Cardinality of Irrational Numbers

Question

I know and I have proved more than once that the set of irrational numbers ($\mathbb{I}$) is uncountable, but now I'm given to solve this problem:

Show that $|\mathbb{I}|=|\mathbb{R}|$,

How can I do that?

Do I need to assume the Continuum Hypothesis in order to that statement to be true?

Thanks

Answer 1

Fix an enumeration of $\Bbb Q$, $q_n$, and find a countably infinite subset of $\Bbb I$, $r_n$.

Now find a map which fixes all the points which are not $r_n$'s, and maps the union $\{q_n,r_n\mid n\in\Bbb N\}$ into $\{r_n\mid n\in\Bbb N\}$.

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The above can be translated quite neatly to cardinal arithmetic. Write $\Bbb I$ as $A\cup B$ where $A$ is countably infinite, and $B\cap A=\varnothing$. Then we have:

$$|\Bbb R|=|\Bbb Q|+|\Bbb I|=\aleph_0+(|A|+|B|)=\aleph_0+(\aleph_0+|B|)=\aleph_0+|B|=|\Bbb I|.$$

Answer 2

Here is how to construct an explicit bijection. First we construct a bijection $\alpha$ from $\mathbb{N}$ to $\mathbb{Q}$. (For definiteness I will take $\mathbb{N}$ to include $0$, this makes no real difference.) For brevity, call $\alpha(i)$ by the name $r_i$.

Now we proceed semi-formally. Map all numbers which are not rational or of the form $r+\sqrt{2}$, where $r$ is rational, to themselves.

Map rational $r_i$ to $r_{2i+1}+\sqrt{2}$. Map $r_i+\sqrt{2}$ to $r_{2i}+\sqrt{2}$.

Answer 3

In what follows, I write $\mathfrak c$ for $|\mathbb R|$, as customary. That $\mathfrak c=2^{\aleph_0}$ can be proved in several ways. For example, the Cantor set $C$, by construction, has size $2^{\aleph_0}$. This shows that $2^{\aleph_0}\le\mathfrak c$. On the other hand, it is easy to see that $\mathbb R$ has the same size as any open interval, and $|(0,1)|\le2^{\aleph_0}$, as we can identify the binary expansion of a number with an infinite sequence of zeros and ones. (I am using that if $A$ and $B$ inject into each other, then there is a bijection between them. This is the Bernstein-Schröder theorem.)

To see the relevance of this, your question is asking for a very particular case of a result discussed here, namely (without assuming the axiom of choice) if $\mathfrak m$ and $\mathfrak n$ are cardinalities, $\mathfrak m+\mathfrak m=\mathfrak m$, and $\mathfrak m+\mathfrak n=2^{\mathfrak m}$, then $\mathfrak n=2^{\mathfrak m}$.

Your question is the case where $\mathfrak m=|\mathbb Q|=\aleph_0$, and $\mathfrak n=|\mathbb I|$, precisely because $|\mathbb R|=2^{\aleph_0}$.

There are other approaches, of course. As discussed here and here, there are reals $r$ such that the set $C+r=\{x+r\mid x\in C\}$ is contained in $\mathbb I$. This shows that $\mathfrak c\le|\mathbb I|$. Since clearly $|\mathbb I|\le\mathfrak c$, we again gave equality.

For yet another variant of this idea, note that ($|\mathbb I|\le\mathfrak c$ and) obviously, $2^{\aleph_0}\le{\aleph_0}^{\aleph_0}$, and the latter is the size of $\mathbb I$, as discussed for example here, where it is shown that not only there is a bijection betwen the space $\mathbb I$ and Baire space $\mathcal N=\mathbb N^{\mathbb N}$ but, in fact, the two spaces are homeomorphic (where the latter has the product topology of countably many copies of the discrete set $\mathbb N$).

Answer 4

$\mathbb{R} = \mathbb{Q} \cup \mathbb{I}$, so $|\mathbb{R}| = |\mathbb{Q}| + |\mathbb{I}| = \max(|\mathbb{Q}|, |\mathbb{I}|) = |\mathbb{I}|$, as the latter is uncountable.