I'm trying to find how to express the following:

enter image description here

I think it would be something like:

$$(A - (B \cup C)) \cup (B - (A \cup C)) \cup (C - (A \cup B))$$

But does this have a name? A simpler way to express it? Since this expression grows fast while one add more sets.

  • $\begingroup$ Isn't that just the symmetric difference $A\triangle B\triangle C$? $\endgroup$ – Clement C. Sep 4 '17 at 1:54
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    $\begingroup$ @ClementC. No, the symmetric difference has $A\cap B\cap C$ in it. $\endgroup$ – Mark S. Sep 4 '17 at 1:55
  • $\begingroup$ @MarkS. You're right, my bad. $\endgroup$ – Clement C. Sep 4 '17 at 1:56
  • $\begingroup$ @ClementC. I initially thought so, but it doesn't. Check the diagram for $ A \Delta B \Delta C $ in this wikipedia article: en.wikipedia.org/wiki/Symmetric_difference $\endgroup$ – stefanmaric Sep 4 '17 at 1:57
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    $\begingroup$ In plain English, “The elements in one and only one of the sets” or “in exactly one set.” You could express it quasi-formally as {$x$ | $x$ is in exactly one of $A$, $B$ or $C$.} $\endgroup$ – Davislor Sep 4 '17 at 7:43

I don't know of a name for this, but one way to express it would be

$$\bigcup\limits_i A_i \setminus \bigcup\limits_{i\ne j} \left(A_i \cap A_j\right)$$

You don't need to specify more complicated intersections, because they're already included in what's being subtracted here. This captures the idea that it's intersections that you don't want.

This also provides for a succinct description of it in words: "the union minus the pairwise intersections".

  • $\begingroup$ And it works for more than 3 given sets. +1. $\endgroup$ – CiaPan Sep 4 '17 at 9:08
  • $\begingroup$ You probably intended $\bigcup\limits_i A_i \setminus \bigcup\limits_{i\ne j} (A_i\cap A_j)$ but leaving out parentheses assumes precedences that I don't think are universally agreed upon (I think most people would either parse $X\setminus Y\cap Z$ as $(X\setminus Y)\cap Z$, or more likely insist on using parentheses whatever way it is intended). $\endgroup$ – Marc van Leeuwen Sep 4 '17 at 13:54
  • $\begingroup$ I thought it was clear from the relative sizes of the cup and the cap, but I can add parentheses if it will help. $\endgroup$ – G Tony Jacobs Sep 4 '17 at 15:02
  • $\begingroup$ Also, taking a union over all $i\ne j$ makes little sense unless both $i$ and $j$ are inside... $\endgroup$ – G Tony Jacobs Sep 4 '17 at 15:14
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    $\begingroup$ A side concern, but if you use that in a paper or a proof: when you first use or introduce it, please describe in plain English what it means as well! Formal notation is great, but a reader will always benefit from a 12-word easy side description... "The subset of elements which appear in exactly one of the sets; that is:" $\endgroup$ – Clement C. Sep 4 '17 at 15:30

This is similar to, but distinct from the symmetric difference.

If your collection is $\mathcal{C}=\{A,B,C,\ldots\}$, then you could write this set with cumbersome notation along the lines of $\left\{a\in\bigcup \mathcal{C}\left|\,\left|\left\{A\in\mathcal{C}\mid a\in A\right\}\right|=1\right.\right\}$. It would probably be clearer to use words, like "$\nabla \mathcal{C}$ will denote the set of all elements that are in exactly one set in the collection $\mathcal{C}$.

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    $\begingroup$ I guess if you define $F$ as the sum of the characteristic functions of each set, then we're looking at $F^{-1}(1)$. $\endgroup$ – G Tony Jacobs Sep 4 '17 at 2:47
  • $\begingroup$ @GTonyJacobs I already upvoted your answer, but consider adding that alternative to your answer; it's a tidy approach. $\endgroup$ – Mark S. Sep 4 '17 at 2:51
  • $\begingroup$ Looks like someone just beat me to it... $\endgroup$ – G Tony Jacobs Sep 4 '17 at 2:52
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    $\begingroup$ +1 Words. I hate it when people forget that we have words, and try to write arcane mathematical sentences for simple concepts. $\endgroup$ – Kevin Sep 4 '17 at 15:51

Since the OP's picture didn't contain a square 'holding' the circles, we pedantically construct our own universal set $U$,

$\tag 1 U = \bigcup\limits_{i=1}^n A_i$

and let $\chi_{A_i}$ be the corresponding characteristic (indicator) functions.


$\tag 2 C = \sum_{i=1}^n \chi_{A_i}$.

It is easy to see that the the simple function $C$ can only take on values contained in $\{1,2,\cdots,n\}$, so

$\tag 3 C:U \to \{1,2,\cdots,n\}$

The inverse image $C^{-1}(\{1\})$ consists of the elements in $U$ that belong to exactly one of the sets $A_i$ in the union (1).

Observe that in general, this inverse image can be the empty set. Consider for example these three sets,

$\qquad A_1 = \{1,2\}$, $A_2 = \{2,3\}$ and $A_3 = \{3,1\}$.

Note: If you choose a larger universal set containing all the $A_i$, then $C(x)$ can take on the value $0$. That would happen whenever $x \notin \bigcup A_i$. Practically, it would be best when looking at any simple function expressed as in (2) to write,

$\tag 4 C:U \to \{0,1,2,\cdots,n\}$

That is the common methodology.

  • $\begingroup$ Why would a universal set be expressed as a union? To define characteristic functions you need a universe that contains all of the $A_i$, but it certainly does not have to be their union. Correspondingly you should allow $0$ in the image of $C$. (The answer remains the inverse image by $C$ of $\{1\}$). $\endgroup$ – Marc van Leeuwen Sep 4 '17 at 13:58
  • $\begingroup$ Yes, it was done to get the OP ask the same question that you came up with. After some thought, we should simply say that there is some $U$ out there in set land and $C$ can take on the value of $0$. $\endgroup$ – CopyPasteIt Sep 4 '17 at 14:20

I don't know of a specific name for this type of expression, but a simpler way to write it would be to use countable unions:

Instead of labeling your three (or in larger cases, $n$) sets by $A$, $B$, and $C$, label them as $A_1$, $A_2$, and $A_3$.

Then for $n$ sets, you could write:

$$\bigcup\limits_{i=1}^n \left(A_i - \bigcup\limits_{1 \leq i \leq n; \, j \neq i} A_j\right)$$

or, alternatively:

$$\bigcup\limits_{i=1}^n\bigcap\limits_{1 \leq i \leq n; \, j \neq i} \left(A_i - A_j\right)$$


Can be written as $(A\bigtriangleup B\bigtriangleup C)-(A \cap B\cap C)$ where $\bigtriangleup$ is the symmetric difference operator.

  • $\begingroup$ This sounded so good, but it is not universal, it only works for a collection of 3 sets. $\endgroup$ – stefanmaric Sep 4 '17 at 19:51

"The set of elements unique to each set."


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