Union of Relative Complements of Nested Subsets
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Theorem
Let $R \subseteq S \subseteq T$ be sets with the indicated inclusions.
Then:
- $\relcomp T S \cup \relcomp S R = \relcomp T R$
where $\complement$ denotes relative complement.
Phrased via Set Difference as Intersection with Relative Complement:
- $\paren {T \setminus S} \cup \paren {S \setminus R} = T \setminus R$
where $\setminus$ denotes set difference.
Proof
From Union with Set Difference:
- $T = T \setminus S \cup S$
and therefore by Set Difference is Right Distributive over Union:
- $T \setminus R = \paren {\paren {T \setminus S} \setminus R} \cup \paren {S \setminus R}$
Now, by Set Difference with Union and Union with Superset is Superset:
- $\paren {T \setminus S} \setminus R = T \setminus \paren {S \cup R} = T \setminus S$
Combining the above yields:
- $T \setminus R = \paren {T \setminus S} \cup \paren {S \setminus R}$
$\blacksquare$
Also see
- Set Difference is Subset of Union of Differences is the inclusion retained when the subset conditions are dropped