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| Title of object: |
symmetric difference |
| Canonical Name: |
SymmetricDifference |
| Type: |
Definition |
| Created on: |
2001-11-16 20:02:26 |
| Modified on: |
2004-02-27 10:38:25 |
| Classification: |
msc:03E20 |
| Keywords: |
set, union, intersection |
| Defines: |
symmetric difference operator |
| Synonyms: |
symmetric difference=set symmetric difference
symmetric difference=symmetric set difference
symmetric difference=symmetric difference between sets |
Revision comment (for changes between this and next version):
| Changes for correction #3729 ('minor typos'). |
Preamble:
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{graphicx}
\usepackage{xypic}
\newcommand{\symd}{\ \triangle} |
Content:
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. It is equal to $(A-B) \cup (B-A)$ and $(A \cup B) - (A \cap B)$.
Note that for any set $A$, we symmetric difference satisfies $A\Delta A=\emptyset$ and $A\Delta\emptyset=A$.
The symmetric difference operator is commutative since $A \symd\ B=(A-B) \cup (B-A) = (B-A) \cup (A-B) = B \symd\ A$.
The operation is also associative. To see this, consider three sets $A, B,$ and $C$. Any given element $x$ is in zero, one, two, or all three of these sets.
\begin{description}
\item If $x$ is not in any of $A, B,$ or $C$, then it is not in the symmetric difference of the three sets no matter how it is computed.
\item If $x$ is in one of the sets, let that set be $A$; then $x \in A \symd\ B$ and $x \in (A \symd\ B) \symd\ C$; also, $x \notin (B \symd\ C)$ and therefore $x \in A \symd\ (B \symd\ C)$.
\item If $x$ is in two of the sets, let them be $A$ and $B$; then $x \notin A \symd\ B$ and $x \notin (A \symd\ B) \symd\ C$; also, $x \in B \symd\ C$, but because $x$ is in $A$, $x \notin A \symd\ (B \symd\ C)$.
\item If $x$ is in all three, then $x \notin A \symd\ B$ but $x \in (A \symd\ B) \symd\ C$; similarly, $x \notin B \symd\ C$ but $x \in A \symd\ (B \symd\ C)$. Thus, $A \symd\ (B \symd\ C) = (A \symd\ B) \symd\ C$.
\end{description}
In general, an element will be in the symmetric difference of several sets iff it is in an odd number of the sets.
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 fixed some set. Finally, we no
The operation is also associative. To see this, consider three sets $A, B,$ and $C$. Any given elemnet $x$ is in zero, one, two, or all three of these sets. If $x$ is not in any of $A, B,$ or $C$, then it is not in the symmetric difference of the three sets no matter how it is computed. If $x$ is in one of the sets, let that set be $A$; then $x \in A \symd\ B$ and $x \in (A \symd\ B) \symd\ C$; also, $x \notin (B \symd\ C)$ and therefore $x \in A \symd\ (B \symd\ C)$. If $x$ is in two of the sets, let them be $A$ and $B$; then $x \notin A \symd\ B$ and $x \notin (A \symd\ B) \symd\ C$; also, $x \in B \symd\ C$, but because $x$ is in $A$, $x \notin A \symd\ (B \symd\ C)$. If $x$ is in all three, then $x \notin A \symd\ B$ but $x \in (A \symd\ B) \symd\ C$; similarly, $x \notin B \symd\ C$ but $x \in A \symd\ (B \symd\ C)$. Thus, $A \symd\ (B \symd\ C) = (A \symd\ B) \symd\ C$.
In general, an element will be in the symmetric difference of several sets iff it is in an odd number of the sets. |
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