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