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Mostly I see a partition of a set $A$ defined as a collection of non-empty disjoint sets whose union is $A$.

I see one reference that allows empty sets to be included in the partition: (Potter, M. Set theory and its philosophy, 2004, Oxford University Press, p. 130). Definition. A collection $B$ of subsets of $A$ is a "partition" of $A$ if each element of $A$ belongs to exactly one element of $B$.

Is there some commonly used terminology to refer to a "partition" which includes empty set(s)?

(The context: a set of injections on the sets of a "partition" of $A$ mapping to the set $C$ comprises a bijection iff their images form a "partition" of $C$. Clearly this is true if empty sets map to empty sets, but it gets somewhat unwieldy to keep adding this to the argument.)

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    $\begingroup$ Do you mean to allow for more than one empty set? Wouldn't that also be a nonstandard definition of set? $\endgroup$
    – Will Sawin
    Commented Oct 20, 2013 at 18:41
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    $\begingroup$ Good question. I’m looking at proofs of Cantor-Bernstein theorem. With injections f:A->B and g:B->A, it seems that in creating a bijection A is “divided” into three elements Af which is mapped to B by f; Ag by inverse of g, and Agf which can be mapped by either. Provided the images are disjoint and give B as a union the bijection is proven. However, any one or two of Af, Ag, and Agf could be empty. Can I call this division a “partition” or what else ? $\endgroup$
    – Tom Collinge
    Commented Oct 21, 2013 at 7:20
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    $\begingroup$ Actually, if there is more than one empty element of the "partition" I don't think it would invalidate the partition as a set. The set is defined by its extension and so { {}, {}, A, B} = { {}, A, B} is a set - yes ? $\endgroup$
    – Tom Collinge
    Commented Oct 24, 2013 at 9:05
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    $\begingroup$ There is only one empty set (one of the ZF axioms says that two sets with the same elements are equal). Unless you mean partitions as indexed partitions $(A_i)_{i\in I}$, in which case indeed $A_i$ could be equal to the empty set for several $i$. $\endgroup$
    – YCor
    Commented Dec 11, 2022 at 10:07
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    $\begingroup$ Question to the OP: are you looking for ordered or unordered partitions? If not specified, I think partition usually means unordered (= set of sets), not ordered (= sequence of sets). $\endgroup$
    – Jukka Kohonen
    Commented Dec 11, 2022 at 10:37

4 Answers 4

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In Bourbaki's terminology the elements of a partition may be empty (cf. E.II.4.7).

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    $\begingroup$ Thanks, but is there any term to differentiate the two possibilities. If a partition is generally understood to have non-empty elements, must one continually note condition that the some elements may be empty ? $\endgroup$
    – Tom Collinge
    Commented Oct 20, 2013 at 12:09
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    $\begingroup$ In view of my answer "generally understood" seems to be not really appropriate. But anyway, if you use Bourbaki terminology throughout (and of course also say that you do so) then you only have to specify if you consider partitions by nonempty sets. $\endgroup$
    – Fred Rohrer
    Commented Oct 20, 2013 at 14:21
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In German you see the term "disjunkte Zerlegung" for this concept, which roughly means "disjoint decomposition".

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We propose the notation $ñ(m,k)$ for the number of partitions of $m$ points into $k$ sets that can be empty. We needed a notation since we proved (in this paper) some relations for this kind of partitions.

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I would say that is a colouring of $A$. While it's often understood to be a function $c\colon X\to A$, it's easily and equivalently understood as the family $(A_x\mid x\in X)$ where $A_x=c^{-1}(x)$.

This has the additional advantage that if you know the size (or indexing set, if we prefer) of the partition it's easily specified as an $X$-colouring.

Note that this terminology will not be satisfactory for all fields; see Jukka Kohonen's comments about the chromatic polynomial of a graph.

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    $\begingroup$ Note that colouring is usually understood to fix the identity of the colours, i.e. colouring $1$ red and $2$ blue is different from the opposite colouring. In contrast, a partition of a set $A$ is often (typically?) understood to be an unordered set of subsets of $A$. $\endgroup$
    – Jukka Kohonen
    Commented Dec 11, 2022 at 10:35
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    $\begingroup$ @JukkaKohonen That's true, though I've found that being able to permute the codomain without changing anything meaningfully is the soft line for when a function becomes a colouring (at least in my experience). Do you have any examples of places where the colouring language is used but the difference between (e.g.) $\{\langle 1,\text{red}\rangle,\langle 2,\text{blue}\rangle\}$ and $\{\langle 1,\text{blue}\rangle,\langle 2,\text{red}\rangle\}$ is meaningful? $\endgroup$
    – Calliope Ryan-Smith
    Commented Dec 11, 2022 at 18:02
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    $\begingroup$ Well, for starters, the chromatic polynomial of a graph. E.g. mathworld.wolfram.com/ChromaticPolynomial.html explicitly says: "(where colorings are counted as distinct even if they differ only by permutation of colors)." $\endgroup$
    – Jukka Kohonen
    Commented Dec 12, 2022 at 3:09

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