intermediate value theorem for extended real numbers

Theorem 1.

Let $\overline{\mathrm{R}}$ be the extended real numbers, and suppose $f\mathrm{:}\overline{\mathrm{R}}\mathrm{\to }\overline{\mathrm{R}}$ is a continuous function. Suppose are such that $f\mathit{}\mathrm{(}{x}_{\mathrm{1}}\mathrm{)}\mathrm{\ne }f\mathit{}\mathrm{(}{x}_{\mathrm{2}}\mathrm{)}$. If $y\mathrm{\in }\mathrm{(}f\mathit{}\mathrm{(}{x}_{\mathrm{1}}\mathrm{)}\mathrm{,}f\mathit{}\mathrm{(}{x}_{\mathrm{2}}\mathrm{)}\mathrm{)}$, then for some $c\mathrm{\in }\mathrm{(}{x}_{\mathrm{1}}\mathrm{,}{x}_{\mathrm{2}}\mathrm{)}$ we have

$$f(c)=y.$$

Proof.

As $\overline{ℝ}$ is homeomorphic to $[0,1]$, we can assume that $f$ is a function $f:[0,1]\to \overline{ℝ}$. For simplicity, let us also assume that $x_1=0$,$x_2=1$, and . Then for some $\epsilon >0$ we have

Let $g:[0,1]\to ℝ$ be the continuous function

$$g(x)=\max \{\min \{f(x),y+\epsilon \},y-\epsilon \}.$$

Now $g(0)=y-\epsilon$ and $g(1)=y+\epsilon$, so for some $c\in (0,1)$, we have $g(c)=y$, and thus $f(c)=y$. ∎

Title	intermediate value theorem for extended real numbers
Canonical name	IntermediateValueTheoremForExtendedRealNumbers
Date of creation	2013-03-22 15:35:15
Last modified on	2013-03-22 15:35:15
Owner	matte (1858)
Last modified by	matte (1858)
Numerical id	6
Author	matte (1858)
Entry type	Theorem
Classification	msc 26A06
Related topic	ExtendedRealNumbers