%! TEX root = TA.tex
% vim: tw=50
% 24/02/2024 12PM

\subsubsection*{Liouville's Transcendentals}

We will always be working over the field $\RR$ in
this section.

We call a number \emph{algebraic} if it is the
root of a polynomial with integer coefficients
(and transcendental otherwise).
Cantor's proof of the uncountability of $\RR$ also
implies that transcendentals exist.

Let $\mathcal{E}_n$ be the collection of all real
roots of polynomials $P(t) = \sum_{j = 0}^m a_j
t^j$ with $a_m \neq 0$, $m \ge 1$, $a_j \in \ZZ$,
$\forall 0 \le j \le n$, $|a_j| \le m$.

The collection of such polynomials is finite and
such polynomial only has finitely many roots, so
$\mathcal{E}_n$ is finite, so $\bigcup_n
\mathcal{E}_n$ the collection of algebraic numbers
is countable. The reals are not countable, so not
all reals are algebraic. Cantor's proof is
non-constructive.

Liouville gave another proof and his proof
exhibits explicit transcendentals.

The proof depends on the following lemma:

\begin{flashcard}[liouville-irrational-alg-approx-lemma]
\begin{lemma*}[Liouville's theorem on approximating
  irrational algebraic numbers]
  \cloze{If $\xi$ is an irrational root of a
  polynomial $P$
  of degree $n$ or less with integer coefficients
  then there exists an $A$ such that
  \[ \left| \xi - \frac{p}{q} \right| \ge
  \frac{A}{q^n} \]
  whenever $p, q$ are integers with $q \ge 1$
  (note that $A$ depends on $\xi$).}
\end{lemma*}

\begin{proof}
  \cloze{First remark is that it is sufficient to prove
  that there exists $N$ and $B > 0$ such that
  \[ \left| \xi - \frac{p}{q} \right| \ge
  \frac{B}{q^n} \qquad \forall p, q \in \ZZ, q \ge
  N .\]
  Let $P$ be the polynomial in question. $P$ has
  only finitely many roots so we can find $\eta >
  0$ such that $\eta$ is the only root of $P$ in
  $[\xi - \eta, \xi + \eta]$. Since $P'$ is
  continuous on $[\xi - n, \xi + n]$, it is
  bounded, i.e. $\exists M$ such that $|P'(t)| \le
  M$ for all $t \in [\xi - \eta, \xi + \eta]$, so
  if $\frac{p}{q} \in [\xi - \eta, \xi + \eta]$
  then
  \[ \left|P(\xi) - P \left( \frac{p}{q}
  \right)\right| \le M \left| \xi - \frac{p}{q} \right| \]
  (MVT). $P(\xi) = 0$, $0 \neq q^n P \left(
  \frac{p}{q} \right) \in \ZZ$. So $M \left| \xi -
  \frac{p}{q} \right| \ge \frac{1}{q^n}$, so
  $\left| \xi - \frac{p}{q} \right| \ge
  \frac{M^{-1}}{q^n}$. If $\frac{p}{q} \in [\xi -
  \eta, \xi + \eta]$, then $\left| \xi - \frac{p}{q}
  \right| \ge \eta \ge \frac{M^{-1}}{q^n}$ for $n$
  large.}
\end{proof}
\end{flashcard}

Liouville's number is an example of an application
of this.
\[ \xi = \sum_{n = 0}^\infty 10^{-n!} \]
If $q_m = 10^{m!}$, $p_m = 10^{m!} \sum_{r = 0}^m
10^{-r!}$. Then
\begin{align*}
  \left| \left( \xi - \sum_{r = 0}^\infty 10^{-n!}
  \right) - \frac{p_m}{q_m} \right|
  &= \sum_{m + 1}^\infty 10^{-n!} \\
  &\le 10^{-(m + 1)!} \left( 1 + \frac{1}{10} +
  \frac{1}{10^2} + \cdots \right) \\
  &\le 2 \cdot 10^{-(m + 1)!}
\end{align*}
$\xi$ irrational since it has a non-periodic
decimal expansion. Then note $\frac{2 \cdot
10^{-(m + 1)!}}{q_m^k} \to 0$ as $m \to \infty$.
So $\xi$ is transcendental by the above lemma.

If we take $\eps_r \in \{1, 2\}$, then we have
that
\[ \sum_r \eps_r 10^{-r!} \]
is transcendental by the same argument. This gives
uncountably many such examples.

\newpage
\section{Baire's Category Theorem}

\begin{flashcard}[baire-category-thm]
\begin{theorem*}[Baire category theorem]
  \label{baire_category}
  \cloze{If $(X, d)$ is a \emph{complete} metric space
  and $u_1, u_2, u_3, \ldots$ are open in $X$ and are such
  that $u_j^c$ has empty interior (i.e. there does
  not exist $V$ open, $V \neq \emptyset$ such that
  $V \subseteq u_j^c$), then $\bigcap_{j =
  1}^\infty u_j \neq \emptyset$.}
\end{theorem*}
\end{flashcard}

The following is the complement (so equivalent):

\begin{theorem*}
  If $F_1, F_2$ are closed and have empty interior
  then $\bigcup F_j \neq X$ (i.e. there exists $x
  \notin \bigcup_{j = 1}^\infty F_x$).
\end{theorem*}