Let $A,A',B,B'$ such that: $A\subseteq B$, $A'\subseteq B'$, $\left|B\right|=\left|B'\right| \gt \aleph_0$, $\left|A\right|=\left|A'\right| = \aleph_0$.
Show that $\left|B-A\right| = \left|B'-A'\right|$
As a start I know that $\left|B-A\right| > \aleph_0$. Otherwise,
$$\left|B\right| = \left|B-A\right| + \left|A\right| \le \aleph_0 + \aleph_0 \le \aleph_0$$
In contrary to $\left|B\right| > \aleph_0$.
For same reason, $\left|B'-A'\right| > \aleph_0$.
How to proceed? I'm kinda stuck at this point.