If ${\displaystyle f}$ is a real-valued function that is defined and continuous on a closed interval ${\displaystyle [a, b]}$ , then for any ${\displaystyle y}$ between ${\displaystyle f(a)}$ and ${\displaystyle f(b)}$ there exists at least one ${\displaystyle c\in[a,b]}$ such that ${\displaystyle f(c)=y}$ .

In Topology

Let ${\displaystyle X}$ be a connected topological space, ${\displaystyle Y}$ a ordered space, and ${\displaystyle f: X \to Y}$ a continuous function. Then for any points ${\displaystyle a,b\in X}$ and point ${\displaystyle y \in Y}$ between ${\displaystyle f(a)}$ and ${\displaystyle f(b)}$ , there exists a point ${\displaystyle c\in X}$ such that ${\displaystyle f(c)=y}$ .

Proofs

Proof (1)

We shall prove the first case, ${\displaystyle f(a)<y<f(b)}$ . The second case is similar.

Let us define set ${\displaystyle A={\big \{}x\in [a,b]:f(x)<y{\big \}}}$ .

Then ${\displaystyle A}$ is none-empty since ${\displaystyle a \in A}$ , and the set is bounded from above by ${\displaystyle b}$ . Hence, by completeness, the supremum ${\displaystyle c=\sup A}$ exists. That is, ${\displaystyle c}$ is the lowest number that for all ${\displaystyle x\in A:x\leq c}$ .

We claim that ${\displaystyle f(c)=y}$ .

Let us assume that ${\displaystyle f(c)>y}$ . We get ${\displaystyle f(c)-y>0}$ .

Since ${\displaystyle f}$ is continuous, we know that for all ${\displaystyle \varepsilon > 0}$ there exists a ${\displaystyle \delta > 0}$ such that ${\displaystyle {\Big |}f(x)-f(c){\Big |}<\varepsilon }$ whenever ${\displaystyle |x-c|<\delta }$ .

For the case ${\displaystyle \varepsilon =f(c)-y}$ there exists a ${\displaystyle \delta > 0}$ s.t. for all ${\displaystyle x\in (c-\delta ,c+\delta )}$ we get

${\displaystyle {\Big |}f(x)-f(c){\Big |}<f(c)-y}$

${\displaystyle -(f(c)-y)<f(x)-f(c)<f(c)-y}$

${\displaystyle {\color {red}y<f(x)}<2f(c)-y}$

By the supremum properties, there exists an ${\displaystyle x\in (c-\delta ,c)}$ that is also ${\displaystyle x \in A}$ .

So there exists such ${\displaystyle x\in (c-\delta ,c)}$ for which we get ${\displaystyle f(x)>y}$ , but ${\displaystyle A}$ is defined for ${\displaystyle f(x)<y}$ . Contradiction.

So now let us assume that ${\displaystyle f(c)<y}$ . We get ${\displaystyle y-f(c)>0}$ .

In the same way, for the case ${\displaystyle \varepsilon =y-f(c)}$ there exists a ${\displaystyle \delta > 0}$ s.t. for all ${\displaystyle x\in (c-\delta ,c+\delta )}$ we get

${\displaystyle {\Big |}f(x)-f(c){\Big |}<y-f(c)}$

${\displaystyle -(y-f(c))<f(x)-f(c)<y-f(c)}$

${\displaystyle 2f(c)-y<{\color {red}f(x)<y}}$

By the supremum properties, all ${\displaystyle x\in (c,c+\delta )}$ are definitely ${\displaystyle x \notin A}$ .

So there exists such ${\displaystyle x\in (c,c+\delta )}$ for which we get ${\displaystyle f(x)<y}$ , but ${\displaystyle x>c=\sup A}$ . Contradiction.

Hence, it must be ${\displaystyle f(c)=y}$ . ${\displaystyle \blacksquare }$