Union Distributes over Intersection/General Result/Proof
Theorem
Let $S$ and $T$ be sets.
Let $\powerset T$ be the power set of $T$.
Let $\mathbb T$ be a subset of $\powerset T$.
Then:
- $\ds S \cup \bigcap \mathbb T = \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X}$
Proof
Union Subset of Intersection
Let $\ds x \in S \cup \bigcap \mathbb T$.
We need to show that:
- $\ds x \in \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X}$
and then by definition of subset we will have shown that:
- $\ds S \cup \bigcap \mathbb T \subseteq \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X}$.
So, we have that $\ds x \in S \cup \bigcap \mathbb T$.
By definition of set union, $x \in S$ or $\ds x \in \bigcap \mathbb T$.
So there are two cases to consider:
$(1): \quad$ Suppose $x \in S$.
Then by definition of set union, $\forall X \in \mathbb T: x \in S \cup X$.
So:
- $\ds x \in \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X}$
$(2): \quad$ Suppose $\ds x \in \bigcap \mathbb T$.
Then by definition of set intersection:
- $\forall X \in \mathbb T: x \in X$
So by definition of set union:
- $\forall X \in \mathbb T: x \in S \cup X$
So:
- $\ds x \in \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X}$
In both cases we see that:
- $\ds x \in \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X}$
so by Proof by Cases, we have that:
- $\ds S \cup \bigcap \mathbb T \subseteq \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X}$
$\Box$
Intersection Subset of Union
Let $\ds x \in \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X}$.
We need to show that:
- $\ds x \in S \cup \bigcap \mathbb T$
and then by definition of subset we will have shown that:
- $\ds \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X} \subseteq S \cup \bigcap \mathbb T$.
So, we have that:
- $\ds x \in \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X}$
By definition of set intersection:
- $(A): \quad \forall X \in \mathbb T: x \in S \cup X$
There are two cases to consider:
- $(1): \quad \forall X \in \mathbb T: x \in X$
Then by definition of set intersection:
- $\ds x \in \bigcap_{X \mathop \in \mathbb T} X$
and so by definition of set union:
- $\ds x \in S \cup \bigcap \mathbb T$
- $(2): \quad \exists X \in \mathbb T: x \notin X$
From $(A)$ we have that $x \in S \cup X$.
But as $x \notin X$ it follows that $x \in S$.
Then by definition of set union:
- $\ds x \in S \cup \bigcap \mathbb T$
In both cases we see that:
- $\ds x \in S \cup \bigcap \mathbb T$
so by Proof by Cases, we have that:
- $\ds \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X} \subseteq S \cup \bigcap \mathbb T$
$\Box$
So we have that:
- $S \cup \ds \bigcap \mathbb T \subseteq \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X}$
and
- $\ds \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X} \subseteq S \cup \bigcap \mathbb T$
and so by definition of set equality:
- $S \cup \ds \bigcap \mathbb T = \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X}$
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text I$: Algebraic Structures: $\S 3$: Unions and Intersections of Sets: Exercise $3.6 \ \text{(b)}$
- 1967: George McCarty: Topology: An Introduction with Application to Topological Groups ... (previous) ... (next): Chapter $\text{I}$: Sets and Functions: Exercise $\text{C}$