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I always thought that $\oplus$ was an operator meaning "xor" in logic. Maybe it does, but how does it work for sets? I've got a question on an assignment due in an hour that asks me to define the set $A\oplus B^{c}$ where

$$\begin{align}U=\{1,2,3,4,5,6,7\}\\A=\{3,4,5,6\}\\B=\{1,3,5,7\}\end{align}$$

I only know of the intersect, union, and compliment operators currently. Not sure why he's asking this question when I don't recall him mentioning this at all in class.

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I tend to think your understanding of "xor" is correct, considering the example given. – Tunococ Jan 28 at 14:36
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"a xor b" would probably mean items in either a or b but not in both (by definition, xor is (a\b)u(b\a)). – Guest 86 Jan 28 at 14:40
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I sort of agree with @Tunococ, though in set theory, it is usually called symmetric difference, and denoted with the symbol $\triangle$. – Harald Hanche-Olsen Jan 28 at 14:41
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This is a good example of why it is desirable to begin assignments well before the deadline, so that questions about notations/definitions/etc. can be asked with sufficient time left over to understand the concept once a clarification is received. – Michael Joyce Jan 28 at 14:59
If this was in a course on algebra, the book/instructor might have wanted to emphasize the "ring" structure of sets, and hence chose $\oplus$ for that reason. (In turning a "Boolean algebra" into a "Boolean ring," the symmetric difference is the additive function of the ring...) – Thomas Andrews Jan 28 at 15:46

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The operation is often denoted by $\triangle$ and called symmetric difference, i.e. $$ A\triangle B=(A\setminus B)\cup (B\setminus A)=\{x: (x\in A)\oplus (x\in B)\}. $$

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Wikipedia shows "Symmetric Difference" using $\ominus$ though, as an alternative - not $\oplus$. – agent154 Jan 28 at 14:45
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Yes, but it also shows a definition using $\oplus$ (the right-most definition in my answer), and hence this could be what your lecturer meant. – Stefan Hansen Jan 28 at 14:50

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