cannot be in the relation since pi is irrational. $\endgroup$
    – GinKin
    Dec 20, 2013 at 20:44
  • 2
    $\begingroup$ @GinKin: You’re misunderstanding the relation. Let $a=\pi+1$ and $b=\pi$; then $a-b=1\in\Bbb Q$, so $a\in[b]_S=[\pi]_S$. For that matter, $b-b=0\in\Bbb Q$, so $b\in[b]_S=[\pi]_S$. There are two members of $[\pi]_S$ already. $\endgroup$ Dec 20, 2013 at 21:08
  • $\begingroup$ Thanks @BrianM.Scott. About the second part Asaf , you mean I need to describe a set of 'representatives' for each equivalence class and show that the union of those is equal to the reals right ? So I suppose this can be $[0,1)$ which is uncountable. But how do I prove that ? (it isn't as easy as you say). $\endgroup$
    – GinKin
    Dec 20, 2013 at 21:49
  • 1
    $\begingroup$ @GinKin: You don’t actually have to specify a set of representatives; you just have to prove that each equivalence class is countable. If $S$ had only countably many equivalence classes, and each of those equivalence classes is countable, what could you say about the cardinality of $\Bbb R$? $\endgroup$ Dec 20, 2013 at 21:52
  • 1
    $\begingroup$ @GinKin: In fact you can show very easily that for any $x\in\Bbb R$, $$[x]_S=x+\Bbb Q=\{x+q:q\in\Bbb Q\}\;,$$ so that the map $q\mapsto x+q$ does give you a bijection. $\endgroup$ Dec 20, 2013 at 22:11

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