# Is $A^c \cap B^c$ finite if $A,B$ are disjoint finite subsets of an uncountable set? [closed]

If $$\Omega$$ is an uncountable set, and $$A$$ and $$B$$ are two finite sets such that $$A, B \subset \Omega$$ and $$A \cap B = \varnothing$$. Is $$A^c \cap B^c$$ finite?

I think it is (for example, $$\Omega = \mathbb R$$, $$A = \{1\}$$, $$B = \{2\}$$) but don't know how to prove it.

• No, it is not finite in your eaxmple and it cannot be finite in general. – Kavi Rama Murthy Apr 13 at 23:46
• In your example, $A^c = \mathbb{R} \setminus \{1\}$ and $B^c = \mathbb{R} \setminus \{2\}$, so $A^c \cap B^c = \mathbb{R} \setminus \{1,2\}$, which is certainly not finite. Two hints: either make a proof from my example, or use DeMorgan's Law. – William Apr 13 at 23:46
• $\mathbb{R}$ is uncountable. so is $\mathbb{R} \setminus \{1,2\}$ – Gopal Anantharaman Apr 13 at 23:51

Note that $$A^c \cap B^c = (A \cup B)^c$$ by De Morgan's law and since $$A \cup B$$ is finite then $$(A \cup B)^c$$ must be infinite.

You can think of cofinite topology on an uncountable set. And then apply that finite intersection of the open sets are open. Obviously, $$A^c$$ and $$B^c$$ are open. Thus, $$A^c\cap B^c$$ is open. Therefore, $$A^c\cap B^c$$ cannot be finite.

If $$A=\{1\}$$ and $$B=\{2\}$$ and $$\Omega=\mathbb R,$$ and complementation is relative to $$\Omega,$$ then $$A^c \cap B^c$$ is the set of all members of $$\mathbb R$$ except $$1$$ and $$2.$$ That is not finite.

Note that by De Morgan’s Law $$(A\cup B)^c = A^c \cap B^c.$$ Because $$A$$ and $$B$$ are finite, $$A\cup B$$ is finite too, and thus $$(A\cup B)^c = \Omega\setminus (A\cup B)$$ is uncountable.

This problem can be viewed in another way "topologically":

If you consider cofinite topology on the uncountable $$\Omega$$: Open sets are those whose complements are finite.

Then $$A$$ and $$B$$ are closed set with respect to the cofinite topology. Clearly, then $$A^c$$ and $$B^c$$ are open. And so is $$A^c \cap B^c$$. Then $$A^c\cap B^c$$ can never be finite, rather complement of the $$A^c\cap B^c$$, i.e., $$A\cup B$$ is finite.