From bf3358629e28f8206b2376273ddee11ce13b9f47 Mon Sep 17 00:00:00 2001 From: Joshua Potter Date: Mon, 22 May 2023 12:52:23 -0600 Subject: [PATCH] Enderton, add prompts to be proven from "Axioms and Operations." --- Bookshelf/Apostol.tex | 6 +- Bookshelf/Enderton/Set.tex | 432 ++++++++++++++++++++++++++++++++++++- preamble.tex | 1 + 3 files changed, 429 insertions(+), 10 deletions(-) diff --git a/Bookshelf/Apostol.tex b/Bookshelf/Apostol.tex index 4aa6c6a..95d1847 100644 --- a/Bookshelf/Apostol.tex +++ b/Bookshelf/Apostol.tex @@ -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} diff --git a/Bookshelf/Enderton/Set.tex b/Bookshelf/Enderton/Set.tex index e376759..be7ccd7 100644 --- a/Bookshelf/Enderton/Set.tex +++ b/Bookshelf/Enderton/Set.tex @@ -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} diff --git a/preamble.tex b/preamble.tex index abdfb77..8e16cf5 100644 --- a/preamble.tex +++ b/preamble.tex @@ -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$\square$}