Enderton. Chapter 2 exercises, 11-25.

finite-set-exercises
Joshua Potter 2023-06-04 17:34:42 -06:00
parent 4624d58924
commit 2dff3f5794
5 changed files with 663 additions and 66 deletions

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@ -1,5 +1,8 @@
\documentclass{report}
\usepackage{graphicx}
\graphicspath{{./Set/images/}}
\input{../../preamble}
\makeleancommands{../..}
@ -1720,7 +1723,7 @@ Show that $A + (B + C) = (A + B) + C$.
\end{proof}
\subsection{\unverified{Exercise 4.16}}%
\subsection{\verified{Exercise 4.16}}%
\label{sub:exercise-4.16}
Simplify:
@ -1728,11 +1731,26 @@ Simplify:
\begin{proof}
TODO
\lean{Bookshelf/Enderton/Set/Chapter\_2}
{Enderton.Set.Chapter\_2.exercise\_4\_16}
Let $A$, $B$, and $C$ be arbitrary sets.
Then
\begin{align*}
[(A \cup B \cup C) \cap (A \cup B)] & - [(A \cup (B - C)) \cap A] \\
& = [A \cup B] - [A] \\
& = \{ x \mid x \in (A \cup B) \land x \not\in A \} \\
& = \{ x \mid x \in \{ y \mid y \in A \lor y \in B \} \land x \not\in A \} \\
& = \{ x \mid (x \in A \lor x \in B) \land x \not\in A \} \\\
& = \{ x \mid (x \in A \land x \not\in A) \lor (x \in B \land x \not\in A) \} \\
& = \{ x \mid F \lor (X \in B \land x \not\in A) \} \\
& = \{ x \mid x \in B \land x \not\in A \} \\
& = B - A.
\end{align*}
\end{proof}
\subsection{\unverified{Exercise 4.17}}%
\subsection{\verified{Exercise 4.17}}%
\label{sub:exercise-4.17}
Show that the following four conditions are equivalent.
@ -1746,11 +1764,58 @@ Show that the following four conditions are equivalent.
\begin{proof}
TODO
\statementpadding
\lean*{Bookshelf/Enderton/Set/Chapter\_2}
{Enderton.Set.Chapter\_2.exercise\_4\_17\_i}
\lean*{Bookshelf/Enderton/Set/Chapter\_2}
{Enderton.Set.Chapter\_2.exercise\_4\_17\_ii}
\lean*{Bookshelf/Enderton/Set/Chapter\_2}
{Enderton.Set.Chapter\_2.exercise\_4\_17\_iii}
\lean{Bookshelf/Enderton/Set/Chapter\_2}
{Enderton.Set.Chapter\_2.exercise\_4\_17\_iv}
Let $A$ and $B$ be arbitrary sets.
We show that (i) $(a) \Rightarrow (b)$, (ii) $(b) \Rightarrow (c)$, (iii)
$(c) \Rightarrow (d)$, and (iv) $(d) \Rightarrow (a)$.
\paragraph{(i)}%
Suppose $A \subseteq B$.
That is, $\forall t, t \in A \Rightarrow t \in B$.
Then there is no element such that $t \in A$ and $t \not\in B$.
By definition of the relative complement, this immediately implies
$A - B = \emptyset$.
\paragraph{(ii)}%
Suppose $A - B = \emptyset$.
By definition of the relative complement,
$$A - B = \emptyset = \{ x \mid x \in A \land x \not\in B \}.$$
Then, for all $t$,
$\neg(t \in A \land t \not\in B) = t \not\in A \lor t \in B$.
This implies, by definition of the subset, that $A \subseteq B$.
It then immediately follows that $A \cup B = B$.
\paragraph{(iii)}%
Suppose $A \cup B = B$.
Then there is no member of $A$ that is not a member of $B$.
In other words, $A \subseteq B$.
This immediately implies $A \cap B = A$.
\paragraph{(iv)}%
Suppose $A \cap B = A$.
Then every member of $A$ is a member of $B$.
This immediately implies $A \subseteq B$.
\end{proof}
\subsection{\unverified{Exercise 4.18}}%
\subsection{\partial{Exercise 4.18}}%
\label{sub:exercise-4.18}
Assume that $A$ and $B$ are subsets of $S$.
@ -1759,11 +1824,30 @@ List all of the different sets that can be made from these three by use of the
\begin{proof}
TODO
We can reason about this diagrammatically:
\begin{figure}[ht]
\includegraphics[width=0.6\textwidth]{venn-diagram}
\centering
\end{figure}
In the above diagram, we assume the left circle corresponds to set $A$ and the
right circle corresponds to $B$.
The the possible sets we can make via the specified operators are:
\begin{itemize}
\item $A - B$, the left circle excluding the overlapping region.
\item $A \cap B$, the overlapping region.
\item $B - A$, the right circle excluding the overlapping region.
\item $(A \cup B) \cap A$, the left circle.
\item $(A \cup B) \cap B$, the right circle.
\item $(A - B) \cup (B - A)$, the symmetric difference.
\item $A \cup B$, the entire diagram.
\end{itemize}
\end{proof}
\subsection{\unverified{Exercise 4.19}}%
\subsection{\verified{Exercise 4.19}}%
\label{sub:exercise-4.19}
Is $\powerset{(A - B)}$ always equal to $\powerset{A} - \powerset{B}$?
@ -1771,11 +1855,35 @@ Is it ever equal to $\powerset{A} - \powerset{B}$?
\begin{proof}
TODO
\lean{Bookshelf/Enderton/Set/Chapter\_2}
{Enderton.Set.Chapter\_2.exercise\_4\_19}
Let $A$ and $B$ be arbitrary sets.
We show (i) that $\emptyset \in \powerset{(A - B})$ and (ii)
$\emptyset \not\in \powerset{A} - \powerset{B}$.
\paragraph{(i)}%
\label{par:exercise-4.19-i}
By definition of the \nameref{ref:power-set},
$$\powerset{(A - B)} = \{ x \mid x \subseteq A - B \}.$$
But $\emptyset$ is a subset of \textit{every} set.
Thus $\emptyset \in \powerset{(A - B)}$.
\paragraph{(ii)}%
By the same reasoning found in \nameref{par:exercise-4.19-i},
$\emptyset \in \powerset{A}$ and $\emptyset \in \powerset{B}$.
But then, by definition of the relative complement,
$\emptyset \not\in \powerset{A} - \powerset{B}$.
\paragraph{Conclusion}%
By the \nameref{ref:extensionality-axiom}, the two sets are never equal.
\end{proof}
\subsection{\unverified{Exercise 4.20}}%
\subsection{\verified{Exercise 4.20}}%
\label{sub:exercise-4.20}
Let $A$, $B$, and $C$ be sets such that $A \cup B = A \cup C$ and
@ -1784,22 +1892,95 @@ Show that $B = C$.
\begin{proof}
TODO
\lean{Bookshelf/Enderton/Set/Chapter\_2}
{Enderton.Set.Chapter\_2.exercise\_4\_20}
Let $A$, $B$, and $C$ be arbitrary sets.
By the \nameref{ref:extensionality-axiom}, $B = C$ if and only if for all sets
$x$, $x \in B \iff x \in C$.
We prove both directions of this biconditional.
\paragraph{($\Rightarrow$)}%
Suppose $x \in B$.
Then there are two cases to consider:
\subparagraph{Case 1}%
Assume $x \in A$.
Then $x \in A \cap B$.
By hypothesis, $A \cap B = A \cap C$.
Thus $x \in A \cap C$ immediately implying $x \in C$.
\subparagraph{Case 2}%
Assume $x \not\in A$.
Then $x \in A \cup B$.
By hypothesis, $A \cup B = A \cup C$.
Thus $x \in A \cup C$.
Since $x \not\in A$, it follows $x \in C$.
\paragraph{($\Leftarrow$)}%
Suppose $x \in C$.
Then there are two cases to consider:
\subparagraph{Case 1}%
Assume $x \in A$.
Then $x \in A \cap C$.
By hypothesis, $A \cap B = A \cap C$.
Thus $x \in A \cap B$, immediately implying $x \in B$.
\subparagraph{Case 2}%
Assume $x \not\in A$.
Then $x \in A \cup C$.
By hypothesis, $A \cup B = A \cup C$.
Thus $x \in A \cup B$.
Since $x \not\in A$, it follows $x \in B$.
\end{proof}
\subsection{\unverified{Exercise 4.21}}%
\subsection{\verified{Exercise 4.21}}%
\label{sub:exercise-4.21}
Show that $\bigcup (A \cup B) = \bigcup A \cup \bigcup B$.
\begin{proof}
TODO
\lean{Bookshelf/Enderton/Set/Chapter\_2}
{Enderton.Set.Chapter\_2.exercise\_4\_21}
Let $A$ and $B$ be arbitrary sets.
By the \nameref{ref:extensionality-axiom}, the specified equality holds if and
only if for all sets $x$,
$$x \in \bigcup (A \cup B) \iff x \in \bigcup A \cup \bigcup B.$$
We prove both directions of this biconditional.
\paragraph{($\Rightarrow$)}%
Suppose $x \in \bigcup (A \cup B)$.
By definition of the union of sets, there exists some $b \in A \cup B$ such
that $x \in b$.
If $b \in A$, then $x \in \bigcup A$ and $x \in \bigcup A \cup \bigcup B$.
Alternatively, if $b \in B$, then $x \in \bigcup B$ and
$x \in \bigcup A \cup \bigcup B$.
Regardless, $x$ is in the target set.
\paragraph{($\Leftarrow$)}%
Suppose $x \in \bigcup A \cup \bigcup B$.
Then $x \in \bigcup A$ or $x \in \bigcup B$.
WLOG, suppose $x \in \bigcup A$.
By definition of the union of sets, there exists some $b \in A$ such that
$x \in b$.
But then $b \in A \cup B$ meaning $x$ is also a member of
$\bigcup (A \cup B)$.
\end{proof}
\subsection{\unverified{Exercise 4.22}}%
\subsection{\verified{Exercise 4.22}}%
\label{sub:exercise-4.22}
Show that if $A$ and $B$ are nonempty sets, then
@ -1807,11 +1988,39 @@ Show that if $A$ and $B$ are nonempty sets, then
\begin{proof}
TODO
\lean{Bookshelf/Enderton/Set/Chapter\_2}
{Enderton.Set.Chapter\_2.exercise\_4\_22}
Let $A$ and $B$ be arbitrary, nonempty sets.
By the \nameref{ref:extensionality-axiom}, the specified equality holds if and
only if for all sets $x$,
\begin{equation}
\label{sub:exercise-4.22-eq1}
x \in \bigcap (A \cup B) \iff x \in \bigcap A \cap \bigcap B.
\end{equation}
We prove both directions of this biconditional.
\paragraph{($\Rightarrow$)}%
Suppose $x \in \bigcap (A \cup B)$.
Then for all $b \in A \cup B$, $x \in B$.
In other words, for every member $b_1$ of $A$ and every member $b_2$ of $B$,
$x$ is a member of both $b_1$ and $b_2$.
But that implies $x \in \bigcap A$ and $x \in \bigcap B$.
\paragraph{($\Leftarrow$)}%
Suppose $x \in \bigcap A \cap \bigcap B$.
That is, $x \in \bigcap A$ and $x \in \bigcap B$.
By definition of the intersection of sets, forall sets $b$, if $b \in A$,
then $x \in b$.
Likewise, if $b \in B$, then $x \in b$.
In other words, if $b$ is a member of either $A$ or $B$, $x \in b$.
That immediately implies $x \in \bigcap (A \cup B$.
\end{proof}
\subsection{\unverified{Exercise 4.23}}%
\subsection{\partial{Exercise 4.23}}%
\label{sub:exercise-4.23}
Show that if $\mathscr{B}$ is nonempty, then
@ -1819,46 +2028,186 @@ Show that if $\mathscr{B}$ is nonempty, then
\begin{proof}
TODO
Refer to \nameref{sub:general-distributive-laws}.
\end{proof}
\subsection{\unverified{Exercise 4.24a}}%
\subsection{\verified{Exercise 4.24a}}%
\label{sub:exercise-4.24a}
Show that if $\mathscr{A}$ is nonempty, then
$\powerset{\bigcap A} = \bigcap\; \{\powerset{X} \mid X \in \mathscr{A} \}$.
$\powerset{\bigcap\mathscr{A}} =
\bigcap\; \{\powerset{X} \mid X \in \mathscr{A} \}$.
\begin{proof}
TODO
\lean{Bookshelf/Enderton/Set/Chapter\_2}
{Enderton.Set.Chapter\_2.exercise\_4\_24a}
Suppose $\mathscr{A}$ is a nonempty set.
Then $\bigcap \mathscr{A}$ is well-defined.
Therefore
\begin{align*}
\powerset{\bigcap\mathscr{A}}
& = \{ x \mid x \subseteq \bigcap \mathscr{A} \}
& \textref{ref:power-set} \\
& = \{ x \mid x \subseteq
\{ y \mid \forall X \in \mathscr{A}, y \in X \} \}
& \text{def'n intersection} \\
& = \{ x \mid \forall t \in x,
t \in \{ y \mid \forall X \in \mathscr{A}, y \in X \} \}
& \text{def'n subset} \\
& = \{ x \mid \forall t \in x,
(\forall X \in \mathscr{A}, t \in X) \} \\
& = \{ x \mid \forall X \in \mathscr{A},
(\forall t \in x, t \in X) \} \\
& = \{ x \mid \forall X \in \mathscr{A}, x \subseteq X \} \\
& = \{ x \mid \forall X \in \mathscr{A}, x \in \powerset{X} \}
& \textref{ref:power-set-axiom} \\
& = \{ x \mid
\forall t \in \{ \powerset{X} \mid X \in \mathscr{A} \}, x \in t \} \\
& = \bigcap\; \{\powerset{X} \mid X \in \mathscr{A}\}.
\end{align*}
\end{proof}
\subsection{\unverified{Exercise 4.24b}}%
\subsection{\verified{Exercise 4.24b}}%
\label{sub:exercise-4.24b}
Show that
$$\bigcup\; \{ \powerset{X} \mid X \in \mathscr{A} \} \subseteq
\powerset{\bigcup A}.$$
\begin{equation}
\label{sub:exercise-4.24b-eq1}
\bigcup\; \{ \powerset{X} \mid X \in \mathscr{A} \} \subseteq
\powerset{\bigcup\mathscr{A}}.
\end{equation}
Under what conditions does equality hold?
\begin{proof}
TODO
\lean{Bookshelf/Enderton/Set/Chapter\_2}
{Enderton.Set.Chapter\_2.exercise\_4\_24b}
We first prove \eqref{sub:exercise-4.24b-eq1}.
Let $x \in \bigcup\; \{ \powerset{X} \mid X \in \mathscr{A} \}$.
By definition of the union of sets,
$(\exists X \in \mathscr{A}), x \in \powerset{X}$.
By definition of the \nameref{ref:power-set}, $x \subseteq X$.
By \nameref{sub:exercise-3.3}, $X \subseteq \bigcup \mathscr{A}$.
Therefore $x \subseteq \bigcup \mathscr{A}$, proving
$x \in \powerset{\mathscr{A}}$ as expected.
\suitdivider
\noindent
We show $\powerset{\bigcup A} \subseteq
\bigcup\;\{ \powerset{X} \mid X \in \mathscr{A} \}$ if and only if
$\bigcup\mathscr{A} \in \mathscr{A}$.
\paragraph{($\Rightarrow$)}%
Suppose $\powerset{\bigcup\mathscr{A}} \subseteq
\bigcup\;\{ \powerset{X} \mid X \in \mathscr{A} \}$.
By definition of the \nameref{ref:power-set},
$\bigcup\mathscr{A} \in \powerset{\bigcup\mathscr{A}}$.
By hypothesis, $\bigcup\mathscr{A} \in
\bigcup\;\{ \powerset{X} \mid X \in \mathscr{A} \}$.
By definition of the union of sets, there exists some $X \in \mathscr{A}$
such that $\bigcup\mathscr{A} \in \powerset{X}$.
That is, $\bigcup\mathscr{A} \subseteq X$.
But $\bigcup\mathscr{A}$ cannot be a proper subset of $X$ since
$X \in \mathscr{A}$.
Thus $\bigcup\mathscr{A} = X$.
This proves $\bigcup\mathscr{A} \in
\bigcup\;\{ \powerset{X} \mid X \in \mathscr{A} \}$.
\paragraph{($\Leftarrow$)}%
Suppose $\bigcup\mathscr{A} \in A$.
Let $x \in \powerset{\bigcup\mathscr{A}}$.
Since $\bigcup\mathscr{A} \in \mathscr{A}$, it immediately follows that
$x \in \{\powerset{X} \mid X \in \mathscr{A}\}$.
\paragraph{Conclusion}%
Equality follows immediately from this fact in conjunction with the proof
of \eqref{sub:exercise-4.24b-eq1}.
\end{proof}
\subsection{\unverified{Exercise 4.25}}%
\subsection{\verified{Exercise 4.25}}%
\label{sub:exercise-4.25}
Is $A \cup \bigcup \mathscr{B}$ always the same as
$\bigcup\; \{ A \cup X \mid X \in \mathscr{B} \}$?
$\bigcup\;\{ A \cup X \mid X \in \mathscr{B} \}$?
If not, then under what conditions does equality hold?
\begin{proof}
TODO
\lean{Bookshelf/Enderton/Set/Chapter\_2}
{Enderton.Set.Chapter\_2.exercise\_4\_25}
We prove that
\begin{equation}
\label{sub:exercise-4.25-eq1}
A \cup \bigcup \mathscr{B} =
\bigcup\;\{ A \cup X \mid X \in \mathscr{B} \}
\end{equation}
if and only if $A = \emptyset$ or $\mathscr{B} \neq \emptyset$.
We prove both directions of this biconditional.
\paragraph{($\Rightarrow$)}%
Suppose \eqref{sub:exercise-4.25-eq1} holds true.
There are two cases to consider:
\subparagraph{Case 1}%
Suppose $B \neq \emptyset$.
Then $A = \emptyset \lor \mathscr{B} \neq \emptyset$ holds trivially.
\subparagraph{Case 2}%
Suppose $B = \emptyset$.
Then $$A \cup \bigcup \mathscr{B} = A \cup \bigcup \emptyset = A$$ and
$$
\bigcup\;\{ A \cup X \mid X \in \mathscr{B} \}
= \bigcup \emptyset \\
= \emptyset.
$$
Then by hypothesis \eqref{sub:exercise-4.25-eq1}, $A = \emptyset$.
Then $A = \emptyset \lor \mathscr{B} \neq \emptyset$ holds trivially.
\paragraph{($\Leftarrow$)}%
Suppose $A = \emptyset$ or $\mathscr{B} \neq \emptyset$.
There are two cases to consider:
\paragraph{Case 1}%
Suppose $A = \emptyset$.
Then $A \cup \bigcup \mathscr{B} = \bigcup{\mathscr{B}}$.
Likewise,
$$
\bigcup \{ A \cup X \mid X \in \mathscr{B} \}
= \bigcup \{ X \mid X \in \mathscr{B} \} \\
= \bigcup \mathscr{B}.
$$
Therefore \eqref{sub:exercise-4.25-eq1} holds.
\paragraph{Case 2}%
Suppose $B \neq \emptyset$.
Then
\begin{align*}
A \cup \bigcup\mathscr{B}
& = \{ x \mid x \in A \lor x \in \bigcup\mathscr{B} \} \\
& = \{ x \mid x \in A \lor (\exists b \in \mathscr{B}) x \in b \} \\
& = \{ x \mid (\exists b \in \mathscr{B}) x \in A \lor x \in b \} \\
& = \{ x \mid (\exists b \in \mathscr{B}) x \in A \cup b \} \\
& = \{ x \mid x \in \bigcup \{ A \cup X \mid X \in \mathscr{B} \} \\
& = \bigcup \{ A \cup X \mid X \in \mathscr{B} \}.
\end{align*}
Therefore \eqref{sub:exercise-4.25-eq1} holds.
\end{proof}

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@ -105,8 +105,6 @@ theorem exercise_3_4 (h : A ⊆ B) : ⋃₀ A ⊆ ⋃₀ B := by
Assume that every member of `𝓐` is a subset of `B`. Show that ` 𝓐 ⊆ B`.
-/
theorem exercise_3_5 (h : ∀ x ∈ 𝓐, x ⊆ B) : ⋃₀ 𝓐 ⊆ B := by
unfold Set.sUnion sSup Set.instSupSetSet
simp only
show ∀ y ∈ { a | ∃ t, t ∈ 𝓐 ∧ a ∈ t }, y ∈ B
intro y hy
rw [Set.mem_setOf_eq] at hy
@ -117,9 +115,8 @@ theorem exercise_3_5 (h : ∀ x ∈ 𝓐, x ⊆ B) : ⋃₀ 𝓐 ⊆ B := by
Show that for any set `A`, ` 𝓟 A = A`.
-/
theorem exercise_3_6a : ⋃₀ (Set.powerset A) = A := by
unfold Set.sUnion sSup Set.instSupSetSet Set.powerset
simp only
theorem exercise_3_6a : ⋃₀ (𝒫 A) = A := by
show { a | ∃ t, t ∈ { t | t ⊆ A } ∧ a ∈ t } = A
ext x
apply Iff.intro
· intro hx
@ -136,11 +133,10 @@ theorem exercise_3_6a : ⋃₀ (Set.powerset A) = A := by
Show that `A ⊆ 𝓟 A`. Under what conditions does equality hold?
-/
theorem exercise_3_6b
: A ⊆ Set.powerset (⋃₀ A)
∧ (A = Set.powerset (⋃₀ A) ↔ ∃ B, A = Set.powerset B) := by
: A ⊆ 𝒫 (⋃₀ A)
∧ (A = 𝒫 (⋃₀ A) ↔ ∃ B, A = 𝒫 B) := by
apply And.intro
· unfold Set.powerset
show ∀ x ∈ A, x ∈ { t | t ⊆ ⋃₀ A }
· show ∀ x ∈ A, x ∈ { t | t ⊆ ⋃₀ A }
intro x hx
rw [Set.mem_setOf]
exact exercise_3_3 x hx
@ -156,10 +152,8 @@ theorem exercise_3_6b
Show that for any sets `A` and `B`, `𝓟 A ∩ 𝓟 B = 𝓟 (A ∩ B)`.
-/
theorem exercise_3_7A
: Set.powerset A ∩ Set.powerset B = Set.powerset (A ∩ B) := by
suffices
Set.powerset A ∩ Set.powerset B ⊆ Set.powerset (A ∩ B) ∧
Set.powerset (A ∩ B) ⊆ Set.powerset A ∩ Set.powerset B from
: 𝒫 A ∩ 𝒫 B = 𝒫 (A ∩ B) := by
suffices 𝒫 A ∩ 𝒫 B ⊆ 𝒫 (A ∩ B) ∧ 𝒫 (A ∩ B) ⊆ 𝒫 A ∩ 𝒫 B from
subset_antisymm this.left this.right
apply And.intro
· unfold Set.powerset
@ -178,7 +172,7 @@ theorem exercise_3_7A
Show that `𝓟 A 𝓟 B ⊆ 𝓟 (A B)`.
-/
theorem exercise_3_7b_i
: Set.powerset A Set.powerset B ⊆ Set.powerset (A B) := by
: 𝒫 A 𝒫 B ⊆ 𝒫 (A B) := by
unfold Set.powerset
intro x hx
simp at hx
@ -195,7 +189,7 @@ theorem exercise_3_7b_i
Under what conditions does `𝓟 A 𝓟 B = 𝓟 (A B)`.?
-/
theorem exercise_3_7b_ii
: Set.powerset A Set.powerset B = Set.powerset (A B) ↔ A ⊆ B B ⊆ A := by
: 𝒫 A 𝒫 B = 𝒫 (A B) ↔ A ⊆ B B ⊆ A := by
unfold Set.powerset
apply Iff.intro
· intro h
@ -247,15 +241,15 @@ theorem exercise_3_7b_ii
Give an example of sets `a` and `B` for which `a ∈ B` but `𝓟 a ∉ 𝓟 B`.
-/
theorem exercise_3_9 (ha : a = {1}) (hB : B = {{1}})
: a ∈ B ∧ Set.powerset a ∉ Set.powerset B := by
: a ∈ B ∧ 𝒫 a ∉ 𝒫 B := by
apply And.intro
· rw [ha, hB]
simp
· intro h
have h₁ : Set.powerset a = {∅, {1}} := by
have h₁ : 𝒫 a = {∅, {1}} := by
rw [ha]
exact Set.powerset_singleton 1
have h₂ : Set.powerset B = {∅, {{1}}} := by
have h₂ : 𝒫 B = {∅, {{1}}} := by
rw [hB]
exact Set.powerset_singleton {1}
rw [h₁, h₂] at h
@ -275,7 +269,7 @@ theorem exercise_3_9 (ha : a = {1}) (hB : B = {{1}})
Show that if `a ∈ B`, then `𝓟 a ∈ 𝓟 𝓟 B`.
-/
theorem exercise_3_10 (ha : a ∈ B)
: Set.powerset a ∈ Set.powerset (Set.powerset (⋃₀ B)) := by
: 𝒫 a ∈ 𝒫 (𝒫 (⋃₀ B)) := by
have h₁ := exercise_3_3 a ha
have h₂ := Chapter_1.exercise_1_3 h₁
generalize hb : 𝒫 (⋃₀ B) = b
@ -289,11 +283,7 @@ Show that for any sets `A` and `B`, `A = (A ∩ B) (A - B)`.
-/
theorem exercise_4_11_i {A B : Set α}
: A = (A ∩ B) (A \ B) := by
unfold Union.union Set.instUnionSet Set.union
unfold SDiff.sdiff Set.instSDiffSet Set.diff
unfold Inter.inter Set.instInterSet Set.inter
unfold Membership.mem Set.instMembershipSet Set.Mem setOf
simp only
show A = fun a => A a ∧ B a A a ∧ ¬B a
suffices ∀ x, (A x ∧ (B x ¬B x)) = A x by
conv => rhs; ext x; rw [← and_or_left, this]
intro x
@ -310,10 +300,7 @@ Show that for any sets `A` and `B`, `A (B - A) = A B`.
-/
theorem exercise_4_11_ii {A B : Set α}
: A (B \ A) = A B := by
unfold Union.union Set.instUnionSet Set.union
unfold SDiff.sdiff Set.instSDiffSet Set.diff
unfold Membership.mem Set.instMembershipSet Set.Mem setOf
simp only
show (fun a => A a B a ∧ ¬A a) = fun a => A a B a
suffices ∀ x, ((A x B x) ∧ (A x ¬A x)) = (A x B x) by
conv => lhs; ext x; rw [or_and_left, this x]
intro x
@ -418,4 +405,249 @@ theorem exercise_4_14 : A \ (B \ C) ≠ (A \ B) \ C := by
end
/-- ### Exercise 4.16
Simplify:
`[(A B C) ∩ (A B)] - [(A (B - C)) ∩ A]`
-/
theorem exercise_4_16 {A B C : Set α}
: ((A B C) ∩ (A B)) \ ((A (B \ C)) ∩ A) = B \ A := by
calc ((A B C) ∩ (A B)) \ ((A (B \ C)) ∩ A)
_ = (A B) \ ((A (B \ C)) ∩ A) := by rw [Set.union_inter_cancel_left]
_ = (A B) \ A := by rw [Set.union_inter_cancel_left]
_ = B \ A := by rw [Set.union_diff_left]
/-! ### Exercise 4.17
Show that the following four conditions are equivalent.
(a) `A ⊆ B`
(b) `A - B = ∅`
(c) `A B = B`
(d) `A ∩ B = A`
-/
theorem exercise_4_17_i {A B : Set α} (h : A ⊆ B)
: A \ B = ∅ := by
ext x
apply Iff.intro
· intro hx
exact absurd (h hx.left) hx.right
· intro hx
exact False.elim hx
theorem exercise_4_17_ii {A B : Set α} (h : A \ B = ∅)
: A B = B := by
suffices A ⊆ B from Set.left_subset_union_eq_self this
show ∀ t, t ∈ A → t ∈ B
intro t ht
rw [Set.ext_iff] at h
by_contra nt
exact (h t).mp ⟨ht, nt⟩
theorem exercise_4_17_iii {A B : Set α} (h : A B = B)
: A ∩ B = A := by
suffices A ⊆ B from Set.inter_eq_left_iff_subset.mpr this
exact Set.union_eq_right_iff_subset.mp h
theorem exercise_4_17_iv {A B : Set α} (h : A ∩ B = A)
: A ⊆ B := Set.inter_eq_left_iff_subset.mp h
/-- ### Exercise 4.19
Is `𝒫 (A - B)` always equal to `𝒫 A - 𝒫 B`? Is it ever equal to `𝒫 A - 𝒫 B`?
-/
theorem exercise_4_19 {A B : Set α}
: 𝒫 (A \ B) ≠ (𝒫 A) \ (𝒫 B) := by
intro h
have he : ∅ ∈ 𝒫 (A \ B) := by simp
have ne : ∅ ∉ (𝒫 A) \ (𝒫 B) := by simp
rw [Set.ext_iff] at h
have := h ∅
exact absurd (this.mp he) ne
/-- ### Exercise 4.20
Let `A`, `B`, and `C` be sets such that `A B = A C` and `A ∩ B = A ∩ C`.
Show that `B = C`.
-/
theorem exercise_4_20 {A B C : Set α}
(hu : A B = A C) (hi : A ∩ B = A ∩ C) : B = C := by
ext x
apply Iff.intro
· intro hB
by_cases hA : x ∈ A
· have : x ∈ A ∩ B := Set.mem_inter hA hB
rw [hi] at this
exact this.right
· have : x ∈ A B := Set.mem_union_right A hB
rw [hu] at this
exact Or.elim this (absurd · hA) (by simp)
· intro hC
by_cases hA : x ∈ A
· have : x ∈ A ∩ C := Set.mem_inter hA hC
rw [← hi] at this
exact this.right
· have : x ∈ A C := Set.mem_union_right A hC
rw [← hu] at this
exact Or.elim this (absurd · hA) (by simp)
/-- ### Exercise 4.21
Show that ` (A B) = ( A) ( B)`.
-/
theorem exercise_4_21 {A B : Set (Set α)}
: ⋃₀ (A B) = (⋃₀ A) (⋃₀ B) := by
ext x
apply Iff.intro
· intro hx
have ⟨t, ht⟩ : ∃ t, t ∈ A B ∧ x ∈ t := hx
apply Or.elim ht.left
· intro hA
exact Or.inl ⟨t, ⟨hA, ht.right⟩⟩
· intro hB
exact Or.inr ⟨t, ⟨hB, ht.right⟩⟩
· intro hx
apply Or.elim hx
· intro hA
have ⟨t, ht⟩ : ∃ t, t ∈ A ∧ x ∈ t := hA
exact ⟨t, ⟨Set.mem_union_left B ht.left, ht.right⟩⟩
· intro hB
have ⟨t, ht⟩ : ∃ t, t ∈ B ∧ x ∈ t := hB
exact ⟨t, ⟨Set.mem_union_right A ht.left, ht.right⟩⟩
/-- ### Exercise 4.22
Show that if `A` and `B` are nonempty sets, then `⋂ (A B) = ⋂ A ∩ ⋂ B`.
-/
theorem exercise_4_22 {A B : Set (Set α)}
: ⋂₀ (A B) = ⋂₀ A ∩ ⋂₀ B := by
ext x
apply Iff.intro
· intro hx
have : ∀ t : Set α, t ∈ A B → x ∈ t := hx
show (∀ t : Set α, t ∈ A → x ∈ t) ∧ (∀ t : Set α, t ∈ B → x ∈ t)
rw [← forall_and]
intro t
exact ⟨
fun ht => this t (Set.mem_union_left B ht),
fun ht => this t (Set.mem_union_right A ht)
· intro hx
have : ∀ t : Set α, (t ∈ A → x ∈ t) ∧ (t ∈ B → x ∈ t) := by
have : (∀ t : Set α, t ∈ A → x ∈ t) ∧ (∀ t : Set α, t ∈ B → x ∈ t) := hx
rwa [← forall_and] at this
show ∀ (t : Set α), t ∈ A B → x ∈ t
intro t ht
apply Or.elim ht
· intro hA
exact (this t).left hA
· intro hB
exact (this t).right hB
/-- ### Exercise 4.24a
Show that is `𝓐` is nonempty, then `𝒫 (⋂ 𝓐) = ⋂ { 𝒫 X | X ∈ 𝓐 }`.
-/
theorem exercise_4_24a {𝓐 : Set (Set α)}
: 𝒫 (⋂₀ 𝓐) = ⋂₀ { 𝒫 X | X ∈ 𝓐 } := by
calc 𝒫 (⋂₀ 𝓐)
_ = { x | x ⊆ ⋂₀ 𝓐 } := rfl
_ = { x | x ⊆ { y | ∀ X, X ∈ 𝓐 → y ∈ X } } := rfl
_ = { x | ∀ t ∈ x, t ∈ { y | ∀ X, X ∈ 𝓐 → y ∈ X } } := rfl
_ = { x | ∀ t ∈ x, (∀ X, X ∈ 𝓐 → t ∈ X) } := rfl
_ = { x | ∀ X ∈ 𝓐, (∀ t, t ∈ x → t ∈ X) } := by
ext
rw [Set.mem_setOf, Set.mem_setOf, forall_mem_comm (· ∈ ·)]
_ = { x | ∀ X ∈ 𝓐, x ⊆ X} := rfl
_ = { x | ∀ X ∈ 𝓐, x ∈ 𝒫 X } := rfl
_ = { x | ∀ t ∈ { 𝒫 X | X ∈ 𝓐 }, x ∈ t} := by simp
_ = ⋂₀ { 𝒫 X | X ∈ 𝓐 } := rfl
/-- ### Exercise 4.24b
Show that
```
{𝒫 X | X ∈ 𝓐} ⊆ 𝒫 𝓐.
```
Under what conditions does equality hold?
-/
theorem exercise_4_24b {𝓐 : Set (Set α)}
: (⋃₀ { 𝒫 X | X ∈ 𝓐 } ⊆ 𝒫 ⋃₀ 𝓐)
∧ ((⋃₀ { 𝒫 X | X ∈ 𝓐 } = 𝒫 ⋃₀ 𝓐) ↔ (⋃₀ 𝓐𝓐)) := by
have hS : (⋃₀ { 𝒫 X | X ∈ 𝓐 } ⊆ 𝒫 ⋃₀ 𝓐) := by
simp
exact exercise_3_3
refine ⟨hS, ?_⟩
apply Iff.intro
· intro rS
have rS : 𝒫 ⋃₀ 𝓐 ⊆ ⋃₀ { 𝒫 X | X ∈ 𝓐 } :=
(Set.Subset.antisymm_iff.mp rS).right
have hA : ⋃₀ 𝓐 ∈ ⋃₀ { 𝒫 X | X ∈ 𝓐 } :=
rS Set.self_mem_powerset_self
conv at hA =>
rhs
unfold Set.sUnion sSup Set.instSupSetSet
simp only
have ⟨X, ⟨⟨x, hx⟩, ht⟩⟩ := Set.mem_setOf.mp hA
have : ⋃₀ 𝓐 = x := by
rw [← hx.right] at ht
have hl : ⋃₀ 𝓐 ⊆ x := ht
have hr : x ⊆ ⋃₀ 𝓐 := exercise_3_3 x hx.left
exact Set.Subset.antisymm hl hr
rw [← this] at hx
exact hx.left
· intro hA
suffices 𝒫 ⋃₀ 𝓐 ⊆ ⋃₀ { 𝒫 X | X ∈ 𝓐 } from
subset_antisymm hS this
show ∀ x, x ∈ 𝒫 ⋃₀ 𝓐 → x ∈ ⋃₀ { x | ∃ X, X ∈ 𝓐𝒫 X = x }
intro x hx
unfold Set.sUnion sSup Set.instSupSetSet
simp only [Set.mem_setOf_eq, exists_exists_and_eq_and, Set.mem_powerset_iff]
exact ⟨⋃₀ 𝓐, ⟨hA, hx⟩⟩
/-- ### Exercise 4.25
Is `A ( 𝓑)` always the same as ` { A X | X ∈ 𝓑 }`? If not, then under
what conditions does equality hold?
-/
theorem exercise_4_25 {A : Set α} (𝓑 : Set (Set α))
: (A (⋃₀ 𝓑) = ⋃₀ { A X | X ∈ 𝓑 }) ↔ (A = ∅ Set.Nonempty 𝓑) := by
apply Iff.intro
· intro h
by_cases h𝓑 : Set.Nonempty 𝓑
· exact Or.inr h𝓑
· have : 𝓑 = ∅ := Set.not_nonempty_iff_eq_empty.mp h𝓑
rw [this] at h
simp at h
exact Or.inl h
· intro h
apply Or.elim h
· intro hA
rw [hA]
simp
· intro h𝓑
calc A (⋃₀ 𝓑)
_ = { x | x ∈ A x ∈ ⋃₀ 𝓑} := rfl
_ = { x | x ∈ A (∃ b ∈ 𝓑, x ∈ b) } := rfl
_ = { x | ∃ b ∈ 𝓑, x ∈ A x ∈ b } := by
ext x
rw [Set.mem_setOf, Set.mem_setOf]
apply Iff.intro
· intro hx
apply Or.elim hx
· intro hA
have ⟨b, hb⟩ := Set.nonempty_def.mp h𝓑
exact ⟨b, ⟨hb, Or.inl hA⟩⟩
· intro ⟨b, hb⟩
exact ⟨b, ⟨hb.left, Or.inr hb.right⟩⟩
· intro ⟨b, ⟨hb, hx⟩⟩
apply Or.elim hx
· exact (Or.inl ·)
· intro h
exact Or.inr ⟨b, ⟨hb, h⟩⟩
_ = { x | ∃ b ∈ 𝓑, x ∈ A b } := rfl
_ = { x | ∃ t, t ∈ { y | ∃ X, X ∈ 𝓑 ∧ A X = y } ∧ x ∈ t } := by simp
_ = ⋃₀ { A X | X ∈ 𝓑 } := rfl
end Enderton.Set.Chapter_2

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@ -9,10 +9,23 @@ Additional theorems and definitions related to basic logic.
/--
The de Morgan law that distributes negation across a conjunction.
-/
theorem not_and_de_morgan {a b : Prop} : (¬(a ∧ b)) ↔ (¬ a ¬ b) := by
theorem not_and_de_morgan : (¬(p ∧ q)) ↔ (¬p ¬q) := by
tauto
/--
Renaming of `not_or` to indicate its relationship to de Morgan's laws.
-/
theorem not_or_de_morgan : ¬(p q) ↔ ¬p ∧ ¬q := not_or
/--
Universal quantification across nested set memberships can be commuted in either
order.
-/
theorem forall_mem_comm {X : Set α} {Y : Set β} (p : α → β → Prop)
: (∀ u ∈ X, (∀ v, v ∈ Y → p u v)) = (∀ v ∈ Y, (∀ u, u ∈ X → p u v)) := by
refine propext ?_
apply Iff.intro
· intro h v hv u hu
exact h u hu v hv
· intro h u hu v hv
exact h v hv u hu

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@ -89,6 +89,14 @@ theorem mem_mem_imp_pair_subset {x y : α}
· intro hy'
rwa [hy']
/-! ## Powerset -/
/--
Every `Set` is a member of its own powerset.
-/
theorem self_mem_powerset_self {A : Set α}
: A ∈ 𝒫 A := subset_self A
/-! ## Symmetric Difference -/
/--
@ -116,8 +124,7 @@ This is the contraposition of `mem_symm_diff_iff_exclusive_mem`.
-/
theorem not_mem_symm_diff_inter_or_not_union {A B : Set α}
: x ∉ (A ∆ B) ↔ (x ∈ A ∩ B) (x ∉ A B) := by
unfold symmDiff
simp
show ¬(x ∈ A ∧ ¬x ∈ B x ∈ B ∧ ¬x ∈ A) ↔ x ∈ A ∧ x ∈ B ¬(x ∈ A x ∈ B)
rw [
not_or_de_morgan,
not_and_de_morgan, not_and_de_morgan,
@ -125,16 +132,12 @@ theorem not_mem_symm_diff_inter_or_not_union {A B : Set α}
not_or_de_morgan
]
apply Iff.intro
· intro hx
apply Or.elim hx.left
· intro nx
apply Or.elim nx.left
· intro nA
exact Or.elim hx.right
(fun nB => Or.inr ⟨nA, nB⟩)
(fun hA => absurd hA nA)
exact Or.elim nx.right (Or.inr ⟨nA, ·⟩) (absurd · nA)
· intro hB
apply Or.elim hx.right
(fun nB => absurd hB nB)
(fun hA => Or.inl ⟨hA, hB⟩)
exact Or.elim nx.right (absurd hB ·) (Or.inl ⟨·, hB⟩)
· intro hx
apply Or.elim hx
· intro hy