w23-logic-2/inputs/lecture_02.tex

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\lecture{02}{2023-10-19}{Topology on $\R$}
\begin{definition}
A set $O \subseteq \R$ is called \vocab{open} in $\R$ iff it is the union
of a set of open intervals.
A set $A \subseteq \R$ is called \vocab{closed} in $\R$
iff it is the complement of an open set.
\end{definition}
\begin{remark}
\begin{itemize}
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\item If $\emptyset \neq O \overset{\text{open}}{\subseteq} \R$
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then $O \sim \R$.
\item If $O \subseteq \R$
is open, then $O$ is the union of open intervals
with rational endpoints, since $\Q$ is dense.
\end{itemize}
\end{remark}
\begin{remark}+
$\{O \subseteq \R\} \sim 2^{\aleph_0} < \cP(\R)$.
\end{remark}
\begin{definition}
We call $x \in \R$
an \vocab{accumulation point}
of $A$ iff for all $a < x < b$
there is some $y \in A$, $y \in (a,b)$, $y \neq x$.
We write \vocab{$A'$} for the set of all accumulation points of $A$.
\end{definition}
\begin{example}
$\{\frac{1}{n+1} | n \in \N\}' = \{0\}$.
\end{example}
\begin{lemma}
\label{lem:closedaccumulation}
A set $A \subseteq \R$ is closed iff $A' \subseteq A$.
\end{lemma}
\begin{refproof}{lem:closedaccumulation}
``$\implies$''
Let $A$ be closed. Suppose that $x \in A' \setminus A$.
Then there exists $(a,b) \ni x$
disjoint from $A$. Hence $x \not\in A' \lightning$
``$\impliedby$''
Suppose $A' \subseteq A$.
\begin{claim}
$A \subseteq \R$ is closed iff all Cauchy sequences
in $A$ converge in $A$.
\end{claim}
\begin{subproof}
Let $A$ be closed and $\langle x_n : n \in \omega \rangle$
a Cauchy sequence in $A$.
Suppose that $x = \lim_{n \to \infty} x_n \not\in A$.
Then there is $(a,b) \ni x$ disjoint from $A$.
However $x_n \in (a,b)$ for almost all $n \in \omega$ $\lightning$
On the other hand let $A$ not be closed.
Then there exists a witness $x \in \R \setminus A$
such that $A \cap (a,b) \neq \emptyset$ for all $(a,b) \ni x$.
In particular,
we may pick $x_n \in (x - \frac{1}{n+1}, x + \frac{1}{n+1}) \cap A$
for all $n < \omega$.
\end{subproof}
Now if $A' \subseteq A$ and $A$ were not closed,
there would be some Cauchy-sequence $(x_n)$
in $A$ such that $\lim_{n \to \infty} x_n \not\in A$.
But then $x \in A' \subseteq A \lightning$.
\end{refproof}
\begin{definition}
$P \subseteq \R$ is called \vocab{perfect}
iff $P \neq \emptyset$ and $P = P'$.
\end{definition}
We want to prove two things:
\begin{itemize}
\item If $P$ is perfect, then $P \sim \R$.
\item If $A$ is closed and uncountable then $A$ has a perfect subset.
In particular $A \sim \R$.
\end{itemize}
\begin{lemma}
Let $P \subseteq \R$ be perfect.
Then $P \sim \R$.
\end{lemma}
\begin{proof}
It suffices to find an injection $f\colon \R \hookrightarrow P$.
We have $\underbrace{\{0,1\}^{\omega}}_{\text{infinite 0-1-sequences}} \sim \R$,
hence it suffices to construct $f\colon \{0,1\}^\omega\hookrightarrow P$.
In order to do that, we are going to construct some
$g\colon \underbrace{\{0,1\}^{<\omega}}_{\text{finite 0-1-sequences}} \to P$
with certain properties
by recursion on the length of $s \in \{0,1\}^{<\omega}$.
Let $g(\emptyset)$ be any point in $P$.
Suppose that $g(s) \in P$ has been chosen for all $s$ of
length $\le n$.
For each $s \in \{0,1\}^{n}$ pick
$g(s) \in (a_s, b_s)$
such that $(a_s, b_s) \cap (a_{s'}, b_{s'}) = \emptyset$
for all $s, s'$ of length $n$,
$b_s - a_s \le \frac{1}{n^3}$
and
$(a_{s\defon{n-1}}, b_{s\defon{n-1}}) \subseteq (a_s, b_s)$.
For each such $s$ pick $x_s \in (a_s, b_s) \cap P$
with $x_s \neq f(s)$.
This is possible since $P \subseteq P'$.
Now set $g(s\concat 0) \coloneqq g(s)$
and $g(s \concat 1) \coloneqq g(x_s)$.
This finishes the construction.
If $t \in \{0,1\}^{\omega}$,
then $(g(t\defon{n}), n < \omega)$ is a Cauchy sequence.
By $P' \subseteq P$ we get that this sequence
converges to a point in $P$.
Define $f(t)$ to be this point.
If $t \neq t' \in \{0,1\}^{\omega}$,
then there is some $n$ such that $t\defon{n} \neq t'\defon{n}$,
hence $f(t) \in [a_{t\defon{n}}, b_{t\defon{n}}]$
and $f(t') \in [a_{t\defon{n}}, b_{t\defon{n}}]$
which are disjoint.
Thus $f(t) \neq f(t')$, i.e.~$f$ is injective.
\end{proof}