Set Theory/Examples/A cup (X cap B) = C, (A cup X) cap B = D

Example in Set Theory

Let $A, B, C, D$ be subsets of a set $S$.

Let there exist $X \subseteq S$ such that:

$A \cup \paren {X \cap B} = C$

$\paren {A \cup X} \cap B = D$

Then:

$A \cap B \subseteq D \subseteq B$

and:

$A \cup D = C$

Converse

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

\(\ds D\)

\(=\)

\(\ds \paren {A \cup X} \cap B\)

\(\text {(1)}: \quad\)

\(\ds \leadsto \ \ \)

\(\ds D\)

\(\subseteq\)

\(\ds B\)

Intersection is Subset

\(\ds D\)

\(=\)

\(\ds \paren {A \cup X} \cap B\)

\(\text {(2)}: \quad\)

\(\ds \leadsto \ \ \)

\(\ds D\)

\(=\)

\(\ds \paren {A \cap B} \cup \paren {X \cap B}\)

Distributive Laws of Set Theory

\(\text {(3)}: \quad\)

\(\ds \leadsto \ \ \)

\(\ds A \cap B\)

\(\subseteq\)

\(\ds D\)

Intersection is Subset

\(\ds \leadsto \ \ \)

\(\ds A \cap B\)

\(\subseteq\)

\(\ds D\)

from $(1)$ and $(3)$

\(\ds \)

\(\subseteq\)

\(\ds B\)

Then:

\(\ds A \cup D\)

\(=\)

\(\ds A \cup \paren {A \cap B} \cup \paren {X \cap B}\)

from $(2)$

\(\ds \)

\(=\)

\(\ds A \cup \paren {X \cap B}\)

Union Absorbs Intersection

\(\ds \)

\(=\)

\(\ds C\)

by hypothesis

$\blacksquare$

Sources

• 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $1$: Sets and Logic: Exercise $12$