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\author{Jiri Lebl}
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\textbf{Problem 10998, year 2003, page 240}
\vspace{0.1in}
Solution by Jiri Lebl,
9635 Genesee Ave Apt E1, San Diego, CA 92121.
\vspace{0.1in}
{\em Problem: Let $D$ be a nonempty, open, connected and relatively compact set
in a metric space $X$ with metric $d$. Prove that if $f$ is a continuous map
from $D$ into $D$ such that $f(D)$ is open, then there exists a point
$x_0 \in D$ so that $d(x_0,\partial D) = d(f(x_0),\partial D)$.}
\vspace{0.1in}
We know that $d(x,\partial D)$ is a continuous map (easy to see by
triangle inequality). Then $g(x) = d(f(x),\partial D) - d(x, \partial D)$
is a continuous map of $D$ to ${\mathbb{R}}$. If we can show that $\exists z
\in D \,\text{s.t.}\, g(z) \geq 0$ and $\exists y \in D \,\text{s.t.}\, g(y) \leq 0$, then
$\exists x_0 \,\text{s.t.}\, d(f(x_0),\partial D) = d(x_0,\partial D)$ by
the intermediate value theorem.
As $d(x,\partial D)$ is continuous and $D$ is relatively compact which means
that $\overline{D}$ is compact, then there exists a $z \in \overline{D}$
such that $d(z,\partial D)$ is maximal. Since $D$ is non-empty and open then
any point in $D$ will have some positive distance to the boundary and thus
this maximal $z$ is in $D$. Then since $f(z) \in D$ we have that
$d(f(z),\partial D) \leq d(z,\partial D)$ and so we have shown
that $\exists z \in D \,\text{s.t.}\, g(z) \leq 0$.
To prove that $\exists y \,\text{s.t.}\, g(y) \geq 0$ we consider 2 cases. First let's
suppose that $f(D) = D$. In this case we can use a similar argument as above
to find $y$ such that $d(f(y), \partial D)$ is maximal and then $g(y) \geq 0$.
Now suppose that $f(D) \subsetneq D$. This means that there exists a point
$\alpha \in D \, \text{s.t.}\, \alpha \in \partial f(D)$, this follows by connectedness
of $D$. So take a sequence of $f(y_i)$ so that they converge to $\alpha$.
Since $\overline{D}$ is compact we can pick $y_i$ such that this
is a convergent sequence as well (we could always just pick a convergent
subsequence) and we suppose that $y_i \rightarrow \beta$. Now $\beta$
must be on the boundary of $D$, since if it wasn't, then $f(\beta) = \alpha$
by continuity and so $\alpha$ would be in $f(D)$, but it is also on the
boundary of $f(D)$, and this is not possible since $f(D)$ is open.
Since $\alpha \in D$ it is a positive distance from the boundary since $D$
is open, so say $d(\alpha,\partial D) = \delta$. So now pick a point
$y_j$ from $\{ y_i \}$ such that $d(y_j, \partial D) < \frac{\delta}{2}$
and such that $d(f(y_j),\alpha) < \frac{\delta}{2}$. Then by
triangle inequality we have $d(f(y_j), \partial D) \geq \frac{\delta}{2}$
and so $g(y_j) \geq 0$. QED!
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