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How to prove the cardinality of $\mathbb{R}/ \mathbb Q$ is equal to the cardinality of $\mathbb{R}$

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Sorry, I wrote it wrong, I meant to be the cardinality of $\mathbb{R}$ mod $\mathbb{Q}$ – Alex Dec 26 '12 at 17:52
You mean the quotient group $\mathbb R / \mathbb Q$ ?? – GEdgar Dec 26 '12 at 17:54
Alex, please edit to clarify what you are asking. If you are asking about the cardinality of the set of irrational numbers (as it is currently written), see… and…. – Jonas Meyer Dec 26 '12 at 18:03
This is a consequence of the axiom of choice. How to prove it, depends a bit on your background, that is, what results you can assume, how familiar you are with choice, etc. Could you specify some of it? – Andrés E. Caicedo Dec 26 '12 at 19:27
If it is the quotient, in fact, without the axiom of choice, it turns out that $\mathbb{R}/\mathbb{Q}$ can have strictly larger cardinality than $\mathbb{R}$... – KSackel Dec 26 '12 at 20:02
up vote 1 down vote accepted

To prove equality we need to either find a bijection between the sets, or two injections between them.

As noted in the comments, this cannot be proved without the axiom of choice. So I am going to use it freely.

Assuming the axiom of choice, if so, we have a function $f\colon\mathbb{R/Q\to R}$ which chooses $f(A)\in A$ for every $A\in\mathbb{R/Q}$. This is an injection because if $A\neq A'$ then $f(A)\in A$ and $f(A)\notin A'$, and vice versa, therefore $f(A)\neq f(A')$.

On the other hand, let $V=\operatorname{rng}(f)$, then $\mathbb R$ is a countable union of copies of $V$, namely $\bigcup_{q\in\mathbb Q}q+V$. Therefore $|V|\cdot\aleph_0=2^{\aleph_0}$. Again, using the axiom of choice, we have that $2^{\aleph_0}=|V|\cdot\aleph_0=\max\{|V|,\aleph_0\}=|V|$.

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Remark: This answers the original version of the question.

The following bijection uses Hilbert's infinite hotel.

The rationals can be enumerated as $q_0,q_1,q_2,\dots$ in various explicit ways.

Define $f: \mathbb{R}\to \mathbb{R}\setminus \mathbb{Q}$ as follows.

If $x$ does not have shape $q$, or $\sqrt{3}+q\sqrt{2}$, where $q$ is rational, let $f(x)=x$.

If $x$ is the rational $q_i$, let $f(x)=\sqrt{3}+q_{2i}\sqrt{2}$.

If $x=\sqrt{3}+q_i\sqrt{2}$, let $f(x)=\sqrt{3}+q_{2i+1}\sqrt{2}$.

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Awesome.[filler] – akkkk Dec 26 '12 at 19:28

$\mathbb{R}= (\mathbb{R} \backslash \mathbb{Q}) \sqcup \mathbb{Q}$, where $\mathbb{Q}$ is countable and $\mathbb{R} \backslash \mathbb{Q}$ infinite, so $$|\mathbb{R}|= |\mathbb{R} \backslash \mathbb{Q} | + | \mathbb{Q}|= \max ( |\mathbb{R} \backslash \mathbb{Q} |, |\mathbb{Q}|) = |\mathbb{R} \backslash \mathbb{Q} |$$

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Indeed, $\mathbb R \setminus \mathbb Q$ is infinite. And so is $\mathbb Q$. You want to write "uncountable", I think. – Rudy the Reindeer Dec 26 '12 at 18:57
@MattN. - being infinite suffices since $\aleph_0$ + any infinite cardinal $k$ is $k$ – Belgi Dec 26 '12 at 19:36
@Belgi Right, I misread the question as having to show that $\mathbb R \setminus \mathbb Q$ is uncountable. – Rudy the Reindeer Dec 26 '12 at 20:52

I don't see what I would have expected to be the "standard answer" to this question, so let me leave it in the hopes it will be helpful to someone.

Proposition: Let $G$ be an infinite group, and let $H$ be a subgroup with $\# H < \# G$. Then $\# G/H = \# G$.

Proof: Let $\{g_i\}_{i \in G/H}$ be a system of coset representatives for $H$ in $G$: then every element $x$ in $G$ can be written as $x = g_{i_x} h_x$ for unique $h_x \in H$ and $i_x \in G/H$. (Note that there is no canonical system of coset representatives: getting one is an archetypical use of the Axiom of Choice.) Thus we have defined a bijection from $G$ to $G/H \times H$, so $\# G = \# G/H \cdot \# H$. Since $\# G$ is infinite, so must be at least one of $\# G/H$, $\# H$, and then standard cardinal arithmetic (again AC gets used...) gives that

$\# G = \# G/H \cdot \# H = \max(\#G/H, \# H)$.

Since we've assumed $\# H < \# G$, we conclude $\# G = \#G/H$.

This applies in particular with $G = \mathbb{R}$, $H = \mathbb{Q}$ to give $\# \mathbb{R}/\mathbb{Q} = \# \mathbb{R} = 2^{\aleph_0}$.

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To see that the uses of choice here are unavoidable:… – Andrés E. Caicedo Apr 23 '13 at 5:58
One can even write a general form theorem from cardinal arithmetics. If $f\colon A\to B$ is surjective, and every fiber have the same cardinality, which is less than that of $A$ then $|A|=|B|$. – Asaf Karagila Apr 23 '13 at 6:17

$\mathbb{R}/\mathbb{Q}$ and $\mathbb{R}$ are both size continuum. So by assuming as vector spaces over the $\mathbb{Q}$ they must have continuum size bases. We know that if two vector spaces have bases of the same size then they are isomorphic.

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So why exactly are they both size continuum? – akkkk Dec 26 '12 at 18:45
@akkkk: Because if B be a basis of R over Q (as a vector space, that is). The cardinality of B is c and that R/Q is like removing one basis element, or one copy of Q. Is that wrong? – Babak S. Dec 26 '12 at 18:48
Babak the OP wants to prove that $\mathbb{R}-\mathbb{Q}$ is of size continuum which you assumed in the first line. – Belgi Dec 26 '12 at 19:35
@Belgi: I wonder what should I do here. Firstly, the OP wrote $R-Q$ and some different good answer came to him, and suddenly he changed his mind to quotient group. See what he replied in the comment. – Babak S. Dec 26 '12 at 19:43
@Belgi: Actually, in the comments the OP says that he is looking for the cardinality of the quotient. – Asaf Karagila Dec 29 '12 at 15:02

Well, heuristically it can go as follows.

We can easily construct a bijection between $\mathbb{N}$ and $\mathbb{N}-\{1\}$. Just send $n \rightarrow n+1$.

If we subtract one element $x_1$ from $\mathbb{R}$, we can decompose $\mathbb{R}$ as $(\mathbb{R}-\mathbb{N})\cup\mathbb{N}$, and $\mathbb{R}-\{x_0\}=(\mathbb{R}-\mathbb{N})\cup(\mathbb{N}-{x_0})$. And we apply the above trick to the latter part.

If we subtract countable elements, say, $x_i$ for all $i\in\mathbb{N}$, we can choose a countable familiy of subsets with cardinal equal to that of $\mathbb{N}$, each containing one of $x_i$. Now decompose $\mathbb{R}$ as a countable union and do as above and you will solve the problem.

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But we don't subtract the rationals. We take the quotient. – Asaf Karagila Jan 4 '13 at 14:13

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