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For sets $A, B, C, D, E,$ and $F$ where any two of these sets have a nonempty intersection. Is \begin{align*} & (A \cup B \cup C)\\ {}\cap{} & (A^c \cup D \cup E)\\ {}\cap{} & (B^c \cup D^c \cup F)\\ {}\cap{} & (C^c \cup E^c \cup F^c) \end{align*}

empty?

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    @user46944 : yup – 2017-02-07
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    guys, how about if we add the condition that any two sets have their intersection nonempty? – 2017-02-07

2 Answers 2

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No.

Let $A =\{ a \}$, $B =\{ b \}$, etc.

Then $b \in B$, $b \in A^c$, $b \in D^c$, and $b \in C^c$, and hence $b$ is in all four sets, and hence in intersection of all those 4 sets.

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Consider $X = \{1,2,3,4,5,6\}$, and take $A = \{1\}$, $B= \{2\}$, $C= \{3\}$, $D= \{4\}$, $E= \{5\}$, and $F= \{6\}$.

Then $A \cup B \cup C = \{1,2,3\}$,

$A^{c} \cup D \cup E = \{2,3,4,5,6 \}$,

$B^{c} \cup D^{c} \cup F = \{ 1,2,3,4,5,6 \}$,

and $C^{c} \cup E^{c} \cup F^{c} = \{1,2,3,4,5,6\}$.

Intersecting the above sets gives $\{2,3\}$, which is not empty.

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    Beat me by 16 seconds ... :) – 2017-02-07
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    @Bram28 Yet your answer is equally as important. :) – 2017-02-07
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    i didn't think about a counter-example.. :'( thanks guys. – 2017-02-07
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    @Bram28,, how about if we add the condition that any two sets have their intersection nonempty? – 2017-02-07
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    @user46944 ow about if we add the condition that any two sets have their intersection nonempty? – 2017-02-07
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    @JefferDaveCagubcob OK, then just add some element (say, 1) to all the sets. So in my example, you'd have $A=\{ a,1\}, B =\{b,1\}$, etc. it will still be true that $b$ is in all four sets and hence in the intersection. – 2017-02-07
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    @Bram28 i get it.. thanks – 2017-02-07