Union with Intersection equals Intersection with Union iff Subset
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Theorem
Let $A$, $B$ and $C$ be sets.
Then:
- $\paren {A \cap B} \cup C = A \cap \paren {B \cup C} \iff C \subseteq A$
Proof
Necessary Condition
Let $C \subseteq A$.
We have:
\(\ds C\) | \(\subseteq\) | \(\ds A\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds A \cap C\) | \(=\) | \(\ds C\) | Intersection with Subset is Subset | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {A \cap B} \cup \paren {A \cap C}\) | \(=\) | \(\ds \paren {A \cap B} \cup C\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds A \cap \paren {B \cup C}\) | \(=\) | \(\ds \paren {A \cap B} \cup C\) | Intersection Distributes over Union |
and another way:
\(\ds C\) | \(\subseteq\) | \(\ds A\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds A \cup C\) | \(=\) | \(\ds A\) | Union with Superset is Superset | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {A \cup C} \cap \paren {B \cup C}\) | \(=\) | \(\ds A \cap \paren {B \cup C}\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {A \cap B} \cup C\) | \(=\) | \(\ds A \cap \paren {B \cup C}\) | Intersection Distributes over Union |
Thus:
- $C \subseteq A \implies \paren {A \cap B} \cup C = A \cap \paren {B \cup C}$
$\Box$
Sufficient Condition
Let $\paren {A \cap B} \cup C = A \cap \paren {B \cup C}$.
Aiming for a contradiction, suppose $\exists x \in C: x \notin A$.
From Set is Subset of Union:
- $C \subseteq \paren {A \cap B} \cup C$
and so by definition of subset:
- $x \in \paren {A \cap B} \cup C$
From Intersection is Subset:
- $A \cap \paren {B \cup C} \subseteq A$
and so (indirectly) by definition of subset:
- $x \notin A \cap \paren {B \cup C}$
Hence by definition of subset:
- $\paren {A \cap B} \cup C \nsubseteq A \cap \paren {B \cup C}$
that is:
- $\paren {A \cap B} \cup C \ne A \cap \paren {B \cup C}$
This contradicts our initial assertion.
Hence by Proof by Contradiction:
- $\neg \exists x \in C: x \notin A$
From Assertion of Universality:
- $\forall x \in C: x \in A$
and so $C \subseteq A$ by definition of subset.
$\blacksquare$
Sources
- 1975: T.S. Blyth: Set Theory and Abstract Algebra ... (previous) ... (next): $\S 1$. Sets; inclusion; intersection; union; complementation; number systems: Exercise $11 \ \text{(d)}$