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I am trying to show if the following can be possible or not.

We have $3$ sets, $A$, $B$ and $C$.

$$|A| = 5$$ $$|B| = 4$$ $$|C| = 3$$ $$|A\cup B \cup C| = 10$$ $$|A \cap B| = 2$$ $$|A \cap B \cap C| = 1$$

My Proof:

$$|A \cup B| = |A|+|B|-|A \cap B| = 5 + 4 - 2 = 7$$

$$|A \cup B| + |C| = |A \cup B \cup C| - |A \cap B \cap C| + |A \cap C| + |B \cap C| \\ \implies 7 + 3 = 10 - 1 + |A \cap C| + |B \cap C| \\\implies |A \cap C| + |B \cap C| = 1$$

But if this is impossible because:

$$|A \cap C| >= |A \cap B \cap C| = 1$$ and $$|B \cap C| >= |A \cap B \cap C| = 1$$

Can someone please confirm if the above is correct?

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    Can someone please confirm if the above is correct? – 2017-02-07
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    No, this is incorrect. – 2017-02-07
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    Is this |A∪B|+|C|=|A∪B∪C|−|A∩B∩C|+|A∩C|+|B∩C| actually true? – 2017-02-07
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    Seems okay to me. – 2017-02-07
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    It is correct. I notice that once you have $|A\cup B|=7$, then since $|C|=3,$ and $|A\cup B \cup C|=10,$ you must have $|C\cap (A\cup B)|=0$, ...(because $10=|(A\cup B)\cup C|=|(A\cup B)|+|C|-|(A\cup B)\cap C|=7+3-|(A\cup B)\cap C|.$)... but then $0=|C\cap (A\cup B)|\geq |C\cap (A\cap B)|=1$, so you get a contradiction. – 2017-02-07

2 Answers 2

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Indeed. You can do it in fewer steps, which makes it more clear. Use the principle of inclusion and exclusion, substitute know values and get an expression for the unknowns.

$\lvert A\cup B\cup C\rvert = \lvert A\rvert+\lvert B\rvert+\lvert C\rvert-\lvert A\cap B\rvert-\lvert A\cap C\rvert-\lvert B\cap C\rvert+\lvert A\cap B\cap C\rvert \\ 10 = 5+4+3-2-\lvert A\cap C\rvert-\lvert B\cap C\rvert+ 1 \\ \lvert A\cap C\rvert+\lvert B\cap C\rvert = 1$

Then because $A\cap C\supseteq A\cap B\cap C$ and $B\cap C\supseteq A\cap B\cap C$ it does follow that $\lvert A\cap C\rvert+\lvert B\cap C\rvert \geq 2$, which contradicts the above.

So we conclude that the given values are inconsistent.

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Just to verify:

$|A| = 5$. Let $A = \{a,b,c,d,e\}$

$|A\cap B| = 2$ so let $A\cap B = \{a,b\}$

$|B| = 4$ so let $B = \{a,b,f,g\}$.

So $A \cup B = \{a,b,c,d,e,f,g\}$ and $|A \cup B| = 7$.

$|A \cup B \cup C| = 10$ so $|(A \cup B \cup C) \setminus (A\cup B)| = 3$

So let $(A \cup B \cup C) \setminus (A\cup B) = \{h,i,j\} \subset C$.

$|C| = 3$ so $C = \{h,i,j\}$ so $A\cap B \cap C =\emptyset$ and $|A \cap B \cap C| = 0$.

Yeah, this is impossible.