How to Prove Bolzano's Theorem

(without any epsilons or deltas!)

Scott E. Brodie, 11/2/97

Let f(x) be a continuous function on the closed interval [a,b], with

    f(a) < 0 < f(b).

We are to show that there is a real number c, between a and b, such that f(c) = 0.

CONTINUITY:

We first need to understand what is meant by a continuous function. This is usually taken to mean that for real numbers near the number x, the function values lie near f(x). This is usually stated in terms of a "challenge" and "answer" dialogue:

We say "f is continuous at x" if, for each open interval J containing f(x), we can find an open interval I containing x so that for each point y in I, f(y) lies in the interval J.

We say "f(x) is continuous" if it is continuous in the sense given above at every point x of its domain.

    It is traditional to think of the "challenge intervals" (the J's) as "small", in which case the "answer intervals" (the I's) will usually have to be small, too, but we will not explicitly rely on this.

As a simple application of this notion of continuity, we may note the

"Sign-Preserving property of Continuous Functions":

    If f is continuous at a, and f(a) < 0, then there is an open interval I containing a such that f(x) < 0 for every x in I.

For a proof, simply take the open interval (2f(a),0) for the challenge interval "J" in the definition of continuity. Of course, a similar statement holds for f(b) > 0, with an analogous proof.

COMMENT on COMPLETENESS:

Note that Bolzano's theorem fails over the field of rational numbers:

    the continuous function

    f(x) = x2 - 2

    takes the value -2 for x = 0 and takes the value +2 for x = 2, yet there is no rational number for which x2 - 2 = 0

This example serves as a reminder that a proof of Bolzano's theorem must include some reference to the "completeness" of the set of real numbers -- the property which distinguishes the real numbers from the rational numbers.

The Axiom of Completeness can be stated in many equivalent forms. One common version which is suitable for our purposes is the

"Least Upper Bound" axiom:

    Every non-empty set of real numbers which is bounded above has a least upper bound.

PROOF of BOLZANO's THEOREM:

Let S be the set of numbers x within the closed interval from a to b where f(x) < 0.

Since S is not empty (it contains a) and S is bounded (it is a subset of [a,b]), the Least Upper Bound axiom asserts the existence of a least upper bound, say c, for S. We show that this number c satisfies the requirements of Bolzano's theorem:

There are three possibilities for f(c): either f(c) < 0, f(c) > 0, or f(c) = 0. We show the first two choices lead to contradictions.

Suppose f(c) < 0, so that c is a member of S. Then the Sign Preserving Property of Continuous Functions asserts the existence of an open interval I containing c where f takes only negative values -- that is, I is a subset of S. But I (and thus S) contains points greater than c, so c cannot be an upper bound (let alone the least upper bound) for S. This contradiction forces us to reject the possibility that f(c) < 0.

Suppose instead that f(c) > 0. The argument is similar but not identical to the previous case. Since f is continuous, there is an open interval H containing c where f takes only positive values. But H contains points less than c, which must also be upper bounds for S, since no point of H can lie in S. This contradicts our choice of c as the least upper bound for S, so we must reject the possibility that f(c) > 0.

We conclude that f(c) = 0, QED.

COMMENTS on the PROOF of BOLZANO's THEOREM:

What is "really" going on here?

One way to look at it is to examine some of the properties of open intervals:

An interval is a set of points with the property that if x, y are distinct points of the set, then every point between x and y is also a point of the set.

An open interval is the set of points which lie strictly between two distinct points, called the "endpoints" of the interval.

    (A "closed interval" is the union of an open interval and its endpoints.)

    An open interval cannot be "degenerate" - a single point is not an open interval.

    For each point, c, in an open interval, there are points of the open interval which are greater than c and (other) points of the interval less than c.

    For each point, c, in an open interval, there is a closed interval about c, contained (as a subset) in the open interval.

    An open interval does not contain its least upper bound nor does it contain its greatest lower bound.

    Now consider the union of two open intervals, say U=(a,b), and V=(c,d), where a < d. Comparing b and c, there are only three possibilities:

      If b < c, then the union of U and V omits the closed interval [b,c];

      If b = c, then the union U and V omits the point b;

      If b > c, then the union of U and V forms an open interval (a,d), but the original intervals share the interval (c,b).

    In summary, it is impossible for the union of two disjoint open intervals to form an interval.

Much of this behavior of open intervals can be generalized to the notion of "open sets". A set is "open" if it contains an open interval containing each of its points.

    For example, in our proof of Bolzano's theorem, we showed, in essence, that the set S of points where f takes on negative values is an open set (it contains an interval about each of its points). Similarly, The set T of points where f takes on positive values is an open set.

    As with an open interval, an open set cannot contain its least upper bound, nor its greatest lower bound.

We can also generalize our observation about the union of two disjoint open intervals:

    Theorem: It is impossible for the union of two disjoint, non-empty open sets to form an interval.

      The proof is nearly a paraphrase of our proof of Bolzano's theorem:

      Suppose to the contrary, that the union of disjoint non-empty open sets U, containing a, and V, containing b, forms an interval, I, with a < b. Consider the set S of points x such that the entire closed interval [a,x] lies in U. Since U is open, it contains an open interval about a, so the set S is not empty. Since U and V are disjoint, the set S is bounded above by b. Then the Least Upper Bound axiom asserts the existence of a least upper bound, say, c, for S. Clearly, a is less than c, and c is at most equal to b.

      If c lies in U, then there must be an open interval about c, in U, such that for some point x in this interval, [a,x] is in U (otherwise there would be a smaller upper bound for S than c). But then we may extend the interval [a,x] beyond c without leaving U, in contradiction to our choice of c as an upper bound for S. Thus, c does not lie in U.

      On the other hand, if c lies in V, then there is an interval about c of points of V, which includes upper bounds for S which are smaller than c, contradicting our choice of c as the least upper bound for S. Thus, c does not lie in V. In particular, c cannot coincide with b.

      We have shown that the union of U and V fails to include the point c, which lies between a and b. Thus the union of U and V cannot be an interval, QED.

With these notions at hand, the logic behind Bolzano's theorem is immediately evident:

If the function f is continuous, with f(a) < 0 and f(b) > 0, then the sets S, of points x where f(x) < 0, and T, of points where f(x) > 0 are open sets. But if no point of [a,b] satisfies f(x)=0, then S and T exhaust [a,b], in contradiction to the previous argument.

In the language of "general topology", a set which cannot be decomposed into two disjoint, non-empty open sets is called "connected", and we have shown that Bolzano's theorem is in essence equivalent to the theorem that a closed interval is a connected set.


Related material
Read more...

  • Cavalieri's Principle
  • Derivative of Sine and Cosine
  • Distance From a Point to a Straight Line
  • Estimating Circumference of a Circle
  • Maximum Volume of a Cut Off Box
  • Mistrust Intuition of the Infinite
  • Naturally Discontinuous Functions
  • Rolle's and The Mean Value Theorems
  • Function, Derivative and Integral
  • Area of a Circle by Rabbi Abraham bar Hiyya Hanasi
  • Schwarz Lantern
  • Two Circles and a Limit
  • Deceptive Appearances
  • Problem 4010 from Crux Mathematicorum
  • |Contact| |Front page| |Contents| |Algebra| |Store|

    Copyright © 1996-2017 Alexander Bogomolny

     61172855

    Search by google: