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fix issue #109

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rzach committed Nov 12, 2016
1 parent 1604102 commit a6a70a4aa4a915128fe93eee1bd60d6887564861
Showing with 67 additions and 22 deletions.
  1. +57 −21 content/sets-functions-relations/size-of-sets/comparing-size.tex
  2. +10 −1 open-logic-config.sty
@@ -24,35 +24,37 @@
\end{explain}
\begin{defn}
-$\card{X} \leq \card{Y}$ if and only if there is an !!{injective}
- function~$f \colon X \to Y$.
+$X$ is \emph{no larger than}~$Y$, $\cardle{X}{Y}$, if and only if there
+ is an !!{injective} function~$f \colon X \to Y$.
\end{defn}
\begin{thm}[Schr\"oder-Bernstein]
-Let $X$ and $Y$ be sets. If $\card{X} \leq \card{Y}$ and $\card{Y}
-\leq \card{X}$, then $\card{X} = \card{Y}$.
+ Let $X$ and $Y$ be sets. If $\cardle{X}{Y}$ and $\cardle{Y}{X}$,
+ then $\cardeq{X}{Y}$.
\end{thm}
\begin{explain}
In other words, if there is a total !!{injective} function from $X$ to
-$Y$, and if there is a total !!{injective} function from $Y$ back to $X$,
-then there is a total !!{bijection} from $X$ to $Y$. Sometimes, it can be
+$Y$, and if there is a total !!{injective} function from $Y$ back to~$X$,
+then there is a total !!{bijection} from $X$ to~$Y$. Sometimes, it can be
difficult to think of a !!{bijection} between two equinumerous sets, so
the Schr\"oder-Bernstein theorem allows us to break the comparison
-down into cases so we only have to think of an !!{injection} from the
+down into cases so we only have to think of !!a{injection} from the
first to the second, and vice-versa. The Schr\"oder-Bernstein theorem,
apart from being convenient, justifies the act of discussing the
``sizes'' of sets, for it tells us that set cardinalities have the
familiar anti-symmetric property that numbers have.
\end{explain}
\begin{defn}
-$\card{X} < \card{Y}$ if and only if there is !!a{injective}
- function~$f\colon X \to Y$ but no !!{bijective}~$g\colon X \to Y$.
+$X$ is \emph{smaller than}~$Y$, $\cardless{X}{Y}$, if and only if
+ there is !!a{injective} function~$f\colon X \to Y$ but no
+ !!{bijective}~$g\colon X \to Y$.
\end{defn}
\begin{thm}[Cantor]
-For all $X$, $\card{X} < \card{\Pow{X}}$.
+ \ollabel{thm:cantor}
+For all $X$, $\cardless{X}{\Pow{X}}$.
\end{thm}
\begin{proof}
@@ -61,18 +63,52 @@
\{x\} \neq \{y\} = f(y)$.
There cannot be !!a{surjective} function~$g\colon X \to \Pow{X}$, let
-alone a !!{bijective} one. For assume that a surjective $g\colon X \to
-\Pow{X}$ exists. Then let $Y = \Setabs{x \in X}{x \notin g(x)}$. If
-$g(x)$ is defined for all $x \in X$, then $Y$ is clearly a
-well-defined subset of~$X$. If $g$ is !!{surjective}, $Y$ must be the
-value of~$g$ for some $x_0 \in X$, i.e., $Y = g(x_0)$. Now consider
-$x_0$: it cannot be !!a{element} of $Y$, since if $x_0 \in Y$ then
-$x_0 \in g(x_0)$, and the definition of~$Y$ then would have $x_0
-\notin Y$. On the other hand, it must be !!a{element} of~$Y$, since
-if it were not, then $x_0 \notin Y = g(x_0)$. But then $x_0$
-satisfies the defining condition of~$Y$, and so $x_0 \in Y$. In either
-case, we have a contradiction.
+alone a !!{bijective} one. For suppose that $g\colon X \to \Pow{X}$.
+Since $g$ is total, every $x \in X$ is mapped to a subset $g(x)
+\subseteq X$. We show that $g$ cannot be surjective. To do this, we
+define a subset~$Y \subseteq X$ which by definition cannot be in the
+range of~$g$. Let
+\[
+\overline{Y} = \Setabs{x \in X}{x \notin g(x)}.
+\]
+Since $g(x)$ is defined for all $x \in X$, $\overline{Y}$ is clearly a
+well-defined subset of~$X$. But, it cannot be in the range
+of~$g$. Let $x \in X$ be arbitrary, we show that $\overline{Y} \neq
+g(x)$. If $x \in g(x)$, then it does not satisfy $x \notin g(x)$, and
+so by the definition of~$\overline{Y}$, we have $x \notin
+\overline{Y}$. If $x \in \overline{Y}$, it must satisfy the defining
+property of~$\overline{Y}$, i.e., $x \notin g(x)$. Since $x$ was
+arbitrary this shows that for each $x \in X$, $x \in g(x)$ iff $x
+\notin \overline{Y}$, and so $g(x) \neq \overline{Y}$. So
+$\overline{Y}$ cannot be in the range of~$g$, contradicting the
+assumption that~$g$ is surjective.
\end{proof}
+\begin{explain}
+ It's instructive to compare the proof of \olref{thm:cantor} to that
+ of \olref[nen]{thm-nonenum-pownat}. There we showed that for any
+ list $Z_1$, $Z_2$, \dots, of subsets of~$\Int^+$ one can construct a
+ set~$\overline{Z}$ of numbers guaranteed not to be on the list. It
+ was guaranteed not to be on the list because, for every $n \in
+ \Int^+$, $n \in Z_n$ iff $n \notin \overline{Z}$. This way, there is
+ always some number that is !!a{element} of one of $Z_n$ and
+ $\overline{Z}$ but not the other. We follow the same idea here,
+ except the indices~$n$ are now !!{element}s of~$X$ instead
+ of~$\Int^+$. The set $\overline{Y}$ is defined so that it is
+ different from~$g(x)$ for each $x \in X$, because $x \in g(x)$ iff
+ $x \notin \overline{Y}$. Again, there is always !!a{element} of~$X$
+ which is !!a{element} of one of $g(x)$ and $\overline{Y}$ but not
+ the other. And just as $\overline{Z}$ therefore cannot be on the
+ list $Z_1$, $Z_2$, \dots, $\overline{Y}$ cannot be in the range
+ of~$g$.
+\end{explain}
+
+\begin{prob}
+ Show that there cannot be !!a{injective} function $g\colon \wp(X) \to
+ X$, for any set $X$. Hint: Suppose $g\colon \wp(X) \to X$ is
+ !!{injective}. Then for each $x \in X$ there is at most one $Y \subseteq
+ X$ such that $g(Y) = x$. Define a set $\overline{Y}$ such that for
+ every $x \in X$, $g(\overline{Y}) \neq x$.
+\end{prob}
\end{document}
View
@@ -480,11 +480,20 @@
% - `\card{X}`: cardinality of a set
\DeclareDocumentCommand \card { m } {\left| #1 \right|}
+% - `\cardle{X}{Y}`: X is no larger than Y
+\DeclareDocumentCommand \cardle { m m } {#1 \preceq #2}
+
+% - `\cardless{X}{Y}`: X is smaller than Y
+\DeclareDocumentCommand \cardless { m m } {#1 \prec #2}
+
+% - `\cardle{X}{Y}`: X is equinumerous with Y
+\DeclareDocumentCommand \cardeq { m m } {#1 \approx #2}
+
% - `\tuple{x,y}`: pairs, tuples, sequences
\DeclareDocumentCommand \tuple { m } {\langle #1 \rangle}
% - `\comp{f}{g}`: composition of f with g, defaults to $g \circ f$
-\DeclareDocumentCommand \comp { m m }{g \circ f}
+\DeclareDocumentCommand \comp { m m }{#2 \circ #1}
% - `\pto`: partial function arrow

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