intermediate value theorem for extended real numbers

Theorem 1.

Let $\overline{\mathrm{R}}$ be the extended real numbers, andsuppose $f\mathrm{:}\overline{\mathrm{R}}\mathrm{\to }\overline{\mathrm{R}}$ 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$y\mathrm{\in }\mathrm{(}f\mathit{}\mathrm{(}{x}_{\mathrm{1}}\mathrm{)}\mathrm{,}f\mathit{}\mathrm{(}{x}_{\mathrm{2}}\mathrm{)}\mathrm{)}$, thenfor 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 . Thenfor 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 $f(c)=y$.∎