Problem: Let $A$ be a non-countable set and let $B\subseteq A$ be a countable set. Prove that $A\setminus B$ is non-countable.
Where, $(X \text{ is non-countable})\;\iff (|X|>|\mathbb{N}|)$.
Of course, a way to prove this would be by finding a function $f:A\to A\setminus B$ that's injective to show that $|A|\leq |A\setminus B|$ thus finishing the proof. I can´t seem to be able to find shuch a function.
My guess would be that a function like that should exist due to the following special case:
$$\big|\mathbb{R}\setminus\big(\;]-\infty,-\pi/2]\cup [\pi/2,\infty[\;\big)\big|=\big|(-\pi/2,\pi/2)\big|=\big|\mathbb{R}\big|$$ Thanks to the existence of the function $\tan:(-\pi/2,\pi/2)\to\mathbb{R}$ which is bijective.
How could I find an appropriate function to prove Problem?
Please help adding or removing tags if necessary. Thanks in advance.
Since $B$ is countable, there exists an injection $f : B \rightarrow \mathbb N$
Suppose $A \setminus B$ is countable. Then there exists an injection $g : A \setminus B \rightarrow \mathbb N $.
Now let $h : A \rightarrow \mathbb N$ be defined by
$$ h(x) = \left \{ \begin{array}{l} 2f(x) \quad &\text{if} \quad x \in B \\ 2g(x) + 1 \quad &\text{if} \quad x \in A \setminus B \end{array} \right. $$
$h : A \rightarrow \mathbb N$ is an injection so $A$ is countable, which is a contradiction.