I wish to prove that $|\mathbb{R}-\mathbb{S}|=2^{\aleph_0}$ when $\mathbb{S}\subset \mathbb{R}$ is countable.
I want to say that $|\mathbb{R}-\mathbb{S}|= |\mathbb{R}|-|\mathbb{S}|$ but we haven't studied yet what subtraction of cardinals means (I can guess, though).
How could I prove this using only basic cardinal properties?
$\newcommand\R{\mathbb{R}}$Suppose that $S\subset\R$ is a countable set of reals, and consider the complementary set $\R-S$. Since the unit interval is uncountable and more specifically contains uncountably many disjoint countably infinite sets (e.g. small translations of the rationals in some tiny interval), there is a countable set $T\subset[0,1]$ of the same size as $S$ that is disjoint from $S$. Thus, $\R-T$ is bijective with $\R-S$ by simply swapping elements of $S$ for $T$ and fixing all other reals. But $\R-T$ contains the interval $[2,3]$, and so $\R-S$ is at least as large as $[2,3]$, which has the same size as $\R$. And so $\R-S$ has size continuum.