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I have a series of subsets that form a set however each subset is actually a partition. Currently I use the conventional subset symbol when writing the appropriate notation and then note the subsets are partitions in the following (or preceding) text. However, it would be good to have a symbol that refers to a partition naturally just like any other set theory notion. Yet, despite looking in various sources and I haven't come across anything so far. So is there a conventionally accepted maths notation/symbol for a partition? Personally have no problem allocating a suitable symbol if none currently exists but I don't want to 'reinvent the wheel' if I can avoid it.

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  • $\begingroup$ Maybe write $\dot\bigcup_i X_i=M$ do indicate that the $X_i$ partition $M$. $\endgroup$ – M. Winter Sep 12 '17 at 7:50
  • $\begingroup$ See math.stackexchange.com/questions/117715/notation-of-partition $\endgroup$ – user472341 Sep 12 '17 at 7:57
  • $\begingroup$ Each subset is a partition? Or the collection of subsets is a partition? $\endgroup$ – Asaf Karagila Sep 12 '17 at 8:09
  • $\begingroup$ Each subset is a partition of the parent set $\endgroup$ – Murray B Sep 12 '17 at 8:29
  • $\begingroup$ it would be good if you provide an example of what you want to write using the symbols that you know. Can you? $\endgroup$ – trying Sep 12 '17 at 8:44
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When you don't want to overload your text with long notations, you can define your own notations.

For instance

Let ${\mathbf{Part}}(X)$ be the set of partitions of $X$.

And later,

Let $(X_i)_{i\in I} \in \mathbf{Part}(X)$

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  • $\begingroup$ The empty set can be one of the sets of a partition? $\endgroup$ – William Elliot Sep 12 '17 at 8:30
  • $\begingroup$ Thanks Lærne, I will take your advice and identify a suitable notation for this. Kind of strange, however, that with all the notation there is, a symbol of a partition has not been allocated yet. $\endgroup$ – Murray B Sep 12 '17 at 8:46
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If $(P_i)_{i \in I}$ is a partition of a set $Q$ you can write

$\dot\bigcup_{i \in I} P_i=Q$ to denote

$\bigcup_{i \in I} P_i=Q$ and $P_i \cap P_j= \emptyset$ for $i,j \in I$ with $i \ne j$.

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  • $\begingroup$ Some P_i is allowed to be empty? $\endgroup$ – William Elliot Sep 12 '17 at 8:27
  • $\begingroup$ No. That is included in the definition of "partition". $\endgroup$ – Fred Sep 12 '17 at 8:30
  • $\begingroup$ By that definition both {{1}} and { empty set, {1} } are partitions of {1}. $\endgroup$ – William Elliot Sep 12 '17 at 8:59
  • $\begingroup$ ???? I said, that according to the definition of "partition": if $(P_i)_{i \in I}$ is a partition of $Q$, then $P_i \ne \emptyset$ for all $i \in I$. $\endgroup$ – Fred Sep 12 '17 at 9:02
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I use
P partition S
for
P is a pairwise disjoint collection of non-empty subsets of S whose union is P.

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Any collection $\mathcal{A}$ of partitions of $X$ are such that $$\mathcal{A}\subset\{X/\sim:\forall\sim\}$$

where it is understood that $\sim$ are equivalence relations on $X$.

Alternatively, see next.

Any map $f\colon X\to Y$ defines a partition of its domain in fibers, a fiber of $y\in Y$ under $f$ being $f^{-1}(\{y\})$. (A fiber is also known as inverse image of a singleton or level set). The equivalence relation associated to the partition in fibers of a map $f$ is called kernel of $f$, indicated by $\ker f$ or by $=_f$, that is, $\forall x_1,x_2\in X$ $$x_1=_fx_2\iff f(x_1)=f(x_2)$$ and the partition in fibers is indicated by $$X/=_f$$

Let's focus our attention on $X^X$, the set of endofunctions on $X$. It is clear that $\{X/=_f:f\in X^X\}$ is the set of all the partitions of $X$. (Indeed given any partition of $X$, $\exists f\in X^X$ that is the forward composition of the natural projection of $X$ on the partition with a choice map for the partition.)

So you don't need to devise a new symbol in place of "$\subset$", because you have that any subset $$\mathcal{A}\subset \{X/=_f:f\in X^X\}$$ is a collection of some partitions of $X$.

N.B. restriction to $X^X$ (instead of considering $X^Y$, $\forall Y$) has been chosen to reduce notation complexity, but it not essential.

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  • $\begingroup$ with all due respect you are over thinking the question. I just want to know if there is a conventionally used symbol that denotes a partition. Given the various replies to the question I can safely assume that there is no such symbol, hence I have taken Larne's reply that I should identify a suitable symbol and define it to be the notation that refers to a partiton. Simple. $\endgroup$ – Murray B Sep 12 '17 at 17:01
  • $\begingroup$ Thanks, but don't mind. No overthinking from me. I like questions about notation anyway. This is an answer I gave before I read your comment, that changed things a little. $\endgroup$ – trying Sep 12 '17 at 23:45
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You say in a comment to your question: "Ok, let's say we have a set, $S$ that consists of balls. Lets make it 100 balls as a nice round number. Now each ball is randomly coloured in one colour of the rainbow. If we let each of the colours be a subset of $S$, which we denote as $s_i$ where $i$ is the index of colours. Now of course $s_i\subset S$ is true but it does not provide the information that the subsets are also partitions (as a colour cannot be present in two subsets). Having a partition notation/symbol would provide more information in the notation without having to make a repeating comment in the text"

Let's rephrase your example

Let $S$ and $I$ be sets (whose elements we call balls and colours, respectively). Let $f\colon S\to I$ be a map (that we call colour of a ball of $S$).

You want to know whether there is any standard symbol to represent

a subset of $S$ whose balls are all and only the balls of $S$ of a given colour

or abstractly,

a subset of $S$ whose elements are all and only the elements of $S$ having the same value under $f$

because the set-inclusion symbol encodes only the first part of this statement:

a subset of $S$

In standard symbol it becomes:

$$\{b\in S|f(b)=i\}$$

that is, by definition

$$f^{-1}(i)$$

or a little more properly $f^{-1}(\{i\})$. It has also a name: the fiber (or level set or inverse image or preimage) of $i$ under $f$.

If you want but it is not essential you can write $f^{-1}(i)\subset S$.

You can also see that $\{f(i)|i\in I\}$ is the set of partition of $S$ under $f$, usually written as $S/{=}_f$.

Moreover if you want you can define a map $s\colon I\to S/{=}_f$ such that (using index notation, instead of functional notation) $s_i=f^{-1}(i)$. You can now write:

$$s_i\subset S$$

So you don't need to substitute a new symbol for the set-inclution symbol, but you need only to define once and for all the partition criterion, through the map $s$ or $f$.

Still one more thing. If you meant that instead of $f$ you have a map $g\colon B\to I$, where $B$ is the set of all balls in the universe and $S\subset B$, $g$ is called colour of a ball (whereby the previous $f$ is given by $f=g|_S$), your statement becomes in symbols $g|_S^{-1}(i)$ or also:

$$g^{-1}(i)\cap S$$

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