AlternativeDefinitionOfFaceOfAConvexSet 2007-05-02 16:24:15 2007-05-02 16:24:15 Definition face of a convex set supporting hyperplane % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here The following definition of a face of a convex set in a real vector space is sometimes useful. Let $C$ be a convex subset of $\mathbb{R}^n$. Before we define faces, we introduce oriented hyperplanes and supporting hyperplanes. Given any vectors $n$ and $p$ in $\mathbb{R}^n$, define the hyperplane $H(n,p)$ by \[ H(n,p) = \{ x \in \mathbb{R}^n \colon n \cdot (x - p) = 0 \}; \] note that this is the degenerate hyperplane $\mathbb{R}^n$ if $n=0$. As long as $H(n,p)$ is nondegenerate, its removal disconnects $\mathbb{R}^n$. The \emph{upper halfspace} of $\mathbb{R}^n$ determined by $H(n,p)$ is \[ H(n,p)^+ = \{ x \in \mathbb{R}^n \colon n \cdot (x - p) \ge 0 \}. \] A hyperplane $H(n,p)$ is a \emph{supporting hyperplane} for $C$ if its upper halfspace contains $C$, that is, if $C\subset H(n.p)^+$. Using this terminology, we can define a \emph{face} of a convex set $C$ to be the intersection of $C$ with a supporting hyperplane of $C$. Notice that we still get the empty set and $C$ as improper faces of $C$. \textbf{Remarks.} Let $C$ be a convex set. \begin{itemize} \item If $F_1 = C\cap H(n_1,p_1)$ and $F_2 = C\cap H(n_2,p_2)$ are faces of $C$ intersecting in a point $p$, then $H(n_1+n_2,p)$ is a supporting hyperplane of $C$, and $F_1\cap F_2 = C\cap H(n_1+n_2,p)$. This shows that the faces of $C$ form a meet-semilattice. \item Since each proper face lies on the base of the upper halfspace of some supporting hyperplane, each such face must lie on the relative boundary of $C$. \end{itemize} \PMlinkescapeword{base}