Is $A^c \cap B^c$ finite if $A,B$ are disjoint finite subsets of an uncountable set? 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.
 A: 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.
A: 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.
A: 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.
A: 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.
A: 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.
