some small changes

inputs/lecture_17.tex

@@ -24,7 +24,7 @@ $X^{X}$ is a compact Hausdorff space.
             is continuous:
 
             Consider $\{f : f \circ f_0 \in U_{\epsilon}(x,y)\}$.
-            We have $ff_0 \in U_{\epsilon}(x,y)$
+            We have $f \circ f_0 \in U_{\epsilon}(x,y)$
             iff $f \in U_\epsilon(x,f_0(y))$.
         \item Fix $x_0 \in X$.
             Then  $f \mapsto f(x_0)$ is continuous.
@@ -43,12 +43,15 @@ and take the closure in  $X^X$.
 \end{definition}
 $E(X,T)$ is compact and Hausdorff,
 since $X^X$ has these properties.
+% TODO THINK ABOUT THIS
 
+\gist{
     Properties of $(X,T)$ translate to properties of $E(X,T)$:
     \begin{goal}
         We want to show that if  $(X,T)$ is distal,
         then $E(X,T)$ is a group.
     \end{goal}
+}{}
 
 \begin{proposition}
     $E(X,T)$ is a semigroup,
@@ -59,11 +62,14 @@ Properties of $(X,T)$ translate to properties of $E(X,T)$:
     Take $t \in T$. We want to show that $tG \subseteq G$,
     i.e.~for all $h \in G$ we have $th \in G$.
 
+    \gist{
         We have that $t^{-1}G$ is compact,
         since $t^{-1}$ is continuous
         and $G$ is compact.
+    }{$t^{-1}G$ is compact.}
+
+    It is $T \subseteq t^{-1}G$ since $T \ni s = t^{-1}\underbrace{(ts)}_{\in G}$.
     
-    Then $T \subseteq t^{-1}G$ since $T \ni s = t^{-1}\underbrace{(ts)}_{\in G}$.
     So $G = \overline{T} \subseteq  t^{-1}G$.
     Hence $tG \subseteq G$.
 
@@ -91,8 +97,8 @@ Properties of $(X,T)$ translate to properties of $E(X,T)$:
 
 \begin{definition}
     A \vocab{compact semigroup} $S$
-    is a nonempty semigroup with a compact
-    Hausdorff topology,
+    is a nonempty semigroup\footnote{may not contain inverses or the identity}
+    with a compact Hausdorff topology,
     such that $S \ni x \mapsto xs$ is continuous for all  $s$.
 \end{definition}
 \begin{example}
@@ -125,8 +131,8 @@ Properties of $(X,T)$ translate to properties of $E(X,T)$:
 \end{proof}
 
 The \yaref{lem:ellisnumakura} is not very interesting for $E(X,T)$,
-since we already know that it has an identity,
-in fact we have chosen $R = \{1\}$ in the proof.
+since we already know that it has an identity.
+%in fact we might have chosen $R = \{1\}$ in the proof.
 But it is interesting for other semigroups.
 
 

inputs/tutorial_12.tex

@@ -59,6 +59,8 @@
     and since $B$ is Hausdorff, compact subsets of $B$ are closed.
 \end{subproof}
 
+\nr 1
+
 Let $(X,T)$ be a flow and $G = E(X,T)$ its Ellis semigroup.
 Let $d$ be a compatible metric on $X$.
 \begin{enumerate}[(a)]