2

I give the definition of symmetric difference of two sets in my article.

  • The symmetric difference of two sets is the set of elements that are in either of the sets but not in their intersection.

A more formal definition:

  • The symmetric difference of two sets $A$ and $B$ is $ A\,\triangle \,B=\left(A\setminus B\right)\cup \left(B\setminus A\right)$.

The definition of symmetry difference only considers two sets, so I feel somewhat uncomfortable when I want to say the sentence below.

The set $F$ is obtained by taking symmetric difference on the several sets $F_1, F_2, \cdots ,F_i$.

(That is to say: $F$ is obtained by $F_1 \triangle F_2 \triangle F_3 \cdots \triangle F_i$)

I don't know if there is any ambiguity or lack of rigor in this sentence. I would love to know a better expression of the above sentence.

PS: I always feel that I need to define the symmetric difference of multiple sets in my article, but it feels a little verbose.

licheng
  • 2,687
  • 1
  • 10
  • 25
  • 1
    that are in either of the sets but not in their intersection --- Regarding writing style issues, I have two comments. [1] If you're going to use the less-mathy expression "in either of the sets", then you should probably end with "but not in both sets" rather than "but not in their intersection". [2] It is probably better to say "that are in at least one of the sets" than "that are in either of the sets", since the former is a bit colloquial whereas the latter is more exactly stated. – Dave L. Renfro Sep 21 '22 at 09:02
  • Thanks! Indeed your definition of the symmetric difference between two sets is more readable. I still have a question: Could my statement "The set $F$ is obtained by taking symmetric difference on the several sets $F_1,F_2,\cdots,F_i.$" be further improved. I am not quite sure if there is a problem with this sentence. – licheng Sep 21 '22 at 09:26
  • 1
    To begin with, it's probably best not to say "is obtained by" unless the set $F$ has already been defined/constructed by another means and at this point you want to say it can also be obtained by symmetric difference operations. So if you're simply defining $F$ at this point, then probably just say "The set $F$ is defined by ..." or "The set $F$ can be defined by ...", where most of the time the former would be used but for some contexts one might use the latter (and I can't think of examples for the latter now). (continued) – Dave L. Renfro Sep 21 '22 at 13:48
  • 1
    As for how to say what you want to for finitely many sets, for me this would depend on what you've stated/proved up to this point about the symmetric difference operation. Maybe say something like "The symmetric difference of a finite sequence $F_1,$ $F_2, \ldots$ $F_i$ of sets is defined by $F_1 \triangle F_2 \triangle F_3 \cdots \triangle F_i,$ which is unambiguous because the operation $\triangle$ is associative" (see these MSE questions for details when more than three sets are involved). (continued) – Dave L. Renfro Sep 21 '22 at 14:00
  • 1
    FYI, many of the properties of the symmetric difference operation were first singled out and proved in The modular difference of classes by Daniell (1917). Two undergraduate expository papers of possible interest are The star product by Chui (The Pentagon, Spring 1965) and Group operations on the power set by Bean (1976; described here). (continued) – Dave L. Renfro Sep 21 '22 at 14:11
  • 1
    Both Chui's paper (which uses the nonstandard term "star product" for the symmetric difference operation) and Bean's paper prove a result (well known, but off-hand I don't know its origin) of possible interest to you: "An element is a member of the symmetric difference of a finite number of sets if and only if it is a member of an odd number of the sets" (quoted -- with removal of one word -- from middle of p. 13 of Bean's paper). (continued) – Dave L. Renfro Sep 21 '22 at 14:19
  • 1
    This result is probably proved somewhere in MSE (I haven't searched), but in addition to the two papers I cited, it is proved in An odd inductive proof by David (1984; JSTOR and freely available copy -- last URL may not work a few years from now, hence the reason for giving the JSTOR link). – Dave L. Renfro Sep 21 '22 at 14:21
  • 1
    By the way, I said "a finite sequence" because I assume you want to define $F_1 \triangle F_2 \triangle F_3 \triangle \cdots \triangle F_i$ for a sequence (ordered) of sets (some of which might be identical to others). However, if you want to discuss the symmetric difference of a finite collection (or finite set) of sets, or maybe a finite multiset of sets, then commutativity of the operation $\triangle$ also needs to be mentioned, and in this case you probably should not subscript the sets in a way that suggests the sets are assumed to be arranged in a finite (linearly ordered) sequence. – Dave L. Renfro Sep 21 '22 at 16:48

1 Answers1

3

The symmetric difference is associative and commutative. That means: $$(F_1\triangle F_2)\triangle F_3)=F_1\triangle(F_2\triangle F_3)=(F_1\triangle F_2)\triangle F_3$$And so on for more sets. You can efficiently define it as you would any iterate of an associative operation - repeat the operation $n$ times by applying successfully applying the operation on the output from the previous step. In the same way, if you have a definition for $a+b$, $a+b+c+d+e$ is defined to be the four-fold iteration of "$+$" and this is unambiguous as $(a+b)+c=a+(b+c)$, so the order is irrelevant.

FShrike
  • 46,840
  • 3
  • 35
  • 94
  • Thank you very much! Did you mean that once I use the symmetric difference multiple times on 3 or more sets, I have to define the symmetric difference for multiple sets first? The reason is that the symmetric difference is associative and commutative. – licheng Sep 21 '22 at 08:45
  • 1
    @licheng I would argue you don’t need to define it at all, just mention that it is an associative operation. – FShrike Sep 21 '22 at 09:01
  • 1
    Because, throughout maths we define the repetition of associative operations in exactly the same way – FShrike Sep 21 '22 at 09:01
  • 1
    You have an incomplete grouping symbol on the first expression. You might want to edit that. It's trivial, so just do it when there is a significant amount of content to add/remove. – soupless Sep 21 '22 at 09:41
  • Is this not simply the set of all $x$ that belong to exactly one member of the given collection? If so, you could also express it quantitatively in terms of characteristic functions as follows: define the multiplicity function (the number of times each $x\in\cup_k F_k$ is covered by the collection) by setting $M(x)= \sum_k \chi_{F_k} (x)$. Then consider the set of $x$ for which $M(x) =1$. – MathFont Sep 21 '22 at 13:54
  • It's important to note that 3-set associativity implies arbitrary associativity for finitely many sets, and the latter is more complex than what is suggested by what you have written. For example, the number of ways to "associatively evaluate" when you have $4,$ $5,$ $6,$ and $n$ many sets is $5,$ $14,$ $42,$ and $\frac{1}{n}\cdot C(2n-2,,n-1)$ many ways. For more details, see this 12 September 2006 sci.math post. – Dave L. Renfro Sep 21 '22 at 14:38