Power Set is Closed under Union

Theorem

Let $S$ be a set.

Let $\powerset S$ be the power set of $S$.

Then:

$\forall A, B \in \powerset S: A \cup B \in \powerset S$

Proof

Let $A, B \in \powerset S$.

Then by the definition of power set, $A \subseteq S$ and $B \subseteq S$.

We also have $A \cup B \subseteq S \iff A \subseteq S \land B \subseteq S$ from Union is Smallest Superset.

Thus $A \cup B \in \powerset S$, and closure is proved.

$\blacksquare$

Also see

• Power Set is Closed under Intersection
• Power Set is Closed under Set Difference

Sources

• 1965: J.A. Green: Sets and Groups ... (previous) ... (next): $\S 4.1$. Binary operations on a set: Example $60$
• 2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics (5th ed.) ... (previous) ... (next): algebra of sets
• 2021: Richard Earl and James Nicholson: The Concise Oxford Dictionary of Mathematics (6th ed.) ... (previous) ... (next): algebra of sets