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Another way to think of this: let $(X, d)$ be
complete. If $P_j$ is a property of a point and:
\begin{enumerate}[(1)]
  \item $P_j$ is stable, i.e. given $x$ with
    property $P_j$, there exists $\delta > 0$ such
    that $y \in B(x, \delta) \implies y$ has
    property $P_j$.
  \item Not $P_j$ is unstable, i.e. given $x \in
    X$ and $\delta > 0$, there exists $y \in B(x,
    \delta)$ such that $y$ has propery $P_j$.
\end{enumerate}
Then there exists $x^*$ with property $P_j$ for
all $j$ (set $F_j$ to be the points not satisfying
$P_j$).

\begin{flashcard}[proof-of-baire-category]
\prompt{Proof of \nameref{baire_category}?}

\begin{proof}[Proof of \nameref{baire_category}]
  \cloze{Choose $x_0 \in X$ and take $\delta_0 = 1$. We
  define $x_j, \delta_j$ revursively (i.e. by
  induction). If $x_j \in X$ and $\delta_j > 0$
  constructed, then we know that there exists
  $x_{j + 1} \notin F_{j + 1}$ such that $d(x_{j +
  1}, x_j) < \frac{\delta_j}{4}$ (by the first
  hypothesis) and $0 <
  \delta_{j + 1} < \frac{\delta_j}{4}$ such that
  $y \in B(x_{j + 1}, \delta_{j + 1}) \implies y
  \notin F_{j + 1}$ (by the second hypothesis).
  Now $\delta_{j + k} \le 4^{-k} \delta_j$ so
  \[ \sum_{k = 0}^\infty d(x_{j + k}, x_{j + k + 1})
  \le \sum_{r = j + 1}^\infty 4^{-r} \delta_j \le
  \frac{\delta_j}{2} \]
  so $(x_j)$ is Cauchy and $x_j \to x$ with
  $d(x_j, x) < \frac{\delta_j}{2}$. Then $x \notin
  F_j$ for every $j$.}
\end{proof}
\end{flashcard}

The result comes with some unsatisfactory but
traditional nomenclature.

\begin{flashcard}[first-category-defn]
\begin{definition*}[First category]
  \cloze{$E$ is of first category if $E \subseteq
  \bigcup_{j = 1}^\infty F_j$ with $F_j$ closed
  and ``nowhere dense'', i.e. with empty
  interior.}
\end{definition*}
\end{flashcard}

\vspace{-1em}
Following immediate consequences are much used:
\begin{enumerate}[(1)]
  \item If $(X, d)$ is complete and $E$ is first
    category then $E \neq X$.
  \item $(X, d)$ complete. The countable union of
    sets of first category is of first category.
    \begin{proof}
      If $\mathcal{F}_j$ is a countable collection
      of nowhere dense closed sets then
      $\bigcup_{j = 1}^\infty \mathcal{F}_j$ is
      the countable union of nowhere dense closed
      sets.
    \end{proof}
\end{enumerate}

\begin{flashcard}[isolated-point-defn]
\begin{definition*}[Isolated point]
  \cloze{Let $(X, d)$ be a metric space. A point $x$ is
  \emph{isolated} if there exists a $\delta > 0$
  such that $B(x, \delta) = \{x\}$.}
\end{definition*}
\end{flashcard}

\begin{flashcard}[complete-no-isolated-is-countable]
\begin{theorem*}
  A complete metric space without \cloze{isolated
  points is uncountable.}
\end{theorem*}

\begin{proof}
  \cloze{Suppose $(X, d)$ is complete and countable. Let
  $X = \{x_1, x_2, \ldots\}$ and no $x_j$
  isolated. Then $\{x_j\}$ is closed and $\{x_j\}$
  is not open since $B(x_j, \delta) \neq \{x_j\}$.
  So $\Int\{x_j\} = \emptyset$. So $X \neq \bigcup
  \{x_j\}$, a contradiction.}
\end{proof}
\end{flashcard}

\begin{corollary*}
  $\RR$ is uncountable.
\end{corollary*}

\vspace{-1em}
Banach's proof that nowhere differentiable
functions exist:

Work on $C([0, 1])$ with uniform norm.

\begin{flashcard}[anywhere-diff-first-category-thm]
\begin{theorem*}[Baire]
  The set of anywhere differentiable continuous
  functions \cloze{is of first category.}
\end{theorem*}

\begin{proof}
  \cloze{Banach shows that
  \[ E_n = \{f \in C([0, 1]) : \exists x \in [0,
  1], |f(x) - f(y)| \le n|x - y| ~\forall y \in
  [0, 1]\} \]
  is closed and nowhere dense. If we can prove
  this then $\bigcup E_j$ is first category. Claim
  is that if $f \notin \bigcup_j E_j$ then $f$ is nowhere
  differentiable.

  Subproof: Suppose $f$ is differentiable at $x$.
  Then there exists $\delta > 0$ such that
  \[ \left| \frac{f(x) - f(y)}{x - y} - f'(x)
  \right| \le 1 \qquad \forall |y| \le \delta \]
  so $|f(x) - f(y)| \le (|f'(x)| + 1)|x - y|$. If
  $|y| \ge \delta$, then
  \[ \left| \frac{f(x) - f(y)}{x - y} \right| \le
  \frac{2\|f\|_\infty}{\delta} \]
  so
  \[ |f(x) - f(y)| \le
  \frac{2\|f\|_\infty}{\delta} |x - y| .\]
  Thus $f \in E_n$ for some $n$.

  Thus all we have to do is to show that $E_n$ is
  closed and nowhere dense (since $C([0, 1])$ is
  complete).

  Closed: Suppose $f_m \in E_n$, $f_m \to f$
  uniformly. Since $f_m \in E_n$, there exists
  $x_m$ such that $|f_m(x_m) - f(y)| \le n|x_m -
  y|$ for all $y \in [0, 1]$. So there exists
  $x^*$ and $m(j) \to \infty$ such that $x_{m(j)}
  \to x^*$. By extracting to a subsequence we may
  assume $x_m \to x^*$. Then:
  \begin{align*}
    |f(x^*) - f(y)|
    &\le |f(x^*) - f(x_m)| + |f(x_m) - f_m(x_m)| +
    |f_m(x_m) - f_m(y)| + |f_m(y) - f(y)| \\
    &\le |f(x^*) - f(x_m)| + \|f - f\|_\infty +
    n|x - y| + \|f_m - f\|_\infty \\
    &\to 0 + 0 + n|x^* - y| + 0
  \end{align*}
  so $|f(x^*) - f(y)| \le n|x^* - y|$.

  Now we show that $E_n$ is nowhere dense. So we
  must show if $f \in C([0, 1])$ and $\delta > 0$
  then there exists $g$ such that $\|f -
  g\|_\infty < \delta$ and $g \notin E_n$. Words
  of wisdom: trying to construct a ``nasty
  function'' from $g$ is hard -- if $g$ happened
  to already be nasty in a weird way, then we might
  accidentally make a ``nice'' function out of it.
  So the first step is we find a nearby ``nice''
  function, and uses this to construct a
  ``nasty'' one. Let $f \in C([0, 1])$. By
  Weierstrass approximation theorem, there exists
  a polynomial $P$ such that $\|f - P\|_\infty <
  \frac{\delta}{4}$. Now look at $g(x) = f(x) +
  \frac{\delta}{4} \sin Nx$ with $N$ to be chosen
  later. Observe that $\|g - f\|_\infty < \delta$
  and we must show that for large $N$, $g - f
  \notin E_n$. Also observe that $P$ is
  continuously differentiable, so $|P'(t)| \le M$
  for some $M$. Suppose $x \in [0, 1]$. Without
  loss of generality $x \in [0, 1)$ (just
  reflect). Consider $N$ large, and
  \[ \left( r\pi + \frac{\pi}{2} \right)
  \frac{1}{N}, \left( r\pi + \frac{3\pi}{2}
  \right) \frac{1}{N} \]
  with $x \le \left( r\pi + \frac{\pi}{2} \right)
  \frac{1}{N}$ and $\left(\frac{r\pi}{2} +
  \frac{\pi}{2} \right) \frac{1}{N}
  < x + \frac{2\pi}{N}$. Then we have that:
  \[ |g(x) - g(y)| \ge |\sin Nx - \sin Ny| - |P(x)
  - P(y)| \ge |\sin N x - \sin N y| - M|x - y| .\]
  So choosing $y = \left( r\pi + \frac{\pi}{2}
  \right) \frac{1}{N}$ or $y = \left( r\pi +
  \frac{3\pi}{2} \right) \frac{1}{N}$ and $N$
  large enough, we get
  \[ |g(x) - g(y)| > (n + M)|x - y| - M|x - y| \ge
  n|x - y| . \qedhere \]}
\end{proof}
\end{flashcard}