diff --git a/inputs/lecture_08.tex b/inputs/lecture_08.tex
index 09efe3f..4288b7f 100644
--- a/inputs/lecture_08.tex
+++ b/inputs/lecture_08.tex
@@ -58,7 +58,7 @@
 \end{proof}
 
 \subsection{Parametrizations}
-\todo{choose better title}
+%\todo{choose better title}
 
 
 Let $\Gamma$ denote a collection of sets in some space.
@@ -109,9 +109,9 @@ where $X$ is a metrizable, usually second countable space.
     put $(y,x) \in \cU$ iff
     $x \in \bigcup \{V_n : y_n = 1\}$.
     $\cU$ is open.
-    Let $V = \bigcup \{V_n : V_n \subseteq V\}$.
-    Pick $y \in 2^\omega$
-    and let $y_n = 1$ iff $V_n \subseteq  V$.
+    For any $V \overset{\text{open}}{\subseteq} X$,
+    define $y \in 2^\omega$
+    by $y_n = 1$ iff $V_n \subseteq  V$.
     Then $\cU_y = V$.
 
 
@@ -125,7 +125,7 @@ where $X$ is a metrizable, usually second countable space.
     Recall that $\eta_1 \le  \eta_2 \implies \Pi^0_{\eta_1}(X) \subseteq \Pi^0_{\eta_2}(X)$.
 
     Note that if $A = \bigcup_n A_n$,  with $A_n \in \Pi^0_{\eta_n}(X)$
-    some $\eta_n < \xi$,
+    for some $\eta_n < \xi$,
     we also have
     $A = \bigcup_n A_n'$ with  $A'_n \in \Pi^0_{\xi_n}(X)$.
 
@@ -146,7 +146,7 @@ where $X$ is a metrizable, usually second countable space.
     Since $2^{\omega}$ embeds
     into any uncountable polish space $Y$
     such that the image is closed,
-    we can  $2^{\omega}$ by $Y$
+    we can replace $2^{\omega}$ by $Y$
     in the statement of the theorem.%
     \footnote{By definition of the subspace topology
     and transfinite induction, $\Sigma^0_\xi(Y)\defon{2^\omega} = \Sigma^0_\xi(2^\omega)$.}
diff --git a/inputs/tutorial_12b.tex b/inputs/tutorial_12b.tex
index acf7793..dca0440 100644
--- a/inputs/tutorial_12b.tex
+++ b/inputs/tutorial_12b.tex
@@ -10,19 +10,18 @@ Let $x = (0)$ and  $y = (\delta_{0,i})_{i \in \Z}$.
 Let $t_n \to \infty$.
 Then $t_n y \to (0) = t_n x$.
 
-The skew shift flow is distal:
-This is tedious but probably not too hard.
-
-The skew shift flow is not equicontinuous:
+% The skew shift flow is distal:
+% This is tedious but probably not too hard.
+% 
+% The skew shift flow is not equicontinuous:
 
 
 
 
 
 
-\subsection{Sheet 11}
 
-We did \yaref{fact:isometriciffequicontinuous}.
+\begin{refproof}{fact:isometriciffequicontinuous}.
 
 $d$ and $d'(x,y) \coloneqq  \sup_{t \in T} d(tx,ty)$
 induce the same topology.
@@ -32,3 +31,4 @@ $\tau \subseteq  \tau'$ easy,
 $\tau' \subseteq \tau'$ : use equicontinuity.
 
 
+\end{refproof}