Prove that if A is an uncountable set and B is a countable set, then A∖B must be uncountable.
I understand I am asking the same question as Uncountable minus countable set is uncountable
but I am trying a different approach that I have not seen in the answers. Is this still a valid proof or am I missing something.
Let ${S\subset{U}}$ and $S$ and $U$ are countable and uncountable respectively. Then we have $|S|$ = $|\mathbb{N}|$ and $|U| > |\mathbb{N}|$.
Let's assume $|U-S|$ is countable so $|U-S| = |\mathbb{N}|$.
$$|U-S| = |U| – |S|$$
$$|\mathbb{N}| = |U| – |\mathbb{N}|$$
$$|2\mathbb{N}| = |U|$$
$$|\mathbb{N}| = |U|$$
This is a contradiction because U is uncountable. So we have that $|U-S|$ is uncountable.
Best Answer
It's not true that $\vert X-Y\vert=\vert X\vert-\vert Y\vert$. In fact, the right hand side doesn't even make sense: we can't subtract cardinals in any good way, since addition of cardinals isn't cancellative (since $$\aleph_0+0=\aleph_0+17=\aleph_0+\aleph_0=\aleph_0,$$ should $\aleph_0-\aleph_0$ be $0$, $17$, $\aleph_0$, or something else?).
Instead, we have to reason purely additively: we do always have $$\vert X\vert +\vert Y\vert=\vert X\cup Y\vert$$ for disjoint $X,Y$ (and if either $X$ or $Y$ is infinite we don't need disjointness). Now take $X=B$ and $Y=A\setminus B$. If both $X$ and $Y$ were countable we would have $$\vert A\vert =\vert X\cup Y\vert=\vert X\vert +\vert Y\vert=\aleph_0+\aleph_0=\aleph_0,$$ contradicting the uncountability of $A$; so if we take a countable set away from an uncountable set, we can't be left with an uncountable set.