Enderton exercises 2.
parent
ead87350b8
commit
1da6e31581
|
@ -49,22 +49,22 @@ If $A$ and $B$ are sets such that for every object $t$,
|
|||
\section{Baby Set Theory}%
|
||||
\label{sec:baby-set-theory}
|
||||
|
||||
\subsection{\verified{Exercise 1}}%
|
||||
\label{sub:baby-set-theory-1}
|
||||
\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 1a}}%
|
||||
\label{ssub:baby-set-theory-1a}
|
||||
\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\_1a}
|
||||
{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 "$\in$".
|
||||
|
@ -75,15 +75,15 @@ $\{\emptyset\} \_\_\_\_ \{\emptyset, \{\emptyset\}\}$.
|
|||
|
||||
\end{proof}
|
||||
|
||||
\subsubsection{\verified{Exercise 1b}}%
|
||||
\label{ssub:baby-set-theory-1b}
|
||||
\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\_1b}
|
||||
{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 "$\in$".
|
||||
|
@ -93,15 +93,15 @@ $\{\emptyset\} \_\_\_\_ \{\emptyset, \{\{\emptyset\}\}\}$.
|
|||
|
||||
\end{proof}
|
||||
|
||||
\subsubsection{\verified{Exercise 1c}}%
|
||||
\label{ssub:baby-set-theory-1c}
|
||||
\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\_1c}
|
||||
{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 "$\in$".
|
||||
|
@ -111,15 +111,15 @@ $\{\{\emptyset\}\} \_\_\_\_ \{\emptyset, \{\emptyset\}\}$.
|
|||
|
||||
\end{proof}
|
||||
|
||||
\subsubsection{\verified{Exercise 1d}}%
|
||||
\label{ssub:baby-set-theory-1d}
|
||||
\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\_1d}
|
||||
{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 "$\in$".
|
||||
|
@ -130,15 +130,15 @@ $\{\{\emptyset\}\} \_\_\_\_ \{\emptyset, \{\{\emptyset\}\}\}$.
|
|||
|
||||
\end{proof}
|
||||
|
||||
\subsubsection{\verified{Exercise 1e}}%
|
||||
\label{ssub:baby-set-theory-1e}
|
||||
\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\_1e}
|
||||
{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 "$\in$".
|
||||
|
@ -149,8 +149,8 @@ $\{\{\emptyset\}\} \_\_ \{\emptyset, \{\emptyset, \{\emptyset\}\}\}$.
|
|||
|
||||
\end{proof}
|
||||
|
||||
\subsection{\verified{Exercise 2}}%
|
||||
\label{sub:baby-set-theory-2}
|
||||
\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.
|
||||
|
@ -158,7 +158,7 @@ Show that no two of the three sets $\emptyset$, $\{\emptyset\}$, and
|
|||
\begin{proof}
|
||||
|
||||
\lean{Bookshelf/Enderton/Set/Chapter\_1}
|
||||
{Enderton.Set.Chapter\_1.exercise\_2}
|
||||
{Enderton.Set.Chapter\_1.exercise\_1\_2}
|
||||
|
||||
By the \nameref{ref:principle-extensionality}, $\emptyset$ is only equal to
|
||||
$\emptyset$.
|
||||
|
@ -171,43 +171,176 @@ Show that no two of the three sets $\emptyset$, $\{\emptyset\}$, and
|
|||
|
||||
\end{proof}
|
||||
|
||||
\subsection{\verified{Exercise 3}}%
|
||||
\label{sub:baby-set-theory-3}
|
||||
\subsection{\verified{Exercise 1.3}}%
|
||||
\label{sub:exercise-1.3}
|
||||
|
||||
Show that if $B \subseteq C$, then $\mathscr{P} B \subseteq \mathscr{P} C$.
|
||||
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\_3}
|
||||
{Enderton.Set.Chapter\_1.exercise\_1\_3}
|
||||
|
||||
Let $x \in \mathscr{P} B$.
|
||||
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
|
||||
$x \in \mathscr{P} C$.
|
||||
$x \in \powerset{C}$.
|
||||
|
||||
\end{proof}
|
||||
|
||||
\subsection{\verified{Exercise 4}}%
|
||||
\label{sub:baby-set-theory-4}
|
||||
\subsection{\verified{Exercise 1.4}}%
|
||||
\label{sub:exercise-1.4}
|
||||
|
||||
Assume that $x$ and $y$ are members of a set $B$.
|
||||
Show that $\{\{x\}, \{x, y\}\} \in \mathscr{P}\mathscr{P} B.$
|
||||
Show that $\{\{x\}, \{x, y\}\} \in \powerset{\powerset{B}}.$
|
||||
|
||||
\begin{proof}
|
||||
|
||||
\lean{Bookshelf/Enderton/Set/Chapter\_1}
|
||||
{Enderton.Set.Chapter\_1.exercise\_4}
|
||||
{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 $\mathscr{P} B$.
|
||||
Then $\{\{x\}, \{x, y\}\}$ is a subset of $\mathscr{P} B$.
|
||||
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 $\mathscr{P}\mathscr{P} B$.
|
||||
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}
|
||||
|
||||
|
|
|
@ -9,7 +9,7 @@ Introduction
|
|||
|
||||
namespace Enderton.Set.Chapter_1
|
||||
|
||||
/-! ### Exercise 1
|
||||
/-! ### Exercise 1.1
|
||||
|
||||
Which of the following become true when "∈" is inserted in place of the blank?
|
||||
Which become true when "⊆" is inserted?
|
||||
|
@ -21,19 +21,19 @@ The `∅` does not equal the singleton set containing `∅`.
|
|||
lemma empty_ne_singleton_empty (h : ∅ = ({∅} : Set (Set α))) : False :=
|
||||
absurd h (Ne.symm $ Set.singleton_ne_empty (∅ : Set α))
|
||||
|
||||
/-- #### Exercise 1a
|
||||
/-- #### Exercise 1.1a
|
||||
|
||||
`{∅} ___ {∅, {∅}}`
|
||||
-/
|
||||
theorem exercise_1a
|
||||
theorem exercise_1_1a
|
||||
: {∅} ∈ ({∅, {∅}} : Set (Set (Set α)))
|
||||
∧ {∅} ⊆ ({∅, {∅}} : Set (Set (Set α))) := ⟨by simp, by simp⟩
|
||||
|
||||
/-- #### Exercise 1b
|
||||
/-- #### Exercise 1.1b
|
||||
|
||||
`{∅} ___ {∅, {{∅}}}`
|
||||
-/
|
||||
theorem exercise_1b
|
||||
theorem exercise_1_1b
|
||||
: {∅} ∉ ({∅, {{∅}}}: Set (Set (Set (Set α))))
|
||||
∧ {∅} ⊆ ({∅, {{∅}}}: Set (Set (Set (Set α)))) := by
|
||||
refine ⟨?_, by simp⟩
|
||||
|
@ -41,19 +41,19 @@ theorem exercise_1b
|
|||
simp at h
|
||||
exact empty_ne_singleton_empty h
|
||||
|
||||
/-- #### Exercise 1c
|
||||
/-- #### Exercise 1.1c
|
||||
|
||||
`{{∅}} ___ {∅, {∅}}`
|
||||
-/
|
||||
theorem exercise_1c
|
||||
theorem exercise_1_1c
|
||||
: {{∅}} ∉ ({∅, {∅}} : Set (Set (Set (Set α))))
|
||||
∧ {{∅}} ⊆ ({∅, {∅}} : Set (Set (Set (Set α)))) := ⟨by simp, by simp⟩
|
||||
|
||||
/-- #### Exercise 1d
|
||||
/-- #### Exercise 1.1d
|
||||
|
||||
`{{∅}} ___ {∅, {{∅}}}`
|
||||
-/
|
||||
theorem exercise_1d
|
||||
theorem exercise_1_1d
|
||||
: {{∅}} ∈ ({∅, {{∅}}} : Set (Set (Set (Set α))))
|
||||
∧ ¬ {{∅}} ⊆ ({∅, {{∅}}} : Set (Set (Set (Set α)))) := by
|
||||
refine ⟨by simp, ?_⟩
|
||||
|
@ -61,11 +61,11 @@ theorem exercise_1d
|
|||
simp at h
|
||||
exact empty_ne_singleton_empty h
|
||||
|
||||
/-- #### Exercise 1e
|
||||
/-- #### Exercise 1.1e
|
||||
|
||||
`{{∅}} ___ {∅, {∅, {∅}}}`
|
||||
-/
|
||||
theorem exercise_1e
|
||||
theorem exercise_1_1e
|
||||
: {{∅}} ∉ ({∅, {∅, {∅}}} : Set (Set (Set (Set α))))
|
||||
∧ ¬ {{∅}} ⊆ ({∅, {∅, {∅}}} : Set (Set (Set (Set α)))) := by
|
||||
apply And.intro
|
||||
|
@ -81,12 +81,12 @@ theorem exercise_1e
|
|||
have nh := h {∅}
|
||||
simp at nh
|
||||
|
||||
/-- ### Exercise 2
|
||||
/-- ### Exercise 1.2
|
||||
|
||||
Show that no two of the three sets `∅`, `{∅}`, and `{{∅}}` are equal to each
|
||||
other.
|
||||
-/
|
||||
theorem exercise_2
|
||||
theorem exercise_1_2
|
||||
: ∅ ≠ ({∅} : Set (Set α))
|
||||
∧ ∅ ≠ ({{∅}} : Set (Set (Set α)))
|
||||
∧ {∅} ≠ ({{∅}} : Set (Set (Set α))) := by
|
||||
|
@ -99,22 +99,22 @@ theorem exercise_2
|
|||
simp at h
|
||||
exact empty_ne_singleton_empty h
|
||||
|
||||
/-- ### Exercise 3
|
||||
/-- ### Exercise 1.3
|
||||
|
||||
Show that if `B ⊆ C`, then `𝓟 B ⊆ 𝓟 C`.
|
||||
-/
|
||||
theorem exercise_3 (h : B ⊆ C) : Set.powerset B ⊆ Set.powerset C := by
|
||||
theorem exercise_1_3 (h : B ⊆ C) : Set.powerset B ⊆ Set.powerset C := by
|
||||
unfold Set.powerset
|
||||
simp
|
||||
intro x hx
|
||||
exact Set.Subset.trans hx h
|
||||
|
||||
/-- ### Exercise 4
|
||||
/-- ### Exercise 1.4
|
||||
|
||||
Assume that `x` and `y` are members of a set `B`. Show that
|
||||
`{{x}, {x, y}} ∈ 𝓟 𝓟 B`.
|
||||
-/
|
||||
theorem exercise_4 (x y : α) (hx : x ∈ B) (hy : y ∈ B)
|
||||
theorem exercise_1_4 (x y : α) (hx : x ∈ B) (hy : y ∈ B)
|
||||
: {{x}, {x, y}} ∈ Set.powerset (Set.powerset B) := by
|
||||
unfold Set.powerset
|
||||
simp only [Set.mem_singleton_iff, Set.mem_setOf_eq]
|
||||
|
|
|
@ -1,4 +1,4 @@
|
|||
\usepackage{amsfonts, amsmath, amsthm}
|
||||
\usepackage{amsfonts, amsmath, amssymb, amsthm}
|
||||
\usepackage{bigfoot}
|
||||
\usepackage{comment}
|
||||
\usepackage[shortlabels]{enumitem}
|
||||
|
@ -131,7 +131,11 @@
|
|||
\newcommand{\ico}[2]{\left[#1, #2\right)}
|
||||
\newcommand{\ioc}[2]{\left(#1, #2\right]}
|
||||
\newcommand{\ioo}[2]{\left(#1, #2\right)}
|
||||
\newcommand{\powerset}[1]{\mathscr{P}\;#1}
|
||||
\newcommand{\ubar}[1]{\text{\b{$#1$}}}
|
||||
|
||||
\let\oldemptyset\emptyset
|
||||
\let\emptyset\varnothing
|
||||
|
||||
% Close off "private" namespace.
|
||||
\makeatother
|
||||
|
|
Loading…
Reference in New Issue