fixing student-submitted typos

Author: rzach

first-order-logic/completeness/compactness.tex

@@ -18,7 +18,7 @@
 \ollabel{thm:compactness} 
 The following hold for any sentences $\Gamma$ and $!A$:
 \begin{enumerate}
-  \item If $\Gamma \Entails !A$ iff there is a finite $\Gamma_0
+  \item $\Gamma \Entails !A$ iff there is a finite $\Gamma_0
     \subseteq \Gamma$ such that $\Gamma_0 \Entails !A$.
   \item $\Gamma$ is satisfiable if and only if it is finitely
     satisfiable. 

first-order-logic/completeness/henkin-expansions.tex

@@ -30,7 +30,7 @@
 \end{explain}
 
 \begin{lem}
-\ollabel{defn:lang-exp}.
+\ollabel{defn:lang-exp}
 If $\Gamma$ is consistent in $\Lang L$ and $\Lang L'$ is obtained from
 $\Lang L$ by adding !!a{denumerable} set of new !!{constant}s $\Obj d_1$,
 $\Obj d_2$, \dots, then $\Gamma$ is consistent in~$\Lang L'$.

first-order-logic/completeness/identity.tex

@@ -20,8 +20,7 @@
 term is that term itself. Hence, if $t$ and $t'$ are different terms,
 their values in the term model---i.e., $t$ and $t'$,
 respectively---are different, and so $\eq[t][t']$ is false.  We can
-fix this, however, using a construction known as ``factoring.''  This
-construction works generally for so-called congruence relations.
+fix this, however, using a construction known as ``factoring.''
 \end{explain}
 
 \begin{defn}

first-order-logic/completeness/outline.tex

@@ -21,7 +21,7 @@
 is possible to directly construct a !!{derivation}, but we will take
 a slightly different tack.  The shift in perspective required is this:
 completeness can also be formulated as: ``if $\Gamma$ is consistent, it
-has a model.''  Perhaps we can use the information in $\Gamma$ together
+has a model.''  Perhaps we can use the information in~$\Gamma$ together
 with the hypothesis that it is consistent to construct a model.  After
 all, we know what kind of model we are looking for: one that is as
 $\Gamma$ describes it!

sets-functions-relations/functions/function-basics.tex

@@ -22,8 +22,8 @@
 \end{explain}
 
 \begin{defn}
-A \emph{function} $f \colon X \to Y$ is a mapping of each !!{element}s
-of~$X$ to an !!{element}s of~$Y$. We call $X$ the \emph{domain} of $f$
+A \emph{function} $f \colon X \to Y$ is a mapping of each !!{element}
+of~$X$ to an !!{element} of~$Y$. We call $X$ the \emph{domain} of $f$
 and $Y$ the \emph{codomain} of $f$. The \emph{range} of $f$ is the
 subset of the codomain that is actually output by $f$ for some input.
 \end{defn}

sets-functions-relations/sets/pairs-and-products.tex

@@ -18,7 +18,7 @@
 
 \begin{defn} 
 Given sets $X$ and $Y$, their \emph{Cartesian product} $X \times Y$ is
-$\Setabs{\tuple{x, y}}{x \in A \text{ and } y \in B}$.
+$\Setabs{\tuple{x, y}}{x \in X \text{ and } y \in Y}$.
 \end{defn}
 
 \begin{ex}

sets-functions-relations/sets/subsets.tex

@@ -33,7 +33,7 @@
 \end{ex}
 
 \begin{explain}
-Note that a set may contain other sets! In particular, a set may
+Note that a set may contain other sets!{} In particular, a set may
 happen to \emph{both} be an !!{element} and a subset of another, e.g.,
 $\{0\} \in \{0, \{0\}\}$ and also $\{0\} \subseteq \{0, \{0\}\}$.
 \end{explain}

sets-functions-relations/size-of-sets/enumerability.tex

@@ -58,17 +58,17 @@
 
 \begin{explain}
 Let's convince ourselves that the formal definition and the informal
-definition using an possibly gappy, possibly infinite list are
+definition using a possibly gappy, possibly infinite list are
 equivalent. !!^a{surjective} function (partial or total) from $\Nat$ to
 a set $X$ enumerates~$X$. Such a function determines an enumeration as
 defined informally above. Then an enumeration for $X$ is the list
 $f(0)$, $f(1)$, $f(2)$, \dots. Since $f$ is !!{surjective}, every
-!!{element} of $X$ is guaranteed to be the value of $f(n)$ for some $n
+!!{element} of $X$ is guaranteed to be the value of $f(n)$ for some~$n
 \in \Nat$.  Hence, every !!{element} of $X$ appears at some finite
 place in the list. Since the function may be partial or not !!{injective},
 the list may be gappy or redundant, but that is acceptable (as noted
 above). On the other hand, given a list that enumerates all
-!!{element}s of~$X$, we can define an !!{surjective} function $f\colon
+!!{element}s of~$X$, we can define a !!a{surjective} function $f\colon
 \Nat \to X$ by letting $f(n)$ be the $(n+1)$st member of the list, or
 undefined if the list has a gap in the $(n+1)$st spot.
 \end{explain}