Union of Power Sets not always Equal to Powerset of Union
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Theorem
The union of the power sets of two sets $S$ and $T$ is not necessarily equal to the power set of their union.
Proof
Let $S = \set {1, 2, 3}, T = \set {2, 3, 4}, X = \set {1, 2, 3, 4}$.
\(\ds S \cup T\) | \(=\) | \(\ds \set {1, 2, 3, 4}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds X\) | \(\subseteq\) | \(\ds S \cup T\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds X\) | \(\in\) | \(\ds \powerset {S \cup T}\) |
But note that $X \nsubseteq S \land X \nsubseteq T$.
Thus:
\(\ds X\) | \(\nsubseteq\) | \(\ds S \land X \nsubseteq T\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds X\) | \(\notin\) | \(\ds \powerset S \land X \notin \powerset T\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \neg (X\) | \(\in\) | \(\ds \powerset S \lor X \in \powerset T)\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds X\) | \(\notin\) | \(\ds \powerset S \cup \powerset T\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \powerset {S \cup T}\) | \(\nsubseteq\) | \(\ds \powerset S \cup \powerset T\) |
So:
- $\powerset {S \cup T} \ne \powerset S \cup \powerset T$
$\blacksquare$
Sources
- 1975: T.S. Blyth: Set Theory and Abstract Algebra ... (previous) ... (next): $\S 2$. Sets of sets: Exercise $5 \ \text{(b)}$
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $1$: Sets and Logic: Exercise $7$