some small changes

inputs/lecture_02.tex

@@ -91,7 +91,6 @@
             because $\tilde{f_U}$ is continuous.
             It is closed}{} in $X \times \R$ \gist{because
             $\tilde{f_U} \to \infty$ for $d(x, U^c) \to 0$}{}.
-            \todo{Make this precise}
 
             Therefore we identified $U$ with a closed subspace of
             the Polish space $(X \times  \R, d_1)$.

inputs/lecture_11.tex

@@ -96,7 +96,7 @@
     Polish space and $\cC = 2^\omega$ the Cantor space,
     then they are Borel isomorphic.
     There is $2^\omega \hookrightarrow X$ Borel
-    (continuous wrt.~to the topology of $X$)
+    (continuous wrt.~the topology of $X$)
     On the other hand
     \[
     X \hookrightarrow\cN \overset{\text{continuous embedding\footnotemark}}{\hookrightarrow}\cC

inputs/tutorial_05.tex

@@ -119,5 +119,3 @@
         Then $f(B) = U_{(x_0,\ldots,x_{n-1})}$ is open,
         hence $f$ is open.
 \end{enumerate}
-
-

inputs/tutorial_08.tex

@@ -95,7 +95,7 @@ for some $B_i \in \cB(Y_i)$.
         Then $\bigcap A_i$ is the image of  $D$
         under $Z \xrightarrow{(y_n) \mapsto f_0(y_0)} X$.
         
-        \paragraph{Other solution}
+        \emph{Other solution:}
 
         Let $F_n \subseteq X \times  \cN$ be closed,
         and $C \subseteq  X \times \cN^{\N}$ defined by