# Set theory vocabulary: the unique subsets which compose a set?

I'm struggling for the correct vocabulary/notation for the following situation. Let's say S is a (very simple) set with the following composition:

S = {1, 2, 3, 4, 5, 6}


Now, there are two subsets which compose this set, A and B:

A = {1, 2}
B = {3, 4, 5, 6}


Obviously, A is a subset of S, $A \subseteq S$, and B is a subset of S, $B \subseteq S$.

Not only are A and B subsets of S, these are subsets which uniquely compose S, i.e.

$A \cup B = S$ and $A \cap B = \varnothing$

(1) What is the vocabulary used to describe these type of subsets? A "unique" subset?

(2) If there is no special terminology, how would one best convey this concept notationally?

e.g.

$A \subseteq S$ and $B \subseteq S$ s.t. $A \cap B = \varnothing$

maybe be one possibility, but it doesn't show that the union of A and B are the set S...

• A and B form a partition of S? – Air Conditioner Apr 8 '18 at 1:34
• The word "partition" is what you want. An obvious caveat is that there will be many different possible partitions as one shouldn't think. Let $A, B$ be partions of $C$ mean that $A$ and $B$ have actually been specified. It hasn't be specified if A = {1,2,3} B ={4,5} or if A = {4} and B={1,2,3,5} or ... whatever. – fleablood Apr 8 '18 at 1:43
• It's also worth being aware that equivalence relations define partitions of sets. en.wikipedia.org/wiki/Equivalence_relation – samerivertwice Apr 9 '18 at 17:53
• You should probably add to your question $A\cup B=S$ which you intend but don't write. The requirements are $A\cup B=S$ and $A\cap B=\emptyset$. – samerivertwice Apr 9 '18 at 17:55
• @ProducerofBS Edited – ShanZhengYang Apr 9 '18 at 18:43

I think you are looking for a partition:

A partition of a set X is a set of nonempty subsets of X such that every element x in X is in exactly one of these subsets...

Equivalently, a family of sets P is a partition of X if and only if all of the following conditions hold:

• The family P does not contain the empty set
• The union of the sets in P is equal to X ...
• The intersection of any two distinct sets in P is empty

[Quoted from Wikipedia]

The standard way to say this is that $\{A,B\}$ is a partition of $S$. In general, a partition of a set $S$ is a set of nonempty subsets of $S$ which are pairwise disjoint (the intersection of any two of them is empty) and whose union is $S$.

Another way to say this is that $A$ is a subset of $S$ and $B$ is the complement of $A$; that is, $B$ consists of exactly those elements of $S$ that are not in $A$. Some commonly used notations for the complement of $A$ (in $S$) are $S\setminus A$, $S-A$, or $A^c$. So your condition could be stated as $A\subseteq S$ and $B=S\setminus A$.