# Intermediate value theorem

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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) . The second case is similar.

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

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 $x\in A:x\le c$ .

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)

${\color{red}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) . Contradiction.

So now let us assume that $f(c) . 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))

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

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) , but $x>c=\sup A$ . Contradiction.

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