Union Distributes over Intersection

Theorem

Set union is distributive over set intersection:

$R \cup \left({S \cap T}\right) = \left({R \cup S}\right) \cap \left({R \cup T}\right)$

General Result

Let $S$ and $T$ be sets.

Let $\mathcal P \left({T}\right)$ be the power set of $T$.

Let $\mathbb T$ be a subset of $\mathcal P \left({T}\right)$.

Then:

$\displaystyle S \cup \bigcap \mathbb T = \bigcap_{X \in \mathbb T} \left({S \cup X}\right)$

Proof

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle x \in R \cup \left({S \cap T}\right)\)	\(\iff\)	\(\displaystyle x \in R \lor \left({x \in S \land x \in T}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Definitions of Union and Intersection
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\iff\)	\(\displaystyle \left ({x \in R \lor x \in S}\right) \land \left({x \in R \lor x \in T}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Rule of Distribution
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\iff\)	\(\displaystyle x \in \left({R \cup S}\right) \cap \left({R \cup T}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Definitions of Union and Intersection

$\blacksquare$

Proof of General Result

Union Subset of Intersection

Let $\displaystyle x \in S \cup \bigcap \mathbb T$.

We need to show that $\displaystyle x \in \bigcap_{X \in \mathbb T} \left({S \cup X}\right)$ and then by definition of subset we will have shown that $\displaystyle S \cup \bigcap \mathbb T \subseteq \bigcap_{X \in \mathbb T} \left({S \cup X}\right)$.

So, we have that $\displaystyle x \in S \cup \bigcap \mathbb T$.

By definition of set union, $x \in S$ or $\displaystyle x \in \bigcap \mathbb T$.

So there are two cases to consider:

$(1): \quad$ Suppose $x \in S$.

Then by definition of set union, $\displaystyle \forall X \in \mathbb T: x \in S \cup X$.

So:

$\displaystyle x \in \bigcap_{X \in \mathbb T} \left({S \cup X}\right)$

$(2): \quad$ Suppose $\displaystyle x \in \bigcap \mathbb T$.

Then by definition of set intersection, $\displaystyle \forall X \in \mathbb T: x \in X$.

So by definition of set union, $\displaystyle \forall X \in \mathbb T: x \in S \cup X$.

So:

$\displaystyle x \in \bigcap_{X \in \mathbb T} \left({S \cup X}\right)$

In both cases we see that:

$\displaystyle x \in \bigcap_{X \in \mathbb T} \left({S \cup X}\right)$

so by Proof by Cases, we have that:

$\displaystyle S \cup \bigcap \mathbb T \subseteq \bigcap_{X \in \mathbb T} \left({S \cup X}\right)$

$\Box$

Intersection Subset of Union

Let $\displaystyle x \in \bigcap_{X \in \mathbb T} \left({S \cup X}\right)$.

We need to show that $\displaystyle x \in S \cup \bigcap \mathbb T$ and then by definition of subset we will have shown that $\displaystyle \bigcap_{X \in \mathbb T} \left({S \cup X}\right) \subseteq S \cup \bigcap \mathbb T$.

So, we have that $\displaystyle x \in \bigcap_{X \in \mathbb T} \left({S \cup X}\right)$.

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:

$\displaystyle x \in \bigcap_{X \in \mathbb T} X$

and so by definition of set union:

$\displaystyle x \in S \cup \bigcap \mathbb T$

$(2): \quad \exists X \in \mathbb T: x \notin X$

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:

$\displaystyle x \in S \cup \bigcap \mathbb T$

In both cases we see that:

$\displaystyle x \in S \cup \bigcap \mathbb T$

so by Proof by Cases, we have that:

$\displaystyle \bigcap_{X \in \mathbb T} \left({S \cup X}\right) \subseteq S \cup \bigcap \mathbb T$

$\Box$

So we have that:

$S\displaystyle \cup \bigcap \mathbb T \subseteq \bigcap_{X \in \mathbb T} \left({S \cup X}\right)$

and

$\displaystyle \bigcap_{X \in \mathbb T} \left({S \cup X}\right) \subseteq S \cup \bigcap \mathbb T$

and so by definition of Equality of Sets:

$\displaystyle S \cup \bigcap \mathbb T = \bigcap_{X \in \mathbb T} \left({S \cup X}\right)$

$\blacksquare$

Also see

• Intersection Distributes over Union

Sources

• Paul R. Halmos: Naive Set Theory (1960)... (previous)... (next): $\S 4$: Unions and Intersections
• Paul R. Halmos: Naive Set Theory (1960)... (previous)... (next): $\S 9$: Families
• W.E. Deskins: Abstract Algebra (1964): Exercise $1.1: 8 \ \text {(e)}$, Exercise $1.4: 6$
• Steven A. Gaal: Point Set Topology (1964)... (previous)... (next): Introduction to Set Theory: $1$. Elementary Operations on Sets
• J.A. Green: Sets and Groups (1965)... (previous)... (next): $\S 1.5$: Example $18$
• Seth Warner: Modern Algebra (1965)... (previous)... (next): $\S 3$: Theorem $3.1$
• Seth Warner: Modern Algebra (1965)... (previous)... (next): Exercise $3.2$
• Seth Warner: Modern Algebra (1965)... (previous)... (next): Exercise $3.6 \ \text{(b)}$
• George McCarty: Topology: An Introduction with Application to Topological Groups (1967): $\text{I}$: Theorem $2 \ \text{(ii)}$, Exercise $\text{C}$
• A.N. Kolmogorov and S.V. Fomin‎: Introductory Real Analysis (1968): $\S 1.2$
• Allan Clark: Elements of Abstract Algebra (1971)... (previous)... (next): $\S 7 \ \text{(a)}$
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• René L. Schilling: Measures, Integrals and Martingales (2005)... (previous)... (next) $\S 2$