Set Theory/Examples/A cup (X cap B) = C, (A cup X) cap B = D/Converse
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Example in Set Theory
Let $A, B, C, D$ be subsets of a set $S$.
Let the following conditions hold:
- $A \cap B \subseteq D \subseteq B$
and:
- $A \cup D = C$
Then these equations hold:
- $A \cup \paren {X \cap B} = C$
- $\paren {A \cup X} \cap B = D$
if:
- $X = D \setminus A$
Proof
Suppose $X = D \setminus A$.
Then:
\(\ds X \cap B\) | \(=\) | \(\ds \paren {D \setminus A} \cap B\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds B \cap D \cap \relcomp S A\) | Set Difference as Intersection with Relative Complement |
\(\ds D\) | \(\subseteq\) | \(\ds B\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds B \cap D \cap \relcomp S A\) | \(=\) | \(\ds D \cap \relcomp S A\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds X \cap B\) | \(=\) | \(\ds D \cap \relcomp S A\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds A \cup \paren {X \cap B}\) | \(=\) | \(\ds A \cup \paren {D \cap \relcomp S A}\) | |||||||||||
\(\ds \) | \(=\) | \(\ds A \cup D\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds C\) |
\(\ds \paren {A \cup X} \cap B\) | \(=\) | \(\ds \paren {A \cup \paren {D \cap \relcomp S A} } \cap B\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {A \cup D} \cap B\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {A \cap B} \cup \paren {D \cap B}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {A \cap B} \cup D\) | as $D \subseteq B$ | |||||||||||
\(\ds \) | \(=\) | \(\ds D\) |
$\blacksquare$
Sources
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $1$: Sets and Logic: Exercise $12$