4.5. CONTINUITY OF THE INVERSE 105

Proof: First it is shown that φ is either strictly increasing or strictly decreasing on(a,b) .

If φ is not strictly decreasing on (a,b), then there exists x1 < y1, x1,y1 ∈ (a,b) such that(φ (y1)−φ (x1))(y1 − x1)> 0. If for some other pair of points, x2 < y2 with x2,y2 ∈ (a,b) ,the above inequality does not hold, then since φ is 1− 1, (φ (y2)−φ (x2))(y2 − x2) < 0.Let xt ≡ tx1 +(1− t)x2 and yt ≡ ty1 +(1− t)y2. Then xt < yt for all t ∈ [0,1] because

tx1 ≤ ty1 and (1− t)x2 ≤ (1− t)y2

with strict inequality holding for at least one of these inequalities since not both t and(1− t) can equal zero. Now define

h(t)≡ (φ (yt)−φ (xt))(yt − xt) .

Since h is continuous and h(0) < 0, while h(1) > 0, there exists t ∈ (0,1) such thath(t) = 0. Therefore, both xt and yt are points of (a,b) and φ (yt)−φ (xt) = 0 contradictingthe assumption that φ is one to one. It follows φ is either strictly increasing or strictlydecreasing on (a,b) .

This property of being either strictly increasing or strictly decreasing on (a,b) carriesover to [a,b] by the continuity of φ . Suppose φ is strictly increasing on (a,b) . (A similarargument holds for φ strictly decreasing on (a,b) .) If x > a, then let zn be a decreasingsequence of points of (a,x) converging to a. Then by continuity of φ at a,

φ (a) = limn→∞

φ (zn)≤ φ (z1)< φ (x) .

Therefore, φ (a)< φ (x) whenever x ∈ (a,b) . Similarly φ (b)> φ (x) for all x ∈ (a,b).

4.5 Continuity of the InverseThe inverse of a continuous function defined on an open interval is also continuous. Thisis an amazing result. If I is an interval, the notation f (I) will be the set { f (x) : x ∈ I} .

Corollary 4.5.1 Let f : (a,b) → R be one to one and continuous. Then f (a,b) is anopen interval, (c,d) and f−1 : (c,d)→ (a,b) is continuous. If f : [a,b]→ R is one to oneand continuous, then f−1 is also continuous.

Proof: Consider the first part. By Lemma 4.4.3, f is strictly increasing or strictlydecreasing. Hence

(f−1)−1

((x,y))≡ f ((x,y)) is an open interval. By Theorem 4.1.1, f−1

is continuous because inverse images of open intervals are open intervals.As to the second claim, here is a direct proof based on notions of compactness. Say

f (xn)→ f (x) where each xn,x are in [a,b]. Does xn → x? If not, there exists a subsequence{xnk

}and ε > 0 such that

∣∣xnk − x∣∣ ≥ ε > 0. However, by compactness, there is a further

subsequence,{

xnkl

}such that liml→∞ xnkl

= x̂ and so, by continuity,

f (x̂) = liml→∞

f(

xnkl

)= lim

n→∞f (xn) = f (x) .

But |x̂− x| = liml→∞

∣∣∣xnkl− x∣∣∣ ≥ ε and so this violates the assumption that f is one to

one. Hence, xn → x and so f−1 ( f (xn)) → f−1 ( f (x)) and so f−1 is continuous at everyf (x) ∈ f ([a,b]).