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fixing issue #107, replacing \Nat with \Int^+ when discussing enumara…

…tions and formulating everything with total functions
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rzach committed Nov 13, 2016
1 parent a6a70a4 commit 370cb024dfd8515d893088475f883a3d32b4eb31
@@ -10,108 +10,180 @@
\olsection{\printtoken{S}{enumerable} Sets}
+One way of specifying a finite set is by listing its !!{element}s. But
+conversely, since there are only finitely many !!{element}s in a set,
+every finite set can be enumerated. By this we mean: its elements can
+be put into a list (a list with a beginning, where each !!{element} of
+the list other than the first has a unique predecessor). Some
+infinite sets can also be enumerated, such as the set of positive
+integers.
+
\begin{defn}[Enumeration]
-Informally, an \emph{enumeration} of a set $X$ is a list (possibly
-infinite) such that every !!{element} of $X$ appears some finite
-number of places into the list. If $X$ has an enumeration, then $X$ is
-said to be \emph{!!{enumerable}}. If $X$ is !!{enumerable} and
-infinite, we say $X$ is !!{denumerable}.
+Informally, an \emph{enumeration} of a set~$X$ is a list (possibly
+infinite) of !!{element}s of~$X$ such that every !!{element} of $X$
+appears on the list at some finite position. If $X$ has an
+enumeration, then $X$ is said to be \emph{!!{enumerable}}. If $X$ is
+!!{enumerable} and infinite, we say $X$ is !!{denumerable}.
\end{defn}
\begin{explain}
A couple of points about enumerations:
\begin{enumerate}
-\item The order of !!{element}s of $X$ in the enumeration does not
- matter, as long as every !!{element} appears: $4$, $1$, $2$5,
- $16$,~$9$ enumerates the (set of the) first five square numbers just
- as well as $1$, $4$, $9$, $16$,~$25$ does.
+\item We count as enumerations only lists which have a beginning and
+ in which every !!{element} other than the first has a single
+ !!{element} immediately preceding it. In other words, there are
+ only finitely many elements between the first !!{element} of the
+ list and any other !!{element}.
+\item We can have different enumerations of the same set~$X$ which
+ differ by the order in which the !!{element}s appear: $4$, $1$,
+ $25$, $16$,~$9$ enumerates the (set of the) first five square
+ numbers just as well as $1$, $4$, $9$, $16$,~$25$ does.
\item Redundant enumerations are still enumerations: $1$, $1$, $2$,
$2$, $3$, $3$,~\dots{} enumerates the same set as $1$, $2$,
$3$,~\dots{} does.
\item Order and redundancy \emph{do} matter when we specify an
- enumeration: we can enumerate the natural numbers beginning with
+ enumeration: we can enumerate the positive integers beginning with
$1$, $2$, $3$, $1$, \dots{}, but the pattern is easier to see when
enumerated in the standard way as $1$, $2$, $3$, $4$,~\dots
\item Enumerations must have a beginning: \dots, $3$, $2$, $1$ is not
an enumeration of the natural numbers because it has no first
!!{element}. To see how this follows from the informal definition,
- ask yourself, ``at what place in the list does the number 76
+ ask yourself, ``at what position in the list does the number 76
appear?''
-\item The following is not an enumeration of the natural numbers: $1$,
- $3$, $5$, \dots, $2$, $4$, $6$, \dots\@ The problem is that the even
- numbers occur at places $\infty + 1$, $\infty + 2$, $\infty + 3$,
- rather than at finite positions.
+\item The following is not an enumeration of the positive integers:
+ $1$, $3$, $5$, \dots, $2$, $4$, $6$, \dots\@ The problem is that the
+ even numbers occur at places $\infty + 1$, $\infty + 2$, $\infty +
+ 3$, rather than at finite positions.
\item Lists may be gappy: $2$, $-$, $4$, $-$, $6$, $-$, \dots{}
- enumerates the even natural numbers.
-\item The empty set is enumerable: it is enumerated by the empty list!
+ enumerates the even positive integers.
+\item The empty set is enumerable: it is enumerated by the empty list!{}
\end{enumerate}
\end{explain}
-The following provides a more formal definition of an
-enumeration:
+\begin{prop}
+ If $X$ has an enumeration, it has an enumeration without gaps or
+ repetitions.
+\end{prop}
+
+\begin{proof}
+ Suppose $X$ has an enumeration $x_1$, $x_2$, \dots{} in which each
+ $x_i$ is an !!{element} of~$X$ or a gap. We can remove repetitions
+ from an enumeration by replacing repeated !!{element}s by gaps. For
+ instance, we can turn the enumeration into a new one in which $x_i'$
+ is $x_i$ if $x_i$ is !!a{element} of~$X$ that is not among $x_1$,
+ \dots, $x_{i-1}$ or is $-$ if it is. We can remove gaps by closing up
+ the elements in the list. To make precise what ``closing up''
+ amounts to is a bit difficult to describe. For instance, we can
+ generate a new enumeration $x_1''$, $x_2''$, \dots, where each
+ $x_i''$ is the first !!{element} in the enumeration $x_1'$, $x_2'$,
+ \dots{} after $x_{i-1}''$.
+\end{proof}
+
+The last proof shows that in order to give precise definitions and
+proofs about enumerations and !!{enumerable} sets, we need a more
+precise definition. The following provides it.
\begin{defn}[Enumeration]
An \emph{enumeration} of a set $X$ is any !!{surjective} function $f:
-\Nat \rightarrow X$.
+\Int^+ \to X$.
\end{defn}
\begin{explain}
Let's convince ourselves that the formal definition and the informal
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~$n
-\in \Nat$. 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 !!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.
+equivalent. !!^a{surjective} function (partial or total) from $\Int^+$
+to a set $X$ enumerates~$X$. Such a function determines an enumeration
+as defined informally above: the list $f(1)$, $f(2)$, $f(3)$,
+\dots. Since $f$ is !!{surjective}, every !!{element} of $X$ is
+guaranteed to be the value of $f(n)$ for some~$n \in \Int^+$. Hence,
+every !!{element} of $X$ appears at some finite position in the
+list. Since the function may not be !!{injective}, the list may be
+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 !!a{surjective} function $f\colon \Int \to X$ by
+letting $f(n)$ be the $n$th !!{element} of the list that is not a gap,
+or the last !!{element} of the list if there is no $n$th !!{element}.
+There is one case in which this does not produce !!a{surjective}
+function: if $X$ is empty, and hence the lsit is empty. So, every
+non-empty list determines !!a{surjective} function $f\colon \Int^+ \to
+X$.
\end{explain}
+\begin{defn}
+ \ollabel{defn:enumerable}
+ A set~$X$ is !!{enumerable} iff it is empty or has an enumeration.
+\end{defn}
+
\begin{ex}
-A function enumerating the natural numbers ($\Nat$) is
-simply the identity function given by $f(n) = n$.
+A function enumerating the positive integers ($\Int^+$) is simply the
+identity function given by $f(n) = n$. A function enumerating the
+natural numbers $\Nat$ is the function $g(n) = n - 1$.
\end{ex}
+\begin{prob}
+ According to \olref{defn:enumerable}, a set $X$ is enumerable iff $X
+ = \emptyset$ or there is !!a{surjective} $f\colon \Int^+ \to X$. It
+ is also possible to define ``!!{enumerable} set'' precisely by: a
+ set is enumerable iff there is !!a{injective} function $g\colon X
+ \to \Int^+$. Show that the definitions are equivalent, i.e., show
+ that there is !!a{injective} function $g\colon X \to \Int^+$ iff
+ either $X = \emptyset$ or there is !!a{surjective} $f\colon \Int^+
+ \to X$.
+\end{prob}
+
\begin{ex}
-The functions $f\colon \Nat \to \Nat$ and $g \colon \Nat \to \Nat$ given by
+The functions $f\colon \Int^+ \to \Int^+$ and $g \colon \Int^+ \to
+\Int^+$ given by
\begin{align}
f(n) & = 2n \text{ and}\\
g(n) & = 2n+1
\end{align}
-enumerate the even natural numbers and the odd natural numbers,
+enumerate the even natural numbers and the odd positive integers,
respectively. However, neither function is an enumeration of
-$\Nat$, since neither is !!{surjective}.
+$\Int^+$, since neither is !!{surjective}.
\end{ex}
\begin{ex}
-The function $f(n) = \lceil \frac{(-1)^n n}{2}\rceil$ (where $\lceil x
-\rceil$ denotes the \emph{ceiling} function, which rounds $x$ up to
-the nearest integer) enumerates the set of integers~$\Int$. Notice
-how $f$ generates the values of $\Int$ by ``hopping'' back and forth
-between positive and negative integers:
+The function $f(n) = (-1)^{n} \lceil \frac{(n-1)}{2}\rceil$ (where
+$\lceil x \rceil$ denotes the \emph{ceiling} function, which rounds
+$x$ up to the nearest integer) enumerates the set of
+integers~$\Int$. Notice how $f$ generates the values of $\Int$ by
+``hopping'' back and forth between positive and negative integers:
\[
-\begin{array}{c c c c c c c}
-f(1) & f(2) & f(3) & f(4) & f(5) & f(6) & \dots \\ \\
-\lceil - \tfrac{1}{2}\rceil & \lceil \tfrac{2}{2} \rceil & \lceil -
-\tfrac{3}{2} \rceil & \lceil \tfrac{4}{2} \rceil & \lceil -\tfrac{5}{2}
-\rceil & \lceil \tfrac{6}{2} \rceil & \dots \\ \\
+\begin{array}{c c c c c c c c}
+f(1) & f(2) & f(3) & f(4) & f(5) & f(6) & f(7) & \dots \\ \\
+- \lceil \tfrac{0}{2} \rceil & \lceil \tfrac{1}{2}\rceil & - \lceil \tfrac{2}{2} \rceil & \lceil \tfrac{3}{2} \rceil & - \lceil \tfrac{4}{2} \rceil & \lceil \tfrac{5}{2}
+\rceil & - \lceil \tfrac{6}{2} \rceil & \dots \\ \\
0 & 1 & -1 & 2 & -2 & 3 & \dots
\end{array}
\]
+You can also think of $f$ as defined by cases as follows:
+\[
+f(n) = \begin{cases}
+ 0 & \text{if $n = 1$}\\
+ n/2 & \text{if $n$ is even}\\
+ -(n-1)/2 & \text{if $n$ is odd and $>1$}
+ \end{cases}
+\]
\end{ex}
+\begin{prob}
+ Show that if $X$ and $Y$ are !!{enumerable}, so is $X \cup Y$.
+\end{prob}
+
+\begin{prob}
+ Show by induction on $n$ that if $X_1$, $X_2$, \dots, $X_n$ are all
+ !!{enumerable}, so is $X_1 \cup \dots \cup X_n$.
+\end{prob}
+
\begin{explain}
That is fine for ``easy'' sets. What about the set of, say, pairs of
natural numbers?{}
\[
-\Nat^2 = \Nat \times \Nat = \Setabs{\tuple{n,m}}{n,m \in \Nat}
+\Int^+ \times \Int^+ = \Setabs{\tuple{n,m}}{n,m \in \Int^+}
\]
-Another method we can use to enumerate sets is to organize them
+We can organize the pairs of positive integers
in an \emph{array}, such as the following:
\[
\begin{array}{ c | c | c | c | c | c}
@@ -129,11 +201,11 @@
\end{array}
\]
-Clearly, every ordered pair in $\Nat^2$ will appear at least
-once in the array. In particular, $\tuple{n,m}$ will appear in the $n$th
-column and $m$th row. But how do we organize the elements of an
-array into a list? The pattern in the array below demonstrates one
-way to do this:
+Clearly, every ordered pair in $\Int^+ \times \Int^+$ will appear
+exactly once in the array. In particular, $\tuple{n,m}$ will appear in
+the $n$th column and $m$th row. But how do we organize the elements of
+such an array into a one-way list? The pattern in the array below
+demonstrates one way to do this:
\[
\begin{array}{ c | c | c | c | c | c }
& & & & & \\
@@ -149,29 +221,30 @@
& \vdots & \vdots & \vdots & \vdots & \ddots\\
\end{array}
\]
-This pattern is called \emph{Cantor's zig-zag method}. Other
-patterns are perfectly permissible, as long as they ``zig-zag''
-through every cell of the array. By Cantor's zig-zag method, the
-enumeration for $\Nat^2$ according to this scheme would be:
+This pattern is called \emph{Cantor's zig-zag method}. Other patterns
+are perfectly permissible, as long as they ``zig-zag'' through every
+cell of the array. By Cantor's zig-zag method, the enumeration for
+$\Int^+ \times \Int^+$ according to this scheme would be:
\[
\tuple{1,1}, \tuple{1,2}, \tuple{2,1}, \tuple{1,3}, \tuple{2,2},
\tuple{3,1}, \tuple{1,4}, \tuple{2,3}, \tuple{3,2}, \tuple{4,1}, \dots
\]
What ought we do about enumerating, say, the set of ordered triples
-of natural numbers?
+of positive integers?
\[
-\Nat^3 = \Nat \times \Nat \times \Nat = \Setabs{\tuple{n,m,k}}{n,m,k \in \Nat}
+\Int^+ \times \Int^+ \times \Int^+ = \Setabs{\tuple{n,m,k}}{n,m,k \in \Int^+}
\]
-We can think of $\Nat^3$ as the Cartesian product of
-$\Nat^2$ and $\Nat$, that is,
+We can think of $\Int^+ \times \Int^+ \times \Int^+$ as the Cartesian
+product of $\Int^+ \times \Int^+$ and $\Int^+$, that is,
\[
-\Nat^3 = \Nat^2 \times \Nat =
-\Setabs{\tuple{\tuple{n,m},k}}{\tuple{n,m} \in \Nat^2, k \in \Nat }
+(\Int^+)^3 = (\Int^+ \times \Int^+) \times \Int^+ =
+\Setabs{\tuple{\tuple{n,m},k}}{\tuple{n,m} \in \Int^+ \times \Int^+, k
+ \in \Int^+ }
\]
-and thus we can enumerate $\Nat^3$ with an array by
-labelling one axis with the enumeration of $\Nat$, and the
-other axis with the enumeration of $\Nat^2$:
+and thus we can enumerate $(\Int^+)^3$ with an array by labelling one
+axis with the enumeration of $\Int^+$, and the other axis with the
+enumeration of $(\Int^+)^2$:
\[
\begin{array}{ c | c | c | c | c | c}
& \textbf 1 & \textbf 2 & \textbf 3 & \textbf 4 & \dots \\
@@ -188,12 +261,18 @@
\end{array}
\]
Thus, by using a method like Cantor's zig-zag method, we may
-similarly obtain an enumeration of~$\Nat^3$.
+similarly obtain an enumeration of~$(\Int^+)^3$.
\end{explain}
\begin{prob}
-Give an enumeration of the set of all ordered pairs of positive
-rational numbers.
+Give an enumeration of the set of all positive rational numbers. (A
+positive rational number is one that can be written as a fraction
+$n/m$ with $n, m \in \Int^+$).
+\end{prob}
+
+\begin{prob}
+Show that $\Rat$ is !!{enumerable}. (A rational number is one that can
+be written as a fraction $z/m$ with $z \in \Int$, $m \in \Int^+$).
\end{prob}
\begin{prob}
@@ -208,10 +287,6 @@
enumerable set is enumerable.
\end{prob}
-\begin{prob}
-Show that if $X$ and $Y$ are !!{enumerable}, so is $X \cup Y$.
-\end{prob}
-
\begin{prob}
A set of positive integers is said to be \emph{cofinite} iff
it is the complement of a finite set of positive integers. Let
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