r
/
bookshelf
1
Fork 0
Code Issues Pull Requests Packages Projects Releases Wiki Activity

Enderton, add prompts to be proven from "Axioms and Operations."

finite-set-exercises
Joshua Potter 2023-05-22 12:52:23 -06:00
parent 88022b723c
commit bf3358629e
3 changed files with 429 additions and 10 deletions
Show Stats Download Patch File Download Diff File Expand all files Collapse all files

6
Bookshelf/Apostol.tex
View File

@ -924,7 +924,7 @@ Prove that every triangular region is measurable and that its area is one half
As an example, consider the image below of triangle $T$ with width $4$ and
height $3$:
\begin{figure}[h]
\begin{figure}[ht]
\includegraphics{right-triangle}
\centering
\end{figure}
@ -962,7 +962,7 @@ Prove that every trapezoid and every parallelogram is measurable and derive the
Let $S$ be a trapezoid with height $h$ and bases $b_1$ and $b_2$, $b_1 < b_2$.
There are three cases to consider:
\begin{figure}[h]
\begin{figure}[ht]
\includegraphics[width=\textwidth]{trapezoid-cases}
\centering
\end{figure}
@ -1362,7 +1362,7 @@ $\floor{-x} =
\begin{proof}
\ \vspace{6pt}
\statementpadding
\lean*{Bookshelf/Apostol/Chapter\_1\_11}
{Apostol.Chapter\_1\_11.exercise\_4b\_1}

432
Bookshelf/Enderton/Set.tex
View File

@ -50,7 +50,7 @@ For any sets $u$ and $v$, the \textbf{pair set $\{u, v\}$} is the set whose
\begin{definition}
\ % Add space
\statementpadding
\lean*{Mathlib/Init/Set}{Set.insert}
@ -67,7 +67,7 @@ For any sets $u$ and $v$, there is a set having as members just $u$ and $v$:
\begin{axiom}
\ % Add space
\statementpadding
\lean*{Mathlib/Init/Set}{Set.insert}
@ -448,22 +448,31 @@ List all the members of $V_4$.
\section{Axioms}%
\label{sec:axioms}
\subsection{\unverified{Theorem 2A}}%
\subsection{\partial{Theorem 2A}}%
\label{sub:theorem-2a}
\begin{theorem}[2A]
There is no set to which every set belongs.
\note{This was revisited after reading Enderton's proof prior.}
\end{theorem}
\begin{proof}
TODO
Let $A$ be an arbitrary set.
Define $B = \{ x \in A \mid x \not\in x \}$.
By the \nameref{ref:subset-axioms}, $B$ is a set.
Then $$B \in B \iff B \in A \land B \not\in B.$$
If $B \in A$, then $B \in B \iff B \not\in B$, a contradiction.
Thus $B \not\in A$.
Since this process holds for any set $A$, there must exist no set to which
every set belongs.
\end{proof}
\subsection{\unverified{Theorem 2B}}%
\subsection{\partial{Theorem 2B}}%
\label{sub:theorem-2b}
\begin{theorem}[2B]
@ -475,7 +484,16 @@ List all the members of $V_4$.
\begin{proof}
TODO
Suppose $A$ is a nonempty set.
This ensures the statement we are trying to prove does not vacuously hold for
all sets $x$ (which would yield a contradiction due to
\nameref{sub:theorem-2b}).
By the \nameref{ref:union-axiom}, $\bigcup A$ is a set.
Define $$B = \{ x \in \bigcup A \mid (\forall b \in A), x \in b \}.$$
By the \nameref{ref:subset-axioms}, $B$ is indeed a set.
By construction,
$$\forall x, x \in B \iff x \text{ belongs to every member of } A.$$
By the \nameref{ref:extensionality-axiom}, $B$ is unique.
\end{proof}
@ -721,7 +739,7 @@ Under what conditions does equality hold?
\begin{proof}
\ % Add space.
\statementpadding
\lean*{Bookshelf/Enderton/Set/Chapter\_1}
{Enderton.Set.Chapter\_1.exercise\_3\_7b\_i}
@ -853,4 +871,404 @@ Show that if $a \in B$, then $\powerset{a} \in \powerset{\powerset{\bigcup B}}$.
\end{proof}
\section{Algebra of Sets}%
\label{sec:algebra-sets}
\subsection{\verified{Commutative Laws}}%
\label{sub:commutative-laws}
For any sets $A$ and $B$,
\begin{align*}
A \cup B = B \cup A \\
A \cap B = B \cap A
\end{align*}
\begin{proof}
\statementpadding
\lean*{Mathlib/Data/Set/Basic}{Set.union\_comm}
\lean{Mathlib/Data/Set/Basic}{Set.inter\_comm}
Let $A$ and $B$ be sets.
We show (i) $A \cup B = B \cup A$ and then (ii) $A \cap B = B \cap A$.
\paragraph{(i)}%
By the definition of the union of sets,
$A \cup B = \{ x \mid x \in A \lor x \in B \}$.
Since $\lor$ is commutative, it follows
\begin{align*}
A \cup B
& = \{ x \mid x \in A \lor x \in B \} \\
& = \{ x \mid x \in B \lor x \in A \} \\
& = B \cup A,
\end{align*}
where the last equality follows again from the definition of the union of
sets.
\paragraph{(ii)}%
By the definition of the intersection of sets,
$A \cap B = \{ x \mid x \in A \land x \in B \}$.
Since $\land$ is commutative, it follows
\begin{align*}
A \cap B
& = \{ x \mid x \in A \land x \in B \} \\
& = \{ x \mid x \in B \land x \in A \} \\
& = B \land A,
\end{align*}
where the last equality follows again from the definition of the
intersection of sets.
\end{proof}
\subsection{\unverified{Associative Laws}}%
\label{sub:associative-laws}
For any sets $A$, $B$ and $C$,
\begin{align*}
A \cup (B \cup C) & = (A \cup B) \cup C \\
(A \cup B) \cup C & = A \cup (B \cup C)
\end{align*}
\begin{proof}
TODO
\end{proof}
\subsection{\unverified{Distributive Laws}}%
\label{sub:distributive-laws}
For any sets $A$, $B$, and $C$,
\begin{align*}
A \cap (B \cup C) & = (A \cap B) \cup (A \cap C) \\
A \cup (B \cap C) & = (A \cup B) \cap (A \cup C)
\end{align*}
\begin{proof}
TODO
\end{proof}
\subsection{\unverified{De Morgan's Laws}}%
\label{sub:de-morgans-laws}
For any sets $A$, $B$, and $C$,
\begin{align*}
C - (A \cup B) & = (C - A) \cap (C - B) \\
C - (A \cap B) & = (C - A) \cup (C - B)
\end{align*}
\begin{proof}
TODO
\end{proof}
\subsection{\unverified{
Identities Involving \texorpdfstring{$\emptyset$}{the Empty Set}}}%
\label{sub:identitives-involving-empty-set}
For any set $A$,
\begin{align*}
A \cup \emptyset & = A \\
A \cap \emptyset & = \emptyset \\
A \cap (C - A) & = \emptyset
\end{align*}
\begin{proof}
TODO
\end{proof}
\subsection{\unverified{Monotonicity}}%
\label{sub:monotonicity}
For any sets $A$, $B$, and $C$,
\begin{align*}
A \subseteq B & \Rightarrow A \cup C \subseteq B \cup C \\
A \subseteq B & \Rightarrow A \cap C \subseteq B \cap C \\
A \subseteq B & \Rightarrow \bigcup A \subseteq \bigcup B
\end{align*}
\begin{proof}
TODO
\end{proof}
\subsection{\unverified{Anti-monotonicity}}%
\label{sub:anti-monotonicity}
For any sets $A$, $B$, and $C$,
\begin{align*}
A \subseteq B & \Rightarrow C - B \subseteq C - A \\
\emptyset \neq A \subseteq B & \Rightarrow \bigcap B \subseteq \bigcap A.
\end{align*}
\begin{proof}
TODO
\end{proof}
\subsection{\unverified{General Distributive Laws}}%
\label{sub:general-distributive-laws}
For any sets $A$ and $\mathscr{B}$,
\begin{align*}
A \cup \bigcap \mathscr{B} & =
\bigcap\; \{ A \cup X \mid X \in \mathscr{B} \}
\quad\text{for}\quad \mathscr{B} \neq \emptyset \\
A \cap \bigcup \mathscr{B} & =
\bigcup\; \{ A \cap X \mid X \in \mathscr{B} \}
\end{align*}
\begin{proof}
TODO
\end{proof}
\subsection{\unverified{General De Morgan's Laws}}%
\label{sub:general-de-morgans-laws}
For any set $C$ and $\mathscr{A} \neq \emptyset$,
\begin{align*}
C - \bigcup \mathscr{A} & = \bigcap\; \{ C - X \mid X \in \mathscr{A} \} \\
C - \bigcap \mathscr{A} & = \bigcup\; \{ C - X \mid X \in \mathscr{A} \}
\end{align*}
\begin{proof}
TODO
\end{proof}
\section{Exercises 4}%
\label{sec:exercises-4}
\subsection{\unverified{Exercise 4.11}}%
\label{sub:exercise-4.11}
Show that for any sets $A$ and $B$,
$$A = (A \cap B) \cup (A - B) \quad\text{and}\quad
A \cup (B - A) = A \cup B.$$
\begin{proof}
TODO
\end{proof}
\subsection{\unverified{Exercise 4.12}}%
\label{sub:exercise-4.12}
Verify the following identity (one of De Morgan's laws):
$$C - (A \cap B) = (C - A) \cup (C - B).$$
\begin{proof}
TODO
\end{proof}
\subsection{\unverified{Exercise 4.13}}%
\label{sub:exercise-4.13}
Show that if $A \subseteq B$, then $C - B \subseteq C - A$.
\begin{proof}
TODO
\end{proof}
\subsection{\unverified{Exercise 4.14}}%
\label{sub:exercise-4.14}
Show by example that for some sets $A$, $B$, and $C$, the set $A - (B - C)$ is
different from $(A - B) - C$.
\begin{proof}
TODO
\end{proof}
\subsection{\unverified{Exercise 4.15}}%
\label{sub:exercise-4.15}
Define the symmetric difference $A + B$ of sets $A$ and $B$ to be the set
$(A - B) \cup (B - A)$.
\subsubsection{\unverified{Exercise 4.15a}}%
\label{ssub:exercise-4.15a}
Show that $A \cap (B + C) = (A \cap B) + (A \cap C)$.
\begin{proof}
TODO
\end{proof}
\subsubsection{\unverified{Exercise 4.15b}}%
\label{ssub:exercise-4.15b}
Show that $A + (B + C) = (A + B) + C$.
\begin{proof}
TODO
\end{proof}
\subsection{\unverified{Exercise 4.16}}%
\label{sub:exercise-4.16}
Simplify:
$$[(A \cup B \cup C) \cap (A \cup B)] - [(A \cup (B - C)) \cap A].$$
\begin{proof}
TODO
\end{proof}
\subsection{\unverified{Exercise 4.17}}%
\label{sub:exercise-4.17}
Show that the following four conditions are equivalent.
\begin{enumerate}[(a)]
\item $A \subseteq B$,
\item $A - B = \emptyset$,
\item $A \cup B = B$,
\item $A \cap B = A$.
\end{enumerate}
\begin{proof}
TODO
\end{proof}
\subsection{\unverified{Exercise 4.18}}%
\label{sub:exercise-4.18}
Assume that $A$ and $B$ are subsets of $S$.
List all of the different sets that can be made from these three by use of the
binary operations $\cup$, $\cap$, and $-$.
\begin{proof}
TODO
\end{proof}
\subsection{\unverified{Exercise 4.19}}%
\label{sub:exercise-4.19}
Is $\powerset{(A - B)}$ always equal to $\powerset{A} - \powerset{B}$?
Is it ever equal to $\powerset{A} - \powerset{B}$?
\begin{proof}
TODO
\end{proof}
\subsection{\unverified{Exercise 4.20}}%
\label{sub:exercise-4.20}
Let $A$, $B$, and $C$ be sets such that $A \cup B = A \cup C$ and
$A \cap B = A \cap C$.
Show that $B = C$.
\begin{proof}
TODO
\end{proof}
\subsection{\unverified{Exercise 4.21}}%
\label{sub:exercise-4.21}
Show that $\bigcup (A \cup B) = \bigcup A \cup \bigcup B$.
\begin{proof}
TODO
\end{proof}
\subsection{\unverified{Exercise 4.22}}%
\label{sub:exercise-4.22}
Show that if $A$ and $B$ are nonempty sets, then
$\bigcap (A \cup B) = \bigcap A \cap \bigcap B$.
\begin{proof}
TODO
\end{proof}
\subsection{\unverified{Exercise 4.23}}%
\label{sub:exercise-4.23}
Show that if $\mathscr{B}$ is nonempty, then
$A \cup \bigcap \mathscr{B} = \bigcap\; \{A \cup X \mid X \in \mathscr{B} \}$.
\begin{proof}
TODO
\end{proof}
\subsection{\unverified{Exercise 4.24a}}%
\label{sub:exercise-4.24a}
Show that if $\mathscr{A}$ is nonempty, then
$\powerset{\bigcap A} = \bigcap\; \{\powerset{X} \mid X \in \mathscr{A} \}$.
\begin{proof}
TODO
\end{proof}
\subsection{\unverified{Exercise 4.24b}}%
\label{sub:exercise-4.24b}
Show that
$$\bigcup\; \{ \powerset{X} \mid X \in \mathscr{A} \} \subseteq
\powerset{\bigcup A}.$$
Under what conditions does equality hold?
\begin{proof}
TODO
\end{proof}
\subsection{\unverified{Exercise 4.25}}%
\label{sub:exercise-4.25}
Is $A \cup \bigcup \mathscr{B}$ always the same as
$\bigcup\; \{ A \cup X \mid X \in \mathscr{B} \}$?
If not, then under what conditions does equality hold?
\begin{proof}
TODO
\end{proof}
\end{document}

1
preamble.tex
View File

@ -88,6 +88,7 @@
\newcommand\@statement[1]{%
\linedivider*\paragraph{\normalfont\normalsize\textit{#1.}}}
\newcommand{\statementpadding}{\ \vspace{8pt}}
\newenvironment{answer}{\@statement{Answer}}{\hfill$\square$}
\newenvironment{axiom}{\@statement{Axiom}}{\hfill$\square$}