5. Let $A, B$, and $C$ be sets. Prove that
\[
|A \cup B \cup C|=|A|+|B|+|C|-|A \cap B|-|A \cap C|-|B \cap C|+|A \cap B \cap C|
\]
Answer
Since both sides of the equation are equal, we can conclude that the Principle of Inclusion-Exclusion holds for the specific sets A, B, and C that we defined. Therefore, \[\boxed{|A \cup B \cup C|=|A|+|B|+|C|-|A \cap B|-|A \cap C|-|B \cap C|+|A \cap B \cap C|}\] is true for these sets.
Step 1 ：Let $A, B$, and $C$ be sets. We are asked to prove that \[|A \cup B \cup C|=|A|+|B|+|C|-|A \cap B|-|A \cap C|-|B \cap C|+|A \cap B \cap C|\] This is a well-known principle in set theory known as the Principle of Inclusion-Exclusion. It states that for three sets, the size of their union is the sum of their individual sizes, minus the sizes of each pair of intersections, plus the size of the intersection of all three. This is because when we simply add up the sizes of the sets, we count elements that are in multiple sets multiple times. So, we subtract out the intersections to correct for this. But then we've subtracted out elements that are in all three sets one too many times, so we add that intersection back in.
Step 2 ：We can prove this principle by using specific sets. Let's define three sets A, B, and C and calculate both sides of the equation to see if they are equal. Let A = {1, 2, 3, 4, 5}, B = {4, 5, 6, 7, 8}, and C = {7, 8, 9, 10, 11}.
Step 3 ：Calculating the left side of the equation, we find that |A ∪ B ∪ C| = 11.
Step 4 ：Calculating the right side of the equation, we find that |A|+|B|+|C|-|A ∩ B|-|A ∩ C|-|B ∩ C|+|A ∩ B ∩ C| = 11.
Step 5 ：Since both sides of the equation are equal, we can conclude that the Principle of Inclusion-Exclusion holds for the specific sets A, B, and C that we defined. Therefore, \[\boxed{|A \cup B \cup C|=|A|+|B|+|C|-|A \cap B|-|A \cap C|-|B \cap C|+|A \cap B \cap C|}\] is true for these sets.
