Prove that for a subset $A$ of $\mathbb{R}$, if $|A|=\omega$ then $|\mathbb{R}-A|=2^{\aleph_0}$.
Clearly, $|\mathbb{R}-A|\le 2^{\aleph_0}=|\mathbb{R}|$.
It suffices to show that for a countable $A\subseteq \{0,1\}^{\mathbb{N}}$ we have $\{0,1\}^{\mathbb{N}}\le \{0,1\}^{\mathbb{N}}-A$, since $\mathbb{R} \sim \{0,1\}^{\mathbb{N}}.$
The idea: Somehow identify $\mathbb{N}$ in $\{0,1\}^{\mathbb{N}}$. Since $|A|\le \omega$, we can then identify $A$ in $\mathbb{N}$ (all within $\{0,1\}^{\mathbb{N}}$). Thus, we will be able to 'transfer' the elements of $A$ to new elements of $\mathbb{N}$ (in $\{0,1\}^{\mathbb{N}}$). An element not in $A$ is mapped to itself. This way, after we remove $A$, we will still have 'a copy of' $A$ in $\{0,1\}^{\mathbb{N}}-A$, proving that $\{0,1\}^{\mathbb{N}}\le \{0,1\}^{\mathbb{N}}-A$.
Is this idea OK?
This problem is given before the introduction of AC, and therefore, before the proposition that there is always an injection from $\mathbb{N}$ to an infinite set $X$. This means that, in principle, I can't use the fact to identify $\mathbb{N}$ in $\{0,1\}^{\mathbb{N}}$.
Are there more elementary approaches to this problem?
Thanks.
