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(3.2) {\sc Theorem.} {\it Any non-degenerate monotone image of an interval
is homeomorphic with the interval.}
{\it Proof.} Let $f(J) = E$ be monotone, where $J = (0,1)$ and $E$ is non-%
degenerate. 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 $m/2^n$, $0\le m\le
2^n$, so that the length of $I(m/2^n)$ is $\le 1/2^{n - 1}$.
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 $\epsilon > 0$ be given. By
uniform continuity of $f$ there exists a $\delta > 0$ such that for any interval $H$
in $J$ of length $< \delta$, $f(H)$ is of diameter $< \epsilon/2$. Let $n$ be chosen so that
$1/2^{n - 1} < \delta$. 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 $n$th 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 $< \delta$ we have
$$h(t_i) + h(t) \subset f(S_i),\enskip i = 1, 2,\enskip{\rm and}\enskip\delta[f(S_1) + f(S_2)] < \epsilon$$
since $\delta[f(S_i)] < \epsilon/2$, $i = 1, 2$. Accordingly, $\rho[h(t_1), h(t_2)] < \epsilon$ and $h$ is
uniformly continuously on $D$.
Let $h$ be extended continuously to $\bar{D} = J$. 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
$h(t)\ne h(x_1)$. Then if $H_1$ and $H_2$ are the closed intervals into which $J$
is divided by $f^{-1}h(t)$ where $x_1\in H_1$, $f(H_1)\cdot f(H_2) = h(t)$ by monotoneity
of $f$ and thus $h(H_1)\cdot h(H_2) = h(t)$ since $h(H_i)\subset f(H_i)$ by definition of $h$.
Accordingly $h(x_1)\not\in h(H_2)$ so that $h(x_1)\ne h(x_2)$. Thus $h(J) = E$ is a
homeomorphism.
\bye