# Uncountable minus uncountable is at most countable? Cocountable topology

Let $$A$$ be an uncountable set. Let $$B \subseteq A$$ be an uncountable proper subset of $$A$$. Is it true that $$A-B$$ is at most countable?

I think I can come up with a counter example: Let $$B, C$$ be uncountable sets such that $$B \cap C = \emptyset$$. Then let $$A = B \cup C$$. Then $$A - B= C$$ is uncountable.

However, I am trying prove something about the cocountable topology on an uncountable set $$X$$. I am trying to describe its open sets but I am not sure how to find subsets of $$X$$ such that its complement is at most countable.

Thank you!

Your counterexample is fine. However, even though "most" uncountable subsets of $$X$$ will have an uncountable complement, that doesn't mean that the cocountable subsets are difficult to find, once you know how to look for them.
I am not sure how to find subsets of $$X$$ such that its complement is at most countable.
Take any at most countable (most people would just say "countable", as that usually includes finite) subset of $$X$$ and consider its complement. It will be cocountable.
For instance, if $$X=\Bbb R$$, then you have the non-integers, or the irrationals, or any number which isn't expressible as a polynomial of $$\pi$$ with integer coefficients as some first examples.
• @TriNguyen Don't forget that $X$ itself is also closed. – Arthur May 9 at 5:51