Cardinality of Set Union/2 Sets
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Theorem
Let $S_1$ and $S_2$ be finite sets.
Then:
- $\card {S_1 \cup S_2} = \card {S_1} + \card {S_2} - \card {S_1 \cap S_2}$
Proof
We have that Cardinality is Additive Function.
From Additive Function is Strongly Additive:
- $\card {S_1 \cup S_2} + \card {S_1 \cap S_2} = \card {S_1} + \card {S_2}$
from which the result follows.
$\blacksquare$
Sources
- 1965: J.A. Green: Sets and Groups ... (previous) ... (next): $\S 1.4$. Union: Example $16$
- 1975: T.S. Blyth: Set Theory and Abstract Algebra ... (previous) ... (next): $\S 1$. Sets; inclusion; intersection; union; complementation; number systems: Exercise $8$
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 19$: Combinatorial Analysis: Exercise $19.1$
- 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $1$: Mappings: $\S 15 \beta$
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $1$: Sets and Logic: Exercise $8 \ \text{(b)}$
- 1982: P.M. Cohn: Algebra Volume 1 (2nd ed.) ... (previous) ... (next): Chapter $1$: Sets and mappings: $\S 1.2$: Sets: Exercise $3$
- 2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics (5th ed.) ... (previous) ... (next): cardinality