Enderton, Baby Set Theory.
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@ -49,20 +49,23 @@ If $A$ and $B$ are sets such that for every object $t$,
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\section{Baby Set Theory}%
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\section{Baby Set Theory}%
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\label{sec:baby-set-theory}
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\label{sec:baby-set-theory}
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\subsection{\partial{Exercise 1}}%
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\subsection{\verified{Exercise 1}}%
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\label{sub:baby-set-theory-1}
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\label{sub:baby-set-theory-1}
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Which of the following become true when "$\in$" is inserted in place of the
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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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blank?
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Which become true when "$\subseteq$" is inserted?
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Which become true when "$\subseteq$" is inserted?
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\subsubsection{\partial{Exercise 1a}}%
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\subsubsection{\verified{Exercise 1a}}%
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\label{ssub:baby-set-theory-1a}
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\label{ssub:baby-set-theory-1a}
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$\{\emptyset\} \_\_\_\_ \{\emptyset, \{\emptyset\}\}$.
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$\{\emptyset\} \_\_\_\_ \{\emptyset, \{\emptyset\}\}$.
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\begin{proof}
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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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Because the \textit{object} $\{\emptyset\}$ is a member of the right-hand set,
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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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the statement is \textbf{true} in the case of "$\in$".
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@ -72,13 +75,16 @@ $\{\emptyset\} \_\_\_\_ \{\emptyset, \{\emptyset\}\}$.
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\end{proof}
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\end{proof}
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\subsubsection{\partial{Exercise 1b}}%
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\subsubsection{\verified{Exercise 1b}}%
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\label{ssub:baby-set-theory-1b}
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\label{ssub:baby-set-theory-1b}
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$\{\emptyset\} \_\_\_\_ \{\emptyset, \{\{\emptyset\}\}\}$.
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$\{\emptyset\} \_\_\_\_ \{\emptyset, \{\{\emptyset\}\}\}$.
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\begin{proof}
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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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Because the \textit{object} $\{\emptyset\}$ is not a member of the right-hand
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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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set, the statement is \textbf{false} in the case of "$\in$".
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@ -87,13 +93,16 @@ $\{\emptyset\} \_\_\_\_ \{\emptyset, \{\{\emptyset\}\}\}$.
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\end{proof}
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\end{proof}
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\subsubsection{\partial{Exercise 1c}}%
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\subsubsection{\verified{Exercise 1c}}%
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\label{ssub:baby-set-theory-1c}
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\label{ssub:baby-set-theory-1c}
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$\{\{\emptyset\}\} \_\_\_\_ \{\emptyset, \{\emptyset\}\}$.
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$\{\{\emptyset\}\} \_\_\_\_ \{\emptyset, \{\emptyset\}\}$.
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\begin{proof}
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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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Because the \textit{object} $\{\{\emptyset\}\}$ is not a member of the
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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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right-hand set, the statement is \textbf{false} in the case of "$\in$".
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@ -102,13 +111,16 @@ $\{\{\emptyset\}\} \_\_\_\_ \{\emptyset, \{\emptyset\}\}$.
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\end{proof}
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\end{proof}
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\subsubsection{\partial{Exercise 1d}}%
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\subsubsection{\verified{Exercise 1d}}%
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\label{ssub:baby-set-theory-1d}
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\label{ssub:baby-set-theory-1d}
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$\{\{\emptyset\}\} \_\_\_\_ \{\emptyset, \{\{\emptyset\}\}\}$.
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$\{\{\emptyset\}\} \_\_\_\_ \{\emptyset, \{\{\emptyset\}\}\}$.
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\begin{proof}
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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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Because the \textit{object} $\{\{\emptyset\}\}$ is a member of the right-hand
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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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set, the statement is \textbf{true} in the case of "$\in$".
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@ -118,13 +130,16 @@ $\{\{\emptyset\}\} \_\_\_\_ \{\emptyset, \{\{\emptyset\}\}\}$.
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\end{proof}
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\end{proof}
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\subsubsection{\partial{Exercise 1e}}%
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\subsubsection{\verified{Exercise 1e}}%
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\label{ssub:baby-set-theory-1e}
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\label{ssub:baby-set-theory-1e}
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$\{\{\emptyset\}\} \_\_ \{\emptyset, \{\emptyset, \{\emptyset\}\}\}$.
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$\{\{\emptyset\}\} \_\_ \{\emptyset, \{\emptyset, \{\emptyset\}\}\}$.
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\begin{proof}
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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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Because the \textit{object} $\{\{\emptyset\}\}$ is not a member of the
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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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right-hand set, the statement is \textbf{false} in the case of "$\in$".
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@ -134,7 +149,7 @@ $\{\{\emptyset\}\} \_\_ \{\emptyset, \{\emptyset, \{\emptyset\}\}\}$.
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\end{proof}
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\end{proof}
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\subsection{\partial{Exercise 2}}%
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\subsection{\verified{Exercise 2}}%
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\label{sub:baby-set-theory-2}
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\label{sub:baby-set-theory-2}
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Show that no two of the three sets $\emptyset$, $\{\emptyset\}$, and
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Show that no two of the three sets $\emptyset$, $\{\emptyset\}$, and
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@ -142,6 +157,9 @@ Show that no two of the three sets $\emptyset$, $\{\emptyset\}$, and
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\begin{proof}
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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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By the \nameref{ref:principle-extensionality}, $\emptyset$ is only equal to
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By the \nameref{ref:principle-extensionality}, $\emptyset$ is only equal to
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$\emptyset$.
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$\emptyset$.
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This immediately shows it is not equal to the other two.
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This immediately shows it is not equal to the other two.
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@ -153,13 +171,16 @@ Show that no two of the three sets $\emptyset$, $\{\emptyset\}$, and
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\end{proof}
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\end{proof}
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\subsection{\partial{Exercise 3}}%
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\subsection{\verified{Exercise 3}}%
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\label{sub:baby-set-theory-3}
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\label{sub:baby-set-theory-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 $\mathscr{P} B \subseteq \mathscr{P} C$.
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\begin{proof}
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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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Let $x \in \mathscr{P} B$.
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Let $x \in \mathscr{P} B$.
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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:powerset}, $x$ 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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@ -169,7 +190,7 @@ Show that if $B \subseteq C$, then $\mathscr{P} B \subseteq \mathscr{P} C$.
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\end{proof}
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\end{proof}
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\subsection{\partial{Exercise 4}}%
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\subsection{\verified{Exercise 4}}%
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\label{sub:baby-set-theory-4}
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\label{sub:baby-set-theory-4}
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Assume that $x$ and $y$ are members of a set $B$.
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Assume that $x$ and $y$ are members of a set $B$.
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@ -177,6 +198,9 @@ Show that $\{\{x\}, \{x, y\}\} \in \mathscr{P}\mathscr{P} B.$
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\begin{proof}
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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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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:powerset}, $\{x\}$ and $\{x, y\}$ are
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@ -1,4 +1,6 @@
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import Mathlib.Init.Set
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import Mathlib.Init.Set
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import Mathlib.Data.Set.Basic
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import Mathlib.Tactic.LibrarySearch
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/-! # Enderton.Chapter_1
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/-! # Enderton.Chapter_1
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@ -7,7 +9,125 @@ Introduction
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namespace Enderton.Set.Chapter_1
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namespace Enderton.Set.Chapter_1
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/-! ### Exercise 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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-/
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/--
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The `∅` does not equal the singleton set containing `∅`.
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-/
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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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`{∅} ___ {∅, {∅}}`
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-/
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theorem exercise_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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`{∅} ___ {∅, {{∅}}}`
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-/
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theorem exercise_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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intro h
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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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`{{∅}} ___ {∅, {∅}}`
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-/
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theorem exercise_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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`{{∅}} ___ {∅, {{∅}}}`
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-/
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theorem exercise_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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intro h
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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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`{{∅}} ___ {∅, {∅, {∅}}}`
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-/
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theorem exercise_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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· intro h
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simp at h
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rw [Set.ext_iff] at h
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have nh := h ∅
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simp at nh
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exact empty_ne_singleton_empty nh
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· intro h
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simp at h
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rw [Set.ext_iff] at h
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have nh := h {∅}
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simp at nh
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/-- ### Exercise 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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: ∅ ≠ ({∅} : Set (Set α))
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∧ ∅ ≠ ({{∅}} : Set (Set (Set α)))
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∧ {∅} ≠ ({{∅}} : Set (Set (Set α))) := by
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refine ⟨?_, ⟨?_, ?_⟩⟩
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· intro h
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exact empty_ne_singleton_empty h
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· intro h
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exact absurd h (Ne.symm $ Set.singleton_ne_empty ({∅} : Set (Set α)))
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· intro h
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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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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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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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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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: {{x}, {x, y}} ∈ Set.powerset (Set.powerset B) := by
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unfold Set.powerset
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simp only [Set.mem_singleton_iff, Set.mem_setOf_eq]
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rw [Set.subset_def]
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intro z hz
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simp at hz
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apply Or.elim hz
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· intro h
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rwa [h, Set.mem_setOf_eq, Set.singleton_subset_iff]
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· intro h
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rw [h, Set.mem_setOf_eq]
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exact Set.union_subset
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(Set.singleton_subset_iff.mpr hx)
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(Set.singleton_subset_iff.mpr hy)
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end Enderton.Set.Chapter_1
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end Enderton.Set.Chapter_1
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