Cardinality of Set Union/3 Sets
Jump to navigation
Jump to search
Theorem
Let $S_1$, $S_2$ and $S_3$ be finite sets.
Then:
\(\ds \card {S_1 \cup S_2 \cup S_3}\) | \(=\) | \(\ds \card {S_1} + \card {S_2} + \card {S_3}\) | ||||||||||||
\(\ds \) | \(\) | \(\, \ds - \, \) | \(\ds \card {S_1 \cap S_2} - \card {S_1 \cap S_3} - \card {S_2 \cap S_3}\) | |||||||||||
\(\ds \) | \(\) | \(\, \ds + \, \) | \(\ds \card {S_1 \cap S_2 \cap S_3}\) |
Proof
This is a specific example of Cardinality of Set Union: General Case.
$\blacksquare$
Sources
- 1975: T.S. Blyth: Set Theory and Abstract Algebra ... (previous) ... (next): $\S 1$. Sets; inclusion; intersection; union; complementation; number systems: Exercise $8$
- 2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics (5th ed.) ... (previous) ... (next): cardinality