tutorial 10

inputs/lecture_01.tex

@@ -61,7 +61,9 @@ However the converse of this does not hold.
     Then $\{x_0\}$ is dense in $X$, but $X$ is not second countable.
 \end{example}
 \begin{example}[Sorgenfrey line]
-    \todo{Counterexamples in Topology}
+    Consider $\R$ with the topology given by the basis $\{[a,b) : a,b \in \R\}$.
+    This is T3, but not second countable
+    and not metrizable.
 \end{example}
 
 \begin{fact}
@@ -98,6 +100,7 @@ However the converse of this does not hold.
     then $X$ is metrisable.
 \end{theorem}
 
+\begin{absolutelynopagebreak}
 \begin{fact}
     If $X$ is a compact Hausdorff space,
     the following are equivalent:
@@ -107,6 +110,7 @@ However the converse of this does not hold.
         \item $X$ is second countable.
     \end{itemize}
 \end{fact}
+\end{absolutelynopagebreak}
 
 \subsection{Some facts about polish spaces}
 
@@ -175,6 +179,7 @@ suffices to show that open balls in one metric are unions of open balls in the o
 \end{proof}
 
 \begin{definition}[Our favourite Polish spaces]
+    \leavevmode
     \begin{itemize}
         \item $2^{\omega}$ is called the \vocab{Cantor set}.
             (Consider $2$ with the discrete topology)
@@ -185,7 +190,7 @@ suffices to show that open balls in one metric are unions of open balls in the o
     \end{itemize}
 \end{definition}
 \begin{proposition}
-    Let $X$ be a separable, metrisable topological space.
+    Let $X$ be a separable, metrisable topological space\footnote{e.g.~Polish, but we don't need completeness.}.
     Then $X$ topologically embeds into the
     \vocab{Hilbert cube},
     i.e. there is an injective $f: X \hookrightarrow [0,1]^{\omega}$ 
@@ -227,7 +232,9 @@ suffices to show that open balls in one metric are unions of open balls in the o
         $f^{-1}$ is continuous.
     \end{claim}
     \begin{subproof}
-        \todo{Exercise!}
+        Consider $B_{\epsilon}(x_n) \subseteq X$ for some $n \in \N$, $\epsilon > 0$.
+        Then $f(U) = f(X) \cap [0,1]^{n} \times [0,\epsilon) \times [0,1]^{\omega}$
+        is open\footnote{as a subset of $f(X)$!}.
     \end{subproof}
 \end{proof}
 \begin{proposition}

inputs/lecture_02.tex

@@ -13,7 +13,6 @@
  \begin{proof}
      Let $C \subseteq X$ be closed.
      Let $U_{\frac{1}{n}} \coloneqq \{x | d(x, C) < \frac{1}{n}\}$.
-     \todo{Exercise}
      Clearly $C \subseteq \bigcap U_{\frac{1}{n}}$.
      Let $x \in \bigcap U_{\frac{1}{n}}$.
      Then $\forall n .~ \exists x_n\in C.~d(x,x_n) < \frac{1}{n}$.
@@ -54,7 +53,7 @@
          then there exists a complete metric on $Y$.
      \end{claim}
      \begin{refproof}{psubspacegdelta:c1}
-         Let $Y = U$be open in $X$.
+         Let $Y = U$ be open in $X$.
          Consider the map
          \begin{IEEEeqnarray*}{rCl}
              f_U\colon U &\longrightarrow &

inputs/lecture_03.tex

@@ -1,6 +1,6 @@
 \lecture{03}{2023-10-17}{Embedding of the cantor space into polish spaces}
 
-% ? \subsection{Trees}  TODO
+\subsection{Trees}
 
 
 \begin{notation}
@@ -255,9 +255,7 @@
 
 
 
-    [To be continued]
     \phantom\qedhere
-
 \end{refproof}
 
 

inputs/lecture_04.tex

@@ -1,6 +1,7 @@
 \lecture{04}{2023-10-20}{}
+
 \begin{remark}
-        Some of $F_s$ might be empty.
+        Some of the $F_s$ might be empty.
 \end{remark}
 
 \begin{refproof}{thm:bairetopolish}
@@ -29,7 +30,6 @@
     \begin{refproof}{thm:bairetopolish:c1}
         Let $x_n$ be a series in $D$
         converging to $x$ in $\cN$.
-        Then $x \in \cN$.
         \begin{claim}
             $(f(x_n))$ is Cauchy.
         \end{claim}
@@ -40,21 +40,19 @@
             $x_m\defon{N} = x\defon{N}$.
             Then for all $m, n \ge M$,
             we have that $f(x_m), f(x_n) \in F_{x\defon{N}}$.
-            So $d(f(x_m), f(x_n)) < \epsilon$ 
+            So $d(f(x_m), f(x_n)) < \epsilon$, i.e.~$(f(x_n))$ is Cauchy.
-            we have that $(f(x_n))$ is Cauchy.
 
-            Since $(X,d)$ is complete,
-            there exists $y = \lim_n f(x_n)$.
-
-            Since for all $m \ge M$, $f(x_m) \in F_{x\defon{N}}$,
-            we get that $y \in  \overline{F_{x\defon{N}}}$.
-            
-            Note that for $N' > N$ by the same argument
-            we get $y \in \overline{F_{x\defon{N'}}}$.
-            Hence
-            \[y \in \bigcap_{n} \overline{F_{x\defon{n}}} = \bigcap_{n} F_{x\defon{n}},\]
-            i.e.~$y \in D$ and $y = f(x)$.
         \end{subproof}
+        Since $(X,d)$ is complete,
+        there exists $y = \lim_n f(x_n)$.
+        Since for all $m \ge M$, $f(x_m) \in F_{x\defon{N}}$,
+        we get that $y \in  \overline{F_{x\defon{N}}}$.
+        
+        Note that for $N' > N$ by the same argument
+        we get $y \in \overline{F_{x\defon{N'}}}$.
+        Hence
+        \[y \in \bigcap_{n} \overline{F_{x\defon{n}}} = \bigcap_{n} F_{x\defon{n}},\]
+        i.e.~$y \in D$ and $y = f(x)$.
     \end{refproof}
 
     We extend $f$ to $g\colon\cN \to X$ 
@@ -70,7 +68,7 @@
     (i.e.~$r = \id$ on $D$ and $r$ is a continuous surjection).
     Then $g \coloneqq f \circ r$.
     
-    To construct $r$, we will define by induction
+    To construct $r$, we will define 
     $\phi\colon \N^{<\N} \to  S$ by induction on the length
     such that
     \begin{itemize}

inputs/tutorial_01.tex

@@ -13,9 +13,9 @@
 \begin{fact}
     \begin{itemize}
         \item Let $X$ be a topological space.
-            Then $X$ 2nd countable $\implies$ X separable.
+            Then $X$ \nth{2} countable $\implies$ X separable.
         \item If $X$ is a metric space and separable,
-            then $X$ is 2nd countable.
+            then $X$ is \nth{2} countable.
     \end{itemize}
 \end{fact}
 \begin{proof}
@@ -32,18 +32,18 @@
 \begin{fact}
     Let $X$ be a metric space.
     If $X$ is Lindelöf,
-    then it is 2nd countable.
+    then it is \nth{2} countable.
 \end{fact}
 \begin{proof}
     For all $q \in \Q$
-    Consider the cover $B_q(x), x \in X$
+    consider the cover $B_q(x), x \in X$
     and choose a countable subcover.
     The union of these subcovers is
     a countable base.
 \end{proof}
 \begin{fact}
     Let $X$ be a topological space.
-    If $X$ is 2nd countable,
+    If $X$ is \nth{2} countable,
     then it is Lindelöff.
 \end{fact}
 \begin{proof}
@@ -60,7 +60,7 @@
 \end{proof}
 \begin{remark}
     For metric spaces the notions
-    of being 2nd countable, separable
+    of being \nth{2} countable, separable
     and Lindelöf coincide.
 
     In arbitrary topological spaces,

jrpie-math.sty

@@ -50,3 +50,13 @@
 \RequirePackage{mkessler-mathfixes} % Load this last since it renews behaviour
 \DeclareMathOperator{\inter}{int} % interior
 \newcommand{\defon}[1]{|_{#1}} % TODO
+
+\RequirePackage[super]{nth}
+
+% TODO MOVE
+% https://tex.stackexchange.com/a/94702
+\newenvironment{absolutelynopagebreak}
+  {\par\nobreak\vfil\penalty0\vfilneg
+   \vtop\bgroup}
+  {\par\xdef\tpd{\the\prevdepth}\egroup
+   \prevdepth=\tpd}

logic3.tex

@@ -61,6 +61,7 @@
 \input{inputs/tutorial_07}
 \input{inputs/tutorial_08}
 \input{inputs/tutorial_09}
+\input{inputs/tutorial_10}
 
 \section{Facts}
 \input{inputs/facts}