Suppose that $X$ is a set and $V_{0}$, $J$, and $V_{1}$ are pairwise disjoint subsets of $X$ whose union is $X$. If the three subsets were nonempty it would be a partition of $X$. However, I wish to include the possibility that at least one of the sets, perhaps two of them, are empty.

For further context I work with abstract convexity. All three sets are convex and the union of $J$ with either $V_{i}$ is also convex. This is an abstraction of the following situation:

Let $f \colon V \rightarrow \mathbb{R}$ be a linear map from a real vector space to $\mathbb{R}$ and $c \in \mathbb{R}$. Define the three sets as follows:

$V_{0} = \{ v \in V \colon f(v) < c \} $

$J = \{ v \in V \colon f(v) = c \} $

$V_{1} = \{ v \in V \colon f(v) > c \} $

Is there a commonly used name for his situation?

  • $\begingroup$ Something about equivalence classes induced by equality under $\operatorname{sgn}\circ f$? $\endgroup$
    – user856
    Feb 5, 2016 at 16:19
  • $\begingroup$ The system of abstract convexity I am interested in, usually it is called a closure system, has as axioms: The sets $\varnothing$ and $X$ are convex. If $\mathcal{C}$ is a collection of convex sets then so is $\cap \mathcal{C}$. The vector space example is an instance of the situation I am interested in. $\endgroup$
    – Jay
    Feb 5, 2016 at 20:56

1 Answer 1


As you say, "partition" doesn't really fit here since that (almost always) would require that none of the sets are empty. I would use the word "decomposition" instead. One example of this use of "decomposition" is in the Hahn decomposition theorem, where either one of the positive and negative sets may be empty.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .