Intermediate value theorem

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

In Topology

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

Proofs

Proof (1)

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

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

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

We claim that $f(c)=y$ .

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

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

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

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

$-(f(c)-y)<f(x)-f(c)<f(c)-y$

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

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

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

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

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

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

$-(y-f(c))<f(x)-f(c)<y-f(c)$

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

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

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

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