cardinality of the complement of a countable subset of R 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?
 A: 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 $B=\{2\cdot 10^{-j}+d:j\in \mathbb N\}.$ 
A: Here we construct. without recourse to the axiom of choice, a countable subset $B$ that is disjoint from $A$; the OP can then use DanielWainfleet's argument, arriving at the desired solution.
Suppose for some $a$ and $b$ in $\mathbb{R}$ with $a \lt b$ the closed interval $[a,b]$ is disjoint from $A$. Then it is trivial to construct $B$. So, we assume that every interval in $\mathbb{R}$ has a nonempty intersection with $A$.
Let the set $A$ be enumerated with a sequence $a_n$. 
Definition: The $a_n \text{-distance}$ of any interval $[a,b]$ is the smallest integer $n_0$ such that $a_{n_0} \in [a,b]$.
Lemma 1: A real number $\hat b$ can be recursively constructed, belonging to the open interval $(0,1)$, that is distinct from the elements in $A$.
Proof
Starting with the closed interval $[0,1]$, divide it into 5 equal pieces and examine the two 'snug'intervals $[\frac{1}{5},\frac{2}{5}]$ and $[\frac{3}{5},\frac{4}{5}]$, selecting the one with the larger $a_n \text{-distance}$. Repeat this process.
This process gives us a nested sequence of closed intervals. By basic properties of the real numbers, the intersection of this decreasing chain of intervals is a singleton $\{\hat b\}$. But it also easy to see that each new nested interval must have a larger $a_n \text{-distance}$ than the preceding one. This of course means that the $\{\hat b\} \cap A = \emptyset$. $\qquad \blacksquare$
Proposition 2: A countable set $B$ can be constructed such that $A \cap B = \emptyset$.
Proof
Note that $A$ is a subset of a larger countable set $A^{`}$, the 'integer translate set $A + \mathbb{Z}$. If a real number $b$ is not in $A^{`}$, then $b + m \notin A^{`}$ for any integer $m$. So, apply lemma 1 to 
$A^{`}$ getting the number $\hat b$. Next, collect the translations $\hat b + m$ to form a countable set disjoint from $A$.
Note that we have also shown that there is no bijective mapping between a countable set $A$ and $\mathbb{R}$.
A: Let $A$ be an infinite countable set, thus $A=\{a_1,a_3...a_n....\}$
Take a countable set $B=\{b_1,b_2...\}$ such that $B \cap A =\emptyset$
Clearly $\mathbb{R}$ \ $A$ is uncountable.
Take an enumeration of $A\cup B$
such that $A \cup B=\{a_1,b_1,a_2,b_2....a_n,b_n...\}$
Take $f: (\mathbb{R}-A) \to \mathbb{R}$
where 
$f(x)=\begin{cases}\
           x &x \in \mathbb{R}-(A \cup B)\\
           a_{n} & x=b_{2n-1}\\
           b_{n} &x=b_{2n}\\
             \end{cases}$
$\text{$f$ is bijective } \Rightarrow |\mathbb{R}| = |\mathbb{R}-A|$
A: Let A = { a1, a2,.. aj,.. } be a countable subset of the reals R.
B = { a - b | a,b in A } is countable.  Thus some p in R - B.
p + A is countable and disjoint from A.  Otherwise
some a,b in A with p + a = b, p = b - a in B, a contradiction.  
Define f:R -> R - A by
f(x) = x, if x not in A $\cup$ p+A
f(a_j) = p + a_2j, for all a_j in A
f(p + a_j) = p + a_(2j-1), for all p + a_j in p + A
f is a bijection;  R and R - A are equinumerous.  
