Has the cardinal of the real set minus a subset of it the same cardinal? I have to check the cardinal of $\bar{A}$(The complement of A) where 
$A = \{ 1 \le x \le 2 \ | \ x \in \mathbb{R}\}$ 
I thought that a good idea would be to argue it by demonstrating that $\bar{A} \subset \mathbb{R}$ but I think that this is not really mathematical. Can this be correct? In case it isn't where can I find some propierties of cardinal related with $\aleph_1$?
 A: The open interval $(1,2)$ is equinumerous with $\mathbb R$ by using the function $\frac{1}{x-1}+\frac{1}{x-2}$.  Thus $A$ is equinumerous with a subset of $\mathbb R$ and vice versa.  A general result of Cantor allows you to conclude that they are equinumerous.
Similarly, the open interval $(2,\infty)$ can be put into 1-1 correspondence with $(1,2)$ (or with $\mathbb R$ itself) by a suitable rational map. Then a subset of the complement of $A$ is equinumerous with a subset of $\mathbb{R}$ and vice versa, and we conclude much as before.
A: Theorem: Whenever we have a subset $A$ of an infinite set $X$ and an injection from $A$ to its complement $X\setminus A$, then $X\setminus A$ has the same cardinality as $X$.
Proof: Obviously we have
$$|A|+|X\setminus A| = |X|$$
which because $X$ is an infinite set means $|A|=|X|$ or $|X\setminus A|=|X|$. So if $|A|<|X|$ the claim follows immediately. On the other hand, because of the injection we have
$$|A|\le |X\setminus A|\le |X|$$
which for $|A|=|X|$ again implies $|X\setminus A|=|X|$ $\square$
Now $\mathbb R$ clearly is an infinite set, and an injection from $A$ to its complement is given e.g. by $x\mapsto x+2$. Therefore the complement of $A$ in $\mathbb R$ has the same cardinality as $\mathbb R$.
