Box 3.0b: Proof of the Intermediate Value Theorem

Note that the material on this page is optional. First year students can almost always count on NOT being tested on this proof

Theorem:

Again the statement of the intermediate value theorem is: if f(x) is continuous on the closed interval,  a £ x £ b  (that is f(x) is continous at every point on the interval and the interval includes the endpoints), and f(a) is not equal to f(b), then for every value, y, that falls in between f(a) and f(b), there exists at least one point, c, in between a and b such that  f(c) = y.

Outline of Proof

The outline of the proof is that we assume that for some y that lies between f(a) and f(b), there is no value, c, lying between a and b that satisifies  f(c) = y. So y is a "missing point." From that we shall show that there must be a point, x, lying between a and b where f(x) is discontinuous. That is the contradiction that proves the theorem, since we began by saying that f(x) is continuous for all x lying between a and b.

Remember that y is the missing point, that is it is one for which  f(c) = y  has no solution on the interval. The strategy will be to pin the missing point down to smaller and smaller intervals. Then we show that the endpoints of those smaller and smaller intervals form Cauchy sequences, and Cauchy sequences always have limits. The definition of continuity will then demonstrate that the limit of the Cauchy sequences must be the solution, c, of f(c) = y. So the only way y can be missing is for f(x) not to be continuous.

Proof

Notice that on the closed interval,  a £ x £ b, there must be some points, x, for which  f(x) > y  and others for which  f(x) < y . That follows because there can be none for which  f(x) = y. We know that there is at least one of each type of point because we know that y falls in between f(a) and f(b), implying that a is of the opposite type from b.

We take the closed interval  a £ x £ b  and divide it precisely in half -- with both fragments being closed. This gives us one closed interval from a to  (a + b)/2, and another from  (a + b)/2  to b. One of the following three situations must be true:

1. All of the points, x, for which  f(x) < y  fall in one of the fragments and all of the points, x, for which  f(x) > y  fall in the other fragment.
2. The lower fragment contains points of both types.
3. The lower fragment contains points of just one type, but the upper fragment contains points of both types.

Whichever of the three situations was the case, the new interval must have the following two properties:

1. It is half the length of the original.
2. It contains points of both types -- that is x's for which  f(x) < y  and x's for which  f(x) > y .

Since the new interval contains both types of points, we can perform the same operation on it and get yet another interval that is half the length of the new one and completely contained in it. And it too will contain points of both types.

Clearly we can keep applying this operation over and over again, getting a new interval each time, always half the length of the last, always completely contained in the last, and always containing both types of points. There is no limit to how many times you can do this.

This gives us a sequence of closed intervals,  I₀, I₁, I₂, ... , where I₀ is the original a to b interval. Each I_n contains all the intervals that follow, just like Russian nesting dolls.

Let  a₀, a₁, a₂, ... be the lower endpoints of the sequence of intervals and let  b₀, b₁, b₂, ... be the upper endpoints of the sequence of intervals. Notice that both sequences of endpoints are, as the subscripts grow, confined to a smaller and smaller range. In fact the range each can be in is halved with each increase of one in the subscript. And by halving the range each time, you can make the range as close to zero as you'd like by taking a sufficiently high subscript. That means that each sequence of endpoints forms a Cauchy sequence. And every Cauchy sequence of real numbers converges to a limit that is also a real number. Not only that, because the difference between b_n and a_n can be made as close to zero as you like by choosing n large enough, both Cauchy sequences must converge to the same real number. Call that real number c.

Observe that c is contained in every interval, I_n, in the list described above. This means that every interval (and not just the ones in the list) that contains c as an interior point (an interior point of an interval is one that is not an endpoint), also contains some of the I_n's from the list. We only require that it contain one of them. So why do we know that every interval containing c as an interior point contains an I_n? Well, you can find an I_n as small as you like simply by choosing n large enough. That means that you can always find one that is small enough to fit completly into any interval containing c.

The consequence of all that is that every interval containing c as an interior point, no matter how small, must contain both types of points -- those for which  f(x) < y  and those for which  f(x) > y . Why? Because every every interval containing c as an interior point also contains an I_n, and every I_n contains points of both types. We also know, because f(x) is continuous, that f(c) exists.

In the final stage of this proof, we show that it must be true that  f(c) = y, where y is the missing point we talked about at the beginning. This is because f(x) is continuous, and that means,

   f(c)  =   lim   f(x)
            x  > c

For if f(c) differed from y then you could set  0 < e < |f(c) - y|. Because every interval, no matter how small, that contains c has both types of points in it (that is the f(x)'s fall on both sides of y), you could never find a d such that  |f(x) - f(c)| £ e  whenever  |x - c| < d.  Hence  |f(x) - f(c)|  could never be guaranteed to be less than  |f(c) - y|. You would always be fighting against some nearby point, x, that causes f(x) to fall on the opposite side of y from f(c). Why? Because  |x - c| < d  represents an interval that contains c, and every interval containing c has x's in it that make f(x) fall on both sides of y. If one f(x) is on the same side of y as f(c), then another x very close by (indeed as close as you like) would lead to an f(x) that fell on the opposite side of y and was therefore at least  |f(c) - y|  away from the first f(x).

And so if f(c) is not equal to y, the limit can't exist, and that means that f(x) can't be continuous.

We started out assuming f(x) to be continuous and assuming the existence of a missing point, y. Since we know that functions can be continuous, it follows that if a function is continuous over an interval, it can't have a missing point.

Comments:

If you got through all that and understood it, congratulations. There are shorter proofs to this theorem, but they involve concepts from a field of mathematics called topology. This proof, though long, sticks with common concepts that we have of real numbers. It also demonstrates the technique of trapping a real number inside of a sequence of nested intervals. It is a powerful method for proving theorems about the real numbers, and is used in many other proofs.

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