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