# Thread:

1. ## Inclusion/Exclusion

I've been working on this problem for a while. Could someone show me how to connect the last part?

Let A, B, and C be finite sets. Prove:
If $|A \cup B \cup C| = |A| + |B| + |C|$
then A, B, and C must be pairwise disjoint.

Here is what I have:

Suppose A, B, and C are finite sets with $|A \cup B \cup C| = |A| + |B| + |C|$.

By inclusion/exclusion, we know that

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

By cancellation, we have:

$|A \cap B \cap C| - |A \cap B| - |A \cap C| - |B \cap C| = 0$

I'm just not sure how to connect that to "Thus A, B, and C must be pairwise disjoint."

2. Originally Posted by absvalue
I've been working on this problem for a while. Could someone show me how to connect the last part?

Let A, B, and C be finite sets. Prove:
If $|A \cup B \cup C| = |A| + |B| + |C|$
then A, B, and C must be pairwise disjoint.

Here is what I have:

Suppose A, B, and C are finite sets with $|A \cup B \cup C| = |A| + |B| + |C|$.

By inclusion/exclusion, we know that

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

By cancellation, we have:

$|A \cap B \cap C| - |A \cap B| - |A \cap C| - |B \cap C| = 0$

I'm just not sure how to connect that to "Thus A, B, and C must be pairwise disjoint."

Well, $A \cap B \cap C \subset A\cap B$, so rearranging

$<br /> |A \cap C| + |B \cap C| = |A \cap B \cap C| - |A \cap B| \le 0<br />$

since the LHS must be non negative we have $|A \cap C| + |B \cap C| = 0$ so both of those terms must be 0. You can do a similar thing with to get $|A \cap B| = 0$ proving the result.
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