Add definitions/prompts for Enderton "Axioms and Operations."
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@ -21,14 +21,14 @@
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\label{ref:empty-set-axiom}
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\label{ref:empty-set-axiom}
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There is a set having no members:
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There is a set having no members:
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$$\exists\; B, \forall\; x, x \not\in B.$$
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$$\exists B, \forall x, x \not\in B.$$
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\section{\defined{Extensionality Axiom}}%
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\section{\defined{Extensionality Axiom}}%
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\label{ref:extensionality-axiom}
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\label{ref:extensionality-axiom}
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If two sets have exactly the same members, then they are equal:
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If two sets have exactly the same members, then they are equal:
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$$\forall\; A, \forall\; B,
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$$\forall A, \forall B,
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\left[\forall\; x, (x \in A \iff x \in B) \Rightarrow A = B\right].$$
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\left[\forall x, (x \in A \iff x \in B) \Rightarrow A = B\right].$$
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\begin{axiom}
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\begin{axiom}
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@ -36,17 +36,24 @@ If two sets have exactly the same members, then they are equal:
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\end{axiom}
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\end{axiom}
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\section{\partial{Pair Set}}%
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\label{ref:pair-set}
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For any sets $u$ and $v$, the \textbf{pair set $\{u, v\}$} is the set whose
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only members are $u$ and $v$.
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\section{\partial{Pairing Axiom}}%
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\section{\partial{Pairing Axiom}}%
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\label{ref:pairing-axiom}
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\label{ref:pairing-axiom}
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For any sets $u$ and $v$, there is a set having as members just $u$ and $v$:
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For any sets $u$ and $v$, there is a set having as members just $u$ and $v$:
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$$\forall\; u, \forall\; v, \exists\; B, \forall\; x,
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$$\forall u, \forall v, \exists B, \forall x,
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(x \in B \iff x = u \text{ or } x = v).$$
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(x \in B \iff x = u \text{ or } x = v).$$
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\section{\defined{Powerset}}%
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\section{\defined{Power Set}}%
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\label{ref:powerset}
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\label{ref:power-set}
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The \textbf{powerset} of some set $A$ is the set of all subsets of $A$.
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For any set $a$, the \textbf{power set $\powerset{a}$} is the set whose members
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are exactly the subsets of $a$.
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\begin{definition}
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\begin{definition}
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@ -58,14 +65,28 @@ The \textbf{powerset} of some set $A$ is the set of all subsets of $A$.
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\label{ref:power-set-axiom}
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\label{ref:power-set-axiom}
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For any set $a$, there is a set whose members are exactly the subsets of $a$:
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For any set $a$, there is a set whose members are exactly the subsets of $a$:
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$$\forall\; a, \exists\; B, \forall\; x, (x \in B \iff x \subseteq a).$$
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$$\forall a, \exists B, \forall x, (x \in B \iff x \subseteq a).$$
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\section{\partial{Subset Axioms}}%
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\label{ref:subset-axioms}
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For each formula $\phi$ not containing $B$, the following is an axiom:
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$$\forall t_1, \cdots \forall t_k, \forall c,
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\exists B, \forall x, (x \in B \iff x \in c \land \phi).$$
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\section{\partial{Union Axiom}}%
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\label{ref:union-axiom}
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For any set $A$, there exists a set $B$ whose elements are exactly the members
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of the members of $A$:
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$$\forall A, \exists B, \forall x \left[ x \in B \iff (\exists b \in A) x \in b \right]$$
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\section{\partial{Union Axiom, Preliminary Form}}%
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\section{\partial{Union Axiom, Preliminary Form}}%
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\label{ref:union-axiom-preliminary-form}
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\label{ref:union-axiom-preliminary-form}
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For any sets $a$ and $b$, there is a set whose members are those sets belonging
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For any sets $a$ and $b$, there is a set whose members are those sets belonging
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either to $a$ or to $b$ (or both):
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either to $a$ or to $b$ (or both):
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$$\forall\; a, \forall\; b, \exists\; B, \forall\; x,
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$$\forall a, \forall b, \exists B, \forall x,
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(x \in B \iff x \in a \text{ or } x \in b).$$
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(x \in B \iff x \in a \text{ or } x \in b).$$
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\endgroup
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\endgroup
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@ -209,10 +230,10 @@ Show that if $B \subseteq C$, then $\powerset{B} \subseteq \powerset{C}$.
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{Enderton.Set.Chapter\_1.exercise\_1\_3}
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{Enderton.Set.Chapter\_1.exercise\_1\_3}
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Let $x \in \powerset{B}$.
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Let $x \in \powerset{B}$.
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By definition of the \nameref{ref:powerset}, $x$ is a subset of $B$.
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By definition of the \nameref{ref:power-set}, $x$ is a subset of $B$.
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By hypothesis, $B \subseteq C$.
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By hypothesis, $B \subseteq C$.
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Then $x \subseteq C$.
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Then $x \subseteq C$.
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Again by definition of the \nameref{ref:powerset}, it follows
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Again by definition of the \nameref{ref:power-set}, it follows
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$x \in \powerset{C}$.
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$x \in \powerset{C}$.
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\end{proof}
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\end{proof}
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@ -230,11 +251,11 @@ Show that $\{\{x\}, \{x, y\}\} \in \powerset{\powerset{B}}.$
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Let $x$ and $y$ be members of set $B$.
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Let $x$ and $y$ be members of set $B$.
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Then $\{x\}$ and $\{x, y\}$ are subsets of $B$.
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Then $\{x\}$ and $\{x, y\}$ are subsets of $B$.
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By definition of the \nameref{ref:powerset}, $\{x\}$ and $\{x, y\}$ are
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By definition of the \nameref{ref:power-set}, $\{x\}$ and $\{x, y\}$ are
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members of $\powerset{B}$.
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members of $\powerset{B}$.
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Then $\{\{x\}, \{x, y\}\}$ is a subset of $\powerset{B}$.
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Then $\{\{x\}, \{x, y\}\}$ is a subset of $\powerset{B}$.
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By definition of the \nameref{ref:powerset}, $\{\{x\}, \{x, y\}\}$ is a member
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By definition of the \nameref{ref:power-set}, $\{\{x\}, \{x, y\}\}$ is a
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of $\powerset{\powerset{B}}$.
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member of $\powerset{\powerset{B}}$.
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\end{proof}
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\end{proof}
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@ -371,4 +392,190 @@ List all the members of $V_4$.
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\end{proof}
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\end{proof}
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\chapter{Axioms and Operations}%
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\label{chap:axioms-operations}
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\section{Axioms}%
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\label{sec:axioms}
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\subsection{\unverified{Theorem 2A}}%
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\label{sub:theorem-2a}
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\begin{theorem}[2A]
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There is no set to which every set belongs.
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\end{theorem}
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\begin{proof}
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TODO
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\end{proof}
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\subsection{\unverified{Theorem 2B}}%
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\label{sub:theorem-2b}
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\begin{theorem}[2B]
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For any nonempty set $A$, there exists a unique set $B$ such that for any
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$x$, $$x \in B \iff x \text{ belongs to every member of } A.$$
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\end{theorem}
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\begin{proof}
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TODO
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\end{proof}
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\section{Exercises 3}%
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\label{sec:exercises-3}
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\subsection{\unverified{Exercise 3.1}}%
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\label{sub:exercise-3.1}
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Assume that $A$ is the set of integers divisible by $4$.
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Similarly assume that $B$ and $C$ are the sets of integers divisible by $9$ and
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$10$, respectively.
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What is in $A \cap B \cap C$?
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\begin{proof}
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TODO
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\end{proof}
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\subsection{\unverified{Exercise 3.2}}%
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\label{sub:exercise-3.2}
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Give an example of sets $A$ and $B$ for which $\bigcup A = \bigcup B$ but
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$A \neq B$.
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\begin{proof}
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TODO
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\end{proof}
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\subsection{\unverified{Exercise 3.3}}%
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\label{sub:exercise-3.3}
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Show that every member of a set $A$ is a subset of $\bigcup A$.
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(This was stated as an example in this section.)
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\begin{proof}
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TODO
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\end{proof}
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\subsection{\unverified{Exercise 3.4}}%
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\label{sub:exercise-3.4}
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Show that if $A \subseteq B$, then $\bigcup A \subseteq \bigcup B$.
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\begin{proof}
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TODO
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\end{proof}
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\subsection{\unverified{Exercise 3.5}}%
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\label{sub:exercise-3.5}
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Assume that every member of $\mathscr{A}$ is a subset of $B$.
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Show that $\bigcup \mathscr{A} \subseteq B$.
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\begin{proof}
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TODO
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\end{proof}
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\subsection{\unverified{Exercise 3.6a}}%
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\label{sub:exercise-3.6a}
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Show that for any set $A$, $\bigcup \powerset{A} = A$.
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\begin{proof}
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TODO
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\end{proof}
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\subsection{\unverified{Exercise 3.6b}}%
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\label{sub:exercise-3.6b}
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Show that $A \subseteq \powerset{\bigcup A}$.
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Under what conditions does equality hold?
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\begin{proof}
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TODO
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\end{proof}
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\subsection{\unverified{Exercise 3.7a}}%
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\label{sub:exercise-3.7a}
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Show that for any sets $A$ and $B$,
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$$\powerset{A} \cap \powerset{B} = \powerset(A \cap B).$$
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\begin{proof}
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TODO
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\end{proof}
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\subsection{\unverified{Exercise 3.7b}}%
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\label{sub:exercise-3.7b}
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Show that $\powerset{A} \cup \powerset{B} \subseteq \powerset(A \cup B)$.
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Under what conditions does equality hold?
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\begin{proof}
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TODO
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\end{proof}
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\subsection{\unverified{Exercise 3.8}}%
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\label{sub:exercise-3.8}
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Show that there is no set to which every singleton (that is, every set of the
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form $\{x\}$) belongs.
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[\textit{Suggestion}: Show that from such a set, we could construct a set to
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which every set belonged.]
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\begin{proof}
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TODO
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\end{proof}
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\subsection{\unverified{Exercise 3.9}}%
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\label{sub:exercise-3.9}
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Give an example of sets $a$ and $B$ for which $a \in B$ but
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$\powerset{A} \not\in \powerset{B}$.
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\begin{proof}
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TODO
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\end{proof}
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\subsection{\unverified{Exercise 3.10}}%
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\label{sub:exercise-3.10}
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Show that if $a \in B$, then $\powerset{a} \in \powerset{\powerset{\bigcup B}}$.
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[\textit{Suggestion}: If you need help, look in the Appendix.]
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\begin{proof}
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TODO
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\end{proof}
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\end{document}
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\end{document}
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@ -131,7 +131,7 @@
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\newcommand{\ico}[2]{\left[#1, #2\right)}
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\newcommand{\ico}[2]{\left[#1, #2\right)}
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\newcommand{\ioc}[2]{\left(#1, #2\right]}
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\newcommand{\ioc}[2]{\left(#1, #2\right]}
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\newcommand{\ioo}[2]{\left(#1, #2\right)}
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\newcommand{\ioo}[2]{\left(#1, #2\right)}
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\newcommand{\powerset}[1]{\mathscr{P}\;#1}
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\newcommand{\powerset}[1]{\mathscr{P}\,#1}
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\newcommand{\ubar}[1]{\text{\b{$#1$}}}
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\newcommand{\ubar}[1]{\text{\b{$#1$}}}
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\let\oldemptyset\emptyset
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\let\oldemptyset\emptyset
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