26A06-IntermediateValueTheoremForExtendedRealNumbers.tex

\documentclass[12pt]{article}
\usepackage{pmmeta}
\pmcanonicalname{IntermediateValueTheoremForExtendedRealNumbers}
\pmcreated{2013-03-22 15:35:15}
\pmmodified{2013-03-22 15:35:15}
\pmowner{matte}{1858}
\pmmodifier{matte}{1858}
\pmtitle{intermediate value theorem for extended real numbers}
\pmrecord{6}{37498}
\pmprivacy{1}
\pmauthor{matte}{1858}
\pmtype{Theorem}
\pmcomment{trigger rebuild}
\pmclassification{msc}{26A06}
\pmrelated{ExtendedRealNumbers}

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\newtheorem{thm}{Theorem}
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\begin{document}
\begin{thm}
Let $\overline{\R}$ be the extended real numbers, and
suppose $f\colon \overline{\R}\to \overline{\R}$ is a continuous function.
Suppose $x_1<x_2\in \overline{\R}$ are such that $f(x_1)\neq f(x_2)$. If
$y\in(f(x_1),f(x_2))$, then
for some $c\in (x_1,x_2)$ we have 
$$
   f(c)=y.
$$
\end{thm}

\begin{proof}
As $\overline{\R}$ is homeomorphic to $[0,1]$, we can assume that $f$ is a function
$f\colon[0,1]\to \overline{\R}$. For simplicity, 
let us also assume that $x_1=0$,$x_2=1$, and $f(0)<f(1)$. Then
for some $\varepsilon>0$ we have 
$$
  f(0)<y-\varepsilon<y<y+\varepsilon < f(1).
$$
Let $g\colon [0,1]\to \R$ be the continuous function
$$
  g(x) = \operatorname{max}\{ \operatorname{min}\{ f(x), y+\varepsilon\}, y-\varepsilon\}.
$$
Now $g(0)=y-\varepsilon$ and $g(1)=y+\varepsilon$, 
so for some $c\in(0,1)$, we have $g(c)=y$, and thus $f(c)=y$. 
\end{proof}
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