Enderton, Baby Set Theory.

finite-set-exercises
Joshua Potter 2023-05-18 16:02:15 -06:00
parent ed981c3892
commit ead87350b8
2 changed files with 153 additions and 9 deletions

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@ -49,20 +49,23 @@ If $A$ and $B$ are sets such that for every object $t$,
\section{Baby Set Theory}% \section{Baby Set Theory}%
\label{sec:baby-set-theory} \label{sec:baby-set-theory}
\subsection{\partial{Exercise 1}}% \subsection{\verified{Exercise 1}}%
\label{sub:baby-set-theory-1} \label{sub:baby-set-theory-1}
Which of the following become true when "$\in$" is inserted in place of the Which of the following become true when "$\in$" is inserted in place of the
blank? blank?
Which become true when "$\subseteq$" is inserted? Which become true when "$\subseteq$" is inserted?
\subsubsection{\partial{Exercise 1a}}% \subsubsection{\verified{Exercise 1a}}%
\label{ssub:baby-set-theory-1a} \label{ssub:baby-set-theory-1a}
$\{\emptyset\} \_\_\_\_ \{\emptyset, \{\emptyset\}\}$. $\{\emptyset\} \_\_\_\_ \{\emptyset, \{\emptyset\}\}$.
\begin{proof} \begin{proof}
\lean{Bookshelf/Enderton/Set/Chapter\_1}
{Enderton.Set.Chapter\_1.exercise\_1a}
Because the \textit{object} $\{\emptyset\}$ is a member of the right-hand set, Because the \textit{object} $\{\emptyset\}$ is a member of the right-hand set,
the statement is \textbf{true} in the case of "$\in$". the statement is \textbf{true} in the case of "$\in$".
@ -72,13 +75,16 @@ $\{\emptyset\} \_\_\_\_ \{\emptyset, \{\emptyset\}\}$.
\end{proof} \end{proof}
\subsubsection{\partial{Exercise 1b}}% \subsubsection{\verified{Exercise 1b}}%
\label{ssub:baby-set-theory-1b} \label{ssub:baby-set-theory-1b}
$\{\emptyset\} \_\_\_\_ \{\emptyset, \{\{\emptyset\}\}\}$. $\{\emptyset\} \_\_\_\_ \{\emptyset, \{\{\emptyset\}\}\}$.
\begin{proof} \begin{proof}
\lean{Bookshelf/Enderton/Set/Chapter\_1}
{Enderton.Set.Chapter\_1.exercise\_1b}
Because the \textit{object} $\{\emptyset\}$ is not a member of the right-hand Because the \textit{object} $\{\emptyset\}$ is not a member of the right-hand
set, the statement is \textbf{false} in the case of "$\in$". set, the statement is \textbf{false} in the case of "$\in$".
@ -87,13 +93,16 @@ $\{\emptyset\} \_\_\_\_ \{\emptyset, \{\{\emptyset\}\}\}$.
\end{proof} \end{proof}
\subsubsection{\partial{Exercise 1c}}% \subsubsection{\verified{Exercise 1c}}%
\label{ssub:baby-set-theory-1c} \label{ssub:baby-set-theory-1c}
$\{\{\emptyset\}\} \_\_\_\_ \{\emptyset, \{\emptyset\}\}$. $\{\{\emptyset\}\} \_\_\_\_ \{\emptyset, \{\emptyset\}\}$.
\begin{proof} \begin{proof}
\lean{Bookshelf/Enderton/Set/Chapter\_1}
{Enderton.Set.Chapter\_1.exercise\_1c}
Because the \textit{object} $\{\{\emptyset\}\}$ is not a member of the Because the \textit{object} $\{\{\emptyset\}\}$ is not a member of the
right-hand set, the statement is \textbf{false} in the case of "$\in$". right-hand set, the statement is \textbf{false} in the case of "$\in$".
@ -102,13 +111,16 @@ $\{\{\emptyset\}\} \_\_\_\_ \{\emptyset, \{\emptyset\}\}$.
\end{proof} \end{proof}
\subsubsection{\partial{Exercise 1d}}% \subsubsection{\verified{Exercise 1d}}%
\label{ssub:baby-set-theory-1d} \label{ssub:baby-set-theory-1d}
$\{\{\emptyset\}\} \_\_\_\_ \{\emptyset, \{\{\emptyset\}\}\}$. $\{\{\emptyset\}\} \_\_\_\_ \{\emptyset, \{\{\emptyset\}\}\}$.
\begin{proof} \begin{proof}
\lean{Bookshelf/Enderton/Set/Chapter\_1}
{Enderton.Set.Chapter\_1.exercise\_1d}
Because the \textit{object} $\{\{\emptyset\}\}$ is a member of the right-hand Because the \textit{object} $\{\{\emptyset\}\}$ is a member of the right-hand
set, the statement is \textbf{true} in the case of "$\in$". set, the statement is \textbf{true} in the case of "$\in$".
@ -118,13 +130,16 @@ $\{\{\emptyset\}\} \_\_\_\_ \{\emptyset, \{\{\emptyset\}\}\}$.
\end{proof} \end{proof}
\subsubsection{\partial{Exercise 1e}}% \subsubsection{\verified{Exercise 1e}}%
\label{ssub:baby-set-theory-1e} \label{ssub:baby-set-theory-1e}
$\{\{\emptyset\}\} \_\_ \{\emptyset, \{\emptyset, \{\emptyset\}\}\}$. $\{\{\emptyset\}\} \_\_ \{\emptyset, \{\emptyset, \{\emptyset\}\}\}$.
\begin{proof} \begin{proof}
\lean{Bookshelf/Enderton/Set/Chapter\_1}
{Enderton.Set.Chapter\_1.exercise\_1e}
Because the \textit{object} $\{\{\emptyset\}\}$ is not a member of the Because the \textit{object} $\{\{\emptyset\}\}$ is not a member of the
right-hand set, the statement is \textbf{false} in the case of "$\in$". right-hand set, the statement is \textbf{false} in the case of "$\in$".
@ -134,7 +149,7 @@ $\{\{\emptyset\}\} \_\_ \{\emptyset, \{\emptyset, \{\emptyset\}\}\}$.
\end{proof} \end{proof}
\subsection{\partial{Exercise 2}}% \subsection{\verified{Exercise 2}}%
\label{sub:baby-set-theory-2} \label{sub:baby-set-theory-2}
Show that no two of the three sets $\emptyset$, $\{\emptyset\}$, and Show that no two of the three sets $\emptyset$, $\{\emptyset\}$, and
@ -142,6 +157,9 @@ Show that no two of the three sets $\emptyset$, $\{\emptyset\}$, and
\begin{proof} \begin{proof}
\lean{Bookshelf/Enderton/Set/Chapter\_1}
{Enderton.Set.Chapter\_1.exercise\_2}
By the \nameref{ref:principle-extensionality}, $\emptyset$ is only equal to By the \nameref{ref:principle-extensionality}, $\emptyset$ is only equal to
$\emptyset$. $\emptyset$.
This immediately shows it is not equal to the other two. This immediately shows it is not equal to the other two.
@ -153,13 +171,16 @@ Show that no two of the three sets $\emptyset$, $\{\emptyset\}$, and
\end{proof} \end{proof}
\subsection{\partial{Exercise 3}}% \subsection{\verified{Exercise 3}}%
\label{sub:baby-set-theory-3} \label{sub:baby-set-theory-3}
Show that if $B \subseteq C$, then $\mathscr{P} B \subseteq \mathscr{P} C$. Show that if $B \subseteq C$, then $\mathscr{P} B \subseteq \mathscr{P} C$.
\begin{proof} \begin{proof}
\lean{Bookshelf/Enderton/Set/Chapter\_1}
{Enderton.Set.Chapter\_1.exercise\_3}
Let $x \in \mathscr{P} B$. Let $x \in \mathscr{P} B$.
By definition of the \nameref{ref:powerset}, $x$ is a subset of $B$. By definition of the \nameref{ref:powerset}, $x$ is a subset of $B$.
By hypothesis, $B \subseteq C$. By hypothesis, $B \subseteq C$.
@ -169,7 +190,7 @@ Show that if $B \subseteq C$, then $\mathscr{P} B \subseteq \mathscr{P} C$.
\end{proof} \end{proof}
\subsection{\partial{Exercise 4}}% \subsection{\verified{Exercise 4}}%
\label{sub:baby-set-theory-4} \label{sub:baby-set-theory-4}
Assume that $x$ and $y$ are members of a set $B$. Assume that $x$ and $y$ are members of a set $B$.
@ -177,6 +198,9 @@ Show that $\{\{x\}, \{x, y\}\} \in \mathscr{P}\mathscr{P} B.$
\begin{proof} \begin{proof}
\lean{Bookshelf/Enderton/Set/Chapter\_1}
{Enderton.Set.Chapter\_1.exercise\_4}
Let $x$ and $y$ be members of set $B$. Let $x$ and $y$ be members of set $B$.
Then $\{x\}$ and $\{x, y\}$ are subsets of $B$. Then $\{x\}$ and $\{x, y\}$ are subsets of $B$.
By definition of the \nameref{ref:powerset}, $\{x\}$ and $\{x, y\}$ are By definition of the \nameref{ref:powerset}, $\{x\}$ and $\{x, y\}$ are

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@ -1,4 +1,6 @@
import Mathlib.Init.Set import Mathlib.Init.Set
import Mathlib.Data.Set.Basic
import Mathlib.Tactic.LibrarySearch
/-! # Enderton.Chapter_1 /-! # Enderton.Chapter_1
@ -7,7 +9,125 @@ Introduction
namespace Enderton.Set.Chapter_1 namespace Enderton.Set.Chapter_1
/-! ### Exercise 1
Which of the following become true when "∈" is inserted in place of the blank?
Which become true when "⊆" is inserted?
-/
/--
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
`{∅} ___ {∅, {∅}}`
-/
theorem exercise_1a
: {∅} ∈ ({∅, {∅}} : Set (Set (Set α)))
∧ {∅} ⊆ ({∅, {∅}} : Set (Set (Set α))) := ⟨by simp, by simp⟩
/-- #### Exercise 1b
`{∅} ___ {∅, {{∅}}}`
-/
theorem exercise_1b
: {∅} ∉ ({∅, {{∅}}}: Set (Set (Set (Set α))))
∧ {∅} ⊆ ({∅, {{∅}}}: Set (Set (Set (Set α)))) := by
refine ⟨?_, by simp⟩
intro h
simp at h
exact empty_ne_singleton_empty h
/-- #### Exercise 1c
`{{∅}} ___ {∅, {∅}}`
-/
theorem exercise_1c
: {{∅}} ∉ ({∅, {∅}} : Set (Set (Set (Set α))))
∧ {{∅}} ⊆ ({∅, {∅}} : Set (Set (Set (Set α)))) := ⟨by simp, by simp⟩
/-- #### Exercise 1d
`{{∅}} ___ {∅, {{∅}}}`
-/
theorem exercise_1d
: {{∅}} ∈ ({∅, {{∅}}} : Set (Set (Set (Set α))))
∧ ¬ {{∅}} ⊆ ({∅, {{∅}}} : Set (Set (Set (Set α)))) := by
refine ⟨by simp, ?_⟩
intro h
simp at h
exact empty_ne_singleton_empty h
/-- #### Exercise 1e
`{{∅}} ___ {∅, {∅, {∅}}}`
-/
theorem exercise_1e
: {{∅}} ∉ ({∅, {∅, {∅}}} : Set (Set (Set (Set α))))
∧ ¬ {{∅}} ⊆ ({∅, {∅, {∅}}} : Set (Set (Set (Set α)))) := by
apply And.intro
· intro h
simp at h
rw [Set.ext_iff] at h
have nh := h ∅
simp at nh
exact empty_ne_singleton_empty nh
· intro h
simp at h
rw [Set.ext_iff] at h
have nh := h {∅}
simp at nh
/-- ### Exercise 2
Show that no two of the three sets `∅`, `{∅}`, and `{{∅}}` are equal to each
other.
-/
theorem exercise_2
: ∅ ≠ ({∅} : Set (Set α))
∧ ∅ ≠ ({{∅}} : Set (Set (Set α)))
∧ {∅} ≠ ({{∅}} : Set (Set (Set α))) := by
refine ⟨?_, ⟨?_, ?_⟩⟩
· intro h
exact empty_ne_singleton_empty h
· intro h
exact absurd h (Ne.symm $ Set.singleton_ne_empty ({∅} : Set (Set α)))
· intro h
simp at h
exact empty_ne_singleton_empty h
/-- ### Exercise 3
Show that if `B ⊆ C`, then `𝓟 B ⊆ 𝓟 C`.
-/
theorem exercise_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
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)
: {{x}, {x, y}} ∈ Set.powerset (Set.powerset B) := by
unfold Set.powerset
simp only [Set.mem_singleton_iff, Set.mem_setOf_eq]
rw [Set.subset_def]
intro z hz
simp at hz
apply Or.elim hz
· intro h
rwa [h, Set.mem_setOf_eq, Set.singleton_subset_iff]
· intro h
rw [h, Set.mem_setOf_eq]
exact Set.union_subset
(Set.singleton_subset_iff.mpr hx)
(Set.singleton_subset_iff.mpr hy)
end Enderton.Set.Chapter_1 end Enderton.Set.Chapter_1