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Let A, B, and C be arbitrary sets taken from the positive integers. I have to prove or disprove that: A ∩ B ∩ C = ∅, then (A ⊆ ~B) or (A ⊆ ~C) Here is my disproof using a counterexample: If A = {} the empty set, B = {2, 3}, C = {4, 5} With these sets defined for A, B, and C, the intersection includes the disjoint set, and then that would lead to A being a subset of B or A being a subset of C which counteracts that if A ∩ B ∩ C = ∅, then (A ⊆ ~B) or (A ⊆ ~C) Is this a sufficient proof? |
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Let $A=\{2,3\}$, $B=\{1,3\}$, and $C=\{1,2\}$. The intersection of the three sets is empty. But none of them is a subset of the complement of another. By symmetry, it is sufficient for example to show that $A$ is not a subset of the complement $B^c$ of $B$. Note that $B^c$ consists of all integers except $1$ and $3$. Since $A$ does contain $3$, $A$ is not a subset of $B^c$. Comment: As has been pointed out in a comment by @Joe, the empty set is a subset of every set, so setting $A=\emptyset$ cannot give you a counterexample, whatever be the choice of $B$ and $C$. |
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As another counterexample, let $A=\mathbb{N}$, and $B$ and $C$ be disjoint sets with at least one element each. Elaborating: On the one hand, since $B$ and $C$ are disjoint, $$ A \cap B \cap C = \mathbb{N} \cap B \cap C = B \cap C = \emptyset. $$ On the other hand, $A = \mathbb{N}$ is not contained in the complement of $B$, since $B$ is not empty, and the same for $C$. |
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