Uncountable minus countable set is uncountable Problem statement
Prove that if $A$ is an uncountable set and $B$ is a countable set, then $A\setminus B$ must be uncountable.
What I think
The statement does not mention $A$ and $B$ relationship. I think there are two possibilities:


*

*If $A \cap B = \emptyset $, then $A\setminus B$ is trivially uncountable

*If $A \cap B = B$, then $B \subset A$ and as a bijection can not be made between $A\setminus B$ and $\mathbb{N}$, $A\setminus B$ is uncountable.


And there is where I'm stuck. How can I prove that a bijecton can't be made?
TIA.
 A: First, the two cases you mention don't include all possibilities.
For your question, note that the union of two countable sets is again countable.
A: Suppose that $A-B=A-A\cap B$ is countable.  Then you have a bijection between $A-B$ and the set of even natural numbers. Since $B$ is countable then $A\cap B$ is also countable and you have a bijection between $A\cap B$ and the set of odd natural numbers.
Now, taking the union of those bijections, you get the bijection between the set
$A=(A-B)\cup (A\cap B)$ and the set of natural numbers, which is an absurd because $A$ is uncountable. Thus $A-B$ is uncountable. Q.E.D.
If $A\cap B$ is just finite countable then the proof goes similarly with obvious changes, see comments.
A: Maybe this is a way to see it. You can make it more precise yourself. Assume that $A\setminus B$ is countable. $B$ is countable, so that would mean that $(A \setminus B) \cup B$ is countable (finite union of countable sets is clearly countable). But then $A \subseteq (A\setminus B)\cup B$, so $A$ is contained in a countable set, so it must itself be countable.
A: Here's what I came up with (inspired by @Thomas's previous answer)
Proof by contradiction:

*

*Assume $A$ is uncountable, $B$ is countable, and $A \setminus B$ is countable.

*Then $(A \setminus B) \cup B$ is countable because the union of countable sets is countable.

*$(A \setminus B) \cup B = (A \cap \overline{B}) \cup B$

*$=(A \cup B) \cap (B \cup \overline{B})$

*$=(A\cup B) \cap U$ (univeral set)

*$=(A \cup B)$

*However, $(A \cup B)$ is clearly uncountable, because we assumed A was uncountable, and the union of an uncountable set with any other set is uncountable.

*We have reached a contradiction, since we also assumed $A \setminus B$ was countable. Therefore, $A \setminus B$ must instead be uncountable.

