If $|A \cup B \cup C | = |A| + |B| + |C|$, then $A, B$, and $C$ must be pairwise disjoint

Question

Let A, B, and C be finite sets. Prove or disprove:

If $|A \cup B \cup C | = |A| + |B| + |C|$, then $A, B$, and $C$ must be pairwise disjoint.

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I started with inclusion-exclusion formula

$$|A \cup B \cup C | = |A| + |B| + |C| - |A \cap B| - |A \cap C| - |B \cap C| + |A \cap B \cap C|$$

Hypothesis says if $|A \cup B \cup C | = |A| + |B| + |C|$, then expression $|A \cap B \cap C| - |A \cap B| - |A \cap C| - |B \cap C| $, must be equal to zero. So, we have

$$ |A \cap B \cap C| - |A \cap B| - |A \cap C| - |B \cap C| = 0$$ $$ |A \cap B \cap C| = |A \cap B| + |A \cap C| + |B \cap C| \tag{1}$$

Now, I am not sure how finish this proof. I found two ways, but I do not know if are correct.

First way

If $x \in |A \cap B \cap C|$, then $x \in |A \cap B| \wedge x\in |A \cap C| \wedge x \in |B \cap C|$.

So if we count the cardinality of $|A \cap B \cap C| = |A \cap B| + |A \cap C| + |B \cap C|$, then if $x \in |A \cap B \cap C|$ then $x$ is counted once but on RHS $x$ is counted three times. Thus the equation is true if $A = B = C = \emptyset$.

Therefore $A, B,$ and $C$ are pairwise disjoint.

Second way

From inclusion-exclusion formula, I know $$|A \cap B \cap C| = - |A| - |B| - |C| + |A \cap B| + |A \cap C| + |B \cap C| - |A \cup B \cup C |$$ I substitute in equation (1) $$ -|A| - |B| - |C| + |A \cap B| + |A \cap C| + |B \cap C| - |A \cup B \cup C | = |A \cap B| + |A \cap C| + |B \cap C|$$ The expression $|A \cap B| + |A \cap C| + |B \cap C|$ cancels out. $$ |A| + |B| + |C| + |A \cup B \cup C | = 0$$

And this equation is true if $A = B = C = \emptyset$.
Therefore $A, B,$ and $C$ are pairwise disjoint.

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My question is which way is correct and better or if there is a different method to prove the proposition. Thank you. (I am self-studying student, who is reading a textbook about discrete mathematics).

Answer 1

• Your first way is correct unless $A \cap B \cap C=\emptyset.$
• The equation in your second way has to be corrected as $$|A \cap B \cap C| = - |A| - |B| - |C| + |A \cap B| + |A \cap C| + |B \cap C| + |A \cup B \cup C |$$ and substituting this leads to a nothing (as you use the same equation twice).

This is how I would answer this question (using inclusion-exclusion principal):
$A \cap B \cap C\subset A \cap B$ and therefore $|A \cap B \cap C| \le |A \cap B|, |A \cap C|, |B \cap C|.$
Then we have $$3|A \cap B \cap C| \le |A \cap B| + |A \cap C| + |B \cap C|.$$ Now combining this with what you have already found $$|A \cap B \cap C| = |A \cap B| + |A \cap C| + |B \cap C|,$$ we can say $$|A \cap B| =|A \cap C| = |B \cap C|=|A \cap B \cap C|=0 .$$

Answer 2

Suppose $A$, $B$, $C$ are not pairwise disjoint. Then the intersection of two of them, wlog $A$ and $B$, is nonempty. Now $|A \cup B| = |A| + |B| - |A \cap B| < |A| + |B|$, and $|A \cup B \cup C| \le |A \cup B| + |C| < |A| + |B| + |C|$.