auto-update(nvim): 2025-02-21 02:39:13

documents/by-course/math-8/pset-5/main.typ

@@ -709,8 +709,19 @@ Every nonempty subset \( S \) of \(\mathbb{Z}^-\) (the set of negative integers)
 
 #proof[
   We construct a new set, $S' = {-x | x in S}$. Then we note that $S'$ is a set
-  of positive integers, and in fact $S' subset.eq NN$. By the well ordering
-  principle, $S'$ always has a least element. Call this least element $k in
-  S'$. $k$ is the least element in $S'$ if and only if $-k$ is the greatest
-  element in $S$. Thus, $S$ always has a largest element.
+  of positive integers, and in fact $S' subset.eq NN$, and it is nonempty. By
+  the well ordering principle, $S'$ always has a least element. Call this least
+  element $s in S'$. Now we show that $s$ is the least element in $S'$ if and
+  only if $-s$ is the greatest element in $S$.
+
+  We say that $s$ is the least element in $S'$ iff. for every $x in S'$ where
+  $x != s$, we have $s < x$. Note that $-s in S$ and for every $x$, $-x$ is in
+  $S$. Also, these are all of the elements in $S$, that is, it has no other
+  elements besides $-x$ and $-s$ for all possible $x$. So, equivalently, for
+  every $-x$ such that $-x != -s$, $-s > -x$. So if $s$ is the least element in
+  $S'$, then $-s in S$ is the greatest element in $S$.
+
+  Thus, as $s$ is guaranteed to exist by the WOP, $-s$ is also guaranteed to
+  exist, and $S$ always has a greatest element.
+
 ]