Merge pull request #1182 from Bolpat/patch-2

Fix typo: Missing closing `|` for absolute value

Author: mikeshulman

errata.tex

@@ -931,6 +931,10 @@
   & 1270-g3f17b85
   & In the proof, $n : \N$ should be $k : \N$. And the range of $i$ should be $0 \leq i \leq k$. Also in the last equation, $r(\lim x) = \ell$ should be $\lim x = \ell$.\\
   %
+  \cref{ctb-uniformly-continuous-sup}
+  & % merge of 2c2d15123c917c33d7ea21c14ca08f86f8d93b55
+  & In the proof, $|f(x) - f(y_i) < \epsilon$ should be $|f(x) - f(y_i)| < \epsilon$.\\
+  %
   \cref{defn:inductive-cover}
   & 57-g671b000
   & In (\cref{defn:inductive-cover-interval-1}), the order of $r$ and $s$ should be flipped on the right-hand side: $(r, s)$ should be $(s, r)$.\\

reals.tex

@@ -2090,7 +2090,7 @@ \section{Compactness of the interval}
   We claim that $m \defeq \lim x$ is the supremum of~$f$. To prove that $f(x) \leq m$ for
   all $x : M$ it suffices to show $\lnot (m < f(x))$. So suppose to the contrary that $m <
   f(x)$. There is $\epsilon : \Qp$ such that $m + \epsilon < f(x)$. But now merely for
-  some $y_i$ participating in the definition of $x_\epsilon$ we get $|f(x) - f(y_i) <
+  some $y_i$ participating in the definition of $x_\epsilon$ we get $|f(x) - f(y_i)| <
   \epsilon$, therefore $m < f(x) - \epsilon < f(y_i) \leq m$, a contradiction.
 
   We finish the proof by showing that $m$ satisfies the second part of the theorem, because