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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!

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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.

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    $\begingroup$ Oh that's right. I can just describe the closed sets instead (which are countable sets) in which I get a 1-1 correspondence with the open sets. This makes sense. Thank you! $\endgroup$ – Tri Nguyen May 9 at 5:50
  • $\begingroup$ @TriNguyen Don't forget that $X$ itself is also closed. $\endgroup$ – Arthur May 9 at 5:51

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