Enderton exercises 2.
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@ -49,22 +49,22 @@ If $A$ and $B$ are sets such that for every object $t$,
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\section{Baby Set Theory}%
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\label{sec:baby-set-theory}
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\subsection{\verified{Exercise 1}}%
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\label{sub:baby-set-theory-1}
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\subsection{\verified{Exercise 1.1}}%
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\label{sub:exercise-1.1}
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Which of the following become true when "$\in$" is inserted in place of the
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blank?
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Which become true when "$\subseteq$" is inserted?
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\subsubsection{\verified{Exercise 1a}}%
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\label{ssub:baby-set-theory-1a}
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\subsubsection{\verified{Exercise 1.1a}}%
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\label{ssub:exercise-1.1a}
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$\{\emptyset\} \_\_\_\_ \{\emptyset, \{\emptyset\}\}$.
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\begin{proof}
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\lean{Bookshelf/Enderton/Set/Chapter\_1}
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{Enderton.Set.Chapter\_1.exercise\_1a}
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{Enderton.Set.Chapter\_1.exercise\_1\_1a}
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Because the \textit{object} $\{\emptyset\}$ is a member of the right-hand set,
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the statement is \textbf{true} in the case of "$\in$".
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@ -75,15 +75,15 @@ $\{\emptyset\} \_\_\_\_ \{\emptyset, \{\emptyset\}\}$.
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\end{proof}
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\subsubsection{\verified{Exercise 1b}}%
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\label{ssub:baby-set-theory-1b}
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\subsubsection{\verified{Exercise 1.1b}}%
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\label{ssub:exercise-1.11b}
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$\{\emptyset\} \_\_\_\_ \{\emptyset, \{\{\emptyset\}\}\}$.
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\begin{proof}
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\lean{Bookshelf/Enderton/Set/Chapter\_1}
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{Enderton.Set.Chapter\_1.exercise\_1b}
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{Enderton.Set.Chapter\_1.exercise\_1\_1b}
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Because the \textit{object} $\{\emptyset\}$ is not a member of the right-hand
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set, the statement is \textbf{false} in the case of "$\in$".
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@ -93,15 +93,15 @@ $\{\emptyset\} \_\_\_\_ \{\emptyset, \{\{\emptyset\}\}\}$.
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\end{proof}
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\subsubsection{\verified{Exercise 1c}}%
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\label{ssub:baby-set-theory-1c}
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\subsubsection{\verified{Exercise 1.1c}}%
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\label{ssub:exercise-1.1c}
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$\{\{\emptyset\}\} \_\_\_\_ \{\emptyset, \{\emptyset\}\}$.
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\begin{proof}
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\lean{Bookshelf/Enderton/Set/Chapter\_1}
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{Enderton.Set.Chapter\_1.exercise\_1c}
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{Enderton.Set.Chapter\_1.exercise\_1\_1c}
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Because the \textit{object} $\{\{\emptyset\}\}$ is not a member of the
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right-hand set, the statement is \textbf{false} in the case of "$\in$".
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@ -111,15 +111,15 @@ $\{\{\emptyset\}\} \_\_\_\_ \{\emptyset, \{\emptyset\}\}$.
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\end{proof}
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\subsubsection{\verified{Exercise 1d}}%
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\label{ssub:baby-set-theory-1d}
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\subsubsection{\verified{Exercise 1.1d}}%
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\label{ssub:exercise-1.1d}
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$\{\{\emptyset\}\} \_\_\_\_ \{\emptyset, \{\{\emptyset\}\}\}$.
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\begin{proof}
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\lean{Bookshelf/Enderton/Set/Chapter\_1}
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{Enderton.Set.Chapter\_1.exercise\_1d}
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{Enderton.Set.Chapter\_1.exercise\_1\_1d}
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Because the \textit{object} $\{\{\emptyset\}\}$ is a member of the right-hand
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set, the statement is \textbf{true} in the case of "$\in$".
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@ -130,15 +130,15 @@ $\{\{\emptyset\}\} \_\_\_\_ \{\emptyset, \{\{\emptyset\}\}\}$.
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\end{proof}
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\subsubsection{\verified{Exercise 1e}}%
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\label{ssub:baby-set-theory-1e}
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\subsubsection{\verified{Exercise 1.1e}}%
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\label{ssub:exercise-1.1e}
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$\{\{\emptyset\}\} \_\_ \{\emptyset, \{\emptyset, \{\emptyset\}\}\}$.
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\begin{proof}
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\lean{Bookshelf/Enderton/Set/Chapter\_1}
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{Enderton.Set.Chapter\_1.exercise\_1e}
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{Enderton.Set.Chapter\_1.exercise\_1\_1e}
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Because the \textit{object} $\{\{\emptyset\}\}$ is not a member of the
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right-hand set, the statement is \textbf{false} in the case of "$\in$".
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@ -149,8 +149,8 @@ $\{\{\emptyset\}\} \_\_ \{\emptyset, \{\emptyset, \{\emptyset\}\}\}$.
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\end{proof}
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\subsection{\verified{Exercise 2}}%
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\label{sub:baby-set-theory-2}
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\subsection{\verified{Exercise 1.2}}%
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\label{sub:exercise-1.2}
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Show that no two of the three sets $\emptyset$, $\{\emptyset\}$, and
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$\{\{\emptyset\}\}$ are equal to each other.
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@ -158,7 +158,7 @@ Show that no two of the three sets $\emptyset$, $\{\emptyset\}$, and
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\begin{proof}
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\lean{Bookshelf/Enderton/Set/Chapter\_1}
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{Enderton.Set.Chapter\_1.exercise\_2}
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{Enderton.Set.Chapter\_1.exercise\_1\_2}
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By the \nameref{ref:principle-extensionality}, $\emptyset$ is only equal to
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$\emptyset$.
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@ -171,43 +171,176 @@ Show that no two of the three sets $\emptyset$, $\{\emptyset\}$, and
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\end{proof}
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\subsection{\verified{Exercise 3}}%
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\label{sub:baby-set-theory-3}
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\subsection{\verified{Exercise 1.3}}%
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\label{sub:exercise-1.3}
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Show that if $B \subseteq C$, then $\mathscr{P} B \subseteq \mathscr{P} C$.
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Show that if $B \subseteq C$, then $\powerset{B} \subseteq \powerset{C}$.
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\begin{proof}
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\lean{Bookshelf/Enderton/Set/Chapter\_1}
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{Enderton.Set.Chapter\_1.exercise\_3}
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{Enderton.Set.Chapter\_1.exercise\_1\_3}
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Let $x \in \mathscr{P} B$.
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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 hypothesis, $B \subseteq C$.
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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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$x \in \mathscr{P} C$.
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$x \in \powerset{C}$.
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\end{proof}
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\subsection{\verified{Exercise 4}}%
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\label{sub:baby-set-theory-4}
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\subsection{\verified{Exercise 1.4}}%
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\label{sub:exercise-1.4}
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Assume that $x$ and $y$ are members of a set $B$.
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Show that $\{\{x\}, \{x, y\}\} \in \mathscr{P}\mathscr{P} B.$
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Show that $\{\{x\}, \{x, y\}\} \in \powerset{\powerset{B}}.$
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\begin{proof}
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\lean{Bookshelf/Enderton/Set/Chapter\_1}
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{Enderton.Set.Chapter\_1.exercise\_4}
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{Enderton.Set.Chapter\_1.exercise\_1\_4}
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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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By definition of the \nameref{ref:powerset}, $\{x\}$ and $\{x, y\}$ are
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members of $\mathscr{P} B$.
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Then $\{\{x\}, \{x, y\}\}$ is a subset of $\mathscr{P} B$.
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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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By definition of the \nameref{ref:powerset}, $\{\{x\}, \{x, y\}\}$ is a member
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of $\mathscr{P}\mathscr{P} B$.
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of $\powerset{\powerset{B}}$.
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\end{proof}
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\section{Sets - An Informal View}%
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\label{sec:sets-informal-view}
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\subsection{\partial{Exercise 2.1}}%
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\label{sub:exercise-2.1}
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Define the rank of a set $c$ to be the least $\alpha$ such that
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$c \subseteq V_\alpha$.
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Compute the rank of $\{\{\emptyset\}\}$.
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Compute the rank of
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$\{\emptyset, \{\emptyset\}, \{\emptyset, \{\emptyset\}\}\}$.
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\begin{proof}
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We first compute the values of $V_n$ for $0 \leq n \leq 3$ under the
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assumption the set of atoms $A$ at the bottom of the hierarchy is empty.
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\begin{align*}
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V_0 & = \emptyset \\
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V_1 & = V_0 \cup \powerset{V_0} \\
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& = \emptyset \cup \{\emptyset\} \\
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& = \{\emptyset\} \\
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V_2 & = V_1 \cup \powerset{V_1} \\
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& = \{\emptyset\} \cup \powerset{\{\emptyset\}} \\
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& = \{\emptyset\} \cup \{\emptyset, \{\emptyset\}\} \\
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& = \{\emptyset, \{\emptyset\}\} \\
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V_3 & = V_2 \cup \powerset{V_2} \\
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& = \{\emptyset, \{\emptyset\}\} \cup
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\powerset{\{\emptyset, \{\emptyset\}\}} \\
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& = \{\emptyset, \{\emptyset\}\} \cup
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\{\emptyset,
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\{\emptyset\},
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\{\{\emptyset\}\},
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\{\emptyset, \{\emptyset\}\}\} \\
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& = \{\emptyset,
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\{\emptyset\},
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\{\{\emptyset\}\},
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\{\emptyset, \{\emptyset\}\}\}
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\end{align*}
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It then immediately follows $\{\{\emptyset\}\}$ has rank $2$ and
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$\{\emptyset, \{\emptyset\}, \{\emptyset, \{\emptyset\}\}\}$ has rank $3$.
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\end{proof}
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\subsection{\partial{Exercise 2.2}}%
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\label{sub:exercise-2.2}
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We have stated that $V_{\alpha + 1} = A \cup \powerset{V_\alpha}$.
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Prove this at least for $\alpha < 3$.
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\begin{proof}
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Let $A$ be the set of atoms in our set hierarchy.
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Let $P(n)$ be the predicate, "$V_{n + 1} = A \cup \powerset{V_n}$."
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We prove $P(n)$ holds true for all natural numbers $n \geq 1$ via induction.
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\paragraph{Base Case}%
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Let $n = 1$.
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By definition, $V_1 = V_0 \cup \powerset{V_0}$.
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By definition, $V_0 = A$.
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Therefore $V_1 = A \cup \powerset{V_0}$.
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This proves $P(1)$ holds true.
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\paragraph{Induction Step}%
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Suppose $P(n)$ holds true for some $n \geq 1$.
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Consider $V_{n+1}$.
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By definition, $V_{n+1} = V_n \cup \powerset{V_n}$.
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Therefore, by the induction hypothesis,
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\begin{align}
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V_{n+1}
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& = V_n \cup \powerset{V_n}
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\nonumber \\
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& = (A \cup \powerset{V_{n-1}}) \cup \powerset{V_n}
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\nonumber \\
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& = A \cup (\powerset{V_{n-1}} \cup \powerset{V_n})
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\label{sub:exercise-2.2-eq1}
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\end{align}
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But $V_{n-1}$ is a subset of $V_n$.
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\nameref{sub:exercise-1.3} then implies
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$\powerset{V_{n-1}} \subseteq \powerset{V_n}$.
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This means \eqref{sub:exercise-2.2-eq1} can be simplified to
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$$V_{n+1} = A \cup \powerset{V_n},$$
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proving $P(n+1)$ holds true.
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\paragraph{Conclusion}%
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By mathematical induction, it follows for all $n \geq 1$, $P(n)$ is true.
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\end{proof}
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\subsection{\partial{Exercise 2.3}}%
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\label{sub:exercise-2.3}
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List all the members of $V_3$.
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List all the members of $V_4$.
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(It is to be assumed here that there are no atoms.)
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\begin{proof}
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As seen in the proof of \nameref{sub:exercise-2.1},
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$$V_3 = \{
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\emptyset,
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\{\emptyset\},
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\{\{\emptyset\}\},
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\{\emptyset, \{\emptyset\}\}
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\}.$$
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By \nameref{sub:exercise-2.2}, $V_4 = \powerset{V_3}$ (since it is assumed
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there are no atoms).
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Thus
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\begin{align*}
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& V_4 = \{ \\
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& \qquad \emptyset, \\
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& \qquad \{\emptyset\}, \\
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& \qquad \{\{\emptyset\}\}, \\
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& \qquad \{\{\{\emptyset\}\}\}, \\
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& \qquad \{\{\emptyset, \{\emptyset\}\}\}, \\
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& \qquad \{\emptyset, \{\emptyset\}\}, \\
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& \qquad \{\emptyset, \{\{\emptyset\}\}\}, \\
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& \qquad \{\emptyset, \{\emptyset, \{\emptyset\}\}\}, \\
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& \qquad \{\{\emptyset\}, \{\{\emptyset\}\}\}, \\
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& \qquad \{\{\emptyset\}, \{\emptyset, \{\emptyset\}\}\}, \\
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& \qquad \{\{\{\emptyset\}\}, \{\emptyset, \{\emptyset\}\}\}, \\
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& \qquad \{\emptyset, \{\emptyset\}, \{\{\emptyset\}\}\}, \\
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& \qquad \{\emptyset, \{\emptyset\}, \{\emptyset, \{\emptyset\}\}\}, \\
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& \qquad \{\emptyset, \{\{\emptyset\}\}, \{\emptyset, \{\emptyset\}\}\} \\
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& \qquad \{\{\emptyset\}, \{\{\emptyset\}\}, \{\emptyset, \{\emptyset\}\}\}, \\
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& \qquad \{\emptyset, \{\emptyset\}, \{\{\emptyset\}\}, \{\emptyset, \{\emptyset\}\}\} \\
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& \}.
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\end{align*}
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\end{proof}
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@ -9,7 +9,7 @@ Introduction
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namespace Enderton.Set.Chapter_1
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/-! ### Exercise 1
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/-! ### Exercise 1.1
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Which of the following become true when "∈" is inserted in place of the blank?
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Which become true when "⊆" is inserted?
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@ -21,19 +21,19 @@ The `∅` does not equal the singleton set containing `∅`.
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lemma empty_ne_singleton_empty (h : ∅ = ({∅} : Set (Set α))) : False :=
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absurd h (Ne.symm $ Set.singleton_ne_empty (∅ : Set α))
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/-- #### Exercise 1a
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/-- #### Exercise 1.1a
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`{∅} ___ {∅, {∅}}`
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-/
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theorem exercise_1a
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theorem exercise_1_1a
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: {∅} ∈ ({∅, {∅}} : Set (Set (Set α)))
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∧ {∅} ⊆ ({∅, {∅}} : Set (Set (Set α))) := ⟨by simp, by simp⟩
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/-- #### Exercise 1b
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/-- #### Exercise 1.1b
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`{∅} ___ {∅, {{∅}}}`
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-/
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theorem exercise_1b
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theorem exercise_1_1b
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: {∅} ∉ ({∅, {{∅}}}: Set (Set (Set (Set α))))
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∧ {∅} ⊆ ({∅, {{∅}}}: Set (Set (Set (Set α)))) := by
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refine ⟨?_, by simp⟩
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@ -41,19 +41,19 @@ theorem exercise_1b
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simp at h
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exact empty_ne_singleton_empty h
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/-- #### Exercise 1c
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/-- #### Exercise 1.1c
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`{{∅}} ___ {∅, {∅}}`
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-/
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theorem exercise_1c
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theorem exercise_1_1c
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: {{∅}} ∉ ({∅, {∅}} : Set (Set (Set (Set α))))
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∧ {{∅}} ⊆ ({∅, {∅}} : Set (Set (Set (Set α)))) := ⟨by simp, by simp⟩
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/-- #### Exercise 1d
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/-- #### Exercise 1.1d
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`{{∅}} ___ {∅, {{∅}}}`
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-/
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theorem exercise_1d
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theorem exercise_1_1d
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: {{∅}} ∈ ({∅, {{∅}}} : Set (Set (Set (Set α))))
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∧ ¬ {{∅}} ⊆ ({∅, {{∅}}} : Set (Set (Set (Set α)))) := by
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refine ⟨by simp, ?_⟩
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@ -61,11 +61,11 @@ theorem exercise_1d
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simp at h
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exact empty_ne_singleton_empty h
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/-- #### Exercise 1e
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/-- #### Exercise 1.1e
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`{{∅}} ___ {∅, {∅, {∅}}}`
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-/
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theorem exercise_1e
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theorem exercise_1_1e
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: {{∅}} ∉ ({∅, {∅, {∅}}} : Set (Set (Set (Set α))))
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∧ ¬ {{∅}} ⊆ ({∅, {∅, {∅}}} : Set (Set (Set (Set α)))) := by
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apply And.intro
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@ -81,12 +81,12 @@ theorem exercise_1e
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have nh := h {∅}
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simp at nh
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/-- ### Exercise 2
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/-- ### Exercise 1.2
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Show that no two of the three sets `∅`, `{∅}`, and `{{∅}}` are equal to each
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other.
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-/
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theorem exercise_2
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theorem exercise_1_2
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: ∅ ≠ ({∅} : Set (Set α))
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∧ ∅ ≠ ({{∅}} : Set (Set (Set α)))
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∧ {∅} ≠ ({{∅}} : Set (Set (Set α))) := by
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@ -99,22 +99,22 @@ theorem exercise_2
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simp at h
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exact empty_ne_singleton_empty h
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/-- ### Exercise 3
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/-- ### Exercise 1.3
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Show that if `B ⊆ C`, then `𝓟 B ⊆ 𝓟 C`.
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-/
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theorem exercise_3 (h : B ⊆ C) : Set.powerset B ⊆ Set.powerset C := by
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theorem exercise_1_3 (h : B ⊆ C) : Set.powerset B ⊆ Set.powerset C := by
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unfold Set.powerset
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simp
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intro x hx
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exact Set.Subset.trans hx h
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/-- ### Exercise 4
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/-- ### Exercise 1.4
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Assume that `x` and `y` are members of a set `B`. Show that
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`{{x}, {x, y}} ∈ 𝓟 𝓟 B`.
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-/
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theorem exercise_4 (x y : α) (hx : x ∈ B) (hy : y ∈ B)
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theorem exercise_1_4 (x y : α) (hx : x ∈ B) (hy : y ∈ B)
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: {{x}, {x, y}} ∈ Set.powerset (Set.powerset B) := by
|
||||
unfold Set.powerset
|
||||
simp only [Set.mem_singleton_iff, Set.mem_setOf_eq]
|
||||
|
|
|
|||
|
|
@ -1,4 +1,4 @@
|
|||
\usepackage{amsfonts, amsmath, amsthm}
|
||||
\usepackage{amsfonts, amsmath, amssymb, amsthm}
|
||||
\usepackage{bigfoot}
|
||||
\usepackage{comment}
|
||||
\usepackage[shortlabels]{enumitem}
|
||||
|
|
@ -131,7 +131,11 @@
|
|||
\newcommand{\ico}[2]{\left[#1, #2\right)}
|
||||
\newcommand{\ioc}[2]{\left(#1, #2\right]}
|
||||
\newcommand{\ioo}[2]{\left(#1, #2\right)}
|
||||
\newcommand{\powerset}[1]{\mathscr{P}\;#1}
|
||||
\newcommand{\ubar}[1]{\text{\b{$#1$}}}
|
||||
|
||||
\let\oldemptyset\emptyset
|
||||
\let\emptyset\varnothing
|
||||
|
||||
% Close off "private" namespace.
|
||||
\makeatother
|
||||
|
|
|
|||
Loading…
Reference in New Issue