ZeroSetOfATopologicalSpace 2007-04-16 17:49:33 2007-05-14 15:12:26 Definition ring of continuous functions zero set level set cozero set \usepackage{amssymb,amscd} \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} \usepackage{pst-plot} \usepackage{psfrag} % define commands here \newtheorem{prop}{Proposition} \newtheorem{thm}{Theorem} \newtheorem{ex}{Example} \newcommand{\real}{\mathbb{R}} Let $X$ be a topological space and $f\in C(X)$, the ring of continuous functions on $X$. The \emph{level set} of $f$ at $r\in \mathbb{R}$ is the set $f^{-1}(r):=\lbrace x\in X\mid f(x)=r\rbrace$. The \emph{zero set} of $f$ is defined to be the level set of $f$ at $0$. The zero set of $f$ is denoted by $Z(f)$. A subset $A$ of $X$ is called a \emph{zero set} of $X$ if $A=Z(f)$ for some $f\in C(X)$. \textbf{Properties}. Let $X$ be a topological space and, unless otherwise specified, $f\in C(X)$. \begin{enumerate} \item Any zero set of $X$ is closed. The converse is not true. However, if $X$ is a metric space, then any closed set $A$ is a zero set: simply define $f:X\to \mathbb{R}$ by $f(x):=d(x,A)$ where $d$ is the metric on $X$. \item The level set of $f$ at $r$ is the zero set of $f-\hat{r}$, where $\hat{r}$ is the constant function valued at $r$. \item $Z(\hat{r})=X$ iff $r=0$. Otherwise, $Z(\hat{r})=\varnothing$. In fact, $Z(f)=\varnothing$ iff $f$ is a unit in the ring $C(X)$. \item Since $f(a)=0$ iff $|f(a)|<\frac{1}{n}$ for all $n\in \mathbb{N}$, and each $\lbrace x\in X \mid |f(x)|<\frac{1}{n} \rbrace$ is open in $X$, we see that $$Z(f)=\bigcap_{n=1}^{\infty}\lbrace x\in X \mid |f(x)|<\frac{1}{n} \rbrace.$$ This shows every zero set is a \PMlinkname{$G_{\delta}$}{G_deltaSet} set. \item For any $f\in C(X)$, $Z(f)=Z(f^n)=Z(|f|)$, where $n$ is any positive integer. \item $Z(fg)=Z(f)\cup Z(g)$. \item $Z(f)\cap Z(g)=Z(f^2+g^2)=Z(|f|+|g|)$. \item $\lbrace x\in X\mid 0\le f(x)\rbrace$ is a zero set, since it is equal to $Z(f-|f|)$. \item If $C(X)$ is considered as an algebra over $\mathbb{R}$, then $Z(rf)=Z(f)$ iff $r\ne 0$. \end{enumerate} The complement of a zero set is called a \emph{cozero set}. In other words, a cozero set looks like $\lbrace x\in X\mid f(x)\ne 0\rbrace$ for some $f\in C(X)$. By the last property above, a cozero set also has the form $\operatorname{pos}(f):=\lbrace x\in X\mid 0<f(x)\rbrace$ for some $f\in C(X)$. Let $A$ be a subset of $C(X)$. The \emph{zero set} of $A$ is defined as the set of all zero sets of elements of $A$: $Z(A):=\lbrace Z(f)\mid f\in A\rbrace$. When $A=C(X)$, we also write $Z(X):=Z(C(X))$ and call it \emph{the family of zero sets} of $X$. Evidently, $Z(X)$ is a subset of the family of all closed $G_{\delta}$ sets of $X$. \textbf{Remarks}. \begin{itemize} \item By properties 6. and 7. above, $Z(X)$ is closed under set union and set intersection operations. It can be shown that $Z(X)$ is also closed under countable intersections. \item It is also possible to define a zero set of $X$ to be the zero set of some $f\in C^*(X)$, the subring of $C(X)$ consisting of the bounded continuous functions into $\mathbb{R}$. However, this definition turns out to be equivalent to the one given for $C(X)$, by the observation that $Z(f)=Z(|f|\wedge \hat{1})$. \end{itemize} \begin{thebibliography}{7} \bibitem{gj} L. Gillman, M. Jerison: {\em Rings of Continuous Functions}, Van Nostrand, (1960). \end{thebibliography}