I have begun some studying of set theory at a level which could be described as "undergraduate level".

So there is some definition of cardinality and bijections between $\mathbb N$ and $\mathbb Q$, and uncountability of $\mathbb R$ and countability of $\mathbb N$ and $\mathbb Q$.

So, it is really a beginning of some possible further study, with just some basic concepts that I have grasped so far.

But, it seems legal to ask at this level, is there a bijection between the set of all countable subsets of $\mathbb I$ and $\mathbb R$?

Here $\mathbb I$ denotes the set of all irrational numbers, and if $\mathbb I_C$ denotes the set of all countable subsets of $\mathbb I$, this question is: Is there a bijection $b: \mathbb I_C \to \mathbb R$?

I would also like to see a constructed example of such bijection.


I'm going to show that there is an injection in both directions, which shows what you want by Cantor-Schroeder-Bernstein.

First, let's show that there is an injection $f:\mathbb R\to\mathbb I_c$ defined by $$f(x)=\begin{cases}\{x,x+1\}&x\notin\mathbb Q\\\{x+\pi\}&x\in\mathbb Q\end{cases}$$

Clearly, rationals and irrationals map to different subsets, since sets with two elements never equal sets with 1 element. Moreover, $x+\pi$ is irrational is $x$ is rational, and $x+1$ is irrational if $x$ is irrational. It's easy to see why this is injective.

Next, we note that there is an bijection $h:\mathbb R^\mathbb N\to\mathbb R$, since there is a standard bijection from $2^\mathbb N\to\mathbb R$ (binary representation), and hence we simply need a bijection from $(2^\mathbb N)^\mathbb N=2^{\mathbb N\times\mathbb N}\to2^\mathbb N$, but that's easy since there is a bijection from $\mathbb N^2\to\mathbb N$.

It's obvious that there is an injection from $\mathbb I_c\to\mathbb R^\mathbb N$, so we are done!

Edit: You may not like that I haven't given you an explicit construction, but it's likely to be more gross than insightful. You can run through CSB to figure out what actual bijection this would result in.


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