104 CHAPTER 4. CONTINUITY AND LIMITS

It gives the existence of a certain point. You see in the picture there is a horizontalline, y = c and a continuous function which starts off less than c at the point a and ends upgreater than c at point b. The intermediate value theorem says there is some point betweena and b shown in the picture as z such that the value of the function at this point equals c.

The theorem is due to Bolzano in 1817. You might think that this is an obvious theorembut this is not the case. It is not even true if you only had the rational numbers and thisincludes all numbers we work with. Suppose you only had the rational numbers. Thenf (x) = x2 − 2 is continuous. f (0) < 0 and f (2) > 0 but the only point between 0 and 2where this function is 0 is the point

√2 which has been known for thousands of years to be

irrational. You have to use something which rules out holes in the real line. Bolzano’s majorcontribution was to identify the concept of completeness as the reason for this theoremrather than some sort of vague notion based on pictures like the one just drawn. He did thislong before Dedekind gave a way to construct the real numbers.

Proposition 4.4.1 Let f : [a,b]→ R be continuous and suppose f (a) f (b) ≤ 0. Thenthere exists x ∈ [a,b] such that f (x) = 0.

Proof: When we have an interval [an,bn] in this argument, cn+1 will be the mid-point (an +bn)/2. Let a0 = a,b0 = b. If [an−1,bn−1] has been chosen such that f (an−1)f (bn−1)≤ 0, consider [an−1,cn] and [cn,bn−1]. Either

f (an−1) f (cn)≤ 0 or f (cn) f (bn−1)≤ 0

since if both products are positive, then

f (an−1) f (cn) f (cn) f (bn−1) = f (cn)2 f (an−1) f (bn−1)> 0

so f (an−1) f (bn−1) > 0. Pick one of the intervals for which the product is non-positive,either [an−1,cn] r [cn,bn−1]. Denote this interval as [an,bn]. Thus f (an) f (bn) ≤ 0. Con-tinue picking the correct subinterval as just described. These nested intervals have exactlyone point in their intersection because they have diameters converging to 0. Call it x. Then

( f (x))2 = limn→∞

f (an) f (bn)≤ 0

This is by Theorem 4.0.7. Thus f (x) = 0.It is easy to generalize this Proposition.

Theorem 4.4.2 Suppose f : [a,b]→R is continuous and suppose either f (a)< c <f (b) or f (a)> c > f (b) . Then there exists x ∈ (a,b) such that f (x) = 0.

Proof: Apply the above proposition to g(x)≡ f (x)−c obtaining a point x ∈ (a,b) withg(x) = f (x)− c = 0.

Here is another lemma which may seem obvious but when you ask why, you begin tosee that it is not as obvious as you thought. In fact, this is a special case of a general theorywhich says that one to one continuous functions from U , an open set in Rp to Rp take opensets to open sets. This is a very difficult result. The notation 1−1 means one to one. Thatis, if x ̸= y, then f (x) ̸= f (y).

Lemma 4.4.3 Let φ : [a,b] → R be a continuous function and suppose φ is 1− 1 on(a,b). Then φ is either strictly increasing or strictly decreasing on [a,b] .