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Enderton. Begin progressing through natural numbers.

finite-set-exercises
Joshua Potter 2023-07-19 19:25:44 -06:00
parent 4679b66abe
commit 9b8ddd2b0d
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6
Bookshelf/Apostol.tex
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@ -187,7 +187,7 @@ That is to say, for each $k = 1, 2, \ldots, n$, there is a real number $s_k$
\begin{note}
At each of the endpoints $x_{k-1}$ and $x_k$ the function must have some
well-defined value, but this need not be the same as $s_k$.
well-defined value, but this need not be the same as $s_k$.
\end{note}
\begin{definition}
@ -1614,8 +1614,8 @@ If $a$ and $b$ are positive integers with no common factor, we have the formula
When $b = 1$, the sum on the left is understood to be $0$.
\begin{note}
When $b = 1$, the proofs of (a) and (b) are trivial. We continue under the
assumption $b > 1$.
When $b = 1$, the proofs of (a) and (b) are trivial.
We continue under the assumption $b > 1$.
\end{note}
\subsubsection{\pending{Exercise 1.11.7a}}%

240
Bookshelf/Enderton/Set.tex
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@ -83,8 +83,12 @@ The \textbf{composition} of sets $F$ and $G$ is
\begin{definition}
\statementpadding
\lean*{Bookshelf/Enderton/Set/Relation}{Set.Relation.comp}
\lean*{Mathlib/Data/Rel}{Rel.comp}
\end{definition}
\section{\defined{Connected}}%
@ -106,8 +110,12 @@ The \textbf{domain} of set $R$, denoted $\dom{R}$, is given by
\begin{definition}
\statementpadding
\lean*{Bookshelf/Enderton/Set/Relation}{Set.Relation.dom}
\lean*{Mathlib/Data/Rel}{Rel.dom}
\end{definition}
\section{\defined{Empty Set Axiom}}%
@ -215,10 +223,60 @@ The \textbf{image of $A$ under $F$} is the set
\begin{definition}
\statementpadding
\lean*{Bookshelf/Enderton/Set/Relation}{Set.Relation.image}
\lean*{Mathlib/Data/Rel}{Rel.image}
\end{definition}
\section{\defined{Inductive Set}}%
\hyperlabel{ref:inductive-set}
A set $A$ is said to be \textit{inductive} if and only if $\emptyset \in A$ and
it is "closed under \nameref{ref:successor}", i.e.
$$(\forall a \in A) a^+ \in A.$$
\begin{note}
Induction is baked into Lean's type system.
In particular, the $\emptyset$ and "closed under successor" properties are
analagous to base and recursive constructors of an inductive data type
respectively.
\end{note}
\begin{definition}
\statementpadding
\lean*{Prelude}{Nat}
\lean*{Mathlib/Init/Set}{Set.univ}
\end{definition}
\section{\defined{Infinity Axiom}}%
\hyperlabel{ref:infinity-axiom}
There exists an \nameref{ref:inductive-set}:
$$(\exists A)\left[ \emptyset \in A \land (\forall a \in A) a^+ \in A \right].$$
\begin{note}
Since the definition of natural numbers in Lean satisfies the properties
required by this axiom, there is no need to explicitly state the axiom
separately in Lean.
\end{note}
\begin{axiom}
\statementpadding
\lean*{Prelude}{Nat}
\lean*{Mathlib/Init/Set}{Set.univ}
\end{axiom}
\section{\defined{Inverse}}%
\hyperlabel{ref:inverse}
@ -227,8 +285,12 @@ The \textbf{inverse} of a set $F$ is the set
\begin{definition}
\statementpadding
\lean*{Bookshelf/Enderton/Set/Relation}{Set.Relation.inv}
\lean{Mathlib/Data/Rel}{Rel.inv}
\end{definition}
\section{\defined{Irreflexive}}%
@ -260,8 +322,9 @@ A \textbf{linear ordering} on $A$ (also called a \textbf{total ordering} on $A$)
\vspace{6pt}
Trichotomy is equivalent to asymmetry and connectivity and asymmetry is
equivalent to antisymmetry and irreflexivity. Thus a linear order, as defined
by Enderton, is a binary relation with the following four properties:
equivalent to antisymmetry and irreflexivity.
Thus a linear order, as defined by Enderton, is a binary relation with the
following four properties:
\vspace{6pt}
\begin{enumerate}[(i)]
@ -278,6 +341,19 @@ A \textbf{linear ordering} on $A$ (also called a \textbf{total ordering} on $A$)
\end{definition}
\section{\defined{Natural Number}}%
\hyperlabel{ref:natural-number}
A \textbf{natural number} is a set that belongs to every inductive set.
The set of all natural numbers exists by virtue of \nameref{sub:theorem-4a}.
This set is denoted as $\omega$.
\begin{definition}
\lean*{Prelude}{Nat}
\end{definition}
\section{\defined{Ordered Pair}}%
\hyperlabel{ref:ordered-pair}
@ -286,8 +362,12 @@ For any sets $u$ and $v$, the \textbf{ordered pair} $\pair{u, v}$ is
\begin{definition}
\statementpadding
\lean*{Bookshelf/Enderton/Set/OrderedPair}{OrderedPair}
\lean*{Prelude}{Prod}
\end{definition}
\section{\defined{Pair Set}}%
@ -385,8 +465,12 @@ The \textbf{range} of set $R$, denoted $\ran{R}$, is given by
\begin{definition}
\statementpadding
\lean*{Bookshelf/Enderton/Set/Relation}{Set.Relation.ran}
\lean*{Mathlib/Data/Rel}{Rel.codom}
\end{definition}
\section{\defined{Reflexive}}%
@ -397,8 +481,12 @@ A binary relation $R$ is \textbf{reflexive} on $A$ if and only if $xRx$ for all
\begin{definition}
\statementpadding
\lean*{Bookshelf/Enderton/Set/Relation}{Set.Relation.isReflexive}
\lean*{Mathlib/Init/Algebra/Classes}{IsRefl}
\end{definition}
\section{\defined{Relation}}%
@ -408,8 +496,12 @@ A \textbf{relation} is a set of \nameref{ref:ordered-pair}s.
\begin{definition}
\statementpadding
\lean*{Bookshelf/Enderton/Set/Relation}{Set.Relation}
\lean*{Mathlib/Data/Rel}{Rel}
\end{definition}
\section{\defined{Restriction}}%
@ -424,6 +516,23 @@ The \textbf{restriction} of a set $F$ to set $A$ is the set
\end{definition}
\section{\defined{Successor}}%
\label{ref:successor}
For any set $a$, its \textbf{successor} is defined by $$a^+ = a \cup \{a\}.$$
\begin{note}
The corresponding Lean reference refers to the `Nat.succ` constructor.
This is not represented internally as a union of sets, but serves the same
role.
\end{note}
\begin{definition}
\lean*{Prelude}{Nat.succ}
\end{definition}
\section{\defined{Subset Axioms}}%
\hyperlabel{ref:subset-axioms}
@ -469,8 +578,12 @@ A binary relation $R$ is \textbf{transitive} if and only if whenever $xRy$ and
\begin{definition}
\statementpadding
\lean*{Bookshelf/Enderton/Set/Relation}{Set.Relation.isTransitive}
\lean*{Mathlib/Init/Algebra/Classes}{IsTrans}
\end{definition}
\section{\defined{Trichotomous}}%
@ -2690,8 +2803,8 @@ If not, then under what conditions does equality hold?
\lean*{Mathlib/SetTheory/ZFC/Basic}{Set.prod}
\begin{note}
The above Lean proof is a definition (i.e. an axiom). It does not prove
such a set's existence from first principles.
The above Lean proof is a definition (i.e. an axiom).
It does not prove such a set's existence from first principles.
\end{note}
Define $C = A \cup B$.
@ -5806,4 +5919,123 @@ Show that $<_L$ is a linear ordering on $A \times B$.
\end{proof}
\chapter{Natural Numbers}%
\hyperlabel{chap:natural-numbers}
\section{Inductive Sets}%
\hyperlabel{sec:inductive-sets}
\subsection{\unverified{Theorem 4A}}%
\hyperlabel{sub:theorem-4a}
\begin{theorem}[4A]
There is a set whose members are exactly the natural numbers.
\end{theorem}
\begin{proof}
By the \nameref{ref:infinity-axiom}, there exists an
\nameref{ref:inductive-set} $A$.
By the \nameref{ref:subset-axioms}, there exists a set $B$ such that
$$x \in B \iff x \in A \land \left[\forall C,
(\emptyset \in C \land
(\forall c \in C) c^+ \in C) \Rightarrow x \in C\right].$$
In other words, $x \in B$ if and only if $x \in A$ and $x$ is a natural
number.
Thus $B$ is the set whose members are exactly the natural numbers.
\end{proof}
\subsection{\unverified{Theorem 4B}}%
\hyperlabel{sub:theorem-4b}
\begin{theorem}[4B]
$\omega$ is inductive, and is a subset of every other inductive set.
\end{theorem}
\begin{proof}
$\omega$ denotes the set of \nameref{ref:natural-number}s.
We show $\omega$ is an \nameref{ref:inductive-set} by proving (i)
$\emptyset \in \omega$ and (ii) $\omega$ is closed under
\nameref{ref:successor}.
\paragraph{(i)}%
\hyperlabel{par:theorem-4b-i}
By definition, $\emptyset$ is a member of every inductive set.
Thus $\emptyset$ is a natural number, i.e. a member of $\omega$.
\paragraph{(ii)}%
\hyperlabel{par:theorem-4b-ii}
Let $n \in \omega$.
That is, let $n$ be a natural number.
By definition, $n$ is a member of every inductive set.
By definition of an inductive set, $n^+$ is then a member of every inductive
set as well.
Thus $n^+$ is a natural number, i.e. $n^+ \in \omega$.
\paragraph{Conclusion}%
By \nameref{par:theorem-4b-i} and \nameref{par:theorem-4b-ii}, it follows
$\omega$ is inductive.
It follows immediately from the definition of a natural number that $\omega$
is a subset of every other inductive set.
\end{proof}
\subsection{\verified{Theorem 4C}}%
\hyperlabel{sub:theorem-4c}
\begin{theorem}[4C]
Every natural number except $0$ is the successor of some natural number.
\end{theorem}
\begin{proof}
\lean{Bookshelf/Enderton/Set/Chapter\_4}
{Enderton.Set.Chapter\_4.theorem\_4c}
Let $T = \{n \mid n = 0 \lor (\exists m) n = m^+\}$.
It trivially follows that $\emptyset \in T$.
Let $x \in T$.
Then $x^+ \in T$ since $(\exists m) x^+ = m^+$, namely $m = x$.
Therefore $T$ is inductive.
By \nameref{sub:theorem-4b}, $\omega$ is a subset of $T$.
Thus every natural number satisfies the condition written in $T$'s definition.
In other words, every natural number except $0$ is the successor of some
natural number.
\end{proof}
\section{Exercises 4}%
\hyperlabel{sec:exercises-4}
\subsection{\verified{Exercise 4.1}}%
\label{sub:exercise-4.1}
Show that $1 \neq 3$ i.e., that $\emptyset^+ \neq \emptyset^{+++}$.
\begin{proof}
\lean{Bookshelf/Enderton/Set/Chapter\_4}
{Enderton.Set.Chapter\_4.exercise\_4\_1}
By definition,
\begin{align*}
1 & = \{\emptyset\} \\
3 & = \{\emptyset, \{\emptyset\}, \{\emptyset, \{\emptyset\}\}\}.
\end{align*}
By the \nameref{ref:extensionality-axiom}, these two sets are trivially not
equal to one another.
\end{proof}
\end{document}

27
Bookshelf/Enderton/Set/Chapter_4.lean Normal file
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@ -0,0 +1,27 @@
import Mathlib.Data.Set.Basic
/-! # Enderton.Set.Chapter_4
Natural Numbers
-/
namespace Enderton.Set.Chapter_4
/-- #### Theorem 4C
Every natural number except `0` is the successor of some natural number.
-/
theorem theorem_4c (n : ℕ)
: n = 0 ∨ (∃ m : ℕ, n = m.succ) := by
match n with
| 0 => simp
| m + 1 => simp
/-- #### Exercise 4.1
Show that `1 ≠ 3` i.e., that `∅⁺ ≠ ∅⁺⁺⁺`.
-/
theorem exercise_4_1 : 1 ≠ 3 := by
simp
end Enderton.Set.Chapter_4