Union of Relative Complements of Nested Subsets

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