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\begin{definition*}
  Let $(X, d)$, $(Y, e)$ be metric spaces. We say
  $f : X \to Y$ is continuous if given $x \in X$,
  $\eps > 0$ there exists $\delta(\eps, x)$ such
  that
  \[ d(x', x) < \delta \implies e(f(x), f(x')) <
  \eps .\]
\end{definition*}

\begin{proposition*}
  This is equivalent to the ``if $U$ open in $Y$
  then $f^{-1}(U)$ open in $X$'' definition.
\end{proposition*}

\begin{proof}
  We first show that if $f$ satisfies the first
  definition, then it satisfies the second.
  Suppose $U$ open in $Y$. If $x \in f^{-1}(Y)$
  then $f(x) \in U$ so there existst $\eps > 0$
  such that $B_Y(f(x), \eps) \subseteq U$. Hence
  there exists $\delta$ such that $f(B(x, \delta))
  \subseteq B_Y(f(x), \eps)$, so $B(x, \delta)
  \subseteq f^{-1}(U)$. So $f^{-1}(U)$ is open.

  For the other direction: if $x \in X$ then given
  $\eps > 0$, $B(f(x), \eps)$ is open so
  $f^{-1}(B(f(x), \eps))$ is open. $x \in
  f^{-1}(B(f(x), \eps))$ so there exists $\delta$
  such that $B(x, \delta) \subseteq f^{-1}(B(f(x),
  \eps))$ and we recover the $\eps, \delta$
  definition.
\end{proof}

We introduce an idea much used in the course:

\begin{flashcard}[distance-to-set-defn]
\begin{definition*}[Distance to $A$]
  \cloze{If $A \subseteq \RR^n$ is closed and non-empty,
  we define
  \[  d(x, A) = \inf_{a \in A} \|x - a\| .\]}
\end{definition*}
\end{flashcard}

\begin{remark*}
  \begin{enumerate}[(1)]
    \item $d(x, A) = 0 \iff x \in A$: the backwards
      direction is trivial, and for the forwards
      direction, if $d(x, A) = 0$ then there
      exists $a_n \in A$ with $d(x, a_n) \to 0$.
      But $A$ is closed.
    \item $x \mapsto d(x, A)$ is continuous.
      \begin{proof}
        Let $\eps > 0$ then for given $x, y$ there
        exists $a \in A$ such that $\|x - a\| \le
        d(x, A) + \frac{\eps}{2}$. So $\|y - a\|
        \le \|x - a\| + \|y - x\| \le d(x, A) +
        \|x - y\| + \frac{\eps}{2}$. So $d(y, A)
        \le \|y - x\| + d(x, A) + \frac{\eps}{2}$.
        Since $\eps$ is arbitrary,
        \[ d(y, A) \le d(x, A) + \|x - y\| \]
        Similarly, $d(x, A) \le d(y, A) + \|x -
        y\|$, so
        \[ |d(x, A) - d(y, A)| \le \|x - y\| .
        \qedhere \]
      \end{proof}
  \end{enumerate}
\end{remark*}

\vspace{-1em}
Now we move towards more interesting results.

\begin{theorem*}
  If $E \subseteq \RR^n$ is compact and $f : E \to
  \RR^n$ is continuous, then $f(E)$ is compact.
\end{theorem*}

\begin{proof}
  If $y_n \in f(E)$ then $y_n = f(x_n)$ for some
  $x_n \in E$. By \nameref{BW}, there exists $n(j)
  \to \infty$ and $x \in E$ such that $x_{n(j)}
  \to x$. So $y_{n(j)} = f(x_{n(j)}) \to f(x) \in
  E$.
\end{proof}

\begin{corollary*}
  If $E \subseteq \RR^n$ is compact and non-empty
  and $f : E \to \RR$ is continuous then $f$ is
  bounded and attains its bounds.
\end{corollary*}

\begin{proof}
  $f(E)$ is compact in $\RR$ so closed and
  bounded. Further since $f(E)$ is bounded above
  (and non-empty), $f(E)$ has a supremum $\alpha$ so
  there exists $e_n$ such that $f(e_n) \to
  \alpha$. By compactness, there exists $n(j) \to
  \infty$ and $e$ such that $e_{n(j)} \to e$ so
  $f(e) = \alpha$.
\end{proof}

\begin{theorem*}[Fundamental Theorem of Algebra]
  \label{FTA}
  If $P$ is a non-constant polynomial then it has
  a root.
\end{theorem*}

\begin{lemma*}
  If $P$ is a non-constant polynomial then $|P(z)|
  \to \infty$ as $|z| \to \infty$.
\end{lemma*}

\begin{proof}
  \begin{align*}
    P(z)
    &= \sum_{j = 0}^n a_j z^j \\
    &= z^n \left( a_n - \sum_j a_j z^{j - n} \right)
  \end{align*}
  $a_n \neq 0$, $n \ge 1$. Note
  \[ a_n - \sum_j a_j z^{j - n} \to a_n \]
  as $|z| \to \infty$, hence since $|z^n| \to
  \infty$ as $|z| \to \infty$, we have $|P(z)| \to
  \infty$ as $|z| \to \infty$.
\end{proof}

\begin{lemma*}
  In particular we can find an $R$ such that
  $|P(z)| > |P(0)|$ for all $|z| \ge R$.
\end{lemma*}

\begin{proof}
  By compactness there exists $|\alpha| \le R$
  such that
  \[ |P(z)| \ge |P(\alpha)| ~\forall |z| \le R .\]
  But $|P(z)| \ge |P(0)| \ge |P(\alpha)| ~\forall
  |z| \ge R$. So $|P(\alpha)| \le |P(z)| ~\forall
  z$.
\end{proof}

We now show that if $\alpha$ is a minima of
$|P(z)|$ then $P(\alpha) = 0$ (to some extent the
rest of the proof is a matter of personal
preference).

\begin{remark*}
  By considering $P(z + \alpha)$ if necessary, we
  may suppose that $|P(0)| \le |P(z)|$ for all
  $z$. If $P(0) = 0$ we are done (using the
  previous two lemmas, we know that $|P(z)|$ has a
  global minimum, because the minimum on
  $B(0, R)$, which is achieved by compactness,
  must be a global minimum).
\end{remark*}

\begin{proof}[Proof of \nameref{FTA}]
  By the remark, assume $|P(0)| \le |P(z)|$ for
  all $z$. If $P(0) = 0$ we are done. If not,
  then
  \[ P(z) = a_0 + \sum_{j = 1}^n a_j z^j \]
  with $a_0 \neq 0$. By considering
  $\frac{P(z)}{a_0}$ we may suppose $a_0 = 1$ and
  that
  \[ P(z) = 1 + a_k z^k + \sum_{j = k + 1}^n a_j
  z^j \]
  with $a_k \neq 0$.

  By considerng $P(r e^{i\theta} z)$ for suitable
  $r, \theta$ we may suppose $a_k = -1$. So
  \[ P(z) = 1 - z^k + \sum_{j = k + 1}^n z^j \]
  Since $\frac{\sum_{j = k + 1}^n a_j z^j}{z^k} \to
  0$ as $z \to 0$, there exists $\delta > 0$ such
  that
  \[ \left| \sum_{j = k + 1}^n a_j z^j \right| <
  \frac{|z|^k}{2} \]
  for $|z| < \delta$. If $\eta =
  \frac{\delta}{2}$, then
  \[ P(\eta) = 1 - \eta^k + \eps(\eta) \]
  with $|\eps(\eta)| < \frac{\eta^k}{2}$. So
  $P(\eta) < 1 = P(0)$, contradiction. So $P(0) =
  0$, and $P$ has a root as desired.
\end{proof}

\begin{corollary*}
  If $P$ has degree $n \ge 1$ then $P(z) = (z -
  a)Q(z)$ for some polynomial $Q$ of degree $n -
  1$.
\end{corollary*}

\begin{proof}
  By long division (in fact proof by induction).
  If $P$ degree $n$, then $P(z) = (z - a) Q(z) +
  r$ with $\deg Q = n - 1$, $r \in \CC$. If $P(a)
  = 0$ then $0 = 0 + r$ so $r = 0$, so $P(z) = (z
  - a)Q(z)$.
\end{proof}

By induction if $P$ has degree $n$,
\[ P(z) = A \prod_{j = 1}^n (z - a_j) \]
for some $A \neq 0$.

\begin{corollary*}
  If $P$ is a polynomial of degree at most $n$
  which vanishes at $n + 1$ points then $P = 0$.
\end{corollary*}

\begin{proof}
  If $\deg P = k \ge 0$, then $P(z) = A \prod_{j =
  1}^k (z - a_j)$. Then $P(z) \neq 0$ for $z \neq
  a_1, a_2, \ldots, a_k$.
\end{proof}
