MathbbR2SetminusCIsPathConnectedIfCIsCountable 2006-08-09 05:16:21 2006-08-23 08:43:10 Theorem % 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} \newtheorem{theo}{Theorem} \newtheorem{coro}{Corollary} % there are many more packages, add them here as you need them % define commands here \begin{theo} Let $C$ be a countable subset of $\mathbb{R}^2$. Then $\mathbb{R}^2\setminus C$ is path connected.\end{theo} We use $\mathbb{R}^2$ simply as an example; an analogous proof will work for any $\mathbb{R}^n, n>1$. \begin{proof} Fix a point $P$ not in $C$. The strategy of the proof is to construct a path $p_x$ from any $x \in \mathbb{R}^2\setminus C$ to $P$. If we can do this then for any $d, d' \in \mathbb{R}^2\setminus C$ we may construct a path from $d$ to $d'$ by first following $p_d$ and then following $p_{d'}$ in reverse. Fix $x \in \mathbb{R}^2\setminus C$, and consider the set of all (straight) lines through $x$. There are uncountably many of these and they meet in the single point $x$, so not all of them contain a point of $C$. Choose one that doesn't and move along it: your distance from $P$ takes on uncountably many values, and hence at some point this distance $r$ from $P$ is not shared by any point of $C$. The whole of the circle with radius $r$, centre $P$, lies in $\mathbb{R}^2\setminus C$ so we may move around it freely. Consider all lines through $P$: these all intersect this circle, and there are uncountably many of them so we may choose one, say $L$, that contains no point of $C$. Moving around the circle until we meet $L$ and then following it inwards completes our path form $x$ to $P$. \end{proof} \begin{coro} Let $f: \mathbb{R}^2 \rightarrow \mathbb{R}$ be continuous and onto. Then $f^{-1} (0)$ is uncountable. \end{coro} \begin{proof} Suppose that $f^{-1} (0)$ is countable. $\mathbb{R}^2$ can be written as the disjoint union \[ f^{-1} (0) \cup f^{-1} ((-\infty, 0)) \cup f^{-1} ((0, \infty)) \] where the last two sets are open (as $f$ is continuous), non-empty (as $f$ is onto) and disjoint. Since pathwise connected is the same as connected for Hausdorff spaces, we have that $\mathbb{R}^2 \setminus f^{-1} (0)$ is not path connected, contradicting the theorem. \end{proof}