Disjoint
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Definition
Let [ilmath]A[/ilmath] and [ilmath]B[/ilmath] be sets. We say "[ilmath]A[/ilmath] and [ilmath]B[/ilmath] are disjoint" if:
 [ilmath]A\cap B=[/ilmath] [ilmath]\emptyset[/ilmath] where [ilmath]A\cap B[/ilmath] denotes the intersection of [ilmath]A[/ilmath] and [ilmath]B[/ilmath]
Disjoint in a set
Let [ilmath]Z[/ilmath] be a set and let [ilmath]A[/ilmath] and [ilmath]B[/ilmath] be sets (with no other requirements), then we say "[ilmath]A[/ilmath] and [ilmath]B[/ilmath] are disjoint in [ilmath]Z[/ilmath]" if:
 [ilmath]A\cap B\cap Z=\emptyset[/ilmath]
Comments on "disjoint in a set"
There are 2 ways to think about it that show intuitively what we mean by "disjoint in a set":
 [ilmath](A\cap Z)\cap(B\cap Z)=\emptyset[/ilmath] is probably the most natural, we're saying that the parts of [ilmath]A[/ilmath] and [ilmath]B[/ilmath] actually in [ilmath]Z[/ilmath] must be disjoint
 This is easily seen to be equivalent to the above definition
 [ilmath]A\cap B\subseteq Z^\complement[/ilmath]  where [ilmath]Z^\complement[/ilmath] denotes the set complement of [ilmath]Z[/ilmath], (which may not always be defined/make sense) and why we have the other form
TODO: I've proved [ilmath](A\cap B\cap Z=\emptyset)\iff(A\cap B\subseteq Z^\complement)[/ilmath] on paper, It seems that we infact have: [ilmath](A\cap B=\emptyset)\iff(A\subseteq B^\complement)[/ilmath], maybe they should get pages....
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See also
 Pairwise disjoint
 Nonempty
 Disjoint in a set (links to above section for "disjoint in a set")
References
