bookshelf/Bookshelf/Enderton/Set.tex

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\documentclass{report}
\input{../../preamble}
\makeleancommands{../..}
\begin{document}
\header{Elements of Set Theory}{Herbert B. Enderton}
\tableofcontents
\begingroup
\renewcommand\thechapter{R}
\setcounter{chapter}{0}
\addtocounter{chapter}{-1}
\chapter{Reference}%
\label{chap:reference}
\section{\partial{Empty Set Axiom}}%
\label{ref:empty-set-axiom}
There is a set having no members:
$$\exists\; B, \forall\; x, x \not\in B.$$
\section{\defined{Extensionality Axiom}}%
\label{ref:extensionality-axiom}
If two sets have exactly the same members, then they are equal:
$$\forall\; A, \forall\; B,
\left[\forall\; x, (x \in A \iff x \in B) \Rightarrow A = B\right].$$
\begin{axiom}
\lean{Mathlib/Init/Set}{Set.ext}
\end{axiom}
\section{\partial{Pairing Axiom}}%
\label{ref:pairing-axiom}
For any sets $u$ and $v$, there is a set having as members just $u$ and $v$:
$$\forall\; u, \forall\; v, \exists\; B, \forall\; x,
(x \in B \iff x = u \text{ or } x = v).$$
\section{\defined{Powerset}}%
\label{ref:powerset}
The \textbf{powerset} of some set $A$ is the set of all subsets of $A$.
\begin{definition}
\lean{Mathlib/Init/Set}{Set.powerset}
\end{definition}
\section{\partial{Power Set Axiom}}%
\label{ref:power-set-axiom}
For any set $a$, there is a set whose members are exactly the subsets of $a$:
$$\forall\; a, \exists\; B, \forall\; x, (x \in B \iff x \subseteq a).$$
\section{\partial{Union Axiom, Preliminary Form}}%
\label{ref:union-axiom-preliminary-form}
For any sets $a$ and $b$, there is a set whose members are those sets belonging
either to $a$ or to $b$ (or both):
$$\forall\; a, \forall\; b, \exists\; B, \forall\; x,
(x \in B \iff x \in a \text{ or } x \in b).$$
\endgroup
\chapter{Introduction}%
\label{chap:introduction}
\section{Baby Set Theory}%
\label{sec:baby-set-theory}
\subsection{\verified{Exercise 1.1}}%
\label{sub:exercise-1.1}
Which of the following become true when "$\in$" is inserted in place of the
blank?
Which become true when "$\subseteq$" is inserted?
\subsubsection{\verified{Exercise 1.1a}}%
\label{ssub:exercise-1.1a}
$\{\emptyset\} \_\_\_\_ \{\emptyset, \{\emptyset\}\}$.
\begin{proof}
\lean{Bookshelf/Enderton/Set/Chapter\_1}
{Enderton.Set.Chapter\_1.exercise\_1\_1a}
Because the \textit{object} $\{\emptyset\}$ is a member of the right-hand set,
the statement is \textbf{true} in the case of "$\in$".
Because the \textit{members} of $\{\emptyset\}$ are all members of the
right-hand set, the statement is also \textbf{true} in the case of
"$\subseteq$".
\end{proof}
\subsubsection{\verified{Exercise 1.1b}}%
\label{ssub:exercise-1.11b}
$\{\emptyset\} \_\_\_\_ \{\emptyset, \{\{\emptyset\}\}\}$.
\begin{proof}
\lean{Bookshelf/Enderton/Set/Chapter\_1}
{Enderton.Set.Chapter\_1.exercise\_1\_1b}
Because the \textit{object} $\{\emptyset\}$ is not a member of the right-hand
set, the statement is \textbf{false} in the case of "$\in$".
Because the \textit{members} of $\{\emptyset\}$ are all members of the
right-hand set, the statement is \textbf{true} in the case of "$\subseteq$".
\end{proof}
\subsubsection{\verified{Exercise 1.1c}}%
\label{ssub:exercise-1.1c}
$\{\{\emptyset\}\} \_\_\_\_ \{\emptyset, \{\emptyset\}\}$.
\begin{proof}
\lean{Bookshelf/Enderton/Set/Chapter\_1}
{Enderton.Set.Chapter\_1.exercise\_1\_1c}
Because the \textit{object} $\{\{\emptyset\}\}$ is not a member of the
right-hand set, the statement is \textbf{false} in the case of "$\in$".
Because the \textit{members} of $\{\{\emptyset\}\}$ are all members of the
right-hand set, the statement is \textbf{true} in the case of "$\subseteq$".
\end{proof}
\subsubsection{\verified{Exercise 1.1d}}%
\label{ssub:exercise-1.1d}
$\{\{\emptyset\}\} \_\_\_\_ \{\emptyset, \{\{\emptyset\}\}\}$.
\begin{proof}
\lean{Bookshelf/Enderton/Set/Chapter\_1}
{Enderton.Set.Chapter\_1.exercise\_1\_1d}
Because the \textit{object} $\{\{\emptyset\}\}$ is a member of the right-hand
set, the statement is \textbf{true} in the case of "$\in$".
Because the \textit{members} of $\{\{\emptyset\}\}$ are not all members of the
right-hand set, the statement is \textbf{false} in the case of
"$\subseteq$".
\end{proof}
\subsubsection{\verified{Exercise 1.1e}}%
\label{ssub:exercise-1.1e}
$\{\{\emptyset\}\} \_\_ \{\emptyset, \{\emptyset, \{\emptyset\}\}\}$.
\begin{proof}
\lean{Bookshelf/Enderton/Set/Chapter\_1}
{Enderton.Set.Chapter\_1.exercise\_1\_1e}
Because the \textit{object} $\{\{\emptyset\}\}$ is not a member of the
right-hand set, the statement is \textbf{false} in the case of "$\in$".
Because the \textit{members} of $\{\{\emptyset\}\}$ are not all members of the
right-hand set, the statement is \textbf{false} in the case of
"$\subseteq$".
\end{proof}
\subsection{\verified{Exercise 1.2}}%
\label{sub:exercise-1.2}
Show that no two of the three sets $\emptyset$, $\{\emptyset\}$, and
$\{\{\emptyset\}\}$ are equal to each other.
\begin{proof}
\lean{Bookshelf/Enderton/Set/Chapter\_1}
{Enderton.Set.Chapter\_1.exercise\_1\_2}
By the \nameref{ref:extensionality-axiom}, $\emptyset$ is only equal to
$\emptyset$.
This immediately shows it is not equal to the other two.
Now consider object $\emptyset$.
This object is a member of $\{\emptyset\}$ but is not a member of
$\{\{\emptyset\}\}$.
Again, by the \nameref{ref:extensionality-axiom}, these two sets must be
different.
\end{proof}
\subsection{\verified{Exercise 1.3}}%
\label{sub:exercise-1.3}
Show that if $B \subseteq C$, then $\powerset{B} \subseteq \powerset{C}$.
\begin{proof}
\lean{Bookshelf/Enderton/Set/Chapter\_1}
{Enderton.Set.Chapter\_1.exercise\_1\_3}
Let $x \in \powerset{B}$.
By definition of the \nameref{ref:powerset}, $x$ is a subset of $B$.
By hypothesis, $B \subseteq C$.
Then $x \subseteq C$.
Again by definition of the \nameref{ref:powerset}, it follows
$x \in \powerset{C}$.
\end{proof}
\subsection{\verified{Exercise 1.4}}%
\label{sub:exercise-1.4}
Assume that $x$ and $y$ are members of a set $B$.
Show that $\{\{x\}, \{x, y\}\} \in \powerset{\powerset{B}}.$
\begin{proof}
\lean{Bookshelf/Enderton/Set/Chapter\_1}
{Enderton.Set.Chapter\_1.exercise\_1\_4}
Let $x$ and $y$ be members of set $B$.
Then $\{x\}$ and $\{x, y\}$ are subsets of $B$.
By definition of the \nameref{ref:powerset}, $\{x\}$ and $\{x, y\}$ are
members of $\powerset{B}$.
Then $\{\{x\}, \{x, y\}\}$ is a subset of $\powerset{B}$.
By definition of the \nameref{ref:powerset}, $\{\{x\}, \{x, y\}\}$ is a member
of $\powerset{\powerset{B}}$.
\end{proof}
\section{Sets - An Informal View}%
\label{sec:sets-informal-view}
\subsection{\partial{Exercise 2.1}}%
\label{sub:exercise-2.1}
Define the rank of a set $c$ to be the least $\alpha$ such that
$c \subseteq V_\alpha$.
Compute the rank of $\{\{\emptyset\}\}$.
Compute the rank of
$\{\emptyset, \{\emptyset\}, \{\emptyset, \{\emptyset\}\}\}$.
\begin{proof}
We first compute the values of $V_n$ for $0 \leq n \leq 3$ under the
assumption the set of atoms $A$ at the bottom of the hierarchy is empty.
\begin{align*}
V_0 & = \emptyset \\
V_1 & = V_0 \cup \powerset{V_0} \\
& = \emptyset \cup \{\emptyset\} \\
& = \{\emptyset\} \\
V_2 & = V_1 \cup \powerset{V_1} \\
& = \{\emptyset\} \cup \powerset{\{\emptyset\}} \\
& = \{\emptyset\} \cup \{\emptyset, \{\emptyset\}\} \\
& = \{\emptyset, \{\emptyset\}\} \\
V_3 & = V_2 \cup \powerset{V_2} \\
& = \{\emptyset, \{\emptyset\}\} \cup
\powerset{\{\emptyset, \{\emptyset\}\}} \\
& = \{\emptyset, \{\emptyset\}\} \cup
\{\emptyset,
\{\emptyset\},
\{\{\emptyset\}\},
\{\emptyset, \{\emptyset\}\}\} \\
& = \{\emptyset,
\{\emptyset\},
\{\{\emptyset\}\},
\{\emptyset, \{\emptyset\}\}\}
\end{align*}
It then immediately follows $\{\{\emptyset\}\}$ has rank $2$ and
$\{\emptyset, \{\emptyset\}, \{\emptyset, \{\emptyset\}\}\}$ has rank $3$.
\end{proof}
\subsection{\partial{Exercise 2.2}}%
\label{sub:exercise-2.2}
We have stated that $V_{\alpha + 1} = A \cup \powerset{V_\alpha}$.
Prove this at least for $\alpha < 3$.
\begin{proof}
Let $A$ be the set of atoms in our set hierarchy.
Let $P(n)$ be the predicate, "$V_{n + 1} = A \cup \powerset{V_n}$."
We prove $P(n)$ holds true for all natural numbers $n \geq 1$ via induction.
\paragraph{Base Case}%
Let $n = 1$.
By definition, $V_1 = V_0 \cup \powerset{V_0}$.
By definition, $V_0 = A$.
Therefore $V_1 = A \cup \powerset{V_0}$.
This proves $P(1)$ holds true.
\paragraph{Induction Step}%
Suppose $P(n)$ holds true for some $n \geq 1$.
Consider $V_{n+1}$.
By definition, $V_{n+1} = V_n \cup \powerset{V_n}$.
Therefore, by the induction hypothesis,
\begin{align}
V_{n+1}
& = V_n \cup \powerset{V_n}
\nonumber \\
& = (A \cup \powerset{V_{n-1}}) \cup \powerset{V_n}
\nonumber \\
& = A \cup (\powerset{V_{n-1}} \cup \powerset{V_n})
\label{sub:exercise-2.2-eq1}
\end{align}
But $V_{n-1}$ is a subset of $V_n$.
\nameref{sub:exercise-1.3} then implies
$\powerset{V_{n-1}} \subseteq \powerset{V_n}$.
This means \eqref{sub:exercise-2.2-eq1} can be simplified to
$$V_{n+1} = A \cup \powerset{V_n},$$
proving $P(n+1)$ holds true.
\paragraph{Conclusion}%
By mathematical induction, it follows for all $n \geq 1$, $P(n)$ is true.
\end{proof}
\subsection{\partial{Exercise 2.3}}%
\label{sub:exercise-2.3}
List all the members of $V_3$.
List all the members of $V_4$.
(It is to be assumed here that there are no atoms.)
\begin{proof}
As seen in the proof of \nameref{sub:exercise-2.1},
$$V_3 = \{
\emptyset,
\{\emptyset\},
\{\{\emptyset\}\},
\{\emptyset, \{\emptyset\}\}
\}.$$
By \nameref{sub:exercise-2.2}, $V_4 = \powerset{V_3}$ (since it is assumed
there are no atoms).
Thus
\begin{align*}
& V_4 = \{ \\
& \qquad \emptyset, \\
& \qquad \{\emptyset\}, \\
& \qquad \{\{\emptyset\}\}, \\
& \qquad \{\{\{\emptyset\}\}\}, \\
& \qquad \{\{\emptyset, \{\emptyset\}\}\}, \\
& \qquad \{\emptyset, \{\emptyset\}\}, \\
& \qquad \{\emptyset, \{\{\emptyset\}\}\}, \\
& \qquad \{\emptyset, \{\emptyset, \{\emptyset\}\}\}, \\
& \qquad \{\{\emptyset\}, \{\{\emptyset\}\}\}, \\
& \qquad \{\{\emptyset\}, \{\emptyset, \{\emptyset\}\}\}, \\
& \qquad \{\{\{\emptyset\}\}, \{\emptyset, \{\emptyset\}\}\}, \\
& \qquad \{\emptyset, \{\emptyset\}, \{\{\emptyset\}\}\}, \\
& \qquad \{\emptyset, \{\emptyset\}, \{\emptyset, \{\emptyset\}\}\}, \\
& \qquad \{\emptyset, \{\{\emptyset\}\}, \{\emptyset, \{\emptyset\}\}\} \\
& \qquad \{\{\emptyset\}, \{\{\emptyset\}\}, \{\emptyset, \{\emptyset\}\}\}, \\
& \qquad \{\emptyset, \{\emptyset\}, \{\{\emptyset\}\}, \{\emptyset, \{\emptyset\}\}\} \\
& \}.
\end{align*}
\end{proof}
\end{document}