tutorial 10
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@ -61,7 +61,9 @@ However the converse of this does not hold.
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Then $\{x_0\}$ is dense in $X$, but $X$ is not second countable.
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\end{example}
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\begin{example}[Sorgenfrey line]
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\todo{Counterexamples in Topology}
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Consider $\R$ with the topology given by the basis $\{[a,b) : a,b \in \R\}$.
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This is T3, but not second countable
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and not metrizable.
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\end{example}
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\begin{fact}
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@ -98,6 +100,7 @@ However the converse of this does not hold.
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then $X$ is metrisable.
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\end{theorem}
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\begin{absolutelynopagebreak}
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\begin{fact}
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If $X$ is a compact Hausdorff space,
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the following are equivalent:
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@ -107,6 +110,7 @@ However the converse of this does not hold.
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\item $X$ is second countable.
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\end{itemize}
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\end{fact}
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\end{absolutelynopagebreak}
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\subsection{Some facts about polish spaces}
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@ -175,6 +179,7 @@ suffices to show that open balls in one metric are unions of open balls in the o
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\end{proof}
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\begin{definition}[Our favourite Polish spaces]
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\leavevmode
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\begin{itemize}
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\item $2^{\omega}$ is called the \vocab{Cantor set}.
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(Consider $2$ with the discrete topology)
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@ -185,7 +190,7 @@ suffices to show that open balls in one metric are unions of open balls in the o
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\end{itemize}
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\end{definition}
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\begin{proposition}
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Let $X$ be a separable, metrisable topological space.
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Let $X$ be a separable, metrisable topological space\footnote{e.g.~Polish, but we don't need completeness.}.
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Then $X$ topologically embeds into the
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\vocab{Hilbert cube},
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i.e. there is an injective $f: X \hookrightarrow [0,1]^{\omega}$
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@ -227,7 +232,9 @@ suffices to show that open balls in one metric are unions of open balls in the o
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$f^{-1}$ is continuous.
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\end{claim}
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\begin{subproof}
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\todo{Exercise!}
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Consider $B_{\epsilon}(x_n) \subseteq X$ for some $n \in \N$, $\epsilon > 0$.
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Then $f(U) = f(X) \cap [0,1]^{n} \times [0,\epsilon) \times [0,1]^{\omega}$
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is open\footnote{as a subset of $f(X)$!}.
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\end{subproof}
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\end{proof}
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\begin{proposition}
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@ -13,7 +13,6 @@
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\begin{proof}
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Let $C \subseteq X$ be closed.
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Let $U_{\frac{1}{n}} \coloneqq \{x | d(x, C) < \frac{1}{n}\}$.
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\todo{Exercise}
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Clearly $C \subseteq \bigcap U_{\frac{1}{n}}$.
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Let $x \in \bigcap U_{\frac{1}{n}}$.
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Then $\forall n .~ \exists x_n\in C.~d(x,x_n) < \frac{1}{n}$.
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@ -1,6 +1,6 @@
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\lecture{03}{2023-10-17}{Embedding of the cantor space into polish spaces}
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% ? \subsection{Trees} TODO
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\subsection{Trees}
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\begin{notation}
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@ -255,9 +255,7 @@
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[To be continued]
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\phantom\qedhere
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\end{refproof}
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@ -1,6 +1,7 @@
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\lecture{04}{2023-10-20}{}
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\begin{remark}
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Some of $F_s$ might be empty.
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Some of the $F_s$ might be empty.
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\end{remark}
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\begin{refproof}{thm:bairetopolish}
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@ -29,7 +30,6 @@
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\begin{refproof}{thm:bairetopolish:c1}
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Let $x_n$ be a series in $D$
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converging to $x$ in $\cN$.
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Then $x \in \cN$.
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\begin{claim}
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$(f(x_n))$ is Cauchy.
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\end{claim}
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@ -40,12 +40,11 @@
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$x_m\defon{N} = x\defon{N}$.
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Then for all $m, n \ge M$,
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we have that $f(x_m), f(x_n) \in F_{x\defon{N}}$.
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So $d(f(x_m), f(x_n)) < \epsilon$
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we have that $(f(x_n))$ is Cauchy.
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So $d(f(x_m), f(x_n)) < \epsilon$, i.e.~$(f(x_n))$ is Cauchy.
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\end{subproof}
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Since $(X,d)$ is complete,
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there exists $y = \lim_n f(x_n)$.
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Since for all $m \ge M$, $f(x_m) \in F_{x\defon{N}}$,
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we get that $y \in \overline{F_{x\defon{N}}}$.
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@ -54,7 +53,6 @@
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Hence
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\[y \in \bigcap_{n} \overline{F_{x\defon{n}}} = \bigcap_{n} F_{x\defon{n}},\]
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i.e.~$y \in D$ and $y = f(x)$.
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\end{subproof}
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\end{refproof}
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We extend $f$ to $g\colon\cN \to X$
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@ -70,7 +68,7 @@
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(i.e.~$r = \id$ on $D$ and $r$ is a continuous surjection).
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Then $g \coloneqq f \circ r$.
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To construct $r$, we will define by induction
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To construct $r$, we will define
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$\phi\colon \N^{<\N} \to S$ by induction on the length
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such that
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\begin{itemize}
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@ -13,9 +13,9 @@
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\begin{fact}
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\begin{itemize}
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\item Let $X$ be a topological space.
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Then $X$ 2nd countable $\implies$ X separable.
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Then $X$ \nth{2} countable $\implies$ X separable.
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\item If $X$ is a metric space and separable,
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then $X$ is 2nd countable.
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then $X$ is \nth{2} countable.
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\end{itemize}
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\end{fact}
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\begin{proof}
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@ -32,18 +32,18 @@
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\begin{fact}
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Let $X$ be a metric space.
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If $X$ is Lindelöf,
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then it is 2nd countable.
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then it is \nth{2} countable.
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\end{fact}
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\begin{proof}
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For all $q \in \Q$
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Consider the cover $B_q(x), x \in X$
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consider the cover $B_q(x), x \in X$
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and choose a countable subcover.
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The union of these subcovers is
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a countable base.
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\end{proof}
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\begin{fact}
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Let $X$ be a topological space.
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If $X$ is 2nd countable,
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If $X$ is \nth{2} countable,
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then it is Lindelöff.
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\end{fact}
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\begin{proof}
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@ -60,7 +60,7 @@
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\end{proof}
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\begin{remark}
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For metric spaces the notions
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of being 2nd countable, separable
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of being \nth{2} countable, separable
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and Lindelöf coincide.
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In arbitrary topological spaces,
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@ -50,3 +50,13 @@
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\RequirePackage{mkessler-mathfixes} % Load this last since it renews behaviour
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\DeclareMathOperator{\inter}{int} % interior
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\newcommand{\defon}[1]{|_{#1}} % TODO
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\RequirePackage[super]{nth}
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% TODO MOVE
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% https://tex.stackexchange.com/a/94702
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\newenvironment{absolutelynopagebreak}
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{\par\nobreak\vfil\penalty0\vfilneg
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\vtop\bgroup}
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{\par\xdef\tpd{\the\prevdepth}\egroup
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\prevdepth=\tpd}
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\input{inputs/tutorial_07}
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\input{inputs/tutorial_08}
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\input{inputs/tutorial_09}
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\input{inputs/tutorial_10}
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\section{Facts}
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\input{inputs/facts}
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