The excercise is to prove that for any countable set of $\mathbb{R}$ , let s call it A, its complement $\mathbb{R}/A$ has the same cardinality as $\mathbb{R}$. The solution I am aware of is straightforward, assuming countable choice, making a second countable subset, taking the union of those two sets, assuming a bijection and so on, but i wanted to try something diffirent, without assuming choice, and i don't know if it s correct.
My thought is that for any countable set, i can write it down as $A=\{a_0, a_1,….\}$ but not necceserily in increasing order. If it was in increasing order, i could simply say that $(a_0, a_1) \in \mathbb{R}-A$ and thus $|\mathbb{R}-A|>=\mathbb{R}$, so i have the equality. I am stuck at proving that there are $a_i, a_j \in A $ such that $a_n \notin (a_i,a_j ) $for any $a_n \in A$. The problem is that this does not hold for any choice of $a_i$. For example, if $A= \{0\} \cup \{ 1/n \}$, then obviously for $0 $ this does not hold, although it holds for any other element. Is there any easy and fast way to get past this?
[Math] cardinality of the complement of a countable subset of R
elementary-set-theory
Best Answer
For countable (finite or infinite) $A\subset \mathbb R$ find a countably infinite $B\subset \mathbb R$ that is disjoint from $A.$ Show there is a bijection $f:B\to B\cup A.$ Extend $f$ to the domain $\mathbb R$ \ $A$ by letting $f(x)=x$ for $x\in \mathbb R \setminus (B\cup A).$ Then $f:(\mathbb R$ \ $A) \to \mathbb R$ is a bijection.
If you want, I will show how to find $B$ without the Axiom of Choice and by the most elementary methods.
Appendix. To find $B$: If $A$ is empty let $B=\mathbb N.$ If $A\ne\phi$ let $A=\{a_n:n\in \mathbb N\}.$ It does not matter whether $a_m=a_n$ for some $m\ne n.$ Let $(a_{n,j})_{j\in \mathbb N}$ be the sequence of decimal digits of $a_n$ to the right of the decimal point, in base $10,$ where $a_{n,j}\ne 9$ for infinitely many $j.$ Let $a^*_j=1$ if $a_{j,j}$ is even and let $a^*_j=2$ if $a_{j,j}$ is odd. Let $d=\sum_{n\in \mathbb N}a^*_j10^{-j}.$ Let $B=\{2\cdot 10^{-j}+d:j\in \mathbb N\}.$