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

Proof by Counterexample:


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