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\begin{proof}
  \phantom{}
  \begin{enumerate}[(i) $\implies$ (ii)]
    \item[(i) $\implies$ (ii)] Let $U \ni 0$
      be a compact neighbourhood of $0$ ($0 \in U
      \subseteq Z$ with $U$ open, $Z$ compact).
      Then there exists $x \in \O_K$ such that
      $x\O_K \subseteq U$. Since $x\O_K$ is
      closed, $x\O_K$ is compact. Hence $\O_K$ is
      compact ($x \O_K \stackrel{\cdot x^{-1}}{\to}
      \O_K$ is a homeomorphism).
    \item[(ii) $\implies$ (i)] $\O_K$ compact
      implies $a + \O_K$ is compact for all $a \in
      K$. So $K$ is locally compact.
    \item[(ii) $\implies$ (iii)] Let $x \in m$,
      and $A_x \subseteq \O_K$ be a set of coset
      representatives for $\O_K / x\O_K$. Then
      $\O_K = \bigcup_{y \in A} y + x\O_K$ is a
      disjoint open cover. So $A_x$ is finite by
      compactness of $\O_K$. So $\O_K / x\O_K$ is
      finite, hence $\O_K / m\O_K$ is finite.o

      Suppose $v$ is not \gls{disc}. Then let
      $x_1, x_2, \ldots$ such that
      \[
        v(x_1) > v(x_2) > \cdots > 0
      .\]
      Then $x\O_K \subsetneq x_2 \O_K \subsetneq
      x_3 \O_K \subsetneq \cdots \subsetneq \O_K$.
      But $\O_K / x\O_K$ is finite so can only
      have finitely many subgroups, contradiction.
    \item[(iii) $\implies$ (ii)] Since $\O_K$ is a
      metric space, it suffices to prove $\O_K$ is
      sequentially compact.

      Let $(x_n)_{n = 1}^\infty$ be a sequence in
      $\O_K$, and fix $\pi \in \O_K$ a \gls{unif}.
      Since $\pi^i \O_K / \pi^{i + 1} \O_K \cong
      k$, $\O_K / \pi^i \O_K$ is finite for all
      $i$ ($\O_K \supseteq \pi\O_K \supseteq
      \cdots \supseteq \pi^i \O_K$). Since $\O_K /
      \pi\O_K$ is finite, there exists $a_1 \in \O_K
      / \pi\O_K$ and a subsequence $(x_n)_{n =
      1}^{\infty}$ such that $x_{1n} \equiv a
      \bmod \pi$ for all $n$.

      Since $\O_K / \pi^2 \O_K$ is finite, there
      exists $a_2 \in \O_K / \pi^2 \O_K$ and a
      subsequence $(x_{2n})_{n = 1}^{\infty}$ of
      $(x_{1n})_{n = 1}^{\infty}$ such that
      $x_{2n} \equiv a_2 \pmod \pi^2 \O_K$.
      Continuing, this, we obtain sequences
      $(x_{in})_{n = 1}^{\infty}$ for $i = 1, 2,
      \ldots$ such that
      \begin{enumerate}[(1)]
        \item $(x_{(i + 1)n})_{n = 1}^\infty$ is a
          subsequence of $(x_{in})_{n = 1}^{\infty}$
        \item For any $i$, there exists $a_i \in
          \O_K / \pi^i \O_K$ such that $x_{in}
          \equiv a_i \bmod \pi^i$ for all $n$.
      \end{enumerate}
      Then necessarily $a_i \equiv a_{i + 1}
      \bmod \pi^i$ for all $i$.

      Now choose $y_i = x_{ii}$. This defines a
      subsequence of $(x_n)_{n = 1}^{\infty}$.
      Moreover, $y_i \equiv a_i \equiv a_{i + 1}
      \equiv y_{i + 1} \bmod \pi^i$. Thus $y_i$ is
      Cauchy, hence converges by completeness.
      \qedhere
  \end{enumerate}
\end{proof}

\begin{example*}
  \phantom{}
  \begin{enumerate}[(i)]
    \item $\Qp$ is a \gls{lf}.
    \item $\Fp((t))$ is a \gls{lf}.
  \end{enumerate}
\end{example*}

More on \glspl{invlim}.

Let $(A_n)_{n = 1}$ a sequence of sets / groups /
rings and $\varphi_n : A_{n + 1} \to A_n$
homeomorphisms.

\begin{fcdefn}[Profinite topology]
  \glsnoundefn{proft}{profinite topology}{NA}%
  % Definition 7.3
  Assume $A_n$ is finite. The \emph{profinite
  topology} on $A \defeq \invlim n A_n$ is the
  weakest topology on $A$ such that $\theta_n : A
  \to A_n$ is continuous for all $n$, where $A_n$
  is equipped with the discrete topology.
\end{fcdefn}

\textbf{Fact:} $A = \invlim n A_n$ with the
profinite topology is compact, totally
disconnected and Hausdorff.

\begin{fcprop}[]
  % Proposition 7.4
  Assuming:
  - $K$ is a \gls{nonarch} \gls{lf}
  Then:
  under the isomorphism $\O_K \cong \invlim n
  \O_K / \pi^n \O_K$ ($\pi \in \O_K$ a
  \gls{unif}), the topology on $\O_K$ coincides
  with the \gls{proft}.
\end{fcprop}

\begin{proof}
  One checks that the sets
  \[
    B \defeq \{a + \pi^n \O_K \st n \in \Nbb_{\ge
    1}, a \in \O_K\}
  \]
  is a basis of open sets in both topologies.

  For $\absval|\bullet|$: clear.

  For \gls{proft}: $\O_K / \O_K / \pi^n \O_K$ is
  continuous if and only if $a + \pi^n \O_K$ is
  open for all $a \in \O_K$.
\end{proof}

Goal: Classify all \glspl{lf}.

\begin{fclemma}[]
  % Lemma 7.5
  Assuming:
  - $K$ is a \gls{nonarch} \gls{lf}
  - $L / K$ a finite extension
  Then:
  $L$ is a \gls{lf}.
\end{fclemma}

\begin{proof}
  \cref{thm_6_1} implies that $L$ is complete and
  \gls{dlyv}. Suffices to show $k_L \defeq \O_L /
  m_L$ is finite. Let $\alpha_1, \ldots, \alpha_n$
  be a basis for $L$ as a $K$ vector space.

  $\|\bullet\|_{\sup}$ (sup norm) equivalent to
  $|\bullet|_L$ implies that there exists $r > 0$
  such that
  \[
    \O_L \subseteq \{x \in L : \|x\|_{\sup} \le r\}
  .\]
  Take $a \in K$ such that $|a| \ge r$, then
  \[
    \O_L \subseteq \bigoplus_{i = 1}^n a\alpha_i
    \O_K \le L
  .\]
  Then $\O_L$ is finitely generated as a module
  over $\O_K$, hence $k_L$ is finitely generated
  over $k$.
\end{proof}

\begin{fcdefn}[Equal characteristic]
  \glsadjdefn{eqc}{equal characteristic}{field}%
  \glsadjdefn{mixc}{mixed characteristic}{field}%
  % Deifnition 7.6
  A \gls{nonarch} \gls{valf} $(K,
  \absval|\bullet|)$ has \emph{equal
  characteristic} if $\char(K) = \char(k)$.
  Otherwise it has \emph{mixed characteristic}.
\end{fcdefn}

\begin{example*}
  $\Qp$ has \gls{mixc}.
\end{example*}

\begin{fcthm}[]
  Assuming:
  - $K$ is a \gls{nonarch} \gls{lf} of \gls{eqc}
  $p > 0$
  Then:
  $K \cong \Fbb_{p^n}((t))$ for some $n \ge 1$.
\end{fcthm}

\begin{proof}
  $K$ complete \gls{dlyv}, $\char K > 0$.
  Moreover, $k \cong \Fbb_{p^n}$ is finite, hence
  perfect.

  By \cref{thm_5_7}, $K \cong \Fbb_{p^n}((t))$.
\end{proof}

\begin{fclemma}[]
  \label{lemma_7_8}
  % Lemma 7.8
  Assuming:
  - $K$ a field
  Then:
  an \gls{absval} $\absval|\bullet|$ is
  \gls{nonarch} if and only if $\absval|n|$ is
  bounded for all $n \in \Zbb$.
\end{fclemma}

\begin{proof}
  \phantom{}
  \begin{enumerate}[$\Rightarrow$]
    \item[$\Rightarrow$] Since $|-1| = 1$, $|-n| =
      |n|$, it suffices to show that $|n|$ bounded
      for $n \ge 1$. Then note that
      \[
        |n| = |1 + 1 + \cdots + 1| \le 1
      .\]
    \item[$\Leftarrow$] Suppose $|n| \le B$ for
      all $n \in \Zbb$. Let $x, y \in K$ with $|x|
      \le |y|$. Then we have
      \begin{align*}
        |x + y|^m
        &= \left| \sum_{i = 0}^{m} {m \choose i}
        x^i y^{m - i} \right| \\
        &\le \sum_{i = 0}^{m} \left| {m \choose i}
        x^i y^{m - i} \right| \\
        &\le |y|^m B(m + 1)
      \end{align*}
      Taking $m$-th roots gives
      \[
        |x + y| \le |y| [B(m + 1)]^{\frac{1}{m}}
      .\]
      The right hand side tends to $|y|$ as $m \to
      \infty$, hence
      \[
        |x + y| \le |y| = \max(|x|, |y|)
        \qedhere
      .\]
  \end{enumerate}
\end{proof}
