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Last time we tried to introduce a metric on
non-empty compact sets in $\RR^n$. We tried
\[ \inf_{e \in E} d(e, F) \]
and failed. We now try
\[ \sigma(E, F) = \sup_{e \in E} d(e, F) .\]

Note $\sigma(E, F) \ge 0$. Now we check:
\begin{align*}
  \sigma(E, F) = 0
  &\implies \sup_{e \in F} d(e, F) = 0 \\
  &\implies d(e, F) = 0 \qquad \forall e \in E \\
  &\implies e \in F \qquad \forall e \in E \\
  &\implies F \supseteq E
\end{align*}
so $\sigma(E, F) \iff E \subseteq F$ (the
backwards direction is easy). Immediately we see
that
\[ \sigma(E, F) \stackrel{?}{=} \sigma(F, E) \]
may fail:
\[ E = \{1\}, F = \{1, 2\} \]
then $\sigma(E, F) \neq \sigma(F, E)$.

However, the triangle inequality
\[ \sigma(E, G) \le \sigma(E, F) + \sigma(F, G) \]
works.

\begin{proof}
  Let $e \in E$, $g g \in G$. Choose $f_e \in F$
  such that $d(e, f_e) = d(e, F)$. Then
  \begin{align*}
    d(e, g)
    &= d(e, f_e) + d(f_e, g) \\
    &= d(e, F) + d(f_e, g)
  \end{align*}
  Now take a supremum over $g \in G$, and we get
  \begin{align*}
    d(e, G)
    &\le d(e, F) + d(f_e, G) \\
    &\le \sigma(E, F) + \sigma(F, G)
  \end{align*}
  so taking sup over $e \in E$ we get
  \[ \sigma(E, G) \le \sigma(E, F) + \sigma(F, G)
  . \qedhere \]
\end{proof}

Now we set
\[ \rho(E, F) = \sigma(E, F) + \sigma(F, E) \]
Now:
\begin{itemize}
  \item $\rho(E, F) \ge 0$
  \item $\rho(E, F) = \rho(F, E)$
  \item $\rho(E, F) = 0$ if and only if $\sigma(E,
    F) = 0$, $\sigma(F, E) = 0$, which happens if
    and only if $E \subseteq F$, $F \subseteq E$,
    which happens if and only if $F = E$.
  \item Finally,
    \[ \rho(E, F) + \rho(E, G) = \sigma(E, F) +
    \sigma(F, E) + \sigma(F, G) + \sigma(G, F) \ge
    \sigma(E, G) + \sigma(G, E) = \rho(E, G) \]
\end{itemize}
We call this metric the \emph{Hausdorff metric}.

\begin{hiddenflashcard}[hausdorff-metric]
\begin{definition*}[Hausdorff metric]
  \cloze{
  \[ \rho(E, F) = \sup_{e \in E} d(e, F) + \sup_{f
  \in F} d(f, E) .\]
  }
\end{definition*}
\end{hiddenflashcard}
We can write it as:
\[ d(E, F) = \sup_{f \in F} \inf_{e \in E} \|e -
f\| + \sup_{e \in E} \inf_{f \in F} \|f - e\| .\]
Moreover, the Hausdorff metric is complete.

The proof depends on the following lemma:

\begin{lemma*}
  If $K_1 \supseteq K_2 \supseteq \cdots$, with
  $K_j$ compact and non-empty, then $K =
  \bigcap_j K_j$ is non-empty (and compact).
  Furthermore, $\rho(K_j, K) \to 0$.
\end{lemma*}

\begin{proof}
  Non-empty: Choose $k_j \in K$. Since $k_j \in
  K_1$, $K_1$ compact, there exists $n(j)$ strictly
  increasing, and some $k \in K_1$ with $k_{n(j)}
  \to k$. But $k_{n(j)} \in K_m$ for all $m \le
  n(j)$, so $k \in K_m$ for all $m \le n(j)$, so
  $k \in \bigcap_m K_m = K$. So $K \neq
  \emptyset$.

  Hausdorff metric convergence: If not then there
  exists a $\delta > 0$ such that $\rho(K_{n(j)},
  K) > \delta$ for some $n(j) \to \infty$. But
  $K_1 \supseteq K_2$, so $\rho(K_n, K) \ge
  \delta$ for all $n$. Now choose $k_j \in K_j$
  such that $d(k_j, K) \ge \delta / 2$. By the
  previous argument, there exists $m(j) \to
  \infty$ such that $k_{m(j)} \to k \in K$, which
  gives a contradiction.
\end{proof}

Second lemma:

\begin{lemma*}
  If $A, B$ are compact in $\RR^n$, then so is
  \[ A + B = \{a + b : a \in A, b \in B\} .\]
\end{lemma*}

\begin{proof}
  Suppose $c_n = a_n + b_n$ with $a_n \in A$, $b_n
  \in B$. Then there exists $n(j) \in \infty$, $a
  \in A$ with $a_{n(j)} \to a \in A$ and there
  exists $m(k) \to \infty$, $b \in B$ such that
  $b_{n(m(j))} \to b$. Then $c_{n(m(j))} \to a +
  b$.
\end{proof}

\begin{flashcard}[hausdorff-metric-is-complete]
\begin{proof}[Proof that Hausdorff metric is
  complete]
  \cloze{It is sufficient to show that ``every
  sufficiently fast Cauchy sequence converges''.
  Thus it suffices to show that if $E_n$ are compact
  non-empty, with
  \[ \rho(E_n, E_{n + 1}) \le 8^{-n} \]
  then $E_n \stackrel{\rho}{\to} E$ for some
  compact non-empty set $E$.

  Let $K_n = E_n + \ol{B(0, 2^{-n})}$. Then $K_n$
  is compact (by the previous lemma), and $K_{n +
  1} \subseteq K_n$. This is because if $x \in
  K_{n + 1}$, then $\exists y \in K_n$ such that
  \[ d(x, y) \le 8^{-n} ,\]
  and then $B(x, 2^{-n - 1}) \subseteq B(y,
  2^{-n})$.

  We have $K_1 \supseteq K_n \supseteq \cdots$, so
  there exists $K$ compact non-empty such that
  $K_n \stackrel{\rho}{\to} K$ (we use $K =
  \bigcap_n K_n$ using the previous lemma). But
  $\rho(K_n, E_n) \to 0$. So $\rho(E_n, K) \to
  0$.}
\end{proof}
\end{flashcard}

\newpage
\section{Runge's Theorem}

We start with Weierstrass Theorem: real
polynomials are uniformly dense in $C(I)$ for all
intervals $I$.

Can we approximate continuous functions $f : K \to
\CC$ ($K$ compact in $\CC$) by polynomials
(uniformly)?

The answer is no:

Take $K = \ol{D(0, 2)}$, $f(z) = \ol{z}$. Suppose
$P$ is a polynomial. Then integrating around the
unit disc we have:
\[ \oint f(z) - P(z) \dd z = \oint \ol{z} \dd z =
\int e^{-i\theta} e^{i\theta} i \dd \theta = 2\pi
i .\]
But then
\[ 2\pi \le \oint |f(z) - P(z)| \dd z \le 2\pi \|f
- P\_\infty, \]
so $\|f - P\|_\infty \ge 1$.

Recall (or take my word) that the uniform limit of
analytic functions is analytic (see
\courseref[CA]{Complex Analysis}). Now try:

If $\Omega$ open, and $K$ compact, with $\Omega
\supseteq K$, $f : \Omega \to \CC$ analytic, then
there exists $P_k$ polynomials such that $P_n \to
f$ uniformly on $K$.

This is also false:

Take
\[ \Omega = D(0, 2) \setminus D(0, 1 / 2) \]
with $f(z) = \frac{1}{z}$. Again,
\[ \oint f(z) \dd z = 2\pi i \]
so argument as before shows that we cannot
approximate $f$ uniformly on $K = \{z : |z| = 1\}$
by polynomials.