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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.

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  • $\begingroup$ No, it is not finite in your eaxmple and it cannot be finite in general. $\endgroup$ – Kavi Rama Murthy Apr 13 at 23:46
  • $\begingroup$ 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. $\endgroup$ – William Apr 13 at 23:46
  • $\begingroup$ $\mathbb{R}$ is uncountable. so is $\mathbb{R} \setminus \{1,2\}$ $\endgroup$ – Gopal Anantharaman Apr 13 at 23:51
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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.

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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.

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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.

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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.

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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.

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