Union equals Intersection iff Sets are Equal

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 $1$
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