Set Difference Relative Complement
From ProofWiki
Theorem
Let $A, B \subseteq S$.
Then the set difference between $A$ and $B$ can be expressed as the intersection with the relative complement with respect to $S$:
- $A \setminus B = A \cap \complement_S \left({B}\right)$
Proof
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle A \setminus B\) | \(=\) | \(\displaystyle \left\{ {x \in A: x \notin B}\right\}\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Definition of set difference | ||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \left\{ {x \in A: x \in \complement_S \left({B}\right)}\right\}\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Definition of relative complement | ||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle A \cap \complement_S \left({B}\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Definition of set intersection |
$\blacksquare$
Sources
- Paul R. Halmos: Naive Set Theory (1960)... (previous)... (next): $\S 5$: Complements and Powers
- T.S. Blyth: Set Theory and Abstract Algebra (1975): $\S 1$
- Thomas A. Whitelaw: An Introduction to Abstract Algebra (1978)... (previous)... (next): Exercise $1.5$