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 .


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\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




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




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

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