symmetric differenceSymmetricDifference2001-11-16 20:02:262008-05-01 19:25:44Definitionsymmetric difference operatorsetunionintersection\usepackage{amssymb}
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\newcommand{\symd}{\triangle}The \emph{symmetric difference} between two sets $A$ and $B$, written $A \symd B$, is the set of all $x$ such that either $x \in A$ or $x \in B$ but not both. In other words,
$$A\symd B:= (A\cup B)\setminus (A\cap B).$$
The Venn diagram for the symmetric difference of two sets $A,B$, represented by the two discs, is illustrated below, in light red:
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\begin{pspicture}(0,0)(8,4)
\begin{psclip}
{\pscircle[fillstyle=vlines,hatchcolor=red,hatchwidth=0.1\pslinewidth,hatchsep=1\pslinewidth](3,2){2}}
\pscircle[fillstyle=solid,hatchcolor=white,hatchwidth=0.1\pslinewidth,hatchsep=1\pslinewidth](5,2){2}
\end{psclip}
\begin{psclip}
{\pscircle[fillstyle=vlines,hatchcolor=red,hatchwidth=0.1\pslinewidth,hatchsep=1\pslinewidth](5,2){2}}
\pscircle[fillstyle=solid,hatchcolor=white,hatchwidth=0.1\pslinewidth,hatchsep=1\pslinewidth](3,2){2}
\end{psclip}
\rput(1.25,3.75){$A$}
\rput(6.75,3.75){$B$}
\rput(0,0){$.$}
\rput(8,4){$.$}
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\subsubsection*{Properties}
Suppose that $A$, $B$, and $C$ are sets.
\begin{itemize}
\item $A \symd B=(A\setminus B) \cup (B\setminus A)$.
\item $A \symd B=A^c \symd B^c$, where the superscript $c$ denotes taking complements.
\item Note that for any set $A$, the symmetric difference satisfies $A \symd A=\emptyset$ and $A \symd \emptyset=A$.
\item The symmetric difference operator is commutative since $A \symd B=(A\setminus B) \cup (B\setminus A) = (B\setminus A) \cup (A\setminus B) = B \symd A$.
\item The symmetric difference operation is associative: $(A \symd B) \symd C = A \symd (B \symd C)$. This means that we may drop the parentheses without any ambiguity, and we can talk about the symmetric difference of multiple sets.
\item Let $A_1,\ldots, A_n$ be sets. The symmetric difference of these sets is written $$\substack{n \\ \displaystyle{\symd} \\ i=1} A_i.$$ In general, an element will be in the symmetric difference of several sets iff it is in an odd number of the sets.
\end{itemize}
It is worth noting that these properties show that the symmetric difference operation can be used as a group law to define an abelian group on the power set of some \PMlinkescapetext{fixed} set.
Finally, we note that intersection distributes over the symmetric difference operator: $$A\cap(B\symd C)=(A\cap B)\symd(A\cap C),$$ giving us that the power set of a given fixed set can be made into a Boolean ring using symmetric difference as addition, and intersection as multiplication.