Union equals Intersection iff Sets are Equal
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Theorem
Let $S$ and $T$ be sets.
Then:
- $\paren {S \cup T = S} \land \paren {S \cap T = S} \iff S = T$
where:
- $S \cup T$ denotes set union
- $S \cap T$ denotes set intersection.
Proof
From Intersection with Subset is Subset:
- $S \subseteq T \iff S \cap T = S$
From Union with Superset is Superset:
- $S \subseteq T \iff S \cup T = T$
That is:
- $T \subseteq S \iff S \cup T = S$
Thus:
- $\paren {S \cup T = S} \land \paren {S \cap T = S} \iff S \subseteq T \subseteq S$
By definition of set equality:
- $S = T \iff S \subseteq T \subseteq S$
Hence the result.
$\blacksquare$
Sources
- 1968: A.N. Kolmogorov and S.V. Fomin: Introductory Real Analysis ... (previous) ... (next): $\S 1$: Sets and Functions: Problem $1$