(3.2) Theorem. Any non-degenerate monotone image of an interval is homeomorphic with the interval.

Proof. Let f(J) = E be monotone, where J = (0, 1) and E is nondegenerate. Let I(1/2) be the closure of the interval J −f^{−1}f(0)−f^{−1}f(1) and let x(1/2) be its mid point. Similarly let I(1/2^{2}) and I(3/2^{2}) be the closures of the left and right intervals remaining on the deletion of f^{−1}f[x(1/2)] from I(1/2) and let x(1/2^{2}) and x(3/2^{2}) be their respective mid points. Likewise I(1/2^{3}), I(3/2^{3}), I(5/2^{3}), I(7/2^{3}), are the closures of the intervals into which I(1/2^{2}) and I(3/2^{2}) are divided by removing f^{−1}f[x(1/2^{2})] and f^{−1}f[x(3/2^{2})] ordered from left to right, and so on indefinitely. In this way we define a collection of intervals I(m/2^{n}) and their mid points x(m/2^{n}) for all dyadic rational numbers m/2^{n}, 0 ≤ m ≤ 2^{n}, so that the length of I(m/2^{n}) is ≤ 1/2^{n−1}.

Now for any dyadic rational m/2^{n} on J we define h(m/2^{n}) = f[x(m/2^{n})]. We next show that h is unformly continuous. Let ϵ > 0 be given. By uniform continuity of f there exists a δ > 0 such that for any interval H in J of length < δ, f(H) is of diameter < ϵ/2. Let n be chosen so that 1/2^{n−1} < δ. Then if t_{1} and t_{2} are points of the set D of dyadic rationals with |t_{1} −t_{2}| < 1/2^{n}, there is at least one point t = j/2^{n} such that for each i (i = 1, 2), t is an end point of an interval T_{i} of the nth dyadic subdivision of J containing t_{i}. If S_{1} and S_{2} are the corresponding intervals to T_{1} and T_{2} in the set I(m/2^{n+1}), since each is of length < δ we have

h(t_{i}) + h(t) ⊂ f(S_{i}), i = 1, 2, and δ[f(S_{1}) + f(S_{2})] < ϵ

since δ[f(S_{i})] < ϵ/2, i = 1, 2. Accordingly, ρ[h(t_{1}), h(t_{2})] < ϵ and h is uniformly continuously on D.

Let h be extended continuously to D̄ = J. Then h(J) = f(J) = E, because for each n the union of the intervals I(m/2^{n}) maps onto E under f, so that the images of the mid points of all these intervals for all n, i.e., the set h(D), is dense in E. Finally, h is (1–1). For if x_{1} and x_{2} are distinct points of J, there exists a point t = j/2^{n} between x_{1} and x_{2} with h(t) ≠ h(x_{1}). Then if H_{1} and H_{2} are the closed intervals into which J is divided by f^{−1}h(t) where x_{1} ∈ H_{1}, f(H_{1}) · f(H_{2}) = h(t) by monotoneity of f and thus h(H_{1}) · h(H_{2}) = h(t) since h(H_{i}) ⊂ f(H_{i}) by definition of h. Accordingly h(x_{1}) ∉ h(H_{2}) so that h(x_{1}) ≠ h(x_{2}). Thus h(J) = E is a homeomorphism.