# intermediate value theorem for extended real numbers

###### Theorem 1.

Let $\overline{\mathbbmss{R}}$ be the extended real numbers, and suppose $f\colon\overline{\mathbbmss{R}}\to\overline{\mathbbmss{R}}$ is a continuous function. Suppose $x_{1} 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

 $f(c)=y.$
###### Proof.

As $\overline{\mathbbmss{R}}$ is homeomorphic to $[0,1]$, we can assume that $f$ is a function $f\colon[0,1]\to\overline{\mathbbmss{R}}$. For simplicity, let us also assume that $x_{1}=0$,$x_{2}=1$, and $f(0). Then for some $\varepsilon>0$ we have

 $f(0)

Let $g\colon[0,1]\to\mathbbmss{R}$ 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 $f(c)=y$. ∎

Title intermediate value theorem for extended real numbers IntermediateValueTheoremForExtendedRealNumbers 2013-03-22 15:35:15 2013-03-22 15:35:15 matte (1858) matte (1858) 6 matte (1858) Theorem msc 26A06 ExtendedRealNumbers