complementComplement2002-02-13 03:02:282008-04-30 14:23:41Definition% this is the default PlanetMath preamble. as your knowledge
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\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsfonts}
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%\usepackage{psfrag}
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% define commands here\section{Definition}
Let $A$ be a subset of $X$. The \emph{complement} of $A$ in $X$ (denoted $A^\complement$ when the larger set $X$ is clear from context) is the set difference $X \setminus A$.
The Venn diagram below illustrates the complement of $A$ in red.
\begin{center}
\begin{pspicture}(0,0)(8,6)
\pspolygon[fillstyle=vlines,hatchcolor=red,hatchwidth=0.1\pslinewidth,hatchsep=1\pslinewidth](0,0)(8,0)(8,6)(0,6)
\pscircle[fillstyle=solid,hatchcolor=white,hatchwidth=0.1\pslinewidth,hatchsep=1\pslinewidth](4,3){2}
\rput(4,3){$A$}
\rput(1,5){$X\setminus A$}
\rput(8.5,5.75){$X$}
\rput(0,0){$.$}
\rput(8,6){$.$}
\end{pspicture}
\end{center}
\section{Properties}
\begin{itemize}
\item $(A^{\complement})^\complement=A$
\item $\emptyset^\complement = X$
\item $X^\complement = \emptyset$
\item If $A$ and $B$ are subsets of $X$, then
$A\setminus B = A\cap B^\complement$, where the complement is taken in $X$.
\end{itemize}
\section{de Morgan's laws}
Let $X$ be a set with subsets $A_i \subset X$ for $i\in I$, where
$I$ is an arbitrary index-set. In other words, $I$ can be finite,
countable, or uncountable. Then
\begin{eqnarray*}
\left( \bigcup_{i\in I} A_i \right)^\complement &=& \bigcap_{i\in I} A_i^\complement, \\
\left( \bigcap_{i\in I} A_i \right)^\complement &=& \bigcup_{i\in I} A_i^\complement.
\end{eqnarray*}