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I've been given the relation described in the title and asked to show it is an equivalence relation, finds the size of each equivalence class and find how many equivalence classes there are. I've managed the first 2 tasks but am struggling with the 3rd.

Each equivalence class is of the form $\{a+q:q\in \mathbb Q\}$, so if $a$ is rational the equivalence class is $\mathbb Q$, and I need only to deal with the irrationals. It's pretty obvious to me that there are uncountable equivalence classes but I can't seem to prove it, it essentially boils down to showing that there are uncountable irrationals that differ by an irrational but I can't prove that either. I've been given a hint to use the fact that if $A,B$ are infinite sets, then $|A|+|B|=|A|\cdot |B|=\max\{|A|,|B|\}$, but I can't find any connection between the set of equivalence classes and a set of this form, presumably with $A$ being countable and $B$ uncountable.

Thanks in advance for the help :).

Adgorn
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    Hint: Given a real $\alpha$, show that there are only countably many reals equivalent to $\alpha$. Deduce that, if there were only countably many equivalence classes, then there'd be only countably many reals. – lulu Aug 10 '20 at 16:42
  • @lulu I can do that, however that would only show that the amount of equivalence classes is not countable. We have not shown in my course that there aren't any cardinalities between $\aleph_0$ and $\aleph_1$, so I need to show explicitly that the cardinality of the set of equivalence classes is the same as that of the reals/irrationals. – Adgorn Aug 10 '20 at 16:54
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    I wonder how much detail we are expected to go into. lulu's hint shows there are uncountably many equivalence classes and the cardinality obviously can't be higher than the reals. But do we have to prove it's not a hypothetical cardinality between the naturals and the reals. I assume we can but I'm not sure I'd want to. – fleablood Aug 10 '20 at 16:55
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    I think we can show the that countable union of sets all of a cardinality M will have a cardinality of $\max(\aleph_0, M)$. so if $\aleph_0 < M < \aleph_1$ we'd have a contradiction. .... I think.... But I'm really not sure we have to. You should ask your professor. – fleablood Aug 10 '20 at 17:02
  • "We have not shown in my course that there aren't any cardinalities between ℵ0 and ℵ1" And it never will. But has it not hypothesized the Continuum Hypothesis? I imaging must courses would have. – fleablood Aug 10 '20 at 17:05
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    See this question. I assume you are working with the Axiom of Choice here...otherwise I think all kinds of things are possible. – lulu Aug 10 '20 at 17:07

1 Answers1

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Self-answer:

Alright so the idea is that every equivalence class has cardinality ${\aleph}_0$, and every real number $x\in \mathbb R$ belongs to one and only one equivalence class. Letting $E$ denote the set of all equivalence classes, we can therefore define a bijection $f:\mathbb R \rightarrow E\times \mathbb N$ defined by $$f(x) = (\textrm{equivalence class of } x, \textrm{index of }x \textrm{ in some listing of the equivalence class}).$$ Therefore (since $E$ is an infinite set by lulu's comment) we conclude that $|\mathbb R|=|E \times \mathbb N|=\max\{|E|,|\mathbb N|\}= \max\{|E|,{\aleph}_0\}=|E|$.

Adgorn
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