Theorem 1   Let be the extended real numbers, and suppose is a continuous function. Suppose 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 is homeomorphic to $[0,1]$ , we can assume that $f$ is a function . 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 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 $f(c)=y$ .