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Author SHA1 Message Date
82f0dfd4de
fixed important typo
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Build latex and deploy / checkout (push) Failing after 15m5s
2024-02-07 16:04:28 +01:00
ffcd9cd45f
Merge branch 'main' of https://git.abstractnonsen.se/josia-notes/w23-logic-3 2024-02-07 16:03:42 +01:00
096906056e
fixed typo 2024-02-07 16:03:39 +01:00
2 changed files with 3 additions and 3 deletions

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@ -218,7 +218,7 @@ Consider $K(\bH^2)$.
A flow $\Z \acts X$ corresponds to the graph of A flow $\Z \acts X$ corresponds to the graph of
\begin{IEEEeqnarray*}{rCl} \begin{IEEEeqnarray*}{rCl}
X &\longrightarrow & X \\ X &\longrightarrow & X \\
1&\longmapsto & 1 \cdot x x&\longmapsto & 1 \cdot x
\end{IEEEeqnarray*} \end{IEEEeqnarray*}
and this graph is an element of $K(\bH^2)$. and this graph is an element of $K(\bH^2)$.

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@ -179,7 +179,7 @@ since $X^X$ has these properties.
\begin{proof} \begin{proof}
Let $G \coloneqq E(X,T)$ and let $d$ be a metric on $X$. Let $G \coloneqq E(X,T)$ and let $d$ be a metric on $X$.
\gist{ \gist{
For all $g \in G$ we need to show that $x \mapsto gx$ is bijective. For all $g \in G$ we need to show that $x \mapsto gx$ is injective.
If we had $gx = gy$, then $d(gx,gy) = 0$. If we had $gx = gy$, then $d(gx,gy) = 0$.
Then $\inf_{t \in T} d(tx,ty) = 0$, but the flow is distal, Then $\inf_{t \in T} d(tx,ty) = 0$, but the flow is distal,
hence $x = y$. hence $x = y$.
@ -195,7 +195,7 @@ since $X^X$ has these properties.
It is $g' = g'gg'$, It is $g' = g'gg'$,
so $\forall x .~g'(x) = g'(g g'(x))$. so $\forall x .~g'(x) = g'(g g'(x))$.
Hence $g'$ is bijective Hence $g'$ is injective
and $x = gg'(x)$, and $x = gg'(x)$,
i.e.~$g g' = \id$. i.e.~$g g' = \id$.
}{ }{