From b81d6b4945bd5bbfe809736a5c4be3b52e8e0b1a Mon Sep 17 00:00:00 2001 From: Raimundo Saona <37874270+saona-raimundo@users.noreply.github.com> Date: Tue, 24 Oct 2023 14:38:54 +0200 Subject: [PATCH] Update class_3.tex Fix typos. --- class_3.tex | 14 +++++++------- 1 file changed, 7 insertions(+), 7 deletions(-) diff --git a/class_3.tex b/class_3.tex index 30d62a2..b053ccc 100644 --- a/class_3.tex +++ b/class_3.tex @@ -116,12 +116,12 @@ \chapter{Class 3} A few remarks are in place. \begin{itemize} \item - Inductively defined sets are countable and consists of finite elements. + Inductively defined sets are countable and consist of finite elements. \item Inductively defined sets can be written as rules $x \implies f(x)$ meaning that, if $x \in X$, then $f(x) \in X$. \item Inductively defined sets allow proof by induction. - Consider prove that for all $x \in X$ we have $G(x)$. + Consider proving that for all $x \in X$ we have $G(x)$. This can be proven by showing \begin{enumerate} \item $G(\bottom)$ @@ -137,7 +137,7 @@ \chapter{Class 3} } In the definition of balanced binary sequences, we consider the complete lattice $(\Sigma^\omega, \subseteq)$ and the function on sets given by $f(X) \defas 01X \cup 10X$. -Then, balanced binary sequences corresponds to $\gfp f$. +Then, balanced binary sequences correspond to $\gfp f$. \Definition[Interval {$[0, 1]$}]{ Define the set $S$ as the largest set $X$ such that @@ -149,15 +149,15 @@ \chapter{Class 3} A few remarks are in place. \begin{itemize} \item - Coinductively defined sets are uncountable and consists of infinite elements. + Coinductively defined sets are uncountable and consist of infinite elements. \item Coinductively defined sets can be written as rules $x \implied f(x)$ meaning that, for all $y \in X$, there exists $x$ such that $y = f(x)$ and $x \in X$. \item - Coinductively defined sets allow proof by induction. - Consider prove that for all $x \in X$ we have $G(x)$. + Coinductively defined sets allow proof by coinduction. + Consider proving that for all $x$, if $G(x)$, then $x \in X$. This can be proven by showing \begin{enumerate} - \item For all $x$ and $i$, if $G(f_i(x))$, then $G(x)$ + \item For all $x$ and $i$, if $G(f_i(x))$, then $G(x)$ \,, \end{enumerate} where $\{f_1, \ldots, f_n\}$ is the set of rules that define the set $X$. \end{itemize}