If is a real-valued function that is defined and continuous on a closed interval , then for any between and there exists at least one such that .

## In Topology

Let be a connected topological space, a ordered space, and a continuous function. Then for any points and point between and , there exists a point such that .

## Proofs

### Proof (1)

We shall prove the first case, . The second case is similar.

Let us define set .

Then is none-empty since , and the set is bounded from above by . Hence, by completeness, the supremum exists. That is, is the lowest number that for all .

We claim that .

Let us assume that . We get .

Since is continuous, we know that for all there exists a such that whenever .

For the case there exists a s.t. for all we get

By the supremum properties, there exists an that is also .

So there exists such for which we get , but is defined for . Contradiction.

So now let us assume that . We get .

In the same way, for the case there exists a s.t. for all we get

By the supremum properties, all are definitely .

So there exists such for which we get , but . Contradiction.

Hence, it must be .