Set Difference and Intersection form Partition/Corollary 2

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Corollary to Set Difference and Intersection form Partition

Let $\O \subsetneqq T \subsetneqq S$.


Then:

$\set {T, \relcomp S T}$

is a partition of $S$.


Proof

First we note that:

\(\ds \O \subsetneqq T\) \(\implies\) \(\ds T \ne \O\)
\(\ds T \subsetneqq S\) \(\implies\) \(\ds T \ne S\)


From the definition of relative complement, we have:

$\relcomp S T = S \setminus T$.

It follows from Set Difference with Self is Empty Set that:

$T \ne S \iff \relcomp S T \ne \O$


From Intersection with Relative Complement is Empty:

$T \cap \relcomp S T = \O$

That is, $T$ and $\relcomp S T$ are disjoint.


From Union with Relative Complement:

$T \cup \relcomp S T = S$

That is, the union of $T$ and $\relcomp S T$ forms the whole set $S$.


Thus all the conditions for a partition are satisfied.

$\blacksquare$


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