Prove there's $\left|A-B\right| = \aleph$.

Question

Let A a set such that $\left|A\right| \ge \aleph$. Prove there's a $B\subseteq A$ such that $\left|B\right|\ge \aleph$ and $\left|A-B\right| = \aleph$.

Lets assume there's a $B\subseteq A$ such that $\left|B\right| = \aleph$ (are we allowed to assume that?)

$A-B \subseteq A$ implying $\left|A-B\right| \le \aleph \le \left|A\right|$

I don't know how to continue. There must be some rule or theorem I'm missing..

I'd be glad for help.

Note: $\aleph$ is the cardinality of thecontinuum.

Answer 1

Let $ f : \mathbb{R} \longrightarrow A$ an injective function.

Then consider simply $B= A \setminus f(\mathbb{R})$.