If f is continuous on [a,b] and k is between f(a) and f(b) then there must be a number, c, in [a,b] such that f(c)=k

The Intermediate Value Theorem can be stated in the following equivalent form:
Suppose that I is an interval in the real numbers R and that f : I -> R is a continuous function. Then the image set f ( I ) is also an interval.

This captures an intuitive property of continuous functions: if f(1) = 3 and f(2) = 5 then the value of f must be 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:

First assume f(a) < k < f(b);  the other case f(b) < k < f(a) is similar.

Let S = {x in [a,b] : f(x) £ k}.  Then S is non-empty (as a is in S) and bounded above by b.  Hence by the "completeness property" of the reals, the least upper bound, or supremum c = sup S exists.  We claim that f(c)=k.

Suppose our claim is false.  First, suppose that f(c)>k.  Then f(c)-k > 0, so there is a d > 0 such that |f(x)-f(c)| < f(c)-k whenever 0 < |x-c| < d, since f is continuous.
But then f(x) > f(c) - (f(c)-k) = k whenever |x-c| < d and then f(x)>k for x in (c-d,c+d) and thus c-d is an upper bound for S which is smaller than c, so c is not the least upper bound, a contradiction.

Now suppose that f(c)<k.  Again, by continuity, so there is a d > 0 such that |f(x)-f(c)| < k-f(c) whenever 0 < |x-c| < d.
But then f(x) < f(c) + (k-f(c)) = k whenever x is in (c-d,c+d) and thus f(c+d/2)£k, and c+d/2 is an element of S larger than c, so c is not an upper bound, a contradiction.

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

Now, Proceed to the Next Theorem in this Series:

This proof was just the first in a series of proofs that take us up to the Fundamental Theorem of Calculus.  The next proof in this series is the Bounded Value Theorem.

Generalization

 The intermediate value theorem can be seen as a consequence of the following two statements from topology:

If X and Y are topological spaces, f : X -> Y is continuous, and X is connected, then f(X) is connected.

A subset of R is connected if and only if it is an interval.

A particular case of the Intermediate Value Theorem is this:

If f is continuous on an interval, and f is sometimes positive and sometimes negative then f must have a zero in the interval, which is known as the Weierstrass Intermediate Value Theorem, named for Karl Weierstrass (1815-1897), a German mathematician who is best known for his rigorous mathematical definitions of the Extreme Value Theorem and other results in calculus.

Proceed to the next theorem in this series:

The next proof in this series is the Bounded Value Theorem.

Go back to Calculus Home

Proceed to the next proof in this series, the Bounded Value Theorem or else skip all the details and go straight to the Fundamental Theorem of Calculus!

Arithmetic Rules -- properties of equality, addition, multiplication, and in particular the Axioms of Real Arithmetic, including the completeness axiom.

Upper Bound -- definition of "upper bound" and "least upper bound" of either sets or functions

Definition of Interval -- a subset satisfying certain properties of a totally connected set such as the set of real numbers.  "Totally connected" means there exists a "£" operator that is reflexive (a property in all sets), and antisymmetric, reflexive, and total (three of the thirteen properties listed above).

Source of the proof: Wikipedia