Intermediate value theorem

INTERMEDIATE VALUE THEOREM: Suppose the I is an interval in R and that f : I -> R is continuous. Then the image set f ( I ) is also an interval.

It is frequently stated in the form

Suppose that f : [a , b] -> R is continuous and that u is a real number satisfying f (a) < u < f (b) or f (a) > u > f (b). Then for some c in (a , b), f(c) = u.

This captures an intuitive property of continuous functions: if f (1) = 3 and f (2) = 5 then f must be equal to 4 somewhere between 1 and 2. It represents the idea that the graph of a continuous function can be drawn without lifting your pencil from the paper.

Proof of IVT: We shall prove the first case; the second is similar.

Let S = {x in [a, b] : f(x) <= u}. Then S is non-empty (e.g. a is in S) and bounded above by b. Hence by the continuum property, c = sup S exists. We claim that f (c) = u.

Suppose first that f (c) > u. Then f (c) - u > 0, so there is a d > 0 such that

| f (x) - f (c) | < f (c) - u whenever | x - c | < d, since f is continuous.

But then f (x) > f (c) - ( f (c) - u ) = u whenever | x - c | < d and then

f (x) > u for x in ( c - d , c + d) and thus c - d is an upper bound for S which is smaller than c, a contradiction.

Suppose next that f (c) < u. Again, by continuity, there is an f > 0 such that

| f (x) - f (c) | < u - f (c) whenever | x - c | < f

Then f (x) < f (c) + ( u - f (c) ) = u for x in ( c - f , c + f)

and there are numbers greater than c for which f (x) < u, a contradiction.

We deduce that f (c) = u as stated.